Foundations of Nuclear and Particle Physics By T. William Donnelly, Joseph A. Formaggio, Barry R. Ho

INSTRUCTOR SOLUTION MANUAL

Foundations of Nuclear and Particle Physics: Solutions to Exercises T. W. DONNELLY MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, MA

J. A. FORMAGGIO MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, MA

B. R. HOLSTEIN UNIVERSITY OF MASSACHUSETTS, AMHERST, MA

R. G. MILNER MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, MA

B. SURROW TEMPLE UNIVERSITY, PHILADELPHIA, PA

Preface

The exercises and their solutions here were developed by us as a central pedagogical tool in the use of the book by the student of nuclear and particle physics. We have found that the typical graduate student will find them challenging and their solution will demand a mature level of understanding. To preserve the effectiveness of these exercises, the solutions manual should be retained by the instructor and not be distributed to the students. We welcome all suggestions for improving the solutions.

April 2019

iii

T.W. Donnelly, B.R. Holstein, R.G. Milner

Contents

iv

1 Introduction

page 1

2 Symmetries

5

3 Building Hadrons from Quarks

23

4 The Standard Model

40

5 QCD and Confinement

54

6 Chiral Symmetry and QCD

66

7 Elastic Electron Scattering from the Nucleon

77

8 Elastic Electron Scattering from the Nucleon

108

9 Hadron Structure via Lepton-Nucleon Scattering

118

10 High-Energy QCD

125

11 The Nucleon-Nucleon Interaction

134

12 The Structure and Properties of Few-Body Nuclei

147

13 Overview of Many-Body Nuclei

157

14 Models of Many-Body Nuclei

175

15 Electron Scattering from Discrete States

204

16 Electroexcitation of High-Lying Excitations of the Nucleus

244

17 Beta Decay

267

18 Neutrino Physics

278

Contents

v

19 The Physics of Relativistic Heavy-Ions

293

20 Astrophysics

307

21 Beyond the Standard Model Physics

319

22 Useful information

320

23 Quantum Theory

321

References

331

1

Introduction 1.1 US Energy Production In 2011, the United States required 3,856 billion kW-hours of electricity. About 20% of this power was generated by ∼100 nuclear fission reactors. About 67% was produced by the burning of fossil fuels, which accounted for about one-third of all greenhouse gas emissions in the U.S. The remaining 13% was generated using other renewable energy resources. Consider the scenario where all the fossil fuel power stations are replaced by new 1-GW nuclear fission reactors. How many such reactors would be needed?

Exercise 1.1 Solution: US Energy Production

3.856 × 106 GW-hours × 0.67 ×

Reactor years × = 295 Reactors 8766 hours GW

1.2 Geothermal Heating It is estimated that 20 TW of heating in the earth is due to radioactive decay: 8 TW from 238 U decay, 8 TW from 232 Th decay, and 4 TW from 40 K decay. Estimate the total amount of 238 U, 232 Th, and 40 K present in the Earth in order to produce such heating.

Exercise 1.2 Solution: Geothermal Heating The number of decays in unit time dt is equal to N (1 − e−dt/τ ) ≈ N dt/τ . That means the total mass of an isotope can be found from: Mtot. =

τP × Matom Edecay

1) 238 U i) P = 8 × 1012 W = 4.99 × 1025 MeV/s ii) τ = 6.446 × 109 years = 1.64 × 1017 s iii) Edecay = 4.267 MeV iv) Matom = 238 amu = 3.95 × 10−25 kg Mtot. = 9.39 × 1017 kg

2) 232 Th 1

Introduction

2

i) P = 8 × 1012 W = 4.99 × 1025 MeV/s ii) τ = 2.027 × 101 0 years = 6.39 × 1017 s iii) Edecay = 4.083 MeV iv) Matom = 232 amu = 3.85 × 10−25 kg Mtot. = 3.00 × 1018 kg

3) 40 K i) P = 4 × 1012 W = 2.50 × 1025 MeV/s ii) τ = 1.805 × 109 years = 5.70 × 1016 s iii) Edecay = 1.31 MeV iv) Matom = 40 amu = 6.64 × 10−26 kg Mtot. = 7.23 × 1016 kg

1.3 Radioactive Thermoelectric Generators A useful form of power for space missions which travel far from the sun is a radioactive thermoelectric generator (RTG). Such devices were first suggested by the science fiction writer Arthur C. Clarke in 1945. An RTG uses a thermocouple to convert the heat released by the decay of a radioactive material into electricity by the Seebeck effect. The two Voyager spacecraft have been powered since 1977 by RTGs using 238 Pu. Assuming a mass of 5 kg of 238 Pu, estimate the heat produced and the electrical power delivered. (Do not forget to include the ∼ 5% thermocouple efficiency.)

Exercise 1.3 Solution: Radioactive Thermoelectric Generators Let’s first look at the instantaneous power produced. We know that: P (t) =

N (0)e−t/τ Edecay N (t)Edecay = τ τ

238

Pu has a 5.593 MeV α decay with a lifetime of τ = 126.5 years (3.99×109 s). 5 kg of 238 Pu is 1.27 × 1025 atoms. At t = 0 years (1977), the thermal power produced was 1.77 × 1016 MeV/s = 2.84 kW. At t = 38 years (2015), the power was down to 2.10 kW. A typical RTG efficiency is about 5%, so we can estimate that Voyager’s power budget has been between 140 and 100 W. We can integrate the thermal power to get the total heat produced.

Z t h i Q= P (t)dt = N (0)Edecay 1 − e−t/τ 0

For t = 38 years, we get a total heat Q = 2.95 × 1012 J. 1.4 Fission versus Fusion Energy can be produced by either nuclear fission or nuclear fusion.

Introduction

3

a) Consider the fission of 235 U into 117 Sn and 118 Sn, respectively. Using the mass information from a Table of Isotopes, calculate (i) the energy released per fission and (ii) the energy released per atomic mass of fuel. b) Consider the deuteron-triton fusion reaction 2

H + 3 H → 4 He + n .

Using the mass information from the Periodic Table of the Isotopes, calculate (i) the energy released per fusion and (ii) the energy released per atomic mass unit of fuel.

Exercise 1.4 Solution: Fission versus Fusion Mass defect of 235 U : 40.9218 MeV Mass defect of 117 Sn: −90.3977 MeV Mass defect of 118 Sn: −91.6528 MeV . Per fission: E = (∆M235 U − ∆M117 Sn − ∆M118 Sn ) = 222.97 MeV . Per amu: E = 222.97 MeV/atom ×

atom = 0.949 MeV/amu 235 amu

Mass defect of 2 H: 13.1357 MeV Mass defect of 3 H: 14.9498 MeV Mass defect of 4 He: 2.4249 MeV Mass defect of n: 8.0713 MeV . Per fission: E = (∆M2 H + ∆M3 H − ∆M4 He − ∆Mn ) = 17.59 MeV . Per amu: E = 17.59 MeV/fission ×

fission = 3.518 MeV/amu 5 amu

1.5 Absorption Lengths A flux of particles is incident upon a thick layer of absorbing material. Find the absorption length, the distance after which the particle intensity is reduced by a factor of 1/e ∼ 37% (the absorption length) for each of the following cases: a) when the particles are thermal neutrons (i.e., neutrons having thermal energies), the absorber is cadmium, and the cross section is 24,500 barns, b) when the particles are 2 MeV photons, the absorber is lead, and the cross section is 15.7 barns per atom,

Introduction

4

c) when the particles are anti-neutrinos from a reactor, the absorber is the Earth, and the cross section is 10−19 barns per atomic electron.

Exercise 1.5 Solution: Absorption Lengths The beam intensity will fall exponentially, according to the differential equation: dN N =− dx λ where λ is the absorbtion length. The beam is reduced by factor of 1/e when x = λ.

The reduction of beam intensity is proportional to the number of interactions per distance, i.e. σn, where σ is the interaction cross section and n is the number density in the absorber.

1 N λ = − dN = σn dx In each part of this problem, one must find a plausible value for n and solve for λ.

1. Neutrons in Cadmium: The density of cadmium is 8.65 g/cm3 . The atomic weight of cadmium is 112 amu = 1.86 × 10−22 g, making n = 4.65 × 1022 cm−3 . λ=

1 b × = 8.78 × 10−4 cm = 8.78µm [4.65 × 1022 cm−3 ] [24500 b] 10−24 cm

2. Photons in Lead: The density of lead is 11.3 g/cm3 . The atomic weight of lead is 207 amu = 3.44 × 10−22 g, making n = 3.28 × 1022 cm−3 . λ=

1 b × = 1.94 cm [3.28 × 1022 cm−3 ] [15.7 b] 10−24 cm

3. Anti-neutrinos through the Earth: The average density of the earth is 5.51 g/cm3 . The four most abundant elements on earth, oxygen, magnesium, silicon, and iron, make up more than 93% of the earth’s mass. A mass-weighted average of their values of e per amu comes to 0.488 e/amu = 2.94 ×1023 e/g. The electron density of the earth is thus 1.51 × 1024 e/cm3 . λ=

1 b × = 6.62 × 1018 cm ≈ 7 light-years [1.51 × 1024 cm−3 ] [10−19 b] 10−24 cm

2

Symmetries 2.1 Noether’s Theorem The mathematician Emmy Noether proved a very important theorem to physics which states that for any invariance of the classical action under a continuous field transformation there exists a classical charge Q which is independent of time and which is connected to a conserved current, ∂µ J µ = 0. That is, the existence of a a symmetry requires the validity of a corresponding conservation law. Well known examples in classical physics are the invariance of the equations of motion under spatial translation, time translation, and rotation, which lead to conservation of momentum, energy, and angular momentum respectively. The purpose of this exercise is to prove this theorem.

We begin with an action Z S=

d4 xL(φ, ∂µ φ)

and consider an infinitesimal transformation of the field φ → φ0 = φ + f (φ) , . where f (φ) is some function of the field. a) Find the equation of motion for a constant value of the infinitesimal and show that one finds the Euler-Lagrange equation δL δL − ∂µ =0. δφ δ∂µ φ b) Calculate the change of the action under this field transformation in the case that = (x) and show that Z S → S 0 = S + d4 x∂µ j µ , where jµ =

δL f (φ) . δ∂µ φ

c) Now integrate by parts and show that if the action is invariant we require that ∂µ j µ = 0, up to a total derivative.

5

Symmetries

6

d) Integrate the equation ∂j 0 = −∇ · j = 0 ∂t over all space, and show that for a local “charge” distribution we have Z dQ =0 where Q = dxj 0 , dt which proves Noether’s theorem.

Exercise 2.1 Solution: Noether’s Theorem a) We have Z 0 = δS =

Z

δL δL δL δL 4 d x δφ + δ∂µ φ = d x f (φ) − ∂µ δφ δ∂µ φ δφ δ∂µ φ 4

so δL δL − ∂µ =0 δφ δ∂µ φ b) Now Z 0 = δS =

δL δ∂µ φ d x δLδφ + δ∂µ φ 4

δL δL = d x f (φ) + (f (φ)∂µ + ∂µ f (φ)) δφ δ∂µ φ

Z δL δL δL = d4 x f (φ) − ∂µ + ∂µ f (φ) δφ δ∂µ φ δ∂µ φ Z

4

c) Then, using, result of a), we have Z Z δL 0 = d4 x∂µ f (φ) = d4 x ∂ µ jµ δ∂µ φ with jµ = f (φ)

δL δ∂µ φ

d) Since 0=

δρ +∇·j δt

we have d dt

Z

3

d xρ = −

Z

3

d x∇ · j =

Z

2.2 Rotation Matrices and Finite Rotations

R→∞

dA · j −→ 0

Symmetries

7

The effect on a spin eigenstate |S, Sz i under a rotation by angle χ about an axis n̂ is given by 0

|S, Sz i = RS (n̂, χ) |S, Sz i , where RS (n̂, χ) = exp(−iχS · n̂) , and where S are the (2S + 1) × (2S + 1) component spin matrices constructed from the relations Sz |S, mi = m |S, mi p (Sx ± iSy ) |S, mi = (S ∓ m)(S ± m + 1) |S, m ± 1i . a) Evaluate the rotation matrix for a spin- 21 system and show that 1

1

R 2 (n̂, χ) = exp(−iχS 2 · n̂) = cos

χ χ − iσ · n̂ sin , 2 2

where σ are the Pauli matrices.

0 b) Calculate the rotated spin- 21 state |1/2, mi for the initial state |1/2, mi for m = ± 21 using n̂ = êz and rotation angle χ and demonstrate that 0

|1/2, mi = exp(−imχ) |1/2, mi . c) Verify the commutation relations for the representations of the spin-1 operators in Eq. (2.41), [Sz , S± ] = ±S± [S+ , S− ] = 2Sz . d) Now, evaluate the rotation matrix for a spin-1 system and show that R1 (n̂, χ) = exp(−iχS1 · n̂) = 1 − (S1 · n̂)2 (1 − cos χ) − iS1 · n̂ sin χ , where S1 are the 3 × 3 spin matrices constructed in the text.

0 e) Calculate the rotated spin state |1, mi for the initial state |1, mi for the cases m = 1, 0, −1 using n̂ = êz and rotation angle χ and demonstrate that 0

|1, mi = exp(−imχ) |1, mi .

Exercise 2.2 Solution: Rotation Matrices and Finite Rotations a) We have σ · n̂2 = 1 so

∞ ∞ χ 2n+1 X X 1 χ 2n 1 χ − + iσ · n̂ − exp −i σ · n̂ = 2 (2n)!

2 (2n + 1)!

2 n=0 n=0

= cos

χ χ − iσ · n̂ sin 2 2

Symmetries

8

b) Pick n̂ = ẑ so

1 1 1 χ χ 1 1 R 2 (χ, n̂)| , ± >= cos − iσz sin | ,± > 2 2 2 2 2 2

χ 1 1 χ χ 1 1 | , ± >= exp ∓i | ,± > = cos ∓ i sin 2 2 2 2 2 2 2 c) Have 

1 Sz S+ =  0 0 and

 0 0 0  0 0 −1

0 0 0 √



 1 √0 2  0 0 0

2 0 0

0 Sz S+ =  0 0

√

2 0 0

  0 √0 2 = 0 0 0

0 0 0

  0 0 0 = 0 −1 0

so 

0 [Sz , S+ ] = Sz S+ − S+ Sz =  0 0

Also, we have 

0  S+ S− = 0 0

√

 √0 √0   2 2 0 0

2 0 0

and 

√0 S− S+ =  2 0

 0 0 0  0 0 0 √ 2 0 0

√

2 0 0

√

 2 0 0 0  0 0

0 0 0

 0 √ − 2  0

 √0 2  = S+ 0

  0 0 2   = 0 √0 0 2 0 0

0 2 0

 0 0  0

  0 √0 2 = 0 0 0

0 2 0

 0 0  2

√

2 0 0

so 

2 [S+ , S− ] = S+ S− − S− S+ =  0 0

0 0 0

 0 0  = 2Sz −2

3

d) Since (S · n̂) = S · n̂ we have 2

exp −iχS · n̂ = 1 + (S · n̂)

∞ X

∞ X 1 1 2n 2n+1 (−χ) + iS · n̂ (−χ) (2n)!

(2n + 1)! n=1 n=0 2

= 1 + (S · n̂) (cos χ − 1) − iS · n̂ sin χ e) Pick n̂ = ẑ so

R1 (χ, n̂)|1, m >= 1 − Sz2 (1 − cos χ) − iSz sin χ |1, m >

Symmetries

9

 (cos χ − i sin χ)|11 > m=1 = 1 − m2 (1 − cos χ) − im sin χ |1, m >=  |10 > m=0  (cos χ + i sin χ)|1, −1 > m = −1 

= exp(−imχ)|1, m >

2.3 Symmetry and Dipole Moments Since the only three-vector associated with an elementary particle in its rest frame is the spin, any dipole moment of the particle must be along the spin direction. For example, in the case of a spin- 21 system such as the nucleon, one writes m = gm

e S, 2m

where m is the magnetic dipole moment, gm is the gyromagnetic ratio, and e/2m is called the Bohr magneton. When such a dipole is placed in a magnetic field B, the corresponding interaction energy is UM = −m · B . a) Analyze this interaction from the point of view of parity and time reversal, and demonstrate that UM is even under both.

Now suppose that we define an analogous electric dipole moment p in terms a gyroelectric ratio ge e S. p = ge 2m When placed in an external electric field E there will exist an interaction energy UE = −p · E . b) Analyze this interaction from the point of view of parity and time reversal and demonstrate that UE is odd under both.

Thus, an elementary particle cannot possess an electric dipole moment if parity and/or time reversal are conserved. c) We know that atoms and molecules can have large static electric dipole moments.

Explain why this fact does this not indicate a violation of parity and time reversal invariance.

Exercise 2.3 Solution: Symmetry and Dipole Moments a) Note that vector potential Aµ is a four-vector so transforms like xµ . That P

T

is, under parity Aµ −→ Aµ while under time reversal Aµ −→ −Aµ . Thus B =

Symmetries

10

P

T

∇ × A −→ B and B = ∇ × A −→ −B. On the other hand, spin is a pseudo-vector P T so S =−→ S and S =−→ −S. Thus we have P

T

S · B −→ S · B and S · B −→ S · B

P

∂ b) On the other hand for the electric field we have E = ∂t A − ∇A0 −→ −E and T

P

T

E =−→ E, while for spin S =−→ S and S =−→ −S. Thus we have P

S · E −→ −S · E

T

and S · E −→ −S · E

c) For a non-elementary system such as a molecule, there exist many nearly degenerate states so that a field reversal has virtually no change in the energy of the system.

2.4 Spin Coupling m

m

Consider a state with two spin- 12 particles having spinors χ1/2s1 (1) and χ1/2s2 (2), where the particles are labeled 1 and 2 and the spin projections along some axis of quantization have the values msi = ± 12 . If the spin angular momenta of the two particles are coupled to total spin S with projection MS one has the state

MS S .

AM S (1, 2) ≡ χ1/2 (1) ⊗ χ1/2 (2) S What values can S and MS have? By explicit evaluation of the Clebsch-Gordan coefficients or 3-j symbols (see [Edm74]) show that the familiar answers are obtained (see [Sch55]). Using the properties of the 3-j symbols prove upon interchange of the coordinates of the two particles that one has S+1 MS S AM AS (1, 2) .

S (2, 1) = (−)

These results can trivially be extended to the case of two isospin- 21 particles, for m m instance two nucleons, where one has isospinors ξ1/2t1 (1) and ξ1/2t2 (2) with mti = ± 12 , one has the coupled state

MT BTMT (1, 2) ≡ ξ1/2 (1) ⊗ ξ1/2 (2) T , and obtains the symmetry under interchange BTMT (2, 1) = (−)T +1 BTMT (1, 2) .

What happens upon coordinate interchange for a system of two nucleons (N N ) where one has both spin and isospin?

Exercise 2.4 Solution: Spin Coupling

Symmetries

11

One begins by writing out the coupling of the two spin-spinors in detail: X m m S AM h1/2 ms1 1/2 ms2 | 1/2 1/2 SMS i χ1/2s1 (1)χ1/2s2 (2) S (1, 2) = ms1 ms2 MS

= (−)

[S]

X 1/2 ms1 m m s1

s2

1/2 ms2

S −MS

m

m

χ1/2s1 (1)χ1/2s2 (2),

√ where as usual [S] ≡ 2S + 1. Clearly, using either the Clebsch-Gordan coefficients or the 3-j symbols one sees that the only values of the total spin quantum numbers are S = MS = 0 and S = 1 with MS = −1, 0, +1. The former is easily written out explicitly using the fact (see Edmonds [Edm74]) that

1/2 ms1

1/2 ms2

0 0

1/2−ms1

=

(−)

√

2

δms2 ,−ms1

yielding i 1 h +1/2 −1/2 −1/2 +1/2 A00 (1, 2) = √ χ1/2 (1)χ1/2 (2) − χ1/2 (1)χ1/2 (2) .

2 For the S = 1, MS = 0 case one has from Edmonds

r 2 1/2 1/2 1 1/2−ms1 ms1 δms2 ,−ms1 (−) = ms1 ms2 0 3 yielding i 1 h +1/2 −1/2 −1/2 +1/2 A01 (1, 2) = √ χ1/2 (1)χ1/2 (2) + χ1/2 (1)χ1/2 (2) .

2 For the remaining cases one has

1/2 1/2 1 1/2 = 1/2 1/2 −1 −1/2

1/2 −1/2

1 +1

1 = −√ , 3

again from Edmonds, yielding +1/2

+1/2

−1/2

−1/2

A+1 1 (1, 2) = χ1/2 (1)χ1/2 (2) A−1 1 (1, 2) = χ1/2 (1)χ1/2 (2).

These are all the expected familiar results from standard discussions of quantum mechanics.

The 3-j symbols have the symmetry property under exchange of the magnetic quantum numbers of the two spinors:

1/2 1/2 S 1/2 1/2 S S+1 = (−) . ms2 ms1 −MS ms1 ms2 −MS Upon exchange of the particle coordinates 1 ↔ 2 in the basic expression for the

Symmetries

12

coupled spin-spinor, one has

X 1/2 1/2 S m m χ1/2s1 (2)χ1/2s2 (1) m m −M s1 s2 S m s1 m s2

X 1/2 1/2 S µs2 µs1 M (2) (1)χ1/2 = (−) S [S] χ1/2 µ µ −M s s S 2 1 µs1 µs2

X 1/2 1/2 S µs2 µs1 S+1 MS (2) = (−) × (−) [S] (1)χ1/2 χ1/2 µ µ −M s1 s2 S µ µ

MS S [S] AM S (2, 1) = (−)

s1

S+1

= (−)

s2

S AM S (1, 2),

where in the second line the replacements ms1 → µs2 and ms2 → µs1 have been made, and in the third line the symmetry relation above has been employed. Note that in the second line the spin-spinors have been written in the opposite order: since they operate in different spaces (namely, those for particles 1 and 2), they commute.

Clearly all of the above goes through for isospin and, specifically, one has the cases T = MT = 0 and T = 1 with MT = −1, 0, +1 having analogous expressions to those above. The exchange symmetry for isospin is immediate, T +1

BTMT (2, 1) = (−)

T AM T (1, 2),

and exchange of the coordinates for systems having both spin and isospin yields S+T

MT S AM S (2, 1)BT (2, 1) = (−)

MT S AM S (1, 2)AT (1, 2).

2.5 L-S Coupling and Central Potentials Consider a nonrelativistic system of two particles, which interact via a potential. In general such an interaction can depend on spin and isospin as well as the particle separation. This problem deals with how to deal with such a potential.

If we represent the spatial coordinates of the two particles by r1 and r2 , then one can change variables to a center-of-mass variable R = (r1 + r2 )/2 and a relative coordinate r = r1 −r2 . The two-particle spatial wavefunction may then be written as a product of a plane wave for the center-of-mass (cm) part times a function involving the relative coordinate, Φ(r). If the state has good orbital angular momentum L, with projection ML , where L is an integer and −L ≤ ML ≤ L, then the spatial wavefunction can be written ML L ΦM L (r) =RL (r)YL (Ωr ) ,

where RL (r), the radial wavefunction depends only on r = |r| and the dependence on the polar and azimuthal angles Ωr = (θ, φ) specifying the direction of r is captured in the spherical harmonic YLML (Ωr ). For a general discussion of these

Symmetries

13

basic ideas see one of the standard books on quantum mechanics, such as [Sch55]; see also [Edm74]. a) Prove that under a parity transformation, namely under inversion of coordinates r → −r one has YLML (Ωr ) → (−)L YLML (Ωr ). One may now form the total wavefunction for an N N system in a given partial wave (i.e., with good orbital angular momentum), M

MT J J ;MT ΨM (LS)J;T (1, 2) = R(LS)J;T (r) [YL (Ωr ) ⊗ AS (1, 2)]J BT (1, 2) ,

where here for simplicity the cm plane wave has been omitted. The total state is presumed to have orbital and spin angular momenta coupled to total angular momentum J with projection MJ . What happens upon interchange of all of the coordinates of the two nucleons? Since the Pauli exclusion principle states that the total wavefunction must be antisymmetric, what quantum numbers are allowed for the N N system? b) Suppose that one has a spin-isospin dependent central potential of the form VC (r) = V1 (r) + V2 (r)σ(1) · σ(2) + V3 (r)τ (1) · τ (2) + V4 (r)σ(1) · σ(2)τ (1) · τ (2) ,

where the factors σ(i) and τ (i) are the familiar Pauli matrices discussed in this chapter. For a state of the type introduced in the previous exercise evaluate the spin and isospin matrix elements of the potential. c) If a so-called spin-orbit potential of the form VLS (r) = {V5 (r) + V6 (r) (τ (1) · τ (2))} (L · S) were to be added to the central potential in the previous Exercise, what would be the form of the orbital-spin-isospin matrix elements? d) One might also want to add a so-called tensor potential of the form VT (r) = {V7 (r) + V8 (r) (τ (1) · τ (2))} S12 (1, 2), involving the tensor operator S12 (1, 2) = 3(r · σ(1))(r · σ(2))/r2 − σ(1) · σ(2) .

Find the result of applying the tensor operator to a singlet spin state, i.e., a state having S = 0. For a potential that contains a tensor term one finds that the orbital angular momentum is not a good quantum number and that states with differing values of L must mix; which states are these? Do you know a relatively familiar example of this mixing? Can you think of other types of potentials not included in the types discussed here in Exercises 2.4 and 2.5?

Exercise 2.5 Solution: L-S Coupling and Central Potentials a) Inversion of coordinates r → −r corresponds to making the transformation

Symmetries

14

(r, θ, ϕ) → (r, π − θ, π + ϕ) under which cos θ → − cos θ, sin θ → + sin θ and m eim(π+ϕ) = (−) eimϕ . Employing this with expressions for the spherical harmonics (see Edmonds [Edm74]) gives the desired result: YLML (Ωr ) → (−)L YLML (Ωr ). In Exercise 2.4 we saw that interchange of coordinates for an N N system having both S+T spin and isospin yielded the phase (−) and hence interchange of coordinates for L+S+T space-spin-isospin N N wavefunctions yields the total symmetry (−) . Since the total wavefunction must be antisymmetric, this implies that L + S + T = odd.

So, for example, one has that 3 S1 and 3 D1,2,3 partial waves have isospin T = 0, while 1 S0 and 3 P0,1,2 partial waves have isospin T = 1 (here the notation is, as usual, 2S+1 LJ with J = L + S). All of these developments are used in Chapter 11. b) Starting with the spin operators, one has S(1, 2) = s(1) + s(2), where as usual s(i) = σ(i)/2. Thus S 2 (1, 2) = s2 (1) + s2 (2) + 2s(1) · s(2) having eigenvalues S(S + 1) for S 2 (1, 2) and 1/2 × 3/2 = 3/4 for s2 (1) and s2 (2).

Using this and rewriting the result in terms of Pauli matrix operators one has σ(1) · σ(2) = 2S(S + 1) − 3 ≡ aS

−3 S = 0 spin singlet = +1 S = 1 spin triplet The isospin case is proven in an analogous way, yielding τ (1) · τ (2) = 2T (T + 1) − 3 ≡ bT

−3 T = 0 isospin singlet = +1 T = 1 isospin triplet which then immediately leads to the desired matrix elements of the potential VC (r) = V1 (r) + aS V2 (r) + bT aV3 (r)+aS bT V4 (r). c) Noting that J=L+S one can form J 2 = L2 + S 2 + 2L · S having eigenvalues J(J + 1), L(L + 1) and S(S + 1) for J, L and S, respectively, which leads to the relationship 1 {J(J + 1) − L(L + 1) − S(S + 1)} ≡ cLSJ 2  0 S=0 J =L    − (L + 1) S = 1 J = L − 1 =  −1 S=1 J =L   +L S =1 J =L+1

L·S=

Symmetries

15

This would lead to the additional terms VLS (r) = {V5 (r) + bT V6 (r)} cLSJ . d) The tensor potential involves somewhat more work. We start by writing the tensor operator S12 (1, 2) = 3(r · σ(1))(r · σ(2))/r2 − σ(1) · σ(2) in terms of contributions that operate only in coordinate space or only in spin space, begining by decomposing the vectors involved in terms of spherical tensors X σ(i) = σm (i)e∗m m=0,±1

r r=

X 4π Y m (Ωr )e∗m . r 3 m=0,±1 1

Let us define tensor products in the two spaces: M

K ΣM K (1, 2) ≡ [σ(1) ⊗ σ(2)]K X h1µ1µ0 | 11KMK i σµ (1)σµ0 (2) =

µµ0

MK

= (−)

[K]

X 1 µ

µµ0

1 µ0

K −MK

4π M [Y1 (Ωr ) ⊗ Y1 (Ωr )]L L 3 X 1 1 4π ML = (−) [L] µ µ0 3 0

σµ (1)σµ0 (2)

ZLML (Ωr ) ≡

µµ

K −MK

0

Y1µ (Ωr )Y1µ (Ωr )

with inverses

1 1 K K σm (1)σm0 (2) = (−) [K] ΣM K (1, 2) m m0 −MK KMK

0 3 X 1 1 L ML m m Y1 (Ωr )Y1 (Ωr ) = (−) [L] ZLML (Ωr ), m m0 −ML 4π X

MK

LML

which are easily derived using the orthogonality properties of the 3-j symbols; as √ usual [x] ≡ 2x + 1. By using Eq. (4.6.5) in Edmonds [Edm74] one has that

√ 1 1 L ML ZL (Ωr ) = 4π YLML (Ωr ), 0 0 0 where the 3-j symbol (the so-called parity 3-j symbol) must have the sum of its entries even, implying that only L = 0 and L = 2 enter. By using the fact that X m σ(1) · σ(2) = (−) σm (1)σ−m (2) m

Symmetries

16

one has immediately that √ σ(1) · σ(2) = − 3Σ00 (1, 2), and by direct evaluation that 1 Z00 (Ωr ) = − √ .

3 Now the remainder of the tensor operator involves X 0 m+m0 4π m (−) 3(r · σ(1))(r · σ(2))/r2 = 3 Y1 (Ωr )Y1−m (Ωr ) 3 0 mm

×σ−m (1)σm0 (2) X M MK K (Ωr )ΣM (−) K ZK =3 K (1, 2) KMK

= Z00 (Ωr )Σ00 (1, 2) X M +3 (−) K Z2MK (Ωr )Σ2−MK (1, 2) =−

√

MK

3Σ00 (1, 2) +3

X

MK

(−)

Z2MK (Ωr )Σ2−MK (1, 2),

MK

where again orthogonality of the 3-j symbols has been employed, together with other relations from above. The first term here cancels with σ(1) · σ(2) (in fact, by design: this is why it is called the tensor operator). One is left with X M S12 (1, 2) = 3 (−) K Z2MK (Ωr )Σ2−MK (1, 2) MK

involving the scalar product of two rank-2 tensor operators. The matrix elements of Z2MK (Ωr ) involve doing the integral over the solid angle of the product of three spherical harmonics, Z M dΩYL0 L0 (Ωr )∗ Y2MK (Ωr )YLML (Ωr ), which by Eq. (4.6.3) of Edmonds [Edm74] is proportional to another parity 3-j symbol 0

L 2 L , 0 0 0 and, since then L + L0 must be even, L and L0 must both be either even or odd for mixing to occur. Also, using the Wigner-Eckart theorem for the spin matrix elements, 0

D E S 2 S S 0 −MS 0 −MK 0 0 S MS Σ 2 (1, 2) SMS = (−) −MS 0 −MK MS × hS 0 kΣ2 (1, 2)k Si ,

Symmetries

17

tells us that only when both S 0 and S are equal to 1 (triplet) can mixing occur.

Thus in the sequence of allowed N N partial waves, 3 S1 , 3 P0,1,2 , 3 D1,2,3 , 3 F2,3,4 , · · · , only mixings of the type 3 S1 − 3 D1 , 3 P2 −3 F2 , · · · can mix; the other triplet states are sometimes called isolated triplets; and the singlet states do not mix, as we have seen. A familiar example of this (see Chapter 11 of the text) is the deuteron whose non-relativistic ground state involves both 3 S1 and 3 D1 configurations mixed via the tensor force.

Finally, it is also possible to have additional types of forces. For instance, using only the facts that the charge is conserved and that all terms must be invariant under interchange of identical particles would allow additional contributions proportional to τ3 (1)τ3 (2) and τ3 (1) + τ3 (2). For low-energy N N scattering one does not need such contributions and finds that the N N force is charge independent. If one assumes the force to be only charge symmetric, then terms of the τ3 (1)τ3 (2) type are also needed.

One may also have contributions that are quadratic in orbital angular momentum, 2 for instance terms involving L2 , (σ(1) · σ(2)) L2 and (L · S) times 1 or τ (1) · τ (2).

2.6 Single-Particle Wavefunctions As we shall see later in the book, it is useful to employ a basis of single-particle wavefunctions when discussing the nuclear many-body problem. One starts with some mean-field potential in which the individual nucleons move and, upon coupling the orbital and spin angular momenta to form the total angular momentum, forms the single-particle wavefunctions m

ms ;mt ψ(`s)j;t (r) = R(`s)j;t (r) [Y` (Ωr ) ⊗ χs ]j j ξtmt ,

where s = t = 1/2 and here the spatial coordinate is with respect to the origin of the potential. What values are allowed for the orbital and total angular momentum quantum numbers ` and j? If the potential has a spin-orbit term such as in Exercise 2.5 c), what does one expect for the splitting between states with the same orbital angular momentum, but different values of j?

Exercise 2.6 Solution: Single-Particle Wavefunctions We use a relationship that is very similar to that in Exercise 2.5 c), namely, starting with j=`+s and forming j 2 = `2 + s2 + 2` · s with eigenvalues j(j + 1), `(` + 1) and s(s + 1) = 3/4, for j 2 , `2 and s2 , respectively,

Symmetries

18

one has 1 {j(j + 1) − `(` + 1) − 3/4} ≡ α`j 2

1 − (` + 1) j = ` + 1/2 = × +` j = ` − 1/2 2

`·s=

and hence a spin-orbit potential that has the form Vso (r) ` · s would require evaluating the radial integral of Vso (r) and multiplying by α`j given above. The physical situation is that the integral comes out to be positive and so the so-called “stretched” states with j = ` + 1/2 are found to be lower in energy, whereas the so-called “jack-knifed” states with j = ` − 1/2 lie higher in energy.

2.7 SU (6) Symmetry Employing the rules given in the text for determining the dimensions of representations of SU (N ) groups using the Young tableaux, consider the direct product of three copies of the fundamental in SU (6), 6, 6⊗6⊗6 (in the text the cases of SU (2) and SU (3) were both discussed). Determine the resulting direct sum of the representations that emerge and find their dimensions.

Exercise 2.7 Solution: SU (6) Symmetry We proceed as in the text, first taking the direct product of two copies of the fundamental representation which yields the result shown in Fig. 2.3, and then taking the direct product with a third fundamental representation we obtain the result in Fig. 2.7. Finally, using the hook rules on the individual Young tableaux, inserting the factors N , etc., and adding the hooks, we obtain the dimensions as discussed in the text. The results are shown in the accompanying figures. For case A we add the factors N , N + 1 and N + 2), which yields dimension N (N + 1) (N + 2) , 2·3 namely, 10 as in the text for SU (3) and 56 in the present case of SU (6). Similarly, for tableaux B1,2 in the second figure, upon inserting the factors N − 1, N and N + 1, yielding these as numerator factors for the three hooks involved, we find dimension (N − 1) N (N + 1) DB1,2 = , 3 namely, 8 as in the text for SU (3) and 70 in the case of SU (6). Finally, the remaining Young tableau has factors N , N − 1 and N − 2, as shown in the third figure, and the three hooks as illustrated, yielding DA =

DC =

N (N − 1) (N − 2) , 2·3

Symmetries

19

A N

3

t

Fig. 2.1

N+1 2

N+2 1

Applying the hook rules for Young tableau A. namely, 1 for SU (3) and 20 in the case of SU (6).

2.8 Spontaneous Symmetry Breaking in Classical Mechanics Consider a frictionless bead of mass m free to slide on a hoop of radius R, which is rotating about a vertical axis with angular velocity ω. a) Show that the potential energy is given by 1 V (θ) = − mω 2 R2 sin2 θ − mgR cos θ , 2

Symmetries

20

B 1,2 N

N+1 1

N-1 3

t

Fig. 2.2

1

Applying the hook rules for Young tableaux B1,2 . where θ is the angle of the bead as measured from the bottom of the hoop. g the shape of the potential has a single minimum, so that b) Show that for ω 2 < R the position of stable equilibrium is at θ = 0. g c) Show that for ω 2 > R the shape of the potential has two minima, so that there exist two positions of stable equilibrium at θ = ± cos−1

g . ω2 R

Symmetries

21

C N

N-1

N-2 3

t

Fig. 2.3

2

1

Applying the hook rules for Young tableau C.

Exercise 2.8 Solution: Spontaneous Symmetry Breaking in Classical Mechanics a) Kinetic energy of rotating bead is T = 12 (R2 θ̇2 + R2 ω 2 sin2 θ) while potential energy is V = −mgR cos θ so Lagrangian is

L=T −V =

1 2 2 1 R θ̇ + mR2 ω 2 sin2 θ + mgR cos θ 2 2

Symmetries

22

so effective potential is 1 Vef f (θ) = − mR2 ω 2 sin2 θ − mgR cos θ 2

Find minimum via dVef f 0= = mgR sin θ − mR2 ω 2 sin θ cos θ = mR sin θ(Rω 2 cos θ − g) dθ b) If Rω 2 < g then factor in parentheses is negative definite so equilibrium must be at θ = 0. g c) If Rω 2 > g then two new equilibrium points at θ = ± cos−1 Rω 2 Note can check stability via d2 Vef f = mR cos θ(Rω 2 cos θ − g) − mR2 ω 2 sin2 θ dθ2 = mR2 ω 2 cos 2θ − mgR cos θ Then if Rω 2 < g curvature is negative at θ = 0 and positive at θ = π so θ = 0 is stable equilibrium point. However, if Rω 2 > g then curvature is positive at both θ = 0, π so these are unstable points, but negative at θ = ± cos−1 ω2gR so these are stable.

3

Building Hadrons from Quarks 3.1 Isospin Symmetry and K → 2π Decay The dominant decay modes of the K mesons are the two-pion channels K 0 → π+ π− ,

K 0 → π0 π0 ,

K ± → π± π0 .

a) Using the fact that pions are spinless and are therefore bosons, show that the final two-pion state resulting from the decay of a kaon must have either isospin zero or two. b) Since ūd carries I = 1 while s̄u carries I = 21 the strangeness-changing nonleptonic Hamiltonian G Hw ∼ √ s̄uūd 2 involves a linear combination of I = 12 and I = 23 components, leading to K → 2π decay amplitudes a1 and a3 . Show that the decay amplitude can be written as 3 A(K + → π + π 0 ) = √ a3 2 2 1 A(K 0 → π + π − ) = −a1 + a3 2 A(K 0 → π 0 π 0 ) = a1 + a3 . c) Compare this parametrization with the experimental lifetimes for the following channels τK 0 →π+ π− ' τK 0 →π0 π0 τK ± →π± π0 .

What do you conclude about the amplitudes a1 and a3 ?

Exercise 3.1 Solution: Isospin Symmetry and K → 2π Decay a) Bose symmetry requires that spatial × spin × isospin piece of the wavefunction be symmetric under interchange. In the case of the two-pion state there is no spin and the spatial component must be an S-wave, which is symmetric under interchange. Hence, the isospin piece must also be symmetric. Now in general, the product of two isotriplet states reads (I1 = 1) × (I2 = 1) → I = 0, 1, 2, where the states with I = 0, 2 are symmetric and the state I = 1 is antisymmetric. Hence, only the states with I = 0, 2 are permitted.

23

Building Hadrons from Quarks

24

b) We have, by isotopic spin symmetry < π a π b |Hw |K n >= ã1 < 1, a; 1, b|0, 0; 11 >< 0, 0;

11 1 1 1 | , ; ,n > 22 2 2 2

11 3 1 1 | , ; ,n > 22 2 2 2

+ã3 < 1, a; 1, b|2, a + b; 11 >< 2, a + b; Here r < 1, +; 1, −|0, 0; 11 >=

1 = − < 1, 0; 1, 0|0, 0; 11 > 3 r

1 2

11 1 1 1 1 < 0, 0; | , ; , − >= 22 2 2 2 2

r

< 1, 1; 1, 0|2, 1; 11 >=

r < 1, +; 1, −|2, 0; 11 >= 31 3 1 1 1 < 2, 0; | , ; , − >= 22 2 2 2 2

r

1 2

1 1 = < 1, 0; 1, 0|2, 0; 11 > 6 2

1 = 2

r

31 3 1 1 1 2 < 2, 1; | , ; , > 3 22 2 2 2 2

Then r

1 ã1 + 6

r < π π |Hw |K >= −

1 ã1 + 2 6

+ 0

√ 3 >= √ ã3 2 2

+ +

0

< π π |Hw |K >= 0 0

0

< π π |Hw |K

+

r

1 ã3 12

r

1 ã3 12

√ Result given in the problem results from redefinitions ã ≡ − 6a1 and ã3 ≡ 1 √ 3a3 . c) Since τK + →π+ π0 >> τK 0 →π+ π− ' τK 0 →π0 π0 we must have |a3 | << |a1 |.

3.2 Isospin Invariance Just as in the case of angular momentum invariance where the Hamiltonian is required to be a scalar under arbitrary rotations, isospin invariance requires that the Hamiltonian be a scalar under rotations in isospin space. Spin and isospin invariance have another similarity in that, since the commutation relations of spin and isospin operators have the same structure [Si , Sj ] = i ijk Sk ,

[Ii , Ij ] = i ijk Ik ,

Building Hadrons from Quarks

25

the irreducible representations must be the same — particles must lie in representations labeled by |I, I3 i with I = 0, 12 , 1, 23 , . . . and −I ≤ I3 ≤ I. In the case of the pion-nucleon interaction, the interaction Lagrangian must be of the form Lint = gπN N N̄ τ · πγ5 N , where τ are the Pauli isospin matrices, π is the isovector pion field and the presence of γ5 is required by the feature that the pion is a pseudoscalar particle. a) Since the nucleon has isospin 12 while the pion has isospin 1, demonstrate that under an infinitesimal isospin rotation by angle δχ i N → N + δχ · τ N 2 π → π − δχ × π . b) From these transformation properties show that the interaction Lagrangian given above is invariant: rot

Lint −→ Lint . c) Using the representations r 1 + (π1 + iπ2 ), π =− 2

r 0

π = π3 ,

−

π =

1 (π1 − iπ2 ) 2

of the pions in terms of their Cartesian components demonstrate that the isospin invariance of Lint requires the pion couplings to nucleons to have the form r r 1 1 − g(npπ ) = − g(pnπ + ) = g(ppπ 0 ) = −g(nnπ 0 ) = gπN N .

2 2 d) Compare the results of c) with the requirements of the Wigner-Eckart theorem

1 1 1 g(Na Nb πc ) = ma 1mc | 1 mb hN ||g||πN i , 2 2 2 where

1 1 1 2 ma 1mc | 2 1 2 mb

is a Clebsch-Gordan coefficient (see Chapter 2).

Of course, unlike angular momentum invariance which is exact, these isospin predictions are expected to be broken at the percent level due to the small differences between the n, p and π + , π 0 , π − masses.

Exercise 3.2 Solution: Isospin Invariance a) We have under rotation N → (1 + iτ · δχ) N

Building Hadrons from Quarks

26

Then πi → πi + ijk δχj πk b) Then N̄ τ · πN → N̄ (1 − iτ · δχ) τ · (π − δχ × π) (1 + iτ · δχ) N = N̄ τ · πN + N̄ (i[τ · π, τ · δχ] + τ · δχ × π) N

= N̄ τ · πN + N̄ τ · i2 π × δχ + π × δχ N = N̄ τ · πN c) Since r 0

τ · π = π τ3 − π

+

1 τ− + π − 2

r

1 τ+ 2

we have r 0

0

gπN N = g(p̄pπ ) = −g(n̄nπ ) = −

1 g(p̄nπ − ) = 2

r

1 g(n̄pπ + ) 2

d) Note that r 1 1 1 1 1 1 1 1 1 1 1 − =< , ; 1, 0| , ; 1 >= − < , − ; 1, 0| , − ; 1 > 3 2 2 2 2 2 2 2 2 2 2 r r 1 1 1 1 1 1 1 1 1 1 1 1 = < , − ; 1, 1| , ; 1 >= − < , ; 1, −1| − , ; 1 > 2 2 2 2 2 2 2 2 2 2 2 2 so completely agrees.

3.3 SU (3) Invariance The near degeneracy of the u, d quarks as well as the n, p and π + , π 0 , π − masses indicate that isospin should be a quite good symmetry, with breaking at the percent level. In the case of SU (3) the symmetry is still useful, with breaking, however, at the ∼20% level, as suggested by the difference between the N, ∆ and ρ, K ∗ masses.

Since the lightest pseudoscalar mesons and spin- 12 baryons are both members of SU (3) octet representations, it is interesting to see what the implications of SU (3) symmetry are both for the masses and the couplings of these particles. In order to study these predictions it is useful to represent these octets in the form q   q 1 0 1 + + π + η π K 2 6 8 8   q q X   − M= λj φj =   π − 12 π 0 + 16 η8 K0   q j=1 1 − 0 K K̄ −2 6 η8  q B=

8 X

  λj Bj =   j=1

1 0 2Σ +

q

Σ− Ξ−

1 0 6Λ

Σ+ −

q

1 0 2Σ +

Ξ0

p q

1 0 6Λ



   n  q −2 16 Λ0

Building Hadrons from Quarks

27

where λj are the Gell-Mann matrices introduced in Chapter 2. Then an SU (3) rotation is given by 8

M0 = UMU† ,

B 0 = U BU †

with U = exp (

iX αj λj ) , 2 j=1

and the simplest possible SU (3) invariants are constructed using the trace Tr(B̄B),

1 Tr(B̄{B, M }), 2

1 Tr(B̄[B, M ]) .

2

That such forms are invariant is clear from the properties of the trace Tr(ABC) → Tr(U AU † U BU † U CU † ) = Tr(ABC) .

Thus the Lagrangian describing the pseudoscalar meson-baryon couplings should have the form X L= gabc Ma B̄b γ5 Bc abc

with gabc =

1 1 DTr(λa λb , λc ) + F Tr(λa [λb , λc ]) = Ddabc + iF fabc , 4 4

where dabc and fabc are the SU (3) structure constants and D and F are the corresponding couplings. a) Use these results to write out the various possible meson-baryon couplings in terms of their SU (3) decompositions, e.g., g(pp̄π 0 ) =

1 (D + F ) 2

1 g(pp̄η 0 ) = − √ (D − 3F ) .

2 3 b) From the experimental results gpp̄π0 ∼ = 13.5

and gpp̄η0 ∼ = −3.9

determine the SU (3) couplings D, F .

To the extent that the mass operator can be written in terms of an SU (3) singlet + plus octet forms, we can then fit the pseudoscalar and 21 baryon masses. c) Show that the masses can then be written in the form 2 2 Mij = M02 δij + MD dij8 + iMF2 fij8 Bij = B 0 δij + BD dij8 + iBF fij8

mesons baryons .

2 , MF2 for mesons and B0 , BD , BF for baryons d) Determine the constants M02 , MD from experimental data.

Building Hadrons from Quarks

28

e) Show that the mass predictions obey the Gell-Mann-Okubo sum rules 1 1 (mN + mΞ ) = (3mΛ + mΣ ) 2 4 1 2 1 (m + + m2K 0 ) = (3m2η + m2π ) 2 K 4 and compare with experiment.

Exercise 3.3 Solution: SU (3) Invariance a) Strong couplings are i 1 1 1 Dd4−i5,4+i5,3 − F f4−i5,4+i5,3 = D + F 2 2 2 2 √ 1 i 1 3 gη0 pp = Dd4−i5,4+i5,8 − F f4−i5,4+i5,8 = − √ D + F 2 2 2 2 3 gπ0 pp =

i 1 1 1 Dd4+i5,4−i5,3 − F f4+i5,4−i5,3 = D − F 2 2 2 2 √ 1 i 1 3 gη0 Ξ− Ξ− = Dd4+i5,4−i5,8 − F f4+i5,4−i5,8 = − √ D − F 2 2 2 2 3 gπ0 Ξ− Ξ− =

gπ0 Σ+ Σ+ = gη0 Σ+ Σ+ =

1 i Dd1−i2,1+i2,3 − F f1−i2,1+i2,3 = F 2 2

1 i 1 Dd1−i2,1+i2,8 − F f1−i2,1+i2,8 = √ D 2 2 3

b) We have D + F = 27.2 D − 3F = 13.5 so, subtracting, have 4F = 40.7, or F = 10.2 and D = 27.2 − 10.2 = 17.0. c) We have in general mB = B0 TrB̄B + BD Tr{B̄, B}λ8 + iBF Tr[B̄, B]λ8 2 m2P = M02 TrM̄ M + MD Tr{M̄ , M }λ8 + iMF2 Tr[M̄ , M ]λ8

so 2 2 Mij = M02 δij + MD dij8 + iMF2 fij8 2 Bij = B0 δij + BD dij8 + iBF fij8

Building Hadrons from Quarks

29

d) Calculate masses via m2K + =

√ 1 2 3 2 1 2 i 1 2 M0 δ4−i5,4+i5 + MD MF d4−i5,4+i5,8 + MF2 f4−i5,4+i5,8 = M02 − √ MD − 2 2 2 2 2 3

1 2 1 2 i 1 2 M δ1−i2,1+i2 + MD d1−i2,1+i2,8 + MF2 f1−i2,1+i2,8 = M02 + √ MD 2 0 2 2 3 √ 1 2 3 2 1 i 2 1 2 2 2 mK 0 = M0 δ6−i7,6+i7 + MD d6−i7,6+i7,8 + MF f6−i7,6+i7,8 = M0 − √ MD − MF 2 2 2 2 2 3 m2π+ =

1 2 2 m2η = M02 δ8,8 + MD d8,8,8 + MF2 f8,8,8 = M02 − √ MD 3 Then M02 =

1 2 (m + m2η ) 2 π

1 MF2 = √ (m2K − m2π ) 3 √ 2 MD = 3(m2η − m2π ) √ 1 1 i 1 3 mp = B0 δ4−i5,4+i5 + BD d4−i5,4+i5,8 + BF f4−i5,4+i5,8 = B0 − √ BD − BF 2 2 2 2 2 3 1 i 1 1 B0 δ1−i2,1+i2 + BD d1−i2,1+i2,8 + BF f1−i2,1+i2,8 = B0 + √ BD 2 2 2 3 √ 1 i 1 3 1 BF mΞ+ = B0 δ4+i5,4−i5 + BD d4+i5,4−i5,8 + BF f4+i5,4−i5,8 = B0 − √ BD + 2 2 2 2 2 3 mΣ+ =

1 mΛ = B0 δ8,8 + BD d8,8,8 + f8,8,8 = B0 − √ BD 3 Then B0 =

1 (mΣ + mΛ ) 2

1 BF = √ (mΞ − mN ) 3 √ BD = 3(mΛ − mN ) e) We have then 1 1 1 (mN + mΞ ) = B0 − √ BD = (3mΛ + mΣ ) 2 4 2 3 1 2 1 1 2 = (3m2η + m2π ) (mK + + m2K 0 ) = M02 − √ MD 2 4 2 3

30

Building Hadrons from Quarks

3.4 SU (4) Representations Just as in the case of SU (3) symmetry, where one looks at rotations among u, d, s quarks, one can look at SU (4) symmetry which involves rotations among u, d, s and c (charm) quarks. Of course, since the c quark is very much heavier than the others, SU (4) should be strongly broken as a dynamical symmetry, but nevertheless its irreducible representations in the meson and baryon sectors should be useful in understanding the shape of the particle spectrum, and that is the goal of this exercise.

In the case of SU (2) we know that the irreducible representations can be given in terms of a line along which the particles having the same total isospin but different values of I3 are given. In the case of SU (3) we require a two dimensional picture in which I3 is given along the x-axis and hypercharge Y , defined through the relationship Q = I3 + Y /2, is plotted along the y-axis, leading to the familiar octet and decuplet representations. For SU (4) we will require a three-dimensional plot with charm plotted along the z-axis and a stacking of SU (3) representaions a) In the case of the pseudoscalar mesons, show that we expect a 15-dimensional representation, and plot the result in terms of the quark content in a threedimensional plot along I3 , Y, C axes. b) Analyze this 15-dimensional representation in terms of its SU (3) content and show that we have {15}SU(4) = (8 + 3 + 3̄ + 1)SU(3)

c) In the case of the spin- 12 baryons show that we expect a 20-dimensional representation, and plot the results in terms of the quark content in a three-dimensional plot along I3 , Y, C axes. d) Analyze this 20-dimensional representation in terms of its SU (3) content and show that we have {20}SU(4) = (8 + 6 + 3 + 3̄)SU(3) e) Compare these representations with what has been observed experimentally.

Exercise 3.4 Solution: SU (4) Representations

a) The pseudoscalar mesons have the structure 1 q̄ i qj − δji q̄ k qk 4 with i, j = 1, 2, 3, 4 so the dimensionality of this representation is d = 4×4−1 = 15. A picture of this reprentation in C, Y, I space can be found in Figure 15.1a of the Quark Model review by C. Amsler, T. DeGrand, and B. Krusche in the Particle Data Group Review of Particle Properties. b) The SU(3) representations which make up the 15-dimensional representation of

Building Hadrons from Quarks

31

SU(4) are 4

X 1 q̄ i qj − δji q̄ k qk , + q̄ i q4 + q̄ 4 qi + q̄ k qk 3 k=1

where i, j = 1, 2, 3 which have the dimensionality 8 + 3 + 3̄ + 1. c) The baryons have the structure qi qj qk jk`m where i, j, k = 1, 2, 3, 4,which has the dimensionality d = 4 × 4×3 2 − 2 × 2 = 20. A picture of this representation in C, Y, I space can be found in Figure 15.4a of the Quark Model review by C. Amsler, T. DeGrand, and B. Krusche in the Particle Data Group Review of Particle Properties. d) The SU(3) representations which make up the 20-dimensional representation of SU(4) are 1 qi qj q` j`n − δin kj` qk qj q` + q4 qi qj + q4 qi qj ijk + q4 q4 qi 3 where i, j = 1, 2, 3 which have the dimensionality 8 + 6 + 3̄ + 3. e) The latest PDG summary lists, in the meson case, C = ±1 mesons D± (1870), D̄0 , D0 (1865), DS± (1969) corresponding to the c̄ quark coupled to u, d, s quarks and the c quark cou¯ s̄ quarks. These are the 3 and 3̄ pieces of the 15-dimensional pled to the ū, d, SU(4) representation. The singlet SU(3) representation is associated with the ηC (2984).

In the case of the baryons the singly charmed particles making up the sex0,+,++ 0,+ tet are the ΣC (2455), the ΞC (2577), and the Ω0C (2696); the anti-triplet +,0 ++,+ + is the ΛC (2287) and the ΞC (2469); the triplet is the ΞCC (3620) and the yet + unseen ΩCC 3.5 Symmetry and Weak Nonleptonic Λ Decay The Λ(1115) has spin- 12 and a lifetime of about 200 ps. Its primary decay mode are into a proton and a negatively charged pion or to a neutron and a neutral pion.

The ∼10−10 second lifetime and the fact that strangeness is changed by one unit indicates that this decay is due to the weak interaction. Hence the decay amplitude will possess a parity-violating as well as a parity-conserving component. a) Show that angular momentum conservation requires that the transition amplitude must be either S-wave or P-wave so that the decay amplitude must have the form M = AS χ†p χΛ + AP χ†p σ · p̂π χΛ , where σ is a Pauli spin matrix and p̂π is a unit vector in the direction of the outgoing pion. b) Evaluate the decay distribution from an unpolarized Λ and show that the distribution is isotropic.

Building Hadrons from Quarks

32

c) Evaluate the decay distribution for the case of decay from a Λ with polarization P n̂, and show that it has the form dΓ ∼ 1 + A1 P n̂ · p̂π dΩ with A1 =

2ReA∗S AP . |AS |2 + |AP |2

d) Show that a nonzero value of A1 implies that pions tend to be emitted either parallel or antiparallel to the Λ spin direction according to whether A1 is positive or negative and show, using mirror arguments, that the existence of such an asymmetry requires a violation of parity invariance. Compare with the explicit form for the asymmetry in terms of AS and AP and comment. e) Evaluate the polarization of the final-state nucleon in the direction n̂ × p̂π for the case of decay from a Λ with polarization P n̂ and show that it is of the form P N · n̂ × p̂π 2ImA∗S AP =P |n̂ × p̂π | . |n̂ × p̂π | |AS |2 + |AP |2 + 2P ReA∗S AP n̂ · p̂π f) Analyze this result from the point of view of simple time reversal invariance arguments and show that one might have expected this polarization to vanish. g) What does the fact that the experimental value for this polarization is nonzero say about the weak decay amplitudes?

Exercise 3.5 Solution: Symmetry and Weak Nonleptonic Λ Decay a) Since initial state has spin 12 , this must be the total angular momentum. Since the nucleon is spin 12 the orbital angular momentum can only be 0 or 1, with amplitudes M = AS χ†p χΛ + AP χ†p σ · p̂π χΛ b) Find decay rate via 1 X 1 |M|2 = Tr|AS + AP σ · p̂π |2 = |AS |2 + |AP |2 2 s ,s 2 p

Λ

c) Find 1 Tr(AS + AP σ · p̂π )(1 + P σ · n̂)(A∗S + A∗P σ · p̂π ) 2 = |AS |2 + |AP |2 + 2ReA∗S AP P n̂ · p̂π d) Define asymmetry via number of pions emitted into the hemisphere parallel to the polarization to those emitted into the opposite hemispherel. Find then

#(n̂ · p̂π > 0) − #(n̂ · p̂π < 0) 2ReA∗S AP A1 = = #(n̂ · p̂π > 0) + #(n̂ · p̂π < 0) |AS |2 + |AP |2

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33

Since spin stays invariant under a parity change while momentum reverses, the correlation n̂ · p̂π is odd under parity. Hence it requires an interference of the opposite parity amplitudes AS and AP . e) Calculate final state nucleon polarization via Pp =

Trσ(AS + AP σ · p̂π )(1 + P σ · n̂)(A∗S + A∗P σ · p̂π ) Tr(AS + AP σ · p̂π )(1 + P σ · n̂)(A∗S + A∗P σ · p̂π ) =

|AS

2iImAS A∗P n̂ × p̂π ∗ 2 P | + 2ReAS AP P n̂ · p̂π

|2 + |A

f) Under time reversal both spin and momentum reverse sign so a triple correlation such as sp · p̂π × sΛ is odd and should vanish. g) Since ImA∗S AP 6= 0, it says that the relative phase between the amplitudes is nonvanishing. By the Fermi-Watson theorem then the pπ − scattering phase in the S- and P-channels must be different and nonvanishing.

3.6 Mesonic States Consider the simple mesonic states which can be constructed from binding a quark and antiquark. a) Write down all of the q q̄ states in the pseudoscalar and vector nonets having good G-parity. b) Using Eq. (3.34), obtain the masses of the pseudoscalar and vector nonets shown in Fig. 3.2, and compare with the physical values (note: the source for the masses is [PDG14], with which the reader should become familiar).

Exercise 3.6 Solution: Mesonic States a) For the low-lying non-strange pseudoscalar mesons we have

1 π+ = √ ud + du 2

1 dd − |uui + dd − |uui π0 = 2 1 − π = − √ (|udi + |dui) , 2 which all have negative G-parity, as discussed in the text, while the non-strange vector mesons discussed there are

1 ρ+ = √ ud − du 2

1 ρ0 = dd − |uui − dd − |uui 2 1 − ρ = √ (|udi − |dui) 2

34

Building Hadrons from Quarks

and have positive G-parity. Here the phase conventions of Close [Clo79] have been adopted. For the pseudoscalar mesons containing one strange quark or antiquark we have analogously 1 K + = √ (|usi + |sui) 2 1 K 0 = √ (|dsi + |sdi) 2 E

1 0 K = −√ sd + ds 2 1 − K = − √ (|usi + |sui) , 2 which all have negative G-parity, and for the vector mesons having one strange quark or antiquark 1 K ∗+ = √ (|usi − |sui) 2 1 ∗0 K = √ (|dsi − |sdi) 2 E

1 ∗0 sd − ds K = −√ 2 1 ∗− K = − √ (|usi − |sui) 2 with positive G-parity. Finally, for the pseudoscalar mesons at the center of the nonet we have

1 η80 = √ |uui + dd − 2 |ssi 2 3

+ |uui + dd − 2 |ssi

1 η00 = √ |uui + dd + |ssi 6

+ |uui + dd + |ssi and for the vector mesons at the center of the nonet

1 ω80 = √ |uui + dd − 2 |ssi 2 3

− |uui + dd − 2 |ssi

1 |uui + dd + |ssi ω00 = √ 6

− |uui + dd + |ssi with G-parities minus and plus, respectively. b) We have for the meson masses from Eq. (3.33) meson M meson = M0meson + Mint

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35

Table 3.1 Pseudoscalar nonet Pseudoscalar meson π± π0 K± 0 K0, K η η0

Computed Value (MeV)

Measured Value (MeV)

% Difference

139.0 139.0 486.8 486.8 772.0 772.0

139.57018 134.9766 493.677 497.614 547.862 957.78

+0.4 -2.9 +1.4 +2.2 -29 +24

with M0meson = m1 + m2 from Eq. (3.31). Equation (3.34) provides a model for the interaction,

ξ −3 pseudoscalar meson Mint = × , +1 vector 4m1 m2 where in the text the following values were proposed for the constituent quark masses and the spin-spin interaction strength: mu ' md ≡ m = 308 MeV ms ≡ m0 = 483 MeV

ξ = 4m2 × 159 MeV.

This results in the values in the table given here, where they are compared with the measured values [PDG14], first for the pseudoscalar nonet of mesons (see Fig. 3.2 in the text). Except for the η and η 0 where mixing occurs (see the discussions in Chapter 3 of the text), the constituent quark model with this simple approach does a very good job of accounting for the measured masses of the low-lying pseudoscalar nonet. For the vector nonet (whose weight diagram is also shown in Fig. 3.2) we find the results in the table and again the model works very well except for the ω and φ where again mixing occurs.

3.7 Baryonic States Consider the simple baryonic states which can be constructed from binding three quarks. a) Using the proton and neutron states given in Eqs. (3.60) and (3.61) together with the charge operator, verify that the correct charges are obtained. b) Using the magnetic dipole operator given in Eq. (3.71) with the baryon octet states given in the text, obtain predictions for the magnetic moments. Compare these results with the physical values given in [PDG14].

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36

Table 3.2 Vector nonet Vector meson

Computed Value (MeV)

Measured Value (MeV)

% Difference

ρ± ρ0 K ∗± ∗0 K ∗0 , K ω φ

775.0 775.0 892.4 892.4 1030.7 1030.7

775.26 775.26 891.66 (895.5) 895.81 782.65 1019.461

+0.03 +0.03 -0.08 (+0.3) +0.4 -24 -1.1

c) Using the mass formulas given in the text, obtain predictions for the masses of both the baryon octet and decuplet (see Fig. 3.4). Compare these results with the physical values given in [PDG14].

Exercise 3.7 Solution: Baryonic States a) The one-body charge operator is given by X Q= q (i) i

with charges q (i), i = 1, 2, 3 equal to +2/3 and −1/3 for u- and d-quarks, respectively. As discussed in the text, since the states involve symmetric configurations, one can set i = 1 and multiply by three in computing Qp = hp ↑ |Q| p ↑i = 3 hp ↑ |q (1)| p ↑i .

Upon substituting Eq. (3.60) for the proton state in terms of the six contributions involving two u quarks and one d quark one gets nine terms to add; these are detailed below:

2 2 1 3 2×2× +2×2× +2×2× − Qp = 18 3 3 3

2 2 1 + 1×1× +1×1× +1×1× − ×2 3 3 3 = 1, as expected. For the neutron one interchanges the factors 2/3 and −1/3 above and gets Qn = 0, also as expected. b) The one-body magnetic dipole operator is handled the same way as the charge operator in part a): X µ= µ (i) σ(i), i

where µ (i) = q (i) · e/2m(i), where q (i) is the quark charge as in part a) with +2/3

Building Hadrons from Quarks

37

Table 3.3 Baryon octet magnetic moments State

Model

Computed Value (µN )

Measured Value (µN )

Difference (µN )

p n Σ+ Σ0 Σ− Ξ0 Ξ− Λ0

µp = 13 (4µu − µd ) µn = 13 (4µd − µu ) µΣ+ = 13 (4µu − µs ) µΣ0 = 13 (2µu + 2µd − µs ) µΣ− = 13 (4µd − µs ) µΞ0 = 13 (4µs − µu ) µΞ− = 31 (4µs − µd ) µΛ 0 = µs

2.59 -1.73 2.49 0.77 -0.96 -1.35 -0.49 -0.58

2.792847356 -1.9130427 2.458 – -1.160 -1.250 -0.6507 -0.613

0.21 -0.19 -0.04 – -0.20 -0.10 -0.16 -0.03

for u and −1/3 for d and s quarks, and m(i) is the quark mass, taken here to be mu ' md ≡ m = 363 MeV for u and d quarks and ms ≡ m0 = 538 MeV for s quarks, as discussed in the text. One has for any of the baryons µ ≡ hB ↑ |µ3 | B ↑i = 3µ3 (1) hB ↑ |σ3 (1)| B ↑i , again using the symmetry of the states. For the states in the baryon octet, starting with the proton, one has µp = 3µ3 (1) hp ↑ |σ3 (1)| p ↑i , again with nine terms as in part a) which are detailed below: µp =

3 {2 × 2 × µu + 2 × 2 × µu + 2 × 2 × (−µd ) 18 +1 × 1 × µu + 1 × 1 × µu + 1 × 1 × µd

+1 × 1 × (−µu ) + 1 × 1 × (−µu ) + 1 × 1 × µd } 1 = (4µu − µd ) , 3 namely the result in Eq. (3.74) in the text. The neutron is easily obtained from this by making the interchange µu ↔ µd : µn =

1 (4µd − µu ) , 3

the result in Eq. (3.75) in the text. The other members of the baryon octet are handled similarly using the configurations given in Chapter 3. The results for the entire multiplet are listed in the accompanying table together with the measured values from [PDG14]. Clearly the trends are good, although the differences presented in the last column show offsets of up to roughly 0.2µN . c) The baryon mass formula was derived in Chapter 3 (see the weight diagrams in Fig. 3.4 for the baryon octet and decuplet). One has baryon M baryon = M0baryon + Mint

Building Hadrons from Quarks

38

with M0baryon = m1 + m2 + m3 , where, as in part b) we use the quark masses m and m0 defined there and in the text. For the baryon decuplet from Eq. (3.80) one has simply

ξ0 M baryon [10] = M0baryon 1 + 4m1 m2 m3 with ξ 0 = 4m2 × 50 MeV, yielding

ξ0 = 3m 1 + 4m3

ξ0 baryon [10] MΣ∗ = (2m + m0 ) 1 + 4m2 m0

ξ0 baryon [10] MΞ∗ = (m + 2m0 ) 1 + 4mm02

ξ0 baryon [10] MΩ = 3m0 1 + .

4m03 baryon [10]

M∆

The baryon octet is found to be somewhat more complicated (see Eqs. (3.81, 3.84, 3.85 and 3.86) in the text): baryon [8]

MN

= 3m −

3ξ 0 4m2

1 1 1 − ξ0 4 m2 mm0 3ξ 0 baryon [8] MΛ = 2m + m0 − 2 4m

1 1 1 baryon [8] = m + 2m0 − ξ0.

MΞ 4 m02 mm0 baryon [8]

MΣ

= 2m + m0 +

From these we have the results given in the accompanying table for the baryon decuplet — remarkably good agreement with the measured values. As noted in the text, the constituent quark model was used to predict the mass and properties of the Ω− ; it was subsequently found experimentally just as predicted, a major triumph for the model. For the low-lying baryon octet we find the results in the table, and again excellent agreement with the measured values is observed.

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39

Table 3.4 Baryon decuplet masses State

Computed Value (MeV)

Measured Value (MeV)

% Deviation

∆ Σ∗+ Σ∗0 Σ∗− Ξ∗0 Ξ∗− Ω−

1239.0 1381.5 1381.5 1381.5 1529.2 1529.2 1682.3

1232 1382.80 1383.7 1387.2 1531.80 1535.0 1672.45

-0.6 +0.1 +0.2 +0.4 +0.2 +0.4 -0.6

Table 3.5 Baryon octet masses State

Computed Value (MeV)

Measured Value (MeV)

% Deviation

p n Σ+ Σ0 Σ− Λ0 Ξ0 Ξ−

939.0 939.0 1179.1 1179.1 1179.1 1114.0 1326.8 1326.8

938.272046 939.565379 1189.37 1192.642 1197.449 1115.683 1314.86 1321.71

-0.08 +0.06 +0.9 +1.2 +1.6 +0.2 -0.9 -0.4

4

The Standard Model 4.1 Z 0 Decays The Z 0 boson decays via the weak neutral current to a pair of fundamental fermions: ¯ ...ν̄τ ντ ūu, dd, a) Show that the decay amplitude of the Z 0 boson to a pair of fundamental spin- 12 fermions is given by h i g2 Amp = i µZ ū(p1 )γµ (I3f + 2 sin2 θW Qf ) + I3f γ5 v(p2 ) , cos θW where I3f is the SU (2)L eigenvalue of the fermion, Qf is the fermion charge, and θW is the weak mixing angle. b) Calculate the decay width for each quark and lepton in the approximation that mf << MZ 0 . c) Compare your results with the results from the particle data book [PDG14], Γ(ēe) = Γ(µ̄µ) = Γ(τ̄ τ ) ' 84 MeV Γ(ūu) = Γ(c̄c) ' 300 MeV ¯ = Γ(s̄s)) = Γ(b̄b) ' 380 MeV Γ(dd) d) Calculate the width due to (unseen) neutrino modes.

Exercise 4.1 Solution: Z 0 Decays Consider Z 0 decay to fermion pair. a) From Eq. 4.38 we have L=

h i g µZ ū(p1 )γµ I3f (1 − γ5 ) − 2 sin2 θW Qf v(p2 ) 2 cos θW h i g f ≡ µZ ū(p1 )γµ gVf − gA γ5 v(p2 ) 2 cos θW

f where gVf = I3f − 2 sin2 θW Qf and gA = I3f b) Calculate the differential decay rate

2 Z Nc d3 p1 d3 p2 g 4 4 ¯ dΓ(f f ) = (2π) δ (PZ − p1 − p2 ) 2mZ (2π)3 2E1 (2π)3 2E2 2 cos θW

40

The Standard Model

41

×

1 3

−η µν +

(p1 + p2 )µ (p1 + p2 )ν m2Z

g2 dΩ = Nc 96π 2 cos2 θW

f2 gVf 2 + gA 2mZ

! −η

f f Trγµ (gVf − gA γ5 ) 6 p2 γν (gVf − gA γ5 ) 6 p 1

µν

(p1 + p2 )µ (p1 + p2 )ν + m2Z

(p1µ p2ν + p1ν p2µ − ηµν p1 · p2 )

2

f2 2 g 2 mZ (gVf 2 + gA ) dΩ g2 gV + gA 2p1 · p2 = 2 2 2 96π cos θW 2mZ 6 cos θW √ 2 Using 2 cosg2 θW = 2 2GF m2Z we have = Nc

GF m3 f2 f2 Γ(f¯f ) = Nc √ Z (gVf 2 + gA ) ' (gVf 2 + gA ) × 330 MeV 6 2π c) Now look at various channels: i) e, µ, τ : Nc = 1; gV = − 21 (1 − 4 sin2 θW ); gA = 21 . Then

Γ(ēe) = Γ(µ̄µ) = Γ(τ̄ τ ) ' 83 MeV 1 + (1 − 4 sin2 θW )2 ' 83 MeV ii) u, c: Nc = 3; gV = 21 − 43 sin2 θW ; gA = 12 . Then

8 Γ(ūu) = Γ(c̄c) =' 250 MeV 1 + (1 − sin2 θW )2 ' 290 MeV 3 Nc = 3; gV = − 21 + 32 sin2 θW ; gA = − 12 . Then

4 2 2 ¯ Γ(dd) = Γ(s̄s) = Γ(b̄b) ' 250 MeV 1 + (1 − sin θW ) ' 370 MeV 3

iii) d, s, b:

d) For neutrino modes we have Nc = 1; gV = gA = 21 . Then width is Γ(ν̄e νe ) = Γ(ν̄µ νµ ) = Γ(ν̄τ ντ ) = 83 MeV(1 + 1) = 166 M eV so the total invisible width is Γ(inv.) = 3 × 166 Mev = 498 MeV

4.2 Yang-Mills Theory The Dirac Lagrangian for a spin- 12 particle of mass m L[ψ(x)] = ψ̄(x) [i 6 ∂ − m] ψ(x) is invariant under the U (1) global phase transformation ψ(x) → ψ 0 (x) = exp(iβ)ψ(x) . a) Demonstrate this invariance.

The Standard Model

42

It is also well known that the Dirac Lagrangian for a spin- 12 particle of mass m and charge e can be made invariant under a local U (1) phase transformation ψ(x) → ψ 0 (x) = exp(ieβ(x))ψ(x) ≡ U ψ(x). b) Demonstrate that the Lagrangian L[ψ(x)] is not invariant under a local U (1) transformation.

The invariance can be accomplished if we introduce an additional dynamical vector field Aµ (x) with the transformation property i Aµ (x) → U Aµ (x)U −1 − U ∂µ U −1 = Aµ (x) − ∂µ β(x) e and modify the Lagrangian to become 1 L[ψ(x), Aµ (x)] = ψ̄(x) [i D 6 − m] ψ(x) − Fµν (x)F µν (x), 4 where Dµ = ∂µ + ieAµ (x) is the covariant derivative and has the property Dµ ψ(x) → Dµ0 ψ 0 (x) = U Dµ ψ(x). c) Verify this transformation property of the covariant derivative. d) Demonstrate the invariance of L[ψ(x), Aµ (x)] under a local U (1) transformation.

Yang and Mills showed how to generalize this local invariance to the case of nonAbelian transformations. We consider the case of SU (2) with a doublet of spin- 21 fields

ψ1 (x) N (x) = ψ2 (x) which transforms as 1 N (x) → N 0 (x) = exp(ig τ · β(x))N (x) ≡ U N (x) 2 under a local SU (2) gauge transformation, where τ are the Pauli isospin matrices and satisfy the commutation relations [τi , τj ] = 2i ijk τk . e) Demonstrate that the Lagrangian L[N (x)] is not invariant under a local SU (2) gauge transformation.

Now define a triplet of vector fields Aµ (x) which transform as 1 1 τ · Aµ (x) → U τ · Aµ (x)U −1 − iU ∂µ U −1 2 2 and a covariant derivative 1 Dµ = ∂µ + ig τ · A(x).

2

The Standard Model

43

f) Demonstrate that the covariant derivative has the property Dµ N (x) → Dµ0 N 0 (x) = U Dµ ψ(x) , . where 1 U = exp(i τ · β(x)) 2 g) For an infinitesimal δβ show that Aµ (x) → Aµ (x) − ∂µ δβ(x) + gδβ(x) × Aµ (x). h) Demonstrate that the Lagrangian 1 L[N (x), A(x)](x) = N̄ (x) [i D 6 − m] N (x) − F µν (x) · F µν (x) 4 is invariant under a local SU (2) gauge transformation provided that F µν (x) is defined via 1 [Dµ , Dν ]N (x) = −ig τ · F µν (x)N (x) 2 i.e., F µν (x) = ∂µ Aν (x) − ∂ν Aµ (x) − gAµ (x) × Aν (x) and has the transformation property 1 1 τ · F µν → U τ · F µν U −1 .

2 2

Exercise 4.2 Solution: Yang-Mills Theory a) If ψ(x) → ψ 0 (x) = U ψ(x) with U = exp(iβ) then ψ̄(x) → ψ̄ 0 (x) = ψ̄(x)U −1 with U −1 = U † = exp(−iβ) so L = ψ̄(x)(i 6 ∂ − m)ψ(x) → ψ̄(x)U −1 (i 6 ∂ − m)U ψ(x) = L b) If β → eβ(x) then L = ψ̄(x)(i 6 ∂ − m)ψ(x) → ψ̄(x)U −1 (i 6 ∂ − m)U ψ(x) = L − eψ̄(x)U −1 (6 ∂β(x))U ψ(x) = L − eψ̄(x)(6 ∂β(x))ψ(x) c) Define iDµ = i∂µ − eAµ so iDµ ψ(x) = (i∂µ − eAµ (x))ψ(x) → [i∂µ − eAµ (x) + e∂µ β(x)) U ψ(x) = U (i∂µ − eAµ − e∂µ β(x) + e∂µ β(x)) ψ(x) = U iDµ ψ(x)

The Standard Model

44

d) Then, since Fµν → Fµν − e∂µ ∂ν β(x) + e∂ν ∂µ β(x) = Fmuν we have 1 1 L = ψ̄(x)(i D 6 −m)ψ(x)− Fµν F µν → ψ̄(x)U −1 U (i D 6 −m)ψ(x)− Fµν F µν = L 4 4 e) We find 1 N̄ (i D 6 − m)N → N̄ U −1 (i D 6 − m)U N = N̄ (i D 6 − m)N − g N̄ τ · ∂µ βN 2 f) Then for covariant derivative iDµ N (x) = (i∂µ − gτ · Aµ (x))N (x) → (i∂µ − gτ · Aµ (x) + τ · ∂µ β) U N (x)

1 1 1 = U i∂µ − g τ · Aµ (x) + τ · ∂µ β − τ · ∂µ β N (x) = U iDµ N (x) 2 2 2 g) Since 1 1 τ · Aµ → U τ · Aµ U −1 − iU ∂µ U −1 2 2 we have 1 1 1 1 1 τ · Aµ → τ · Aµ + ig[ τ · δβ, τ · Aµ ] + gτ · ∂µ δβ 2 2 2 2 2 or Aµ → Aµ − gδβ × Aµ + g∂µ δβ h) First look at 1 1 1 1 − F µν · F µν = − Tr τ · F µν · τ · F µν 4 2 2 2 1 1 1 1 → − TrU τ · F µν U −1 · U τ · F µν U −1 = − F µν · F µν 2 2 2 4 Then 1 1 L = N̄ (i D 6 − m)N − F µν · F µν → N̄ U −1 U (i D 6 − m)N − F µν · F µν = L 4 4 Check: 1 1 1 [Dµ , Dν ]N = −ig∂µ τ · Aν + ig∂ν τ · Aµ + ig 2 τ · Aµ × Aν = −igF µν 2 2 2 with F µν = ∂µ Aν − ∂ν Aµ − gAµ × Aν Also, for infinitesimal transformation 1 1 1 1 1 τ · F µν → U τ · F µν U −1 = τ · F µν − ig[ τ · δβ, τ · F µν ] 2 2 2 2 2 =

1 1 τ · F µν + g τ · δβ × F µν 2 2

The Standard Model

45

4.3 Neutrino Oscillations with Two Flavors Consider the standard analysis of neutrino oscillations supposing that there are only two types of neutrinos. Production of these neutrinos in a weak interaction emphasizes the flavor states, νa and νb , while the neutrino states with definite mass are ν1 and ν2 . These two pairs of states are related by a 2 × 2 unitary matrix with a single parameter, the mixing angle θ12 , via

ψa cos θ12 sin θ12 ψ1 = , ψb − sin θ12 cos θ12 ψ2 which implies the states a and b can transform back and forth between each other.

Consider plane-wave states of neutrinos 1 and 2 that have well-defined energies Ei and momenta Pi large compared with their rest masses mi , which propagate essentially at the speed of light. a) For propagation along the x-axis, show that the momenta Pi are

m2 c4 Ei 1− i 2 , Pi ≈ c 2Ei and therefore 2 3

ψi (x, t) = ψi,0 ei(Pi x−Ei t)/~ ≈ ψi,0 eEi (x/c−t)/~ e−imi c x/2Ei ~ . b) The first phase factor can be ignored to write 2 3

ψi (x, t) ≈ ψi,0 e−imi c x/2Ei ~ , where x ≈ ct. This exponential phase factor is slightly different for mass states 1 and 2, which leads to an oscillatory interference term in the spatial/time dependence of an initial single-flavor state. Consider a neutrino created at time t = 0 in a pure flavor state a with ψa,0 = 1, ψb,0 = 0. Find the initial mass states ψ1,0 and ψ2,0 . c) Determine the evolution of flavor states ψa (x) and ψb (x). d) Show that the probability that the initial state flavor a is still a after the neutrino has traveled a distance x is given by Pa→a (x) = 1 − sin2 2θ12 sin2

∆m212 x , 4E

1 2 where 2 E 2= 2 (E1 + E2 ) is the average energy of the two neutrinos and ∆m12 = m1 m2 E1 − E2 E.

e) Explain how neutrino oscillation between different mass eigenstates is consistent with conservation of energy.

Exercise 4.3 Solution: Neutrino Oscillations with Two Flavors We begin by taking the energy to be constant in order that the time development of the system be the same for all components.

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46

a) For a state of mass mi , since the momentum is given in terms of the energy by r q m2 m2 2 2 Pi = E − mi = E 1 − 2i ' E − i + . . .

E 2E the wavefunction must have the form m2 ix

ψ(xt, x) ∼ ei(P x−Et) = eiE(x−t) e−i 2E b) Since

ψ1 ψ2

c −s s c

ψa (0) ψb (0)

1 0

ψ1 (0) ψ2 (0)

c s

=

ψa ψb

the initial state

=

corresponds to

=

where we have defined c = cos θ12 and s = sin θ12 . c) The time evolved state is     m2 m2

−i 2E1 x −i 2E1 x ce ψ1 (t) ψ (0)e 1 iE(x−t) e  eiE(x−t) = = m2 m2 ψ2 (t) −i 2E2 x −i 2E2 x ψ (0)e se 2

Now project back to the flavor basis via

ψa c = ψb −s

s c

ψ1 ψ2

and find

ψa (t, x) ψb (t, x)

=

c −s

s c

 m2 −i 2E1 x ce  eiE(x−t)  m2 −i 2E2 x se 



 m2 m2 2 −i 2E1 x 2 −i 2E2 x c e + s e  eiE(x−t) = m2 m2 2 2 −cse−i 2E x + cse−i 2E x d) Since m2 1

m2 2

2

Pab (t) = c2 s2 e−i 2E x − e−i 2E x 2 2

= 4c s e

−i

2 m2 1 +m2 x 4E

= sin2 2θ12 sin2

1 2i

2 m2 −m2 m2 +m2 −i 14E 2 x i 14E 2 x e −e

∆m2 ∆m2 x ' sin2 2θ12 sin2 t 4E 4E

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47

by unitarity we have Paa (t) = 1 − Pab (t) = 1 − sin2 θ12 sin2

∆m2 t 4E

e) In this picture energy of the two states is assumed to be identical, so no problem with conservation.

4.4 Neutrino Oscillations Using QFT

ΦD,β (k )

ΦP,α (k)

Uα,i

Uβ,i νi (q)

ΦP,α (p)

ΦD,β (p )

In this exercise, we consider neutrino mixing from a diagrammatic perspective.

Consider the case that a nucleus, represented by the wavefunction ΦP,α (p) makes a transition to a final state ΦP,α (k) by emitting a virtual massive neutrino νi . The neutrino propagates and then interacts with a second nucleus, located at some distance L after some time T such that ΦD,β (p0 ) + νi → ΦD,β (k 0 ). Thus p(p0 ) k(k 0 ) represents the production (detection) momentum, and we consider the coupling term at each vertex to have the form Uα,i and Uβ,i respectively. a) Using the function Φ to represent the initial and final wavefunction for the production and detection nucleus, write out the transition amplitude for the reaction assuming that there exist n neutrinos, each with a distinct mass mi , which participate in the interaction. Assume the momentum transfer of the interaction is given by the four-vector q. One can take into account the fact that the production and detection locations are spatially separated by explicitly considering the plane-wave term of the wavefunction e−iq·x , where x = (tP − tD , xP − xD ) = (T, L) represents the spacetime separation of the two interaction points. Show, by rewriting the energy-momentum conservation delta function in terms of its Fourier transform, how the e−iq·x emerges. b) Using complex integration methods show that √ Z AL i√A|L| dp φ(q)eip·L 1 φ( )e . = − (2π)3 A − p2 4π|L| |L|

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48

c) Using the integral from b) and the completeness of neutrino spinors X usj ūsj =6 p + mj s

extract the amplitude and probability for the exchange to occur. Show that the phase term, when expanded in terms of the mass, is consistent with the usual oscillation form.1

Exercise 4.4 Solution: Neutrino Oscillations Using QFT First it is useful to review the conventional way to look at this problem, without wavefunctions— Z Z Z 0 0 1 d4 q iq·(x−y) e Amp = d4 xei(k−p)·x d4 yei(k −p )·y (2π)4 q 2 − m2 Z =

d4 q 1 (2π)4 δ 4 (k − p + q)(2π)4 δ 4 (k 0 − p0 − q) (2π)4 q 2 − m2 = (2π)4 δ 4 (k + k 0 − p − p0 )

1 (k − p)2 − m2

a) Now put in the wavefunctions and couplings. We have Z Z Z 4 i(k−p)·x 4 i(k0 −p0 )·y Amp = d xe ΦP,α Uα,i (x) d ye ΦD,β (y)Uβ,i

1 d4 q iq·(x−y) e 4 2 (2π) q − m2i

Perform the time integrations, yielding Amp = 2πδ(Ek + Ek0 − Ep − Ep0 ) ×

Z

3

i(k0 −p0 )·y

d ye

Z ΦD,β (y)Uβ,i

Z

d3 xe−i(k−p)·x ΦP,α (x)Uα,i

d3 q −iq·(x−y) 1 e (2π)3 q02 − q 2 − m2i

where q0 = Ek − Ep = Ep0 − Ek0 b) We need Z d3 q −iq·(x−y) 1 I(x − y) = e 2 3 (2π) q0 − q 2 − m2i Note that for a neutrino of energy q0 its momentum is |pi | = have Z d3 q −iq·(x−y) 1 I(x − y) = − e (2π)3 q 2 − p2i 1

p

q02 − m2i so we

You may find that your expression remains a function of the nuclear wavefunctions. These functions disappear once the proper normalization is applied. You are free to treat them as a constant for this problem.

The Standard Model

49

which gives the simple Yukawa form I(x − y) =

1 ei|pi ||x−y| 4π|x − y|

c) We have then Ampi = 2πδ(Ek + Ek0 − Ep − Ep0 ) ×

Z

0

0

Z

d3 xe−i(k−p)·x ΦP,α (x)Uα,i

d3 yei(k −p )·y ΦD,β (y)Uβ,i

1 ei|pi ||x−y| 4π|x − y|

' 2πδ(Ek + Ek0 − Ep − Ep0 )ΦP,α (k − p)Uα,i ×ΦD,β (k0 − p0 )Uβ,i

1 ei|pi ||x−y| 4π|x − y|

Now approximate |pi | = Eν −

m2i 2Eν

and sum over neutrino species Amp = 2πδ(Ek + Ek0 − Ep − Ep0 )ΦP,α (k − p)ΦD,β (k0 − p0 ) ×

X

1 eiEν |x−y| 4π|x − y|

m2 i

Uα,i Uβ,i e−i 2Eν |x−y|

i

In the two component mixing case we have

1 (m21 − m22 )|x − y| 2 2 2 2 Prob ∼ Uα,1 Uβ,1 + Uα,2 Uβ,2 + 2Uα,1 Uβ,1 Uα,2 Uβ,2 cos |x − y|2 2Eν Comparing to the massless neutrino case we have Prob − Probm=0 ∼ −

1 (m21 − m22 )|x − y| 4Uα,1 Uβ,1 Uα,2 Uβ,2 sin2 2 |x − y| 4Eν

4.5 Spontaneous Symmetry Breaking: Meissner Effect for Scalar Electrodynamics The Meissner effect refers to the exclusion of a magnetic field from the interior of a superconductor, and will be explored in this exercise.

The Lagrangian density for the classical electrodynamics of charged spinless particles is given by 1 L = − Fµν F µν + (Dµ φ)∗ (Dµ φ) − V (φ) 4

The Standard Model

50

with Dµ = ∂µ − ieAµ being the covariant derivative and V (φ) =

1 2 ∗ λ m (φ φ) + (φ∗ φ)2 2 4

representing the potential energy. Here φ is a complex scalar field. a) Using Noether’s theorem, identity the electromagnetic current of the φ field. b) Find the Hamiltonian density. c) If m2 > 0 show that the ground state of the system has φ = 0, Aµ = 0. (In this case the theory is that of normal electrodynamics.) d) There can exist another phase of the theory if the parameter m2 = µ2 < 0.

In this case show that the lowest energy state has Aµ = 0, φ = const. and calculate φ. What is the mass of the photon? This phase has electromagnetic fields being screened. To see this calculate the potential between two static point charges, ρ(x) = Qδ 3 (x), using Green’s function techniques (as would be done in Jackson [Jac75]), but including the photon mass in the Green’s function equation. What sets the scale of the screening distance? e) Now add an external field to the problem. One way to do this is to add an external current to generate the field by including a new term in the Lagrangian µ L → L0 − Aµ Jext

with µ νµ Jext = ∂µ Fext

If the fields vanish at infinity this is equivalent to 1 µν .

L0 + Fµν Fext 2 In order to see that this does act like an applied field, show that if one disregards µν the scalar field the equations of motion require F µν = Fext f) Show that there exist two simple solutions to the equations of motion in the presence of a constant applied field µν i) φ = 0, F µν = Fext µν ii) φ = const, F = 0

where const is the value of φ calculated in part d) above. Again, these correspond to unscreened and screened phases for the electromagnetic field. µν g) Calculate the energy of the two phases when Fext describes a constant magnetic field. Show that solution ii) is preferred for B < Bcritical and that i) has the lower energy for B > Bcritical . Then, there exists a phase transition from the screened to the unscreened phase as one increases the external magnetic field— this is the simple analogy of the Meissner effect.

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51

Suggestions: In a) and b) remember that φ is a complex scalar field, i.e., φ 6= φ∗ , and that it has two degrees of freedom. These can be chosen as φ = φ1 + iφ2 , with φ1 , φ2 being two real fields. However, it is probably easier (and more common) to treat φ and φ∗ in applying Lagrangian techniques.

The photon mass can be obtained from the equation of motion when φ = const.

In e),f) and g) continue to assume that m2 = −µ2 < 0. µν It is best to work with the second form of the source term, 12 Fµν Fext .

In g), do not forget to include the source term in the Hamiltonian density.

Exercise 4.5 Solution: Spontaneous Symmetry Breaking: Meissner Effect for Scalar Electrodynamics a) Lagrangian is 1 L = − Fµν F µν + Dµ φ∗ Dµ φ − V (φ) 4 Invariance under φ → e−iχ φ ' φ − iχφ means Noether current is jµ =

δL ∗ δL φ + φ = φ∗ Dµ φ − (Dµ φ)∗ φ δ∂µ φ∗ δ∂µ φ

b) We have δL = D0 φ∗ , δ φ̇

δL = D0 φ δ φ̇∗

δL = Fi0 δ Ȧi

δL =0 δ Ȧ0

so, using Ȧ = E + ∇A0 and − 14 Fµν F µν = 21 (B 2 − E)2 ) 1 H = D0 φ∗ φ̇+φ̇∗ D0 φ−D0 φ∗ D0 φ+Dφ∗ ·Dφ+V (φ)+E·(E+∇A0 )+ (B 2 −E)2 ) 2 = ∂0 φ∗ φ0 φ + ieA0 (φ∗ ∂0 φ − (∂0 φ)∗ φ) − ieA0 (φ∗ ∂0 φ − (∂0 φ)∗ φ) − e2 A20 φ∗ φ 1 1 1 +(Dφ)∗ · Dφ + (B 2 + E)2 ) − A0 ρ + m2 φ∗ φ + λ(φ∗ φ)2 2 2 4 where we have used ∇ · E = ρ. Hamiltonian is then

Z 1 H = d3 x ∂0 φ∗ φ0 φ − e2 A20 φ∗ φ + (Dφ)∗ · Dφ + (B 2 + E)2 ) 2 1 1 −A0 ρ + m2 φ∗ φ + λ(φ∗ φ)2 2 4

c) Note if φ = 0, Aµ = 0 we have H = 0, which is a minimum.

The Standard Model

52

d) Set m2 = −µ2 . With Aµ = 0 and static field minimum requires

d 1 2 1 1 2 λ µ2 − µ x + λx = x − µ2 = 0 i.e. x = φ∗ φ = dx 2 4 2 2 λ so Z H=

1 µ2 1 d x − µ2 · + λ 2 λ 4 3

µ2 λ

2 ! =−

Z

d3 x

1 µ4 4 λ

Piece relevant for Green’s function is

Z 2 1 3 2µ 2 H= d x ∇A0 · ∇A0 − e A − ρA0 2 λ 0 and equation of motion is δH δH ∇· − = δ(∇A0 ) δA0

2 2µ

2

∇ − 2e

λ

A0 − ρ = 0

Write ρ = Qδ 3 (x) = Q

Z

d3 k ik·x e (2π)3

so

2

− k + 2e

2 2µ

λ

A0 (k) = Q

and A0 (k) =

2

−Q

2

k + 2e2 µλ

Then Z A0 (x) =

d3 k −ik·x −Q −κ|x| −Q e e 2 = 3 2 µ 2 (2π) 4π|x| k + 2e

with κ2 = 2e2

λ

µ2 λ2

e) We have, disregarding the scalar field, 1 1 µν ext 1 µν ext ext = − (F µν − Fext )(Fµν − Fµν ) + Fext Fµν L −→ L + F µν Fµν 2 4 4 µν ext Here 41 Fext Fµν is a constant and can be discarded, so

1 µν ext L = − (F µν − Fext )(Fµν − Fµν ) 4 f) Pick static solution, so φ̇ = Ȧµ = 0 and Z H = d3 x ∇φ∗ · ∇φ − ieA · ∇φ∗ + ieA · ∇φ − e2 A20 φ∗ φ + e2 A2 φ∗ φ 1 1 1 1 − µ2 φ∗ φ + λ(φ∗ φ)2 + (∇A0 − E ext )2 + (∇ × A − B ext )2 2 4 2 2

The Standard Model

53

Now let φ → φ + δφ, A → A + δA, A0 → A0 + δA0 so Z H −→ H + d3 x [∇φ∗ · ∇δφ + ∇δφ∗ · ∇φ − ieδA · ∇φ∗ −ieA · δ∇φ∗ + ieδA · ∇φ + ieA · δ∇φ − 2e2 δA0 A)0φ∗ φ 1 +2e2 A · δAφ∗ φ − e2 (A20 − A2 )(φ∗ δφ + δφ∗ φ) − µ2 (δφ∗ φ + φ∗ δφ) 2 1 + λφ∗ φ(δφ∗ φ + φ∗ δφ) − ∇δA0 · (−∇A0 − E ext ) + ∇ × δA · (∇ × A − B ext )2 2 We have then δH = −ie∇φ∗ + ie∇φ + 2e2 Aφ∗ φ + ∇ × (∇ × A − B ext ) = 0 δA δH = −2e2 A0 φ∗ φ − ∇ · (∇A0 − E ext ) = 0 δA0 1 1 δH = −∇2 φ + ie∇ · A − e2 (A20 − A2 )φ − µ2 φ + λφ∗ φφ = 0 δφ∗ 2 2

Solution a) has φ = 0 so B = B ext and E = E ext .

2

Solution b) has φ 6= 0 so A0 = A = 0 and φ∗ φ = µλ = 0

R 4 g) Solution a) has Ha = 0, while solution b) has Hb = d3 x − 41 µλ + 12 B 2ext so 4

Hb < Ha until B 2ext > 12 µλ .

5

QCD and Confinement 5.1 Quark-Antiquark Interaction in QCD Consider a quark and antiquark interacting via gluon exchange, as shown in the figure.

3

4

1

2

a) Calculate the interaction amplitude M and show that it has the form g 2 fC M = − s 2 ū3 γ µ u1 v̄4 γµ v2 , q

where fC =

1 † a † a c λ c1 c2 λ c4 4 3

is a color factor.

Note that the amplitude has the same form as the QED amplitude for e+ e− → µ+ µ− scattering except for the color factor fC . b) Show that if the quark-antiquark are in an octet color configuration fC = − 61 , while if they are in a color singlet state fC = 43 . c) If the quark-antiquark are close enough to justify the use of perturbation theory show that we have the potential 4αs

− 3r color singlet Vqq̄ (r) = . + α6rs color triplet We see then that the quark-antiquark interaction is attractive (repulsive) in the color singlet (triplet) configuration, which explains why only color singlet mesons are found.

54

QCD and Confinement

55

However, this form of the potential is only valid at very short distances. At larger separations it is necessary to take gluon-gluon interactions into account. In QED the interaction between a pair of particles, one with charge +e and one with charge −e has the familiar dipole form, as shown in the left-hand side of the figure below.

On the other hand, in QCD the strong gluon-gluon interaction squeezes the lines of force together into a flux tube as shown in the right-hand side of the figure below.

e

q

e

q

d) Show that the existence of a flux tube implies the quark-antiquark potential of the form 4 αs + λr , V (r) = − 3 r as shown in the figure below.

1

αs VQCD =− 4 + λr 3 r

V (r) GeV 0

VQED = − α r

−1

0.5

r fm

1

1.5

It is this flux tube which leads to confinement. The linear potential imples that it requires an infinite amount of energy to completely separate the quark and antiquark, i.e., they are confined. Rather, as the quark and antiquark separate they produce additional q q̄ pairs.

QCD and Confinement

56

Exercise 5.1 Solution: Quark-Antiquark Interaction in QCD) Consider gluon exchange between quarks and anti-quarks. a) Then g 2 fC Amp = − s 2 ūi3 γ µ uj1 v̄4k γµ v2` q where 8

fC = b) Color singlet state is

1X a a λ λ 4 a=1 ij k`

q P 3 1 3

c=1 |c̄c > We have then 8

3

1 XX fC = δjk δi` λaij λak` 12 a=1 i,j=1 3

=

8

1 XX a a λ λ 3 i,j=1 a=1 ij ji

However, 8 X

λaij λaj` = CF δi`

a=1

where CF = 1 + 1 + 1 + 1 + 1 +

16 1 = 3 3

Then fC =

1 i` 1 4 δ δi` CA = CF = 12 4 3

On the other we can use the color q hand if we have a color triplet state then P 8 1 octet state 2 (|11̄ > −|22̄ >). Then, defining S(ijk`) = a=1 λaij λak` we have 1 (S(1111) + S(2222) − S(1221) − S(2112)) 8

1 1 1 1 = 1+ +1+ −1−1−1−1 =− 8 3 3 6

fC =

c) In the low energy limit we have q0 → 0 so Z d3 q iq·r gs2 fC † V (r) = − e χ χ1 χ†4 χ2 (2π)3 q2 3 g 2 fC = − s χ†3 χ1 χ†4 χ2 4πr

57

QCD and Confinement

d) Since energy of flux tube is proportional to the length, the effective long-distance potential must have the form Vef f (r) = −

4gS2 + λr 12πr

5.2 Anomalous Symmetry Breaking Anomalous symmetry breaking is said to occur when the process of quantization destroys a symmetry which was present in the classical theory. This phenomenon generally occurs in quantum field theory, where the trace anomaly, chiral anomaly, etc., are well-known manifestations. However, there exist two examples of anomalous symmetry breaking in ordinary quantum mechanics: the two-dimensional delta function potential and the 1/r2 potential. In this exercise we explore the former.

Note that this is not the familiar delta function potential bound state which is solved in many quantum mechanics courses and is not anomalous. Rather, we deal here with a two-dimensional delta function. It is in two dimensions that classical scale invariance is valid and for which a quantum mechanical anomaly arises, as studied below.

Consider a particle of mass m moving in two dimensions under the influence of a potential V (r) = −λδ 2 (r) .

The time-independent Schrödinger equation then reads

1 2 2 ∇ − λδ (r) ψ(r) = Eψ(r) . − 2m a) Show that the Schrödinger equation is invariant under a scale transformation r → r/ζ

and E → ζ 2 E .

Such an invariance means that there can exist no bound state, since any solution ψE (r) with energy E must have a sister solution ψζ 2 E (r/ζ), implying that if there exists one negative energy bound state solution then there must exist a continuum of solutions with energies ranging from zero to negative infinity. Furthermore, because the system is not bounded from below, a particle which finds itself in a negativeenergy state would cascade wildly to negative infinity, releasing unlimited amounts of energy in the process, so that there can exist no such bound (negative-energy) state. However, this exercise will show that such a state does exist. We begin by finding a positive energy (scattering) solution of the Schrödinger equation ψk+ (r). b) Demonstrate that in momentum space the Schrödinger equation becomes k2 + p2 + + φE (p) − λψE (0) = φ (p) , 2m 2m E

QCD and Confinement

58

where φ+ k (p) =

Z

dre−ip·r ψk+ (r)

is the momentum space wavefunction. c) Solve this equation and show that 2 2 φ+ k (p) = (2π) δ (p − k) +

2mλψk+ (0) . p2 − k2 − i

d) Demonstrate that ψk+ (0) =

1 , 1 − λG+ (0)

where +

dp θ(Λ2 − p2 ) , (2π)2 p2 − k2 − i

Z

G (0) = 2m

and we have been forced to regularize the Green’s function by use of a cutoff Λ, since otherwise there would exist a divergence at large p. e) Solve for the Green’s function and show that G+ reg (0) = f) Using Z 2m

m Λ2 . log π −k2

dp eip·r im 1 = H (kr) (2π)2 p2 − k2 − i 2 0

show that the scattering solution can be written as 2 −1

1 Λ 1 i 1 ik·r − log H (kr) . ψk (r) = e + 2mλ 4π −k 2 4 0 Here J0 (kr), N0 (kr) are the regular, irregular Bessel functions of order zero, (1) while H0 (kr) = J0 (kr) + iN0 (kr) is the related circular Hankel function. g) Find the scattering amplitude and show that r

−1 1 1 1 Λ2 i f (θ) = − log 2 − .

2πk mλ 2π k 2 h) Making the analytic continuation k → iκ, show that f (θ) has a pole at energy Eb.s. = −

Λ2 2π exp(− ), 2m mλ

which corresponds to a bound state.

The existence of this bound state shows that the scale invariance has been broken — there exists an anomaly. The reason for the anomaly is seen to be the breaking of scale invariance due to introduction of a cutoff in order to define the Green’s function, and is necessary in order to regularize the short-distance behavior of the

QCD and Confinement

59

theory.

Exercise 5.2 Solution: Anomalous Symmetry Breaking a) Begin with the time-independent Schrödinger equation

∇2 − − λδ 2 (r) ψ(r) = Eψ(r) 2m and note that under the scaling r → r/ζ we have ∇ → ζ∇ and δ 2 (r) → ζ 2 δ(r) so that

∇2 ∇2 2 2 2 − − λδ (r) → ζ − − λδ (r) 2m 2m Hence TISE is invariant provided that E → ζ 2 E. b) Taking Fourier transform RHS becomes Z k2 + φ (p) RHS → E d2 re−ip·r ψk+ (r) = 2m k while LHS becomes LHS → −

∇2 2m

Z

d2 re−ip·r ψk+ (r) − λ =

Z

d2 re−ip·r ψk+ (r)δ 2 (r)

p2 + φ (p) − λψk+ (0) 2m k

c) Homogeneous solution is δ 2 (p − k), while particular solution of b) is φ+ k (p) = − k2

λψk+ (0)

p2 2m − 2m + i

so general solution is 2 2 φ+ k (p) = (2π) δ (p − k) − k2

λψk+ (0)

p2 2m − 2m + i

d) Solve c) for ψk+ (0) ψk+ (0) =

Z

d2 p + φ (p) = 1 + 2mλψk+ (0) (2π)2 k

Z

so ψk+ (0) =

d2 p 1 (2π)2 p2 − k2 − i

1 1 − λG+ (0)

where +

G (0) = 2m

Z

d2 p 1 (2π)2 p2 − k2 − i

QCD and Confinement

60

e) Find +

G (0) = 2m · 2π

Z Λ 0

pdp 2πm Λ2 − k 2 = ln 2 (2π)2 p2 − k − i −k2 '

f) Find ψk+ (r) = = eik·r −

Z

m Λ2 ln π −k2

imλ 1 d2 p ip·r e φk (p) = eik·r − H0 (kr)ψk (0) (2π)2 2

imλ 1 i H 1 (kr) 2 H0 (kr) = eik·r − 1 4 10 Λ2 mλ Λ2 i 1 − π (ln k2 − iπ) 2mλ − 4π ln k2 − 4

g) Scattering amplitude is defined via π

r→∞

ψk+ −→ eik·r + f (θ)

ei(kr+ 4 ) √ r

Using r→∞ H01 (kr) −→

r

2 ikr e iπkr

we find π 1 f (θ) √ ei(kr+ 4 ) = r

so f (θ) =

r

i eikr 2 1 1 8πk 2mλ − 4π ln Λk2 − 4i

1 1 2 1 1 2πk mλ − 2π ln Λk2 − 2i

h) To find pole set 1 1 Λ2 − ln 2 = 0 mλ 2π κ so ln

κ2 2π =− Λ2 mλ

or Eb.s. = −

κ2 Λ2 2π =− exp − 2m 2m mλ

5.3 Probing Color Experimentally Consider the process of e+ e− annihilation to either quark-antiquark pairs of flavor f or µ+ µ− pairs: e+ + e− → f + f¯. From the Feynman rules for QED, we expect that the cross section for this process should be proportional to Q2f , where Qf is the charge of the fermion.

QCD and Confinement

61

a) Show that R≡

X σ(e+ e− → hadrons) = N Q2f , c σ(e+ e− → µ+ µ− ) f

where Nc is the number of colors of quark and the sum over f is over the number of active flavors (see also Section 10.1). b) Calculate R for (i) f = u, d, s, that is, below charm threshold; (ii) for f = u, d, s, c, that is, below bottom threshold; and (iii) for f = u, d, s, c, b, that is, below top threshold. Compare your results with data from the Particle Data Group [PDG14].

The agreement here between theory and experiment constitutes strong evidence for the existence of color degrees of freedom.

Exercise 5.3 Solution: Probing Color Experimentally Consider photon exchange diagram between e+ e− and f¯f . a) Then Amp = −Nc

e2 Q2f i µ j k v̄ γ u1 v̄4 γµ u`2 q2 3

In high energy limit we have dσCM ∼ Nc2 Q2f α2 (pµ1 pν2 + pν1 pµ2 − η µν p1 · p2 ) (p3µ p4ν + p4µ p3ν − ηµν p3 · p4 ) dΩ = Nc2 Q2f α2 (2p1 · p3 p2 · p4 + 2p1 · p4 p2 · p3 )

= 2Nc2 Q2f α2 E 4 (1 − cos θ)2 + (1 + cos θ)2 = 4Nc2 Q2f α2 E 4 (1 + cos2 θ) Then for ratio R we have dσ + − (e e → f¯f ) R = dσdΩ + − = Nc Q2f + µ− ) (e e → µ dΩ

b) Then if move past threshold with charge Qf the ratio R increases by 3Q2f . Then h 2 2 i 2 i) For u, d, s quarks find R = 3 32 + − 13 + − 13 =2

2 ii) For u, d, s, c quarks find additional ∆R = 3 23 = 43 so R = 10 3 2 iii) For u, d, s, c, b quarks find additional ∆R = 3 − 13 = 13 so R = 11 3 Agreement with experiment is basically good, but should include loop corrections.

5.4 Color Factors for Quarks

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62

The color factor for quark i−quark j interaction → quark k−quark l via single-gluon exchange is X X 1 1 < i|taij |j >< k|takl |l >= taij takl = δil δjk − δij δkl ≡ f (ijkl) .

2 6 a a The color factor f (ijkl) depends on whether pairs of colors are the same or not. a) Show that the color factor for quark-quark scattering with quarks of the same color (i = j = k = l) is f = 31 . b) Show that the color factor for quark-quark scattering with i = j and k = l is f = − 61 . c) Show that the color factor for quark-quark scattering with i = l and j = k is f = 21 . d) Show that f is otherwise zero.

Exercise 5.4 Solution: Color Factors for Quarks Using the relation 1 fijk` = 2

1 δi` δjk − δij δk` 3

1 2

we find a) If i = j = k = ` we find f=

1−

1 ·1 3

=

1 3

b) If i = j and k = ` but i 6= ` we find

1 1 1 f= 0− ·1 =− 2 3 6 c) If i = ` and j = k but i 6= j we find

d) If i 6= j and i 6= ` then

1 f= 2

1 f= 2

1 1− ·0 2

=

1 2

1 0− ·0 =0 3

5.5 Color Factors CF , CA , and TF for Gluons Consider the three gluon processes in the figure.

QCD and Confinement

63

q −→ qg

g −→ gg

CF

CA

g −→ qq

TF

a) Show that the three color factors are 8 X 3 X

taij tajk = δik CF

a=1 j=1 8 X

f abc f abd = δ cd CA

a,b=1

T r(ta tb ) = δ ab TF , where ta are the generators of SU (3) ( 21 λa , where λa are the Gell-Mann matrices) and f abc are the structure constants of SU (3). b) Show that CF = 34 , CA = 3, and TF = 12 .

The color factor ratios CA /CF and TF /CF have been determined experimentally in measurements of angular correlations, of event shapes, and of the differences between gluon and quark multi-jet events in e+ e− collisions. The data are in very good agreement with predictions and directly validate the SU (3) gauge group structure of QCD. Exercise 5.5 Solution: Color Factors CF , CA , and TF for Gluons From the figures we find a) i)

P8

P3 1 a a a=1 j=1 4 λij λjk ≡ δik CF 3 abc abd f f = δ cd CA a,b=1 1 a b ab Tr 4 λ λ = δ TF

ii) iii) b) Now evaluate constants

P8 P3 i) CF = 14 a=1 j=1 λa1j λaj1 = 14 1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 + 13 = 43 P8 ii) CA = a,b=1 f 1ab f 1ab = (f 123 )2 + (f 132 )2 + (f 147 )2 + (f 174 )2 + (f 156 )2 + (f 165 )2 = 2 · 1 + 4 · 14 = 3 iii) TF = Tr 14 λ1 λ1 = 21 5.6 Positronium and Quarkonium

64

QCD and Confinement

The energy levels of the hydrogen atom are

1 me α2 1 = −13.6 2 eV .

En = − 2 2 n n For positronium, we can change to the reduced mass of the e+ e− system, i.e., µr = 12 me to find the energy levels En = −6.8/n2 eV. The lowest energy level of positronium (n=1) is −6.8 eV. The next highest energy level (n=2) is −1.7 eV. Both n=1 and n=2 levels have singlet and triplet level splittings. The lower-energy 11 S0 (singlet) state and 13 S1 (triplet) states are split by 8.4 × 10−4 eV. For n=2, there is also a splitting of the 21 S0 and 23 S1 states as well as of the 2P0,1,2 states.

Take the bound state of the quark-antiquark system where the quarks have the same flavor (quarkonium) and (i) using the positronium level structure described above and (ii) scaling from the Coulomb potential of positronium to the QCD potential of Exercise 5.1, determine the level structure of quarkonium. To a good approximation, this procedure works well for charmonium and bottomonium. Note that toponium does not exist, since the top quark decays through the electroweak interaction before a bound state can form.

Exercise 5.6 Solution: Positronium and Quarkonium This problem is inspired by the article Quarkonium by Elliott D. Bloom and Gary J. Feldman in Scientific American in May 1982. The figure from their paper shows a side-by-side comparison of the spectrum of energy states for positronium compared with charmonium. The energy differences in charmonium are greater by a factor of about 108 . The energy of a state is determined by the principal quantum number n and by the orientation of the particle spins and the orbital angular momentum. In positronium the various combinations of angular momentum cause only tiny shifts in energy, but in charmonium the shifts are much larger. All energies are given with reference to the 13 S1 state. At 6.8 eV positronium dissociates. At 633 MeV above the energy of the J/ψ, charmonium becomes quasi-bound because it can decay into D0 and D0 mesons.

65

QCD and Confinement

6

Chiral Symmetry and QCD 6.1 ππ Scattering Lengths Consider the lowest-order chiral Lagrangian L=

Fπ2 F2 Tr(Dµ U Dµ U † ) + π Tr(2B0 m(U + U † )), 4 4

where m is the quark mass matrix and B0 = m2π /(mu + md ). a) Expand to O(φ2 ) and demonstrate that

3 X 1 µ m2 L2 = ∂ φj ∂µ φj − π φ2j .

2 2 j=1 b) Expand to O(φ4 ) and show that in SU (2) L4 =

1 m2π 1 φ · φφ · φ + (φ · ∂µ φ)2 + (φ · φ)2 .

2 2 6Fπ 2Fπ 24Fπ2

c) Evaluate s-wave scattering and show that the I=0, 2 scattering lengths are given by 7m2π m2π 2 a00 = a = − .

0 32πFπ2 16πFπ2 d) Evaluate p-wave scattering and show that the I=1 scattering length is given by a11 =

m2π .

24πFπ2

These are the so-called Weinberg scattering lengths.

Exercise 6.1 Solution: ππ Scattering Lengths a) Here Dµ U =

i 1 1 i 1 τ · ∂µ φ − ∂µ (φ · φ) − ∂µ (τ · φφ · φ) + . . .

2 Fπ 2! Fπ 3! Fπ3

Dµ U † = −

1 1 i i 1 τ · ∂µ φ − ∂µ (φ · φ) + ∂µ (τ · φφ · φ) + . . .

Fπ 2! Fπ2 3! Fπ3 U + U† = 2 −

66

1 1 φ·φ+ (φ · φ)2 + . . .

2 Fπ 12Fπ4

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67

so Fπ2 1 1 1 φ · ∂µ φφ · ∂ µ φ + φ · φφ · φ + . . .

TrDµ U Dµ U † = ∂µ φ · ∂ µ φ + 4 2 2Fπ2 6Fπ2 Fπ2 m2π m2 (φ · φ)2 + . . .

Trm2π (U + U † ) = − π φ · φ + 4 2 24Fπ2

Then for O(π 2 ) piece L(2) =

1 1 ∂µ π · ∂ µ π − m2π π · π 2 2

[b)] For O(π 4 ) piece L(4) =

1 m2π 1 µ φ · ∂ φφ · ∂ φ + φ · φφ · φ + (φ · φ)2 µ 2Fπ2 6Fπ2 24Fπ2

[c)] Now calculate ππ amplitude. For δ ab δ cd piece we have

m2 1 Amp = δ ab δ cd 8 π 2 − 2 2 (p2a + p2b + p2c + p2d ) 24Fπ 6Fπ

1 +2 2 (pa · pc + pa · pd + pb · pc + pb · pd ) 2Fπ

1 m2π 1 4 s − m2π = δ ab δ cd (p + p ) · (p + p ) + ( − ) = δ ab δ cd a b c d 2 2 Fπ Fπ 3 3 Fπ2 [c)] Have then Amp =

s − m2π ab cd t − m2π ac bd u − m2π ad bc δ δ + δ δ + δ δ Fπ2 Fπ2 Fπ2

Then T I=0 (s, t, u) =

3(s − m2π ) + t − m2π + u − m2π 2s − m2π = Fπ2 Fπ2

T I=2 (s, t, u) =

2m2π − s t − m2π + u − m2π = Fπ2 Fπ2

Project out s-wave and p-wave Z 1 1 2s − m2π s→4m2π 7m2π 0 T0 (s) = d(cos θ)T I=0 (s, t, u) = −→ = a00 64π −1 32πFπ2 32πFπ2 T02 (s) =

1 64π

Z 1 −1

d(cos θ)T I=2 (s, t, u) = −

s − 2m2π s→4m2π 2m2π −→ − = a20 32πFπ2 32πFπ2

[c)] For p-wave T I=1 (s, t, u) =

t−u t − m2π − u + m2π = 2 Fπ Fπ2

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68

so T11 (s) =

1 64π

Z 1

2 s→4m2π k (s) 1 4 2 1 2 k (s) cos θ = k (s) −→ a Fπ2 24πFπ2 m2π 1

d(cos θ) cos θ −1

6.2 Pion Form Factor and Unitarity Verify that the pion form factor as calculated in chiral perturbation theory

2Lr9 (µ) 2 1 q2 m2π q2 2 2 2 2 F1π (q ) = 1 + q + (q − 4mπ )H( 2 ) − q log 2 − , Fπ2 96π 2 Fπ2 mπ µ 3 where r H(x) = 2 +

q  x−4 −1 + x−4 x q log + iπθ(x − 4) , 4 1 + x−4 x 

satisfies the unitarity requirement 2Im F1π (q 2 )(p1 − p2 )µ = −

Z

d3 q1 d3 q2 (2π)4 δ 4 (p1 + p2 − q1 − q2 ) (2π)6 2q10 2q20

×(q1 − q2 )µ < π + (q1 )π − (q2 )|T |π + (p1 )π − (p2 ) > , where T is the two-derivative (tree-level) pion-pion scattering amplitude.

Exercise 6.2 Solution: Pion Form Factor and Unitarity We need the on-shell π + π − scattering amplitude < π + π − |π + π − >=

t − m2π 2m2π − u s − m2π + = 2 2 Fπ Fπ Fπ2

so calculate Z d3 q1 d3 q 2 2m2 − (q1 − p2 )2 Iµ = − (2π)4 δ 4 (q1 + q2 − p1 − p2 )(q1µ − q2µ ) π 0 0 3 3 (2π) 2q1 (2π) 2q2 Fπ2 Work in CM frame, so I0 = 0 and Z q d3 q1 1 0 I=− δ(2p − 2 q 21 + m2π ) 2 (2p02 q10 − 2q 1 · p2 )2q 1 1 0 0 2 (2π) 2q1 2q2 Fπ =−

Z

dΩ1 q 21 1 1 2 |q 1 |3 (2p02 q10 − 2q 1 · p2 )2q 1 = p 02 2 2 2|q | 1 F 4π 4q1 3πFπ2 8q10 2 π 0 q1

2

Set equal to 4p2 ImF1π (q ) so unitarity gives ImF1π (q 2 ) =

1 |q 1 |3 48πFπ2 q10

Chiral Symmetry and QCD

69

while Feynman diagram calculation gives 3

ImF1π (s) =

1 (s − 4m2π ) 2 √ 96πFπ2 2 s

In CM system s = 4q102 so 3

ImF1π (q 2 ) =

1 (4q102 − 4m2π ) 2 1 |q 1 |3 p = 96πFπ2 48πFπ2 q10 2 4q102

6.3 Haag’s Theorem A popular model which is chiral symmetric and renormalizable is the linear sigma model, which involves the interaction of pions with a scalar meson σ and can be written in the “standard” form Lstd =

1 1 ∂µ π · ∂ µ π + [∂µ σ∂ µ σ − m2σ σ 2 ] 2 2

1 −λFπ σ(σ 2 + π 2 ) − λ(σ 2 + π 2 )2 , 4 √ where mσ = 2λFπ . After integrating out the σ this model can be written as an effective Lagrangian involving only the pions Lef f =

Fπ2 Tr(∂µ U ∂ µ U † ), 4

where U = exp

i τ ·π .

Fπ

Calculate π + π 0 scattering at lowest (tree) order using two very different forms of the chiral effective Lagrangian. a) In the standard form of the linear σ-model show that λ Lstd = − (π · π)2 − λFπ σπ · π + . . .

4 and i Mπ+ π0 →π+ π0 = −2iλ + (−2iλFπ )2 2 q − m2σ

= −2iλ 1 +

2λFπ2 q2 = i + ..., q 2 − 2λFπ2 Fπ2

where q = pπ+ − pπ+ is the momentum transfer. f

i

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70

b) In the exponential representation, demonstrate that Lef f =

1 (π · ∂µ π)2 − π · π∂µ π · ∂ µ π + . . .

2 6Fπ

and Mπ+ π0 →π+ π0 = i

q2 + ... .

Fπ2

Note that despite very different looking forms, the physics is the same. This is guaranteed by a deep result in quantum field theory called Haag’s theorem.

Exercise 6.3 Solution: Haag’s Theorem a) Using 1 L1 = − λ(π · π)2 − λFπ σπ · π 4 there are two contributions to π + π 0 scattering. Using π · π = π0 π0 + π+ π− + π− π+ the contact term is Ampa = −i

λ · 8 = −i2λ 4

while the pole term is i Ampb = (−iλFπ · 2) 2 (−iλFπ · 2) q − m2σ Total is

2λFπ2 q2 2λFπ2 4 = −i2λ 1 − (1 + 2 + O(q )) Amptot = −i2λ 1 + 2 q − m2σ m2σ mσ But m2σ = 2Fπ2 λ so Amptot = i2λ

q2 q2 = i + ... m2σ Fπ2

b) We have Dµ U = =

i 1 1 i 1 τ · ∂µ φ − ∂µ (φ · φ) − ∂µ (τ · φφ · φ) + . . .

2 Fπ 2! Fπ 3! Fπ3

1 i i τ · ∂µ φ − 2 φ · ∂µ φ − (τ · ∂µ φφ · φ + 2τ · φφ · ∂µ φ) + . . .

Fπ Fπ 6Fπ3

Then Fπ2 1 1 TrDµ U Dµ U † = ∂µ π · ∂ µ π − ∂µ π · ∂ µ ππ · π 4 2 6Fπ2 +(

1 1 − )π · ∂µ ππ · ∂ µ π · π 2 2Fπ 3Fπ2

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71

so L(4) =

1 1 π · ∂µ ππ · ∂ µ π · π − ∂µ π · ∂ µ ππ · π 6Fπ2 6Fπ2

The π + π 0 scattering amplitude is then Amp = −i = 2i

4p+i · p+f + 4p0i · p0f 2p+i · p0f + 2p+f · p0i − 2p+i · p0i − 2p+f · p0f +i 6Fπ2 6Fπ2

(p+f − p+i )2 (p0f − p0i )2 (p+i − p+f · (p0f − p0i ) q2 + 2i + 2i = i 6Fπ2 6Fπ2 6Fπ2 Fπ2

6.4 Compton Scattering in Effective Field Theory The lowest order Hamiltonian for Compton scattering is given in quantum mechanics by e2 A2 ∗ φ φ 2m where φ is the wavefunction of a spinless particle, where e, m are the particle’s charge, mass respectively.

H0 =

a) Calculate the laboratory frame cross section for unpolarized photons from this spin zero system and show that it takes the form 0 2 α2 ω dσ = (1 + cos2 θ) dΩ 2m2 ω where α = e2 /4π is the fine structure constant and ω, ω 0 are the frequencies of incoming and outgoing photons, respectively.

Now, include the first order corrections due to electric and magnetic polarizabilities αE and βM , which are described by the Hamiltonian 1 H1 = − [4παE E 2 + 4πβM H 2 ]φ∗ φ .

2 b) Calculate the cross section, including the polarizability effects to first order, and show that 2 dσ α2 ω 0 1 = 2 { (1 + cos2 θ) dΩ m ω 2

mωω 0 · (αE + βM )(1 + cos θ)2 + (αE − βM )(1 − cos θ)2 } .

2α Thus, polarizabilities may be measured via experimental study of the Compton cross section, as discussed in the text. −

Exercise 6.4 Solution: Compton Scattering in Effective Field Theory

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72

a) Find Amp0 =

e2 2ˆ i · ˆ f 2m

so 2

1 X e4 1 · 4 1 + cos2 θ |Amp0 |2 = 2 2 4m 2 i,f =1

(1)

(2)

where we have used k̂i = ẑ with polarizations ˆ i = x̂, ˆ i = ŷ and k̂f = (1) (2) cos θẑ + sin θŷ with polarizations ˆ f = cosˆθx − sin θẑ, ˆ f = ŷ. Then 1 dσ = 2ωi

Z

d3 kf 2πδ(m + ωi − ωf − (2π)3 2ωf

2 q 1 X 2 2 m + (ki − kf ) ) |Amp0 |2 2 i,f =1

so e4 dσ 1 ωf 1 (1 + cos2 θ) = 1 2 dΩf 16π ωi 1 + Ef (ωf − ωi cos θ) 2m2 Since p2f = m2 = (pi + ki − kf )2 = m2 + 2m(ωi − ωf ) − 2ωi ωf (1 − cos θ) we have ωf 1 1 = ' 1 ωi 1 + ωmi (1 − cos θ) 1 + Ef (ωf − ωi cos θ) and ωf2 α2 dσ (1 + cos2 θ) = 2 dΩf ωi 2m2 b) Non-zero amplitudes are (1) (1) Amp(ˆ i , ˆf ) =

2 e 0 0 cos θ − 2παE ωω cos θ − 2πβM ωω m

and (2) (2) Amp(ˆ i , ˆ)f =

2

e 0 0 − 2παE ωω − 2πβM ωω cos θ m

Then 2

1 X 1 e4 m m |Amp0 |2 = (1 + cos2 θ)(1 − 2 4παE ωω 0 ) − 2 8πβM ωω 0 cos θ 2 2 2m e e i,f =1

Then ωf2 α2 dσ = 2 2 dΩf ωi m

1 mωω 0 2 2 2 (1 + cos θ) − (αE + βM )(1 + cos θ) + (αE − βM )(1 − cos θ) 2 2α

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73

6.5 Applied Compton Scattering: Why the Sky is Blue Quantum mechanics should be able to give a simple account of two very obvious facts: a) The sky is blue (and the sunset red) b) The skylight is highly polarized.

(If you haven’t had direct experience with the polarization, borrow a pair of polaroid sunglasses and study photons coming from the sun scattered nearly perpendicularly.) For the classical explanation of these phenomena read the Feynman Lectures On Physics, Vol. I, pp. 32–6 to 32–9 [Fey63]. a) Frequency Dependence of the Scattering Cross Section: As a model for the sky, consider a gas of one-electron atoms in the ground state |Ai. Consider elastic (Rayleigh) scattering of photons by one of the atoms in the case that the photon energy is much smaller than the energy separation between |Ai and any excited states |Ii. (Note: ω ≈ 12 eV for visible light compared to (EI − EA )min = 10.4 eV for hydrogen.) Show that, in the long wavelength approximation, the cross section is given by Xh i 1

dσ = m2 r02 ω 4 ˆ∗f · r AI (ˆ i · r)IA + (ˆ i · r)AI ˆ∗f · r IA dΩ ωIA

2

.

I

The strong increase in cross section with frequency (∼ ω 4 ) gives the blue color of the sky and the red color of the sunset. Rayleigh was the first to understand this many years ago.

Hint: Start from the Kramers–Heisenberg formula. Take |Bi = |Ai and ωf = ωi = ω. Expand the energy denominators in powers of ω/ωIA

1 1 ω ω2 − = 1+ + 2 + ...

EA − EI − ω ωIA ωIA ωIA keeping terms up to second order. Show that the lowest order terms combine to give −ˆ ∗f ·ˆ i , and exactly cancel the “seagull term” arising from A·A. Eliminate 1 the ωIA by noting that < p >AI = −imωIA < r >AI ,

< p >IA = imωIA < r >IA .

Why? Use completeness to sum over states, |Ii and note the commutation role of operators r and p. Show that the first order terms give zero (by repeating the previous trick). Finally, demonstrate that the second order terms combine to give the desired result, neglecting higher powers of ω.

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74

b) Polarization of Scattered Light: that

Study the case k̂i = êz , k̂f = êy . If êi = êx show Amp(ˆ f = êx ) 6= 0 .

Amp(ˆ f = êz ) = 0

If ˆi = êy show that there is no scattering in the y direction. c) What can you say about the polarization of photons scattered in the y-direction if the incident photon beam is completely unpolarized? What is the relevance of this result to the effectiveness of polaroid sunglasses? d) Why isn’t the sky violet, since even more violet light is scattered than blue?

Exercise 6.5 Solution: Applied Compton Scattering: Why the Sky is Blue a) Use Kramers-Heisenberg Formula ∗ e2 p ˆ f · ˆ i < B|A > m 2ωf 2ωi

Amp =

ˆ ∗f · pBI ˆ i · pIA ˆ i · pBI ˆ ∗f · pIA 1 X 1 X + + (0) (0) (0) (0) m E − E + ωi + i m E − E − ωf + i I

A

I

I

A

#

I

and write 1 (0)

=−

(0)

=−

(0)

EA − EI + ωi 1 (0)

EA − EI − ωf

so

1 (0)

ωi

− (0)

(0)

(0)

−

(0)

−

EI − EA

(EI − EA )2

1

ωf

(0)

+ (0)

EI − EA

Amp =

(0)

(EI − EA )2

ωi2 (0)

(0)

+ ...

(0)

+ ...

(EI − EA )3 ωf2 (0)

(EI − EA )3

∗ e2 p ˆ f · ˆ i < B|A > m 2ωf 2ωi

" # 1 1 X ∗ ωi ωi2 ˆ f · pBI ˆ i · pIA − + + + ...

(0) (0) (0) (0) (0) (0) m EI − EA (EI − EA )2 (EI − EA )3 I " #) ωf2 1 ωf 1 X ∗ ˆ i · pBI ˆ f · pIA − + + ... − (0) (0) (0) (0) (0) (0) m EI − EA (EI − EA )2 (EI − EA )3 I Write (0)

(0)

(0)

pIA = −im(EI − EA

(0)

and pAI = im(EI − EA

so ˆ ∗f · ˆ i < B|A > −

1 1 X ˆ ∗f · pBI ˆ i · pIA + ˆ i · pBI ˆ ∗f · pIA (0) (0) m E −E I

I

A

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75

= ˆ ∗f · ˆ i < B|A > −i

X I

ˆ ∗f · pBI ˆ i · r IA + ˆ i · pBI ˆ ∗f · r IA

= ˆ ∗f · ˆ i < B|A > −iˆ ∗f a ˆ ib < B|[pa , rb ]|A > = ˆ ∗f · ˆ i < B|A > −ˆ ∗f a ˆ ib δ ab < B|A >= 0 For term linear in ω consider elastic scattering so note

1 X ω ˆ ∗f · pAI ˆ i · pIA − ˆ i · pAI ˆ ∗f · pIA (0) (0) 2 m I (EI − EA ) = mω

X I

ˆ ∗f · r AI ˆ i · r IA − ˆ i · r AI ˆ ∗f · r IA

= mω < A|ˆ i · rˆ ∗f · r − ˆ ∗f · rˆ i · r|A >= 0 Thus term linear in ω vanishes for elastic scattering so

1 e2 ω 2 X ∗ ˆ f · pAI ˆ i · pIA + ˆ i · pAI ˆ ∗f · pIA Amp = 2 (0) (0) 2m ω (E − E )3 I

I

X ∗

ˆ f · r AI ˆ i · r IA + ˆ i · r AI ˆ ∗f · r IA = e2 ω I

A

1 (0) (0) (EI − EA )

Then Z dσ =

X ω2 X d3 kf 2πδ(ωi − ωf ) |Amp|2 = dΩf |Amp|2 3 2 (2π) (2π) , , i

i

f

f

or

1 dσ e4 ω 4 X ∗ ˆ f · r AI ˆ i · r IA + ˆ i · r AI ˆ ∗f · r IA | = |2 2 (0) (0) dΩf 4π (E − E ) I

I

A

b) Choose ˆ i = êx and ˆ f = êz . Then dσ e4 ω 4 X 1 = | |2 [zAI xIA + xAI zIA ] (0) (0) dΩf 4π 2 (E − E ) I

I

A

which vanishes since, by the Wigner-Eckart theorem, ZAI , zIA vanish unless mA = mI while XIA xAI vanish unless mA = mI ± 1.

Choose ˆ i = êx and ˆ f = êx . Then dσ e4 ω 4 X 1 = | [xAI xIA + xAI xIA ] (0) |2 (0) dΩf 4π 2 (E − E ) I

=

I

e4 ω 4 X 1 | |xAI |2 (0) |2 > 0 2 (0) 2π (E − E ) I

I

A

A

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76

Choose ˆ i = êy = k̂f . Then final polarization must be in x or z direction.

But can’t be in z-direction for the same reason vanishes above. In the case of polarization in the x-direction 1 dσ e4 ω 4 X | [xAI yIA + yAI xIA ] (0) = |2 (0) dΩf 4π 2 (EI − EA ) I Then X I

xAI yIA ∝

X X

(O11 − O1−1 )(O11 + O1−1 ) = (O11 )2 − (O1−1 )2 I

I

which vanishes by the Wigner-Eckart theorem. c) Sunlight is unpolarized, meaning an equal mixture of two orthogonal polarizations. Then if we take initial direction to be z-direction and photon scatters into the y-direction, only the original polarization along the x-direction will scatter. But then the scattered photon must be polarized along x-direction, so choose polarizer the block this direction. d) There is more light scattered in the violet but the eye’s ability to detect this scattered lightdrops off rapidly, meaning that it cannot be seen .

7

Elastic Electron Scattering from the Nucleon 7.1 Mott Scattering The Mott cross section for a electron of incident energy E0 scattered through an angle θe is written as dσ (α cos θe /2)2 , = dΩ (2E0 sin2 θe /2)2 where α is the fine structure constant. Plot the Mott cross section for E0 = 1 GeV in units of barns/sr versus θe . Note in particular the sharp rise at small angles, which can be very useful in luminosity measurements.

Exercise 7.1 Solution: Mott Scattering

77

Elastic Electron Scattering from the Nucleon

78

(rad)

7.2 Electron Scattering from a Spinless Target Consider scattering of unpolarized electrons of energy E from a static, spinless charge distribution Zeρ(r) with ∇2 φ(r) = −Zeρ(r) normalized so that

R

ρ(r)dr = 1.

a) From first-order perturbation theory, show thatR the amplitude for scattering from the initial state to the final state Tf i = −i jµf i Aµ d4 x, where Aµ is the electromagnetic field and jµf i = −eūf γµ ui ei(pf −pi )·x , is given by Z 0 Tf i = −i2πδ(Ef − Ei )(−eūf γ ui ) eiq·r φ(r)dr . b) Show that Z

eiq·r φ(r)dr =

ZeF (q) , |q|2

where F (q) is the Fourier transform of the charge distribution.

Elastic Electron Scattering from the Nucleon

79

c) By summing over final and averaging over initial electron spins show that 1 Σs ,s |ūf γ 0 ui |2 = 4E 2 (1 − v 2 sin2 θe /2) , 2 f i where θe is the electron scattering angle. d) Show that

dσ dσ = |F (q)|2 . dΩ dΩ M ott e) Finally, show that if an electron beam is replaced by a beam of pointlike, spinless particles, the only change is that the factor (1 − v 2 sin2 θe /2) in c) above is replaced by unity. Why does the electron spin make no difference in the nonrelativistic limit v → 0? By considering the electron helicity, explain why you would anticipate the cos2 θe /2 behavior of c) in the extreme relativistic limit (ERL).

Exercise 7.2 Solution: Electron Scattering from a Spinless Target a) We need the vector potential, which in momentum space is Z Aµ (q) ' δµ0 e d3 re−iq·r φ(r) Then the T-matrix is Z Z T (q) = d4 xjµ (x)Aµ (x) ' 2πδ(E2 − E1 )e2 ū(p2 )γ 0 u(p1 ) d3 re−iq·r φ(r) b) Fourier transform of equation ∇2 φ(r) = −Zeρ(r) is q2

Z

d3 re−iq·r φ(r) = ZeF (q)

where Z F (q) =

d3 re−iq·r ρ(r)

c) Calculate spin sum, average via 1 1 X |ū(p2 )γ 0 u(p1 )|2 = Tr(6 p2 + m)γ0 (6 p1 + m)γ0 2 s ,s 2 i

=

f

1 θe 4(m2 + (2E 2 − p1 · p2 )) = 2(E 2 − v 2 E 2 + E 2 + v 2 E 2 cos θ) = 4E 2 (1 − v 2 sin2 ) 2 2 d) Calculate cross section via Z 3 E1 1 d p2 1 θe dσ = 2πδ(E2 − E1 )Z 2 e4 |F (q)|2 4 4E12 (1 − v12 sin2 ) 3 p1 4E1 E2 (2π) q 2

Elastic Electron Scattering from the Nucleon

80

= dΩ2

Z 2 e2 E12 θe (1 − v12 sin2 )|F (q)|2 2 4 (2π) q 2

so 1 θe dσ (1 − v12 sin2 )|F (q)|2 = = Z 2 α2 4 θe 2 4 dΩ2 2 4m1 v1 sin 2

dσ dΩ2

M ott

|F (q)|2

e) If incident beam were spinless difference is that 1 X θe |ū(p2 )γ 0 u(p1 )|2 = 4E 2 (1 − v 2 sin2 ) → (p1 + p2 )2 ' 4E 2 2 s ,s 2 i

f

so that dσ 1 |F (q)|2 = = Z 2 α2 2 4 dΩ2 4m1 v1 sin4 θ2e

dσ dΩ2

Ruth

|F (q)|2

The reason that spin does not matter as v → 0 is that plane-wave Dirac solution is r

E1 + m χ1 χ1 p1 →0 −→ u(p1 ) = σ·p1 0 2m E1 +m so that the spin dependence arises only at O(v1 /c).

Factor 1 − v 2 sin2 θ2e in the ERL can be understood by considering helicity. Suppose spin is aligned with momentum. Then at scattering angle θe the projection of a final state spin along direction p̂2 onto initial state spin along direction p̂1 is 2

θe θe θe θe 1 − sin = cos2 = lim 1 − v 2 sin2 cos v→1 0 2 2 2 2

7.3 Longitudinal-Transverse Decomposition of Electron Scattering It is useful to decompose the electron scattering cross section into its longitudinal and transverse components, since the various contributions often depend on rather different aspects of the projections of the EM current. a) Verify the basic expression for the electromagnetic response given in terms of the matrix elements of the EM current, namely, Eqs. (7.62) and (7.63): " #

Q2 2 2 0 2 2 Rf i = ( + ) − q |ρf i (q)| q2 " + 2

1 0 kk sin θe q +

2

√

# 2

+Q

0

2 ( + )

n o 2 2 |Jf i (q; +1)| + |Jf i (q; −1)|

Q2 q2

1 0 kk sin θe q

Elastic Electron Scattering from the Nucleon

81

× {2Re {ρf i (q)∗ (Jf i (q; +1) − Jf i (q; −1))}} "

+ −2

1 0 kk sin θe q

2 #

{2Re {Jf i (q; +1)∗ Jf i (q; −1)}}

≡ v0 vL WfLi + vT WfTi + vT L WfTiL + vT T WfTiT . b) Prove the identity D E E D ω cJM (q) i b f M f LJM (q) i = − q that results from invoking the continuity equation for the EM current and its matrix elements. c) Using angular momentum properties of the matrix elements of the current, provide the details in the proofs that yield the representations E2 X unpol 4π X D cJ Ji WfLi ≡ W L = Jf M 2Ji + 1 J≥0

if

and X if

WfTi ≡ W T

unpol

=

4π X 2Ji + 1

D E2 Jf TbJel (q) Ji

J≥1

E 2 D mag b , + Jf TJ (q) Ji (see Eqs. (7.91), (7.93) and (7.94)).

Exercise 7.3 Solution: Longitudinal-Transverse Decomposition of Electron Scattering a) Equation (7.46) reads Rf i = 2Rµ Jfµi (q)

2

∗

− Q2 (Jf i (q))µ Jfµi (q)

with q µ = k µ − k 0µ 1 Rµ = (k µ + k 0µ ) 2

Jfµ (q) = Jf0i (q) , Jf i (q) Jf0i (q) = ρf i (q) X Jf i (q) = Jf i (m)e∗ (q, m) m=0,±1

Jf i (m) = e (q, m) · Jf i (q) ,

Elastic Electron Scattering from the Nucleon

82

as defined in the text. The convention of having the z-axis (µ = 3) along q is assumed as usual. In momentum space one has for the current conservation condition qµ Jfµi (q) = 0 which immediately yields

ω Jf i (0) = ρf i (q) . q The four-vector scalar product of the current with itself written in terms of its spherical vector projections becomes o n 2 2 2 ∗ 2 (Jf i (q))µ Jfµi (q) = |ρf i (q)| − Jf1i (q) + Jf2i (q) + Jf3i (q) " 2 # n o ω 2 2 2 = 1− |ρf i (q)| − |Jf i (+1)| + |Jf i (−1)| q 2 n o Q 2 2 2 = |ρ (q)| − |J (+1)| + |J (−1)| , f i f i f i q2 where we have substituted for Jf3i (q) = Jf i (0) =

ω ρf i (q) q

and used the facts that 1 Jf1i (q) = − √ [Jf i (+1) − Jf i (−1)] 2 i 2 Jf i (q) = √ [Jf i (+1) + Jf i (−1)] .

2 The initial and final three-momenta of the electrons lie in the 13-plane (the y-axis is then normal to the electron scattering plane) and one can show that 1 0 kk sin θe q k 2 = k 02 = 0 1 k 3 = k (k − k 0 cos θe ) q 1 k 03 = − k 0 (k 0 − k cos θe ) q k 1 = k 01 =

so that 1 0 kk sin θe q R2 = 0

ω 1 2 3 02 k −k = ( + 0 ) , R = q q R1 =

Elastic Electron Scattering from the Nucleon

83

and therefore 2Rµ Jfµi (q) = 2R0 Jf0i (q) − 2R · Jf i (q)

ω ω = ( + 0 ) ρf i (q) − ( + 0 ) ρf i (q) − 2R1 Jf1i (q) q q 2 √ 1 Q ρf i (q) + 2 kk 0 sin θe [Jf i (+1) − Jf i (−1)] . = ( + 0 ) 2 q q This yields for the other term required above 2Rµ Jfµi (q)

2

2 Q2 2 = ( + ) |ρf i (q)| q2

2 √ 1 0 Q + 2 ( + 0 ) kk sin θ e q2 q ×2Re {ρ∗fi (q) [Jfi (+1) − Jfi (−1)]} 2 n

1 0 2 2 kk sin θe |Jf i (+1)| + |Jf i (−1)| +2 q −2Re [J∗fi (+1)Jfi (−1)]} 0 2

and, adding the two derived expressions together, proves the sought-after identity. b) For the Coulomb matrix element one has E Z D c f MJM (q) i = dx jJ (qx)YJM (Ωx )ρf i (x) Z = dx MJM (qx)ρf i (x) whereas, for the longitudinal matrix element one has

D E Z b JM (q) i = dx i ∇MJM (qx) · Jf i (x) f L q Z

i = − dxMJM (qx)∇ · Jf i (x) q D E ω cJM (q) i , =− f M q where the second line above comes from partial integration and the third line from the continuity equation. c) Working from Eq. (7.78) in the text, first we employ the orthogonality properties of the 3-j sysmbols to write X Jf Mi Mf

−Mf

J M

Ji Mi

Jf −Mf

J0 M0

Ji Mi

=

1

2 δJ J δM M 0

[J]

0

Elastic Electron Scattering from the Nucleon

84

with [J] = WL

√

2J + 1. Then for W L

unpol

unpol

we have

2

E 2X D (4π) X 1 cJ Ji YJM (Ωq )YJM (Ωq )∗ Jf M 2Ji + 1 2J + 1 M J≥0 E D X 2 4π cJ (q) Ji = Jf M 2Ji + 1

=

J≥0

using the fact from Edmonds [Edm74] that X

YJM (Ωq )YJM (Ωq )∗ =

M

2J + 1 .

4π

The remaining transverse contribution is handled similarly, first yielding Eq. (7.92) in the text and, noting that no electric/magnetic interferences can occur once both angular momentum and parity symmetries are enforced, finally yielding the desired result

D E 2 E2 unpol 4π X D Jf TbJel (q) Ji . + Jf TbJmag (q) Ji WL = 2Ji + 1 J≥1

7.4 Elastic Scattering of Fast Electrons by an Atom Apply the Born approximation to the scattering of a fast electron by an atom, using as a perturbation, the interaction V (r) = −

Z Zα X α + , r |r − ri | i=1

which represents the Coulomb potential between the incident electron, and the nucleon of charge Z|e|, and the individual electrons which make up the atom.

For the initial-state wavefunction take hr, r1 , . . . rZ |ii = eiki ·r ψ0 (r1 , . . . rZ ) and for the final-state hr, r1 , . . . rZ |f i = eikf ·r ψ0 (r1 , . . . rZ ) , where ψ0 (r1 , . . . rZ ) is the wavefunction of the ground state of the atom, assumed to be the same before and after the collision. (Note: In principle, the electron wavefunction should be properly antisymmetrized. However, at high energies this antisymmetrization can be shown to be relatively unimportant.) a) Demonstrate that the differential cross section takes the form dσ dσ = |1 − F (q)|2 , dΩ dΩ point

Elastic Electron Scattering from the Nucleon

85

where Z 2 α2 dσ = dΩ point 4m2 v 4 sin4 θe /2 is the cross section for an electron of charge |e| and velocity v to scatter from a point-particle target of charge Z|e|; here, F (q) is the “form factor” for the atomic ground state |ψ0 > + * Z X 1 iq·rj ψ0 e F (q) = ψ0 Z j=1 Z

1 Y = Z i=1

Z dri

Z X j=1

eiq·rj |ψ0 (r1 , . . . rZ )|2 .

b) For the specific case of electron scattering from the ground state of a hydrogen atom, show that the differential scattering cross section has the form  2 α2 dσ  = 1 − 4 2 4 dΩ 4m v sin θe /2

1 1+

q2 a20

 2  ,

4

where a0 is the Bohr radius.

In general, a form factor arises in problems in which scattering is due to a composite system made up of identical scatterers (electrons in our case), which are distributed in space. The form factor expresses the feature that the scattering amplitude generated by the scattering centers differ from one another because of purely geometric considerations — variation of the phase of the incident wave at the scattering center and of the scattered wave at the detector. Measurement of the form factor thus yields information about the distribution of scattering centers within the composite system, i.e., the Fourier transform of the spatial distribution. For a given momentum transfer q, the form factor is sensitive to variations in the geometric distribution over distances of order d such that 1 qd ∼ 1, i.e., d ∼ , q and this is the motivation behind the use of scattering experiments to map the distribution of constituent scatterers. c) More quantitatively, consider the situation with a source S and a detector D with scattering centers at O, O0 . The full scattering amplitude is then M=

0 0 1 ikri 1 1 1 e A(ki → kf ) eikrf + 0 eikri A(k0i → k0f ) 0 eikrf , ri rf ri rf

where ri (ri0 ) is the distance from the source to scattering center O (O0 ) and rf (rf0 ) is the distance from scattering center to the detector. Show that, if ri , rf are much larger than the distance ρ between the scattering centers, then

Elastic Electron Scattering from the Nucleon

86

it is reasonable to take the intrinsic scattering amplitude A(ki → kf ) as being the same at both centers, so that M'

1 ikri 1 e A(ki → kf )(1 + eiq·ρ ) eikrf , ri rf

where q = ki − kf is the momentum transfer. Explain your result for atomic scattering obtained in a) of this exercise in light of this form. d) Large Momentum Transfer: Show that if q is much larger than the atomic radius, dσ is essentially the Rutherford cross section for scattering then |F (q)| 1 and dΩ by the Coulomb field. e) Small Momentum Transfer: Assume for simplicity that the atomic ground-state wavefunction ψ0 has total angular momentum zero. Expand F (q) in powers of q and show that + * Z X 1 2 2 ri ψ0 + . . .

ZF (q) = Z − q ψ0 6 i=1 Is the differential scattering cross section finite or infinite at θe = 0? Explain this result. f) Analyze the hydrogen atom cross section calculated in part b) in terms of the large- and small-q limits discussed in d) and e).

Exercise 7.4 Solution: Elastic Scattering of Fast Electrons by an Atom a) In Born approximation Z Amp(q) =

d3 re−iq·r V (r)

Z 4πZα X 4πα −iq·rj =− 2 + e q q2 j=1

where we have used Z

3

−iq·r

d re Then

1 = e−iq·rj |r − rj |

Amp(q) = −

Z

d3 se−iq·s

4πZα [1 − F (q)] q2

where F (q) =

Z X 1 < ψ0 | e−iq·rj |ψ0 > Z j=1

b) For hydrogen atom in ground state the wavefunction is s 1 r ψ1S (r) = exp − 3 πa0 a0

1 |s|

Elastic Electron Scattering from the Nucleon

87

Then Fourier transform is F (q) =< ψ0 |e

−iq·r

1 |ψ0 >= πa30

4 = |q|a30

Z ∞ 0

Z

3

−iq·r− a2r

d re

0

− a2r

drr sin |q|re

0

=

Z ∞

4 = 3 a0

0

!2

1 1+

−2r

drr2 j0 (|q|r)e a0

q2 a20 4

Since T (q) =

Z

d3 r

4πα α = 2 |r| q

we have dσ 1 m α2 1− =|− |T (q)|2 |1 − F (q)|2 = 2 4 θe 2 4 dΩ 2π q2 a2 4m v sin 2 1+ 40

2

c) We have M∼ '

1 A(ki → kf )eik(ri +rf ) 1 + exp ik(ri + rf − ri0 − rf0 ) ri rf

1 A(ki → kf )eik(ri +rf ) (1 + exp(iki · (x1 − x2 ) − ikf · (x1 − x2 ))) ri rf '

1 A(ki → kf )eik(ri +rf ) (1 + exp(−iq · (x1 − x2 )) ri rf

d) If q >> 1/a then phase is large and lage cancellations occur, so F (q) << 1 meaning that the electron scattering piece vanishes so that the scattering is basically the Rutherford scattering from the nucleus. e) If q << 1/a then can expand exponential as 1 exp −iq · ri = 1 − iq · ri + (q · ri )2 + . . .

2 so 1−

=1−

Z X 1 1 1 F (q) ' 1 − < ψ0 |Z − (q · rj )2 + . . . |ψ0 > Z Z 2 j=1 Z Z 1 1 X 2 1 2X 2 < ψ0 |Z − q2 rj + . . . |ψ0 >= q r + ...

Z 6 j=1 6Z j=1 j

Here the q2 cancels that in the Rutherford amplitude so that the cross section is finite. f) We note dσ Z 2 α2 = 1− dΩ 4m2 v 4 sin4 θ2e

2

1 1+

q2 a20 4

2

Elastic Electron Scattering from the Nucleon

88

Then if |q| << 1/a0 1−

1 1+

q2 a20

2 →

1 2 2 q a0 2

4

so 1 1 dσ → (2mZα)2 4 · q4 a40 = 2Z 2 α2 m2 a40 dΩ q 2 If |q| >> 1/a0 then 1−

1 1+

q2 a20

2 → 1 −

4

16 q4 a40

so dσ Z 2 α2 → dΩ 4m2 v 4 sin4 θ2e

7.5 Relativistic Coulomb Scattering In this exercise, we calculate the differential cross section for the scattering of two non-identical spinless particles of charge and mass, −e, m1 and Ze, m2 , respectively.

In the case of electrodynamics, the equation-of-motion of the photon field Aµ (x) is Aµ (x) = jµ (x) , where jµ (x) is the current density, and the term which couples Aµ (x) and j µ (x) is Z Lint = − d4 xj µ (x)Aµ (x) .

For a charged scalar field we have 0

hp01 |jµ (x)|p1 i = ei(p1 −p1 )·x e(p1 + p01 )µ . a) Show that the interaction of the two particles is given by calculating the scattering amplitude via Z S = d4 xLint (x) ×

Z

4

Z

d x

4

d y

Z

d4 q i(p02 −p2 )µ ·x e Ze(p2 + p02 )µ eiq·(x−y) ei(p2 −p1 )·y e(p1 + p01 )µ (2π)4

= (2π)4 δ 4 (p1 + p2 − p01 − p02 )Ze(p1 + p01 ) · e(p2 + p02 ) ≡ (2π)4 δ 4 (p1 + p2 − p01 − p02 )T (q) , where T (q) = 4πZα

(p1 + p01 ) · (p2 + p02 ) . (p1 − p01 )2

1 (p1 − p01 )2

Elastic Electron Scattering from the Nucleon

89

b) Square to find the transition probability: interpret this result via [(2π)4 δ 4 (p1 + p2 − p01 − p02 )]2 = (2π)4 δ 4 (p1 + p2 − p01 − p02 )

Z

Z dt

dx exp(−i(p1 + p2 − p01 − p02 ) · x)

= (2π)4 δ 4 (p1 + p2 − p01 − p02 )T V , where T is the interaction time and V is the volume. Since the cross section is the transition rate per unit volume divided by the incident flux, show that Z 1 dp01 dp0 2 dσ = (2π)4 δ 4 (p1 + p2 − p01 − p02 )|T (q)|2 , Flux 2E10 2E20 where Flux = 2E1 · 2E2 · |v1 − v2 | is the flux, with 2E representing relativistic normalization factors, due to the normalization condition Z d3 xφ∗k (x)i∂0 φk0 (x) − i∂0 φ∗k (x)φk0 (x) = 2Ek δ 3 (k − k0 ) for the Klein-Gordon states c) Calculate the cross section in the laboratory frame (i.e., the frame in which the particle having mass m2 and four-momentum p2 is at rest) and show that 2

1 1 dσ p01 = |T (q)|2 0 0 0 dΩ (2π)2 16m2 p1 E2 p1 + E1 (p01 − p1 cos θ) with T (q) = 4παZ

m2 (E1 + E10 ) .

2 m1 − E1 E10 + p1 p01 cos θ

d) Now transform to the frame where the target particle is very heavy, m2 → ∞ , and the incident particle is moving slowly v1 1, c and compare your result with the Rutherford cross section dσRuth Z 2 α2 . = dΩ (2m1 v12 )2 sin4 θe /2

Elastic Electron Scattering from the Nucleon

90

Exercise 7.5 Solution: Relativistic Coulomb Scattering a) The T-matrix element is 1 T (q) = Aµ (q)e(p1 + p01 )µ = Ze(p2 + p02 )µ 2 e(p1 + p01 )µ q so T (q) = Ze2 p1 + p01 · (p2 + p02 ) (p1 − p01 )2 b) Cross section is, by this reasoning Z 1 d3 p02 d3 p01 dσ = (2π)4 δ 4 (p1 + p2 − p01 − p02 )|T (q)|2 0 3 2E1 2E2 |v1 − v2 | (2π) 2E1 (2π)3 2E20 c) Calculate Z 1 1 d3 p01 1 dσ = 0 2m2 2p1 (2π)2 2E2 2E10 q q q 0 ×δ(m2 + p21 + m21 − (p1 − p01 )2 + m22 − p12 + m21 )|T (q)|2 0

= dΩ01

1 1 1 p12 1 1 2 0 −p cos θ |T (q)| 0 0 0 2 p p 1 1 1 2m2 2p1 (2π) 2E2 2E1 0 + 0 E1

E2

0

=

dΩ01 1 p12 |T (q)|2 0 0 0 2 (2π) 16p1 m2 p1 E2 + (p1 − p1 cos θ)E10

d) Here T (q) = Ze2

2m2 (E1 + E10 ) (p1 + p01 ) · (2p2 + p1 − p01 ) = Ze2 0 2 2 (p1 − p1 ) 2m1 − 2E1 E10 + 2p1 · p01

e) As m2 → ∞ and E1 ' E10 ' m1 we have m21 − E1 E10 + p1 p01 cos θ → m21 − (m1 +

p2 p21 )(m1 + 2 ) + p1 p01 cos θ 2m1 2m1

1 θe = − (p21 + p1 0 2 + p1 p01 cos θ ' −2p21 sin2 2 2 so

m1 m2 T (q) ' Ze2 2 2 θe p1 sin 2

and Z 2 α2 dσ Z 2 e4 1 m21 m22 ' = = 0 2 dΩ1 (2π)2 16m2 p41 sin4 θ2e 4m21 v14 sin4 θ2e

7.6 Gravitational Scattering

dσ dΩ01

Ruth

Elastic Electron Scattering from the Nucleon

91

We can analyze the gravitational scattering of two massive particles in a fashion analogous to that used for Coulomb scattering. In the case of electrodynamics the equation-of-motion of the photon field Aµ (x) is determined by Aµ (x) = jµ (x) , where jµ (x) is the current density and the term which couples Aµ (x) and j µ (x) is Z Lint = − d4 xj µ (x)Aµ (x) .

For a charged scalar field we have 0

hp01 |Jµ (x)| p1 i = ei(p1 −p1 )·x e(p1 + p01 )µ .

Then the interaction of a two particles, one with charge e1 and mass m1 and a second with charge e2 and mass m2 , is given by calculating the scattering amplitude via Z 1 1 p S = − d4 xLint (x) = √ 4E1 E2 4E10 E20 ×

Z

4

d x

Z

4

d y

Z

0 d4 q i(p01 −p1 )·x e e1 (p1 + p01 )eiq·(x−y) ei(p2 −p2 )·y e2 (p02 + p2 )µ (2π)4

= −(2π)4δ 4 (p1 + p2 − p01 − p02 )e1 (p1 + p01 )e2 · (p2 + p02 )

1 (p1 − p01 )2

≡ (2π)4δ 4 (p1 + p2 − p01 − p02 )Tem (q) .

Then the potential energy is given by the Fourier transform of the transition amplitude Z e1 e2 dq −iq·r e Tem (q) = , Vem (r) = (2π)3 4πr which is the familiar Coulomb potential.

A parallel formalism can be used to describe the lowest-order gravitational interaction. In this case, the graviton field, which carries spin 2, is described by a symmetric second-rank tensor, hµν (x), which obeys the equation-of-motion

1 hµν (x) = −16πG Tµν (x) − ηµν Tr(T (x)) , 2 where here G is Newton’s constant, Tµν (x) is the energy-momentum tensor, and Tr (T (x)) = η µν Tµν (x) is its trace. The fields hµν (x) and Tµν (x) are coupled by the interaction term Z 1 Lint = − d4 xTµν (x)hµν (x) , 2 where for a scalar field we have

0 1 < p01 |Tµν (x)|p1 >= ei(p1 −p1 )·x p1µ p01ν + p01µ p1ν − ηµν (p01 · p1 − m2 ) p .

4E1 E10

Elastic Electron Scattering from the Nucleon

92

The relation between the graviton field hµν (x) and the metric tensor gµν (x) is gµν (x) = ηµν + hµν (x) , where 1  0 ηµν =   0 0 

0 −1 0 0

0 0 −1 0

 0 0   0  −1

is the usual Minkowski metric. a) Consider a very massive particle M (which could be the sun, for example) at rest at the origin so that 0

hp02 |Tµν (x)| p2 i = ei(p2 −p2 )·x M δµ0 δν0 .

Using Fourier transform methods, solve the equation-of-motion for hµν (x) and show that h00 (x) = −

2GM , |x|

hij (x) = −

2GM δij |x|

hi0 (x) = 0

when the mass is placed at the origin of coordinates. Compare this result with the Schwarzschild metric in the limit GM/|x| << 1 [?]. b) Consider the scattering by M of a much lighter scalar particle with mass m. Show that the scattering amplitude is given by Z −1 S = d4 xLint (x) = p 4E1 E10 ×

Z

d4 x

Z

d4 y

Z

d4 q i(p01 −p1 )·x 4πG e M (2δµ0 δν0 − ηµν )eiq·(x−y) 2 (2π)4 q

0

×ei(p2 −p2 )·y [p2µ p02ν + p02µ p2ν − ηµν (p2 · p02 − m2 )] = −(2π)4δ 4 (p1 + p2 − p01 − p02 )8πGM 2

1 1 1 p (4E1 E10 − 2m2 ) (p1 − p01 )2 2M 4E1 E10

≡ (2π)4δ 4 (p1 + p2 − p01 − p02 )Tgr (q) . c) Evaluate the Fourier transform of the static limit of the transition amplitude, and show that the mutual interaction of m and M is described by the potential energy Z dq −iq·r GM m Vgr (r) = e Tgr (q) = − , (2π)3 r which is the conventional Newtonian result.

Elastic Electron Scattering from the Nucleon

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d) Calculate the differential scattering cross section using the transition amplitude Tgr (q) and show that Z E1 1 dp2 dσ = 2π(E10 − E1 )|Tgr (q)|2 = (2GM )2 2 (2E12 − m2 )2 dΩ 3 p1 (2π) q = (2GM )2

1 (2E12 − m2 )2 dΩ .

16p41 sin4 θ/2

e) We can connect with the corresponding classical result by using the relation between the impact parameter b and scattering angle θ 2πbdb = dσ .

Show that in the massless (m = 0) limit this relation can be integrated to become 4GM θ= b in the small angle scattering limit.

This is the relation derived by Einstein, and has been tested via the measurement of the deflection of starlight during an eclipse. (These experiments involved photons with spin-one rather than spin-zero particles as considered above. However, general relativity requires that all massless particles must follow the same trajectory, a geodesic, regardless of spin.)

Exercise 7.6 Solution: Gravitational Scattering a) Have −q 2 hµν (q) = −16πGM

1 δµ0 δν0 − ηµν 2

so 16πGM δµ0 δν0 − 12 ηµν hµν (q) = q2

and Z hµν (r) =

d3 q iq·r −16πGM δµ0 δν0 − 21 ηµν e (2π)3 q2

4GM δµ0 δν0 − 12 ηµν =− |r|

Schwarzschild solution is in isotropic coordinates !2 1 − GM 2GM 2|r| g00 = =1− + ...

GM |r| 1 + 2|r|

gij = −δij

GM 1+ 2|r|

2

= −δij

2GM 1− + ... |r|

Elastic Electron Scattering from the Nucleon

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b) The T-matrix element is T (q) = p

=p

1 1 1 1 1 1 8πGM µν p δ δ − h (q)T = η µ0 ν0 µν µν 2 4E1 E10 2M 2 4E1 E10 2M q 2

× p1µ p01ν + p1ν p01µ − ηµν (p1 · p01 − m2 )

1 1 8πGM 2E12 − (p1 · p01 − m2 ) − p1 · p01 + 2(p1 · p01 − m2 ) 2 0 4E1 E1 2M q

1 8πGM 1 2E12 − m2 =p 2 0 4E1 E1 2M q

c) Potential is Vgr (r) = −

Z

GM m d3 q iq·r 4πGM m e =− (2π)3 q2 |r|

d) Cross-section is E1 dσ = |p1 |

Z

2 d3 p2 1 GM (2E12 − m21 ) 2πδ(E1 − E2 ) 4π (2π)3 E1 q2 p2 1 = dΩ2 2 E1 dE dp

2GM (2E12 − m21 ) q2

2

2

i.e., dσ G2 M 2 (2E12 − m2 )2 = dΩ2 4p41 sin4 θ2 e) Then if m = 0 dσ G2 M 2 = dΩ2 sin4 θ2 or dσ =

G2 M 2 dθ 2π sin θdθ ' 32π(GM )2 3 = 2πbdb θ sin4 θ2

Then bdb =

dθ 1 2 (4GM )2 1 db = (4GM )2 3 = − d 2 2 θ 2 θ

or b = 4GM θ 7.7 Small-Angle Electron Scattering Most frequently, the kinematics and leptonic factors such as in Eqs. (7.68–7.71) are given in the extreme relativistic limit, where the electron mass is neglected with respect to its energy or three-momentum. However, this simplification is not necessary, and under some conditions, it is essential to retain the mass.

Elastic Electron Scattering from the Nucleon

95

a) Using the standard expressions for the square of the three-momentum transfer 2

q 2 = k 2 + k 0 − 2kk 0 cos θe and for the energy transfer ω = − 0 , and keeping the electron mass finite, obtain an expression for the four-momentum transfer squared. Here q = |q|, k = |k| and k 0 = |k0 |. b) By adding and subtracting a term 4kk 0 sin2 θe /2 to and from the expression obtained in part a), write the square of the four-momentum transfer in the form Q2 = 4kk 0 sin2 θe /2 + ∆, and obtain an expression for ∆. When considering electron scattering at very small angles, the first term here goes to zero, but, in general, the second term does not. Find an approximate expression for the latter in the small-angle limit for a situation involving the excitation of a nucleus (or any system in general) from its ground state to an excited state at excitation energy E ∗ . In fact, one can see that the four-momentum transfer cannot ever be zero in such a situation and that one always has Q2 > 0 (spacelike). c) Using the results in parts a) and b), find the critical angle θec where, for θe ≤ θec , one must retain the electron mass, while for θe > θec it is safe to employ the ERL. d) Obtain expressions for the analogs of the kinematic factors vL,T,T L,T T (usually denoted VL,T,T L,T T ) when the electron mass is kept finite.

Exercise 7.7 Solution: Small-Angle Electron Scattering a) Using the expressions for q 2 and ω given above it is straightforward to obtain Q = q2 − ω2 :

Q2 = k 2 + k 02 − 2kk 0 cos θe − 2 + 02 − 2 0

= 2 0 − kk 0 cos θe − m2e 2

≡ Q20 + ∆,

where the two contributions are found by adding and subtracting 2kk 0 , namely, Q20 = 4kk 0 sin2 θe /2

∆ = 2 0 − kk 0 − m2e 2

= m2e

2 (k − k 0 ) .

0 + kk 0 + m2e

Using the first of the last two expressions for ∆ one clearly sees that both Q20 and ∆ and hence Q2 are non-negative (Q2 > 0 means spacelike). b) Part a) has the general answer where the electron mass is not assumed to be zero. Clearly one could stop here and use the above expressions for Q2 in any

Elastic Electron Scattering from the Nucleon

96

numerical analysis. However, it is interesting to go a bit further and make some sensible approximations to the results. First, note that the ∆-term is manifestly proportional to m2e : in most situations, namely, except for very low energies, one can invoke the ERL for the expression that multiplies the square of the electron mass. Then one has ω2 . ∆ ' m2e ( − ω) Accordingly, when θe → 0, it can be shown that the first term (Q20 ) goes to zero, but the expression here shows that the second term (∆) goes to ∆0 = m2e

ω02 , ( − ω0 )

where ω0 is the energy transfer obtained when the scattering angle goes to zero.

Energy conservation tells us that q Mi + ω = Mf2 + q 2 with Mi and Mf the initial and final nuclear masses, respectively. The nuclear excitation energy is E ∗ = Mf − Mi , and putting this together we have ω = ω∗ +

Q2 2Mi

where

Mf2 − Mi2 E∗ ∗ =E 1+ ω ≡ .

2Mi 2Mi ∗

Thus, we have found that ω0 = ω ∗ +

∆0 2Mi

and that, since ∆0 is of order m2e , and can therefore be neglected with respect to ω ∗ , we have ω0 ' ω ∗ together with 2 Q θe →0 ' m2e

ω ∗2 , ( − ω ∗ )

where ω ∗ is given in terms of E ∗ in the above. We have then proven that, for inelastic scattering, Q2 is always greater than zero, namely, spacelike and can never reach the lightcone where Q2 = 0.

Note that elastic scattering is different: in that case both Q20 and ∆, and hence Q2 go to zero, reaching the limiting situation where q = ω = Q2 = 0 when θe = 0. This is, however, a trivial circumstance, since any elastic form factor is always known at that special kinematic point. c) A critical angle, θec , can be defined as the place where Q20 and ∆ are equal, that is,

2 ∗2 ∆ 1 me ω ∗ 2 c 2ω sin θe /2 = ' m ' 4kk 0 4kk 0 e 0 2 0

Elastic Electron Scattering from the Nucleon

97

or θec '

me ω ∗ me ω ∗ ' .

0 ( − ω ∗ )

d) Using the expressions in part a) for the three- and four-momentum transfers one has 2 2 Q = ρ2 VL = q2 with ρ ≡ Q2 /q 2 , as usual. Defining

2

v0 ≡ ( + 0 ) − q 2

= 2 0 + kk 0 cos θe + m2e and following procedures similar to those in part a), one has from Eq. (7.62) in the text " # 2 1 kk 0 sin θe /2 2 VT = 2 +Q v0 q " 2 # kk 0 sin θe /2 1 VT T = −2 v0 q 0

√ kk sin θe /2 1 0 .

VT L = − 2 ( + ) ρ v0 q To bring these expressions into the form used earlier in this exercise, one can develop v0 just as was done for Q2 above to find 2

v0 = 4kk 0 cos2 θe /2 + m2e

2 ( + 0 ) 0 + kk 0 + m2e

and again the ERL can be used for the second term. Using the expression for q 2 one can write 2kk 0 cos θe = k 2 + k 02 − q 2 and hence h ih i 2 2 2 (2kk 0 sin θe ) = q 2 − (k − k 0 ) (k + k 0 ) − q 2 , yielding 2

(2kk 0 sin θe ) =

2

1 Q − 2 0 − kk 0 − m2e 4 cos2 θe /2

× v0 − 2 0 − kk 0 + m2e ,

permitting one to write all of the kinematic factors in terms of leading variables plus mass contributions.

7.8 Weisskopf Units

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98

In the measurement of an electromagnetic transition probability within a nucleus, it is useful to compare the measured magnitude of a given quantity with a “natural” size, i.e., with what one might expect from a single nucleon making a transition from one orbital to another within the nucleus — the “single-particle value”. For electric transitions, the B(Eλ) value is the square of the transition matrix element for a single proton, which can be written as Jf Mf erλ Yλµ Ji Mi = e(radial integral) × (angular integral) , where the radial integral is r

λ

=

Z ∞ 0

drr2 Rn∗ f `f (r)rλ Rni `i (e)

and the angular integral yields 4π.

1

a) Assume the nucleus to be a uniformly charged sphere with radius R0 A 3 . Show that λ 3 rλ = R0λ A 3 .

3+λ b) For the present purpose, take e2 λ 2 r 4π and show that the single-particle estimate becomes

2 2λ e2 3 BW (Eλ) = R02λ A 3 , 4π 3 + λ Best (Eλ) =

which is called the Weisskopf unit.

Exercise 7.8 Solution: Weisskopf Units a) We can write < Jf Mf |erλ Yλµ |Ji Mi >= radial integral · angular integral , where the radial integral takes the form Z ∞ < rλ >= Rn∗ f lf rλ Rni li r2 dr . o

For a sphere of uniform density of radius R, we can apply the normalization condition as follows Z ∞ 1= Rn∗ f lf Rni li r2 dr 0

= Rn∗ f lf Rni li

Z R

r2 dr

0

1 = Rn∗ f lf Rni li R3 , 3

Elastic Electron Scattering from the Nucleon

99

which implies that Rn∗ f lf Rni li =

3 .

R3

Hence, Z ∞

Rn∗ f lf rλ Rni li r2 dr Z ∞ ∗ rλ+2 dr = Rnf lf Rni li

< rλ > =

o

0

1 3 = 3 Rλ+3 R λ+3 3 = Rλ λ+3 3 = Rλ Aλ/3 λ+3 0

QED .

b) We have BW (Eλ) = | < Jf Mf |erλ Yλµ |Jf Mf > |2

= e2 (radial integral)2 · (angular integral)2 ,

where the radial and angular integrals have the values < rλ > and √14π , respectively.

Thus, 1 BW (Eλ) = e2 < rλ >2 4π

2 e2 3 = R02λ A2λ/3 4π 3 + λ

QED .

7.9 Effective Field Theory and the EM Multipole Expansion The application of effective field theoretic ideas to the electromagnetic interaction is manifested in a multipole expansion, which represents the interaction in terms of a few constants (multipoles). a) Show that the electromagnetic interaction of a nucleus of spin J can be represented in terms of an expansion ∂Ei 1X + ...

Qij Hem = Zeφ0 − M · B − 6 ij ∂xj with M=µ

e J 2mN J

eQ 3 2 Qij = (Jj Jj + Jj Ji ) − δij J .

J(2J − 1) 2

Here µ is the magnetic dipole moment in units of nucleon magnetons and Q is the electric quadrupole moment of the particle having charge Ze.

Elastic Electron Scattering from the Nucleon

100

b) Show that the matrix element of the electromagnetic current can be written at low momentum transfer as J, Mf , pf |Vµem |J, Mi , pi = δMi ,Mf M k;Mi

f −iCJ1;J

ijk ηµj

Pµ +CJ2;J Mf k; Mi (2M )3

r

Pµ F1 (q 2 ) 2M

qi F2 (q 2 ) 2M

4π k Y (q̂)q2 F3 (q 2 ) + . . . , 5 2

where Pµ = (pf + pi )µ is the total momentum and M = 21 (M1 + M2 ) is the nuclear mass. c) Verify that the current is conserved, in that q µ J, Mf , pf |Vµem |J, Mi , pi = 0 . d) Show that the electromagnetic form factors F1 (q 2 ), F2 (q 2 ), F3 (q 2 ) are connected to the moments via F1 (0) = Ze r F2 (0) = s F3 (0) = −

J +1 µA J

(J + 1)(2J + 3) 2M 2 Q, J(2J − 1) 3

where A is the atomic number.

Exercise 7.9 Solution: Effective Field Theory and the EM Multipole Expansion a) The electromagnetic interaction can be written as Z H = d3 r(ρ(r)φ(r) − j(r) · A(r)) Now expand φ(r) =

∞ X 1

n! n=0

(r · ∇)n φ(r)

= φ(0) + r · ∇φ(0) + A(r) =

∞ X 1

n! n=0

r=0

1 r · ∇r · ∇φ(0) + . . .

2!

(r · ∇)n A(r)

r=0

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101

= A(0) + r · ∇A(0) + In static limit E = −∇ · φ so φ(r) = φ(0) − r · E −

1 r · ∇r · ∇A(0) + . . .

2!

1 1 r : (∇i Ej + ∇j Ei ) : r + . . .

2!

2

Note that ∇2 φ = −∇ · E = 0 in the charge-free region where the expansion is being performed.

Z Z 1 1 HE = φ(0) d3 rρ(r) − E · d3 rrρ(r) − Qij (∇i Ej + ∇j Ei ) + . . .

2!

2 Here Z

d3 rρ(r) = Ze

Z

d3 rrρ(r) = d

is the total charge, and

is the electric dipole moment, which vanishes for an elementary particle by parity arguments, and Z 1 Qij = d3 r(ri rj − δij r2 )ρ(r) 3 is the quadrupole moment Also, since ∇ × A = B we have A(r) = A(0) + r · ∇A(0) +

1 r · ∇r · ∇A(0) + . . .

2!

Then HM = −

Z

d3 rj · A = −A(0) ·

Since for a closed circuit we have then Z HM = −

R

Z

d3 rj −

Z

d4 rr · ∇j · A(0)

d3 rj = 0 and, in free space ∇ · j = − ∂ρ ∂t = 0

d3 r(r × j) · ∇ × A(0) = −M · B

where Z M=

d3 rr × j

is the magnetic dipole moment. b) Form follows simply from Wigner-Eckart theorem and realization that charge carries spin zero, magnetic moment spin 1, and quadrupole moment spin 2. c) Note that P · q = (p1 + p2 ) · (p1 − p2 ) = p21 − p22 = M 2 − M 2 = 0 so the terms in F1 (q 2 ), F3 (q 2 ) vanish, while the term F2 (q 2 ) vanishes since ijk q i q j = (q × q)k = 0.

Elastic Electron Scattering from the Nucleon

102

d) Since the electromagnetic Lagragngian density is L = jµ Aµ and in the nonrelativistic limit Pµ /2M → δµ0 , we must have Z d3 q iq·r e F1 (q2 ) = ρ(r) (2pi)3 Then Z

so F1 (0) =

R

d3 re−iq·r ρ(r) = F1 (q)

d3 rρ(r) = Ze. We also have

< J, J|Q̂33 |J, J >= so J0;J CJ2;J

1 (2M )

eQ < J, J|3Jˆz2 − Ĵ2 |J, J >= eQ J(2J − 1

1 1 2 3 1 2 3 2 2 2 F3 (0)q ( 2 cos θ − 2 ) = 6 q ( 2 cos θ − 2 )eQ

so s (2M )2 1 2M 2 (J + 1)(2J + 3) eQ F3 (0) = − eQ 20;J = − 6 3 J(2J − 1) CJ2;J For magnetic field Jk;J qk CJ1;J F2 (0) =

2M

r

J qz qz =µ J + 1 2M 2mN

so r F2 (0) = µ ·

r J +1 M J +1 = µA J mN J

7.10 The Anapole Moment The electromagnetic interaction becomes more complex when the weak interaction, and its associated parity violation, is included. In addition to the charge, magnetic dipole, electric quadrupole, etc. moments, new effects arise. One such effect, an interaction between the nuclear spin and the electron momentum called the anapole moment, is explored in this exercise.

Consider the matrix element of the electromagnetic current between a pair of nucleons, which can be written as

ν em 2 2 iσµν q N (p2 )|Vµ (0)|N (p1 ) = ūN (p2 ) F1 (q )γµ − F2 (q ) uN (p1 ) , 2mN where q = p1 − p2 is the four-momentum transfer. a) Verify that the current matrix element is conserved, in that q µ N (p2 )|Vµem (0)|N (p1 ) = 0.

103

Elastic Electron Scattering from the Nucleon

b) By going to the nonrelativistic limit,( q0 ' 0, |q| << mN ), show that the matrix element becomes NR

hN (p2 )|V0em (0)|N (p1 )i −→ F1 (0)χ†2 χ1 + . . .

NR

N (p2 )|Vjem (0)|N (p1 ) −→ F1 (0)

(p2 + p1 )j † (F1 (0) + F2 (0)) † χ2 χ1 − χ2 i(σ×q)j χ1 +. . .

2mN 2mN

R c) Equating dxV0em with the electric charge, show that we require F1 (0) = QN , where QN is the nucleon charge. d) Show that the nucleon magnetic moment is given by e (F1 (0) + F2 (0))σ , µN = 2mN so that F2 (0) is identified with the anomalous magnetic moment κN .

In the first part of this exercise, it was assumed that parity is conserved. However, in the presence of the weak interaction, parity is violated, and the general form of the nucleon matrix element of the electromagnetic current requires the addition of two additional form factors. e) Show that in the presence of parity violation, the general form of the nucleon electromagnetic matrix element is

iσµν q ν N (p2 )|Vµem (0)|N (p1 ) = ūN (p2 ) F1 (q 2 )γµ − F2 (q 2 ) 2mN

iσµν q ν γ5 F3 (q 2 ) (γµ q 2 − qµ 6 q)γ5 − F4 (q 2 ) uN (p1 ) . + 2mN 2mN f) Verify that the result in a) still pertains. g) Extend b) and show that NR

hN (p2 )|V0em (0)|N (p1 )i −→ F1 (0)χ†2 χ1 − NR

N (p2 )|Vjem (0)|N (p1 ) −→ F1 (0)

F4 (0) † χ σ · qχ1 + . . .

2mN 2

p2 + p1 † (F1 (0) + F2 (0)) † χ2 χ1 − χ2 i(σ×q)j χ1 2mN 2mN

+F3 (0)(q 2 χ†2 σj χ1 − qj χ†2 σχ1 · (p1 + p2 ) + . . . h) Show that a nonzero value of F4 (0) indicates a violation of both parity and time-reversal invariance and corresponds to the existence of an electric dipole moment of size e σ. dE = F4 (0) 2mN i) Suppose the nucleon interacts with an electron. Show that a nonzero value of F3 (0) indicates a violation of parity, but not time reversal invariance, and signals the existence of a local interaction H=

1 F3 (0)σ N · γ0 γ elec , 2mN

Elastic Electron Scattering from the Nucleon

104

where σ N is the nucleon spin and the subscript elec indicates the electron matrix element. Because of the local nature of the interaction, this sort of term is not part of the typical electromagnetic multipole expansion. This form factor F3 (0) was called the anapole moment by Zeldovich.

Exercise 7.10 Solution: The Anapole Moment a) We have

q µ < N (p2 )|Vµem |N (p1 ) >= ū(p2 ) 6 qF1 (q 2 ) −

i σµν q µ q ν F2 (q 2 ) u(p1 ) = 0 2mN

since ū(p2 ) 6 qu(p1 ) = ū(p2 )(6 p1 − 6 p2 )u(p1 ) = (mN − mN )ū(p2 )u(p1 ) = 0 b) In NR limit s 0

ū(p2 )γ u(p1 ) '

×

χ†2

σ · p2 − χ†2 2mN

2mN

1 0

0 −1

= χ†2 χ1 + O(

1 ) m2N

j

ū(p2 )γ u(p1 ) '

p2

2mN + 2m1N

2mN

s

s

p2

2mN + 2m2N

χ1 σ·p1 2mN

p2

2mN + 2m1N

2mN

σ · p2 0 × χ†2 − χ†2 −σj 2mN = χ†2 χ1

s

p2

2mN + 2m2N

2mN σj 0

χ1 σ·p1 2mN

i 1 p1j + p2j − (χ† σχ1 × (p1 − p2 ))j + O( 2 ) 2mN 2mN 2 mN s ij

ū(p2 )σ u(p1 ) '

p2 2mN + 2m2N

s

p2

2mN + 2m1N

2mN

2mN

σ · p2 σk × χ†2 − χ†2 ijk 0 2mN

0 σk

= ijk χ†2 σk χ1 + O(

1 ) m2N

χ1 σ·p1 2mN

Elastic Electron Scattering from the Nucleon

105

s ū(p2 )σ 0j u(p1 ) '

×i χ†2 = iχ†2 χ1

p2

2mN + 2m2N

2mN

σ · p2 0 − χ†2 σj 2mN

s

p2

2mN + 2m1N 2mN

σj 0

χ1

σ·p1 2mN

p1j − p2j 1 + (χ†2 σχ1 × (p1 + p2 ))j + O( 2 ) 2mN mN

Then < N (p2 )|V0em |N (p1 ) >= F1 (q 2 )ū(p2 )γ 0 u(p1 ) −

i F2 (q 2 )ū(p2 )σ 0j u(p1 )qj 2mN

NR

−→ F1 (0)χ†2 χ1 + . . . < N (p2 )|Vjem |N (p1 ) >= F1 (q 2 )ū(p2 )γ j u(p1 ) − NR

−→ F1 (0)χ†2 χ1

i F2 (q 2 )ū(p2 )σ jα u(p1 )qα 2mN

i p1j + p2j − (F1 (0) + F2 (0)) (χ† σχ1 × q)j + . . .

2mN 2mN 2

c) Since < N (p2 )|V0em (x)|N (p1 ) >= ei(p2 −p1 )·x < N (p2 )|V0em (0)|N (p1 ) > we have QN χ†2 χ1 (2π)2 δ 3 (p2 − p1 ) =

Z

d3 x < N (p2 )|V0em (x)|N (p1 ) >

= (2π)2 δ 3 (p2 − p1 ) < N (p2 )|V0em (0)|N (p1 ) >= (2π)2 δ 3 (p2 − p1 )F1 (0)χ†2 χ1 so F1 (0) = QN or from previous problem find F1 (0) = Ze. d) Now look at < N (p2 )|Vjem |N (p1 ) >= −(F1 (0) + F2 (0))

i (χ† σχ1 × q)j + . . .

2mN 2

and use previous problem. We must translate notation. Define K via χ†2 (↑)σz χ1 (↑) = 1 = K < so for spin 12 we have K = we have

q

3 2 =

q

11 111 ; 1, 0| 1> 22 222

J+1 J . Then, replacing F2 (0) −→ F1 (0) + F2 (0)

e K(F1 (0) + F2 (0)) = 2mN

r

J +1 µ J

i.e., (F1 (0) + F2 (0))

e =µ 2mN

Elastic Electron Scattering from the Nucleon

106

e) Clearly must introduce γ5 for parity violation. Then besides the two terms given, could also have P µ γ5 . However, we have Gordon identities ū(p2 )γµ γ5 u(p1 ) = ū(p2 )

6 p2 1 γµ γ5 u(p1 ) = ū(p2 ) (p2µ + iσµν pν2 ) γ5 u(p1 ) mN mN

ū(p2 )γµ γ5 u(p1 ) = −ū(p2 )γµ

6 p1 1 γ5 u(p1 ) = − ū(p2 ) (p1µ − iσµν pν1 ) γ5 u(p1 ) mN mN

so, subtracting i Pµ ū(p2 )γ5 u(p1 ) − ū(p2 )σµν q ν γ5 u(p1 ) 2mN 2mN

=

so is not independent. f) Current is clearly conserved since σµν q µ q | nu = 0 and q µ ū(p2 )(γµ q 2 − 6 qqµ )γ5 u(p1 ) = ū(p2 ) 6 qq 2 − 6 qq 2 )u(p1 ) = 0 g) Note that s 0

ū(p2 )γ γ5 u(p1 ) '

ū(p2 )(γ 0 q 2 − 6 qq 0 )γ5 u(p1 ) =

j

ū(p2 )γ γ5 u(p1 ) ' ×

2mN σ·p1

0 −1

2mN

χ1

q 2 − q02 † 1 χ σχ1 · (p1 + p2 ) + O( 3 ) 4m2N 2 mN

s

χ†2

p2

2mN + 2m1N

2mN

σ · p2 1 × χ†2 − χ†2 0 2mN

s

p2 2mN + 2m2N

s

p2 2mN + 2m2N

2mN

σ · p2 − χ†2 2mN

p2

2mN + 2m1N 2mN

0 −σj

σ·p1

σj 0

2mN

χ1

ū(p2 )(γ j q 2 − 6 qq j )γ5 u(p1 ) = q 2 χ†2 σj χ1 − qj χ†2 σχ1 · (p1 + p2 ) + O( s ij

ū(p2 )σ γ5 u(p1 ) '

p2

2mN + 2m2N 2mN

σ · p2 σk × χ†2 − χ†2 ijk 0 2mN =

s

1 ) m3N

p2

2mN + 2m1N 2mN

0 σk

σ·p1 2mN

χ1

1 1 ijk (χ†2 χ1 (p1k − p2k ) − i(χ†2 σχ1 × (p1 + p2 )))k + O( 2 ) 2mN mN

Elastic Electron Scattering from the Nucleon

107

s 0j

ū(p2 )σ γ5 u(p1 ) '

×i χ†2

p2

2mN + 2m2N

s

p2

2mN + 2m1N

2mN 2mN

σ·p1 σ · p2 0 σj 2mN − χ†2 σj 0 2mN χ1 = iχ†2 σj χ1 + O(

1 ) m2N

Then < N (p2 )|V0em |N (p1 ) >= F1 (q 2 )ū(p2 )γ 0 u(p1 ) − NR

−→ F1 (0)χ†2 χ1 −

1 F4 (0)χ†2 σχ1 · q + . . .

2mN

< N (p2 )|Vjem |N (p1 ) >= F1 (q 2 )ū(p2 )γ j u(p1 ) − NR

−→ F1 (0)χ†2 χ1 +

i F2 (q 2 )ū(p2 )σ 0j u(p1 )qj 2mN

i F2 (q 2 )ū(p2 )σ jα u(p1 )qα 2mN

i p1j + p2j − (F1 (0) + F2 (0)) (χ† σχ1 × q)j χ1 2mN 2mN 2

1 F3 (0)(q 2 χ†2 σj χ1 − qj χ†2 σχ1 · (p1 + p2 )) + . . .

2mN

h) When dotted with the vector potential, this corresponds to an interaction of the form e e H= σ · ∇φ = − S·E 2mN mN i.e., an electric dipole moment eS/mN . Since an electric field changes sign under parity but not time reversal, while spin changes sign under T but not under P, this interaction violates both T and P. i) Since spin is invariant while momentum changes sign under parity, a term linear in spin and quadratic in momentum does not change sign under a parity transformation. However, the spatial component of a polar vector must change sign, so this must be a pseudovector. The term proportional to q µ can be dropped due to gauge invariance. The factor of q 2 cancels the 1/q 2 of the photon propagator, meaning this is a local interaction proportional to χ†2 σχ1 on the hadronic side, which is dotted into u † γ 0 γu on the electron side.

8

Elastic Electron Scattering from the Nucleon 8.1 Elastic Electron-Proton Scattering Consider the elastic scattering of electrons from a proton target. a) Using the expressions for the virtual photoabsorption cross section, show that the ep elastic cross section may be written as R=

Q2 Q2 1 d2 σ /σM ott = ( 2 )2 RL + 2 RT , 0 dΩdE q 2q E

where q 2 = Q2 + ν 2 q2 θe E = [1 + 2 2 tan2 ]−1 .

Q 2 2

Q b) Further, with τ = 4m show that 2 N

RL = (1 + τ )G2E RT = 2τ G2M .

Exercise 8.1 Solution: Elastic Electron Scattering from the Nucleon (a) We can write the elastic ep cross section as a product of the Mott cross section, a δ-function to conserve energy, and the standard one-photon exchange angular distribution including the Sachs electric and magnetic elastic form-factors

2 4α2 E 0 cos2 ( θ2 ) d2 σ Q2 G2E + τ G2M 2 2 θ = δ(ν − ) + 2τ GM tan ( ) , dΩdE 0 Q4 2m 1+τ 2 where τ ≡ Q2 /4m2N . Further, the cross section can be written in terms of functions W1 (ν, Q2 ) and W2 (ν, Q2 ) as

2 4α2 E 0 cos2 ( θ2 ) d2 σ 2 2 2 θ = W2 (ν, Q ) + 2W1 (ν, Q ) tan ( ) . dΩdE 0 Q4 2 108

Elastic Electron Scattering from the Nucleon

109

Thus, GE and GM are related to W1 (ν, Q2 ) and W2 (ν, Q2 ) by W1elastic (ν, Q2 ) =

Q2 2 Q2 Q2 GM δ(ν − ) = τ G2M δ(ν − ) 2 4mN 2mN 2mN

W2elastic (ν, Q2 ) =

Q2 G2E + τ G2M δ(ν − ).

1+τ 2mN

We express W1 and W2 in terms of longitudinal and transverse cross sections as K σT 4π 2 α K 1 W2 = (σT + σL ) , 2 4π α 1+τ

W1 =

where K is the flux of photons. Substituting into the expression for the elastic ep cross section, we have K 1 2K θ d2 σ /σM ott = (σT + σL ) + 2 σT tan2 ( ) 0 2 dΩdE 4π α 1+τ 4π α 2

K Q2 2K K Q2 2 θ = σ + tan ( ) + σT L 4π 2 α q 2 4π 2 α 2 4π 2 α q 2

K Q2 q2 K Q2 2 θ σL + 2 1 + 2 2 tan ( ) σT . = 4π 2 α q 2 4π α q 2 Q 2 With the definitions K q2 σL 4π 2 α Q2 K q2 σT RT ≡ 2 2 4π α Q2 −1

q2 2 θ , E ≡ 1 + 2 2 tan ( ) Q 2 RL ≡

we have d2 σ Q2 Q2 1 /σM ott = ( 2 )2 RL + 2 RT 0 dΩdE q 2q E

QED.

(b) From (a), we have W1 = τ G2M =

K σT = RT /2 .

4π 2 α

Thus, RT = 2τ G2M

QED .

Also from (a), we have G2E + τ G2M K Q2 Q2 Q2 K K = (σT + σL ) 2 = σL 2 + 2 σT 2 2 2 1+τ 4π α q 4π α q 4π α q 2 2 2 Q Q = RL + RT /2 2 . q2 q

W2 =

110

Elastic Electron Scattering from the Nucleon 2

2

Q Q 2 Note that q 2 = Q2 + q 2 = Q2 + ( 2m )2 = Q2 (1 + 4m 2 ) = Q (1 + τ ). Thus, N N

q2 Q2 = 1 + τ . We can then write

RL =

2 q 1 τ Q2 G2E + G2M − τ G2M 2 ( 2 )2 1+τ 1+τ q Q

= (1 + τ )G2E + τ (τ + 1)G2M − τ (τ + 1)G2M = (1 + τ )G2M .

(8.1) 8.2 Elastic Electron-Proton Scattering Kinematics Consider elastic electron-proton scattering. The electron beam of energy E is incident on a fixed proton target, while the final-state electron has energy E 0 and is scattered through an angle θe with respect to the incident beam direction. a) Show that the final-state electron energy in elastic kinematics is given by E0 =

E

.

2 1 + 2E M sin (θe /2)

b) Show that the angle θq , between the direction of the three-momentum transfer q and the incident beam, is given by E0 sin θq = p sin θe , Q2 (1 + τ ) 2

Q where Q2 is the four-momentum transfer squared and τ ≡ 4M 2. c) Consider an experiment to measure the elastic scattering cross section at an incident energy of 2 GeV as a function of

−1 E ≡ 1 + 2(1 + τ ) tan2 θe /2 ,

the photon’s degree of longitudinal polarization. It is desired to measure the elastically scattered electron and proton in coincidence over E values of 0.3 to 1.0. Over what range in θe and θq must the electron and proton be detected, respectively?

Exercise 8.2 Solution: Elastic Electron-Proton Scattering Kinematics (a) From energy-momentum conservation at the electron vertex, we have θ Q2 = 4EE 0 sin2 ( ) .

2 From energy-momentum at the hadron vertex, we have W 2 = m2N − Q2 + 2mN ν .

For elastic scattering W 2 = m2N so Q2 = 2mN ν. Thus, we have θ 4EE 0 sin2 ( ) == 2mN (E − E 0 ) .

2

Elastic Electron Scattering from the Nucleon

111

Solving for E 0 , we have E0 =

E 2E sin2 ( θ2 ) 1+ m N

.

(b) Applying the law of sines, we have |k0 | |q| = , sin θq sin θ so with |k0 | = E0 and |q| = q sin θq =

E 0 sin θ . q

We have q 2 = ν 2 + Q2 so s s 2 p p Q Q2 )2 + Q2 = Q2 (1 + ) = Q2 (1 + τ ). q = ν 2 + Q2 = ( 2mN 4m2N Hence sin θq = p

E0 Q2 (1 + τ )

sin θ.

(c) We have E = 2 GeV. We can write Q2 4m2N 2mN ν = 4m2N E − E0 = 2mN E − 1+2E/mE sin2 ( θ ) N 2 = .

2mN

τ=

Now τ is expressed as a function of E and θ only. Thus, we can write    −1 E − 1+2E/mE sin2 ( θ ) θ N 2  E = 1 + 2  tan2 ( ) .

2mN 2 For 0.3 ≤ E ≤ 1 and E = 2 GeV, we can solve for θ: it lies between 0◦ and 80◦ .

Next we consider θq . From (b), we have sin θq = p

E0 Q2 (1 + τ )

sin θ E 2E 1+ m sin2 ( θ2 )

N =v # sin θ . u E "

E− 2E θ u 2 1+ sin ( ) mN 2 E t2mN E − 1+ 2mN 1+ 2E sin2 ( θ ) mN

2

Elastic Electron Scattering from the Nucleon

112

For 0◦ ≤ θ ≤ 80◦ , we have 21◦ ≤ θq ≤ 90◦ .

8.3 Dipole Form Factor Consider elastic electron scattering from the proton. a) If the proton charge distribution ρ(r) is assumed to have an exponential form ρ0 e−r/α , show that the form factor has the dipole form F (|q|) =

1 , (1 + α2 q 2 )2

with q 2 = −|q|2 . √ b) Show that the RMS proton charge radius is 12α. c) The Compton wavelength of a particle of mass m is defined as 1 . m It is the approximate length scale at which relativistic quantum field theory becomes necessary for an accurate description. First, write the above equation with appropriate factors ~ and c to make it dimensionally correct. Second, calculate the Compton wavelength of the proton. In electron scattering, the characteristic length scale probed at a given momentum transfer q is λC =

λ∼

1 . q

Again, include factors ~ and c to make the expression dimensionally correct.

Explain why the charge distribution of the proton is not well determined from experiment.

Exercise 8.3 Solution: Dipole Form Factor We have ρ(r) = ρ0 e−r/α .

The total charge Q must be Z Q=

ρ(r)dr Z ∞

ρ(r)4πr2 dr Z ∞ = 4πρ0 e−r/α r2 dr

=

0

0

= 8πρ0 α3 .

Elastic Electron Scattering from the Nucleon

113

(a) We compute Z 4π ρ(r) sin(qr)rdr q Z 4π ∞ = ρ0 e−r/α sin(qr)rdr q 0 4πρ0 2α3 q = q (1 + α2 q 2 )2 1 = 8πρ0 α3 (1 + α2 q 2 )2 1 . =Q· (1 + α2 q 2 )2

F (q 2 ) =

We identify F (|q|) =

1 .

(1 + α2 q 2 )2

(b) We can write Z 4π F (q ) = ρ(r) sin(qr)rdr q

Z 4π 1 3 = ρ(r) qr − (qr) ..... rdr q 3!

Z 4π 1 −r/α 3 = ρ0 e qr − (qr) ..... rdr q 6 Z Z 1 = ρ(r)4πr2 dr − q 2 r2 ρ(r)4πr2 dr + .....

6 Z ∞ 1 = 8πρ0 α3 − q 2 r2 ρ(r)4πr2 dr + .....

6 0

1 2 2 = Q 1 − q < r > + ....... , 6 2

where R ∞ 2 −r/α r ρ0 e 4πr2 dr <r >≡ 0 Q R ∞ 2 −r/α r ρ e 4πr2 dr 0 = R0 ∞ −r/α ρ0 e 4πr2 dr 0 2

4πρo · 24α5 4πρ0 · 2α3 = 12α3 . =

Thus, the RMS proton charge radius is √ √ < r2 > = 12α.

Elastic Electron Scattering from the Nucleon

114

(c) The Compton wavelength for the proton is λC =

197 M eV · f m ~c = = 0.21 f m .

2 mp c 938 M eV

The proton charge radius is experimentally determined to be about 0.85 fm. Thus, for momentum transfers above about 1 (GeV/c)2 we expect that relativistic effects will be large. For low momentum transfers, i.e. distance scales longer than the Compton wavelength, we expect that the Fourier transform of the proton formfactor would be a reasonable approximation to the charge distribution. Note that for nuclei the Compton wavelength will be smaller by a factor of A and thus the Fourier transform of the charge form-factor obtained from electron scattering is valid, and does not have the difficulty for the case of the nucleon.

8.4 Charged Particle Scattering The Compton wavelength of a particle of mass m is defined in the previous exercise. a) Calculate the Compton wavelength for the nucleus 40 Ca. b) As shown in exercise 8.3, in electron scattering the characteristic scale probed at a given momentum transfer q is λ ∼ ~c/q. Compare elastic electron scattering from the proton with 40 Ca. Explain why the charge distribution of 40 Ca is well determined from experiment but the charge distribution of the proton is not.

Exercise 8.4 Solution: Charged Particle Scattering (a) We have ~ mnucleus c 1.05 × 10−34 = 39.96 × 1.66 × 10−27 × 3 × 108 = 5.3 × 10−18 m

λC =

= 5.3 × 10−3 f m .

(b) The characteristic length scale probed experimentally is λ ≈ ~/q. For 40 Ca, the length scale at which the relativistic quantum field theory corrections become applicable is about 40 times smaller than for the proton. Hence, for electron scattering experiments from 40 Ca at energies below 1 GeV, the charge radius determined from electron scattering experiments is valid.

8.5 Continuous Beam Consider an experiment to measure the reaction (e, e0 p) on a nucleus A. Assume that the beam comes in pulses of width T with f beam pulses per second. The

115

Elastic Electron Scattering from the Nucleon

electrons are detected with a time-averaged rate Re in an electron spectrometer.

The protons are detected with a time-averaged rate Rp in a proton spectrometer.

Coincidence events are detected at a time-averaged rate Rcoinc within a coincidence resolving time τ . (a) Show that the fraction of the time the beam is on target, called the duty factor d, is given by d = T f . (b) Prove that the ratio of true events to accidental events can be approximated by

trues Rcoinc d = . accidentals Re Rp τ This shows that the quality of coincidence experiments is greatly improved by using continuous rather than pulsed electron beams. This has driven the technical development of continuous electron beams and has greatly improved the scientific scope of electron scattering experiments.

Exercise 8.5 Solution: Continuous Beam

(a) With f pulses per second, each of width T , the fraction of time the beam is on target is d=T ·f , known as the duty factor. The SLAC End Station A beam had a duty factor of about 10−4 ; the medium energy accelerators at MIT-Bates, Saclay and NIKHEF typically had a duty factor of about 10−2 . Modern electron accelerators at Jefferson Lab and Mainz have duty factors close to unity. (b) The time-averaged singles rates in the electron and proton detectors are Re and Rp , respectively. Thus, when the beam is on, these rates are increased by a factor of 1/d. The instantaneous rates in the electron and proton detectors are Re /d and Rp /d, respectively. The accidental trigger rate Racc within a time window τ when the beam is one is given by Racc =

Re Rp · ·τ . d d

A low duty-factor machine will have a very high rate of singles and consequently a very high rate of accidental coincidences. The instantaneous true coincidence rate when the beam is on is Rtrue = Rcoinc /d. Thus the ratio of true to accidental

116

Elastic Electron Scattering from the Nucleon

coincidences is given by trues Rtrue = accidentals Racc d2 Rcoinc · = d Re Rp τ Rcoinc d = Re Rp τ .

(8.2)

8.6 Measurement of Spin Asymmetries Consider scattering of longitudinally polarized electrons from a polarized proton target. In a single-arm experiment, the final-state scattered electron is detected and its four-momentum is measured. The experimental spin asymmetry is measured by reversing the direction of either the incident electron spin or the target proton spin.

The numbers of events for parallel spins (N+ ) and anti-parallel spin-states (N− ) are measured and their relative luminosity determined. Then, one can form the experimental asymmetry N+ − N− .

Aexp = N+ + N− a) For small asymmetries A 1, show that the statistical uncertainty in the experimental asymmetry is δA = √1N , where N ≡ N+ + N− is the total number of events. b) In a real experiment, the electron beam polarization pB and the target polarization pT are less than unity. Further, there can be extraneous unpolarized material in the polarized target which dilutes the measured asymmetry by a dilution factor f . Show that √ δA ∼ ( N · f · pB · pT )−1 .

A common target material used is ammonia (NH3 ) where f = 3/17.

Exercise 8.6 Solution: Measurement of Spin Asymmetries (a) The measured asymmetry is A=

N+ − N− .

N+ + N−

We assume that N+ and N− are independent with no correlations so s ∂A 2 ∂A 2 δA = ( ) (δN+ )2 + ( ) (δN− )2 . ∂N+ ∂N− Further, we assume that both N+ and N− are normally distributed so that δN+ =

Elastic Electron Scattering from the Nucleon

117

p N+ and δN− = N− . Hence, we have "

2

2 #2 1 1 N+ − N− N+ − N− δA = N+ + N− − − .

N+ + N− (N+ + N− )2 N+ + N− (N+ + N− )2

p

Let N = N+ + N− so we have A=

N+ − N− N+ − N− = 1.

N+ + N− N

Then 1/2

1 1 δA = N+ 2 (1 − A)2 + N− 2 (1 − A)2 N N

1/2 1 1 ≈ N+ 2 + N− 2 N N 1 =√ N .

(8.3)

(b) The measured asymmetry Ameas is related to the physics asymmetry Aphys as follows N+ − N− Ameas = = pB · pT Aphys .

N+ + N− Lower beam and target polarizations reduce the size of the measured asymmetry or conversely increase the experimental uncertainty. For no dilution √ −1 N · pB · pT . δA = If we now add extraneous material to the polarized target so that there is a dilution f , the counting statistics will include scattering from both polarized and extraneous material. Then √ −1 δA = N · f · pB · pT , where we note that N will be increased by scattering from the extraneous material.

Hadron Structure via Lepton-Nucleon Scattering

9

9.1 SU(6) Picture of the Proton Consider the simple nonrelativistic SU (6) wavefunction for the polarized proton r p ↑=

r 2 1 (u ↑ u ↑)d ↓ + (u ↑ u ↓)d ↑ .

3 3

a) Show that u ↑= 35 , u ↓= 13 , d ↑= 13 , and d ↓= 23 . b) Hence, show that in this model the total spin of the proton is carried by the quarks. c) Show that ∆u − ∆d = 53 . This is equivalent to the Bjorken sum rule. Compare with data. d) Determine the asymmetries for spin-dependent DIS from the proton and neutron in this model. Compare with data. e) Determine the ratio of proton to neutron magnetic moment in this model. Compare with data.

Exercise 9.1 Solution: SU(6) Picture of the Proton The proton wavefunction is r |p ↑>=

2 (u ↑ u ↑)d ↓ + 3

r

1 (u ↑ u ↓)d ↑ .

3

(a) The probability to find the u quark as ↑, i.e. P (u ↑), is r r 5 2 2 1 2 ) +( ) = .

P (u ↑) = 2 · ( 3 3 3 Similarly, r

1 2 1 ) = 3 3 r 1 1 2 ) = P (d ↑) = ( 3 3 r 2 2 2 P (d ↓) = ( ) = .

3 3

P (u ↓) = (

(b) We have ∆u ≡ P (u ↑) − P (u ↓) = 43 . Similarly, ∆d ≡ P (d ↑) − P (d ↓) = − 13 .

118

Hadron Structure via Lepton-Nucleon Scattering

119

Thus, ∆u + ∆d = 1. In this model, the valence quarks carry 100% of the proton’s spin. (c) We then have 4 1 5 − (− ) = .

3 3 3 This is the equivalent of the Bjorken Sum Rule. ∆u − ∆d =

(d) Consider now the asymmetries we expect to measure in spin-dependent DIS. We can write P 2 i ei ∆qi .

A= P e2i qi Using the results above, for the proton we expect ( 32 )2 ∆up + (− 13 )2 ∆dp ( 23 )2 · 2 + (− 13 )2 · 1 4 1 = ∆up + ∆dp 9 9 1 4 4 1 = · + · (− ) 9 3 9 3 15 = 27 5 = .

9 Similarly, for the neutron we expect Ap =

An =

( 23 )2 ∆un + (− 31 )2 ∆dn ( 23 )2 · 1 + (− 13 )2 · 2 4

= 9 4

= 9

∆un + 19 ∆dn 6 9

1 4 · −1 3 + 9 · (3) 6 9

=0.

Note we have used isospin symmetry to assume ∆un = ∆dp and ∆dn = ∆up .

The measured data in the valence quark region x ≈ 0.5 are in reasonable agreement with this prediction. (e) The ratio of proton to neutron magnetic moments is given by 2 ∆up − 13 ∆dp µp = 32 1 µn 3 δdp − 3 ∆up 2 4 ( ) − 1 (− 1 ) = 32 3 1 3 1 43 3 (− 3 ) − 3 ( 3 ) 3 =− .

2

Hadron Structure via Lepton-Nucleon Scattering

120

µ

Experimentally, µnp is -1.46, in very good agreement with the valence quark model prediction.

9.2 The Mandelstam Variables In two-body scattering A + B → C + D, it is convenient to introduce the Mandelstam variables: s ≡ (pA + pB )2 /c2 t ≡ (pA − pC )2 /c2

u ≡ (pA − pD )2 /c2 . a) Show that s + t + u = m2A + m2B + m2C + m2D . b) Find the CM energy of A in terms of s, t, u and their masses. c) Find the LAB energy of A, assuming B is at rest. d) Find the total CM energy. e) For elastic scattering of identical particles of mass m A + A → A + A, p is the 3-momentum of the incident particle in the CM frame and θ is the scattering angle. Express s, t, u in terms of p, θ and m.

Exercise 9.2 Solution: The Mandelstam Variables Use Mandelstam variables s = (pA + pB )2 , t = (pA − pC )2 , u = (pA − pD )2 Then a) Expanding and adding, we find s + t + u = 3m2A + m2B + m2C + m2D + 2pA · (pB − pC − pD ) = 3m2A + m2B + m2C + m2D − 2pA · pA = m2A + m2B + m2C + m2D b) In CM the energy is given by CM CM 2 s = (EA + EB )

Thus

with

CM 2 CM 2 p2 = (EA ) − m2A = (EB ) − m2B

q √ CM CM )2 + m2 − m2 s = EA + (EA B A

Squaring, we have √ CM CM 2 CM 2 s + (EA ) − 2 sEA = (EA ) + m2B − m2A so

√ CM EA = (s + m2A − m2B )/2 s

c) In laboratory frame L s = m2A + m2B + 2mB EA

so

L EA = (s − m2A − m2B )/2mB

Hadron Structure via Lepton-Nucleon Scattering

121

√ CM d) Total CM energy is Etot = s e) In the equal mass case we have CM 2 s = 4(EA ) = 4(m2A + p2 ) CM 2 t = 2m2A − 2((EA ) − p2 cos θ) = −2p2 (1 − cos θ) CM 2 u = 2m2A − 2((EA ) + p2 cos θ) = −2p2 (1 + cos θ)

Check: s + t + u = 4m2A + 4p2 − 4p2 = 4m2A 9.3 Inclusive Deep Inelastic Cross Section The cross section for inclusive DIS in terms of the two structure functions is written as 2

d2 σ α~ 2W1 sin2 (θ/2) + W2 cos2 (θ/2) . = 2 0 dE dΩ 2E sin (θ/2) Using the Callan-Gross relation, show that the cross section can be written in terms of one structure function as follows

2

d2 σ F1 (x) α~ 2EE 0 2 = 1+ cos (θ/2) . dE 0 dΩ 2M (E − E 0 )2 E sin2 (θ/2) Using F1 (x) from [PDG14], one can calculate the inclusive DIS cross section as a function of kinematics.

Exercise 9.3 Solution: Inclusive Deep Inelastic Cross Section Beginning with the cross section d2 σ = dE 0 dΩ

α 2E sin2 θ2

!2

2W1 sin2

θ θ + W2 cos2 2 2

we use the Callan-Gross relation 4EE 0 sin2 θ2 W2 M F2 M q2 = = 2x = 2 = W1 ν F1 ν ν (E − E 0 )2 Then d2 σ = W1 dE 0 dΩ

α 2E sin2 θ2

F1 = 2M

α E sin θ2

!2 "

!2

4EE 0 sin2 θ2 θ + 2 sin cos2 2 (E − E 0 )2 2 2 θ

2EE 0 θ 1+ cos2 0 2 (E − E ) 2

#

Hadron Structure via Lepton-Nucleon Scattering

122

9.4 DIS Kinematics Consider DIS lepton scattering. a) On a plot of Q2 versus ν, show the region of deep inelastic scattering from a fixed target for a beam of leptons of energy E0 = (i) 5, (ii) 10, (iii) 20, and (iv) 50 GeV. 2 b) With the Mandelstam variable s = ECM , x is the Bjorken scaling variable , and p·q y = p·k , show that Q2 ≈ s · x · y . c) On a plot of Q2 versus ν (log versus log), plot the accessible DIS region for (i) 11 GeV lepton beam on a proton fixed target, (ii) 27 GeV electron beam colliding head-on with a 940 GeV proton beam (HERA collider which operated from 1993 until 2007 at DESY, Hamburg, Germany), and (iii) 10 GeV electron beam colliding head-on with a 250 GeV proton beam (proposed future electron-ion collider).

Exercise 9.4 Solution: DIS Kinematics (a) In ep scattering W 2 = (p + q)2 = M 2 − Q2 + 2p · q = M 2 − Q2 + 2M ν ≥ M 2 .

Hence, 2M ν ≥ Q2 . The equals sign is obtained for elastic scattering. Deep Inelastic Scattering (DIS) is defined for W 2 ≥ 4 (GeV)2 and Q2 ≥ 1 (GeV)2 , which corresponds to 2M ν + M 2 − 4 > Q2 . This is plotted in Fig. 9.1. (b) We have Q2 Q2 = 2p · q 2M ν p·q Mν y= = . p·k ME

x=

(9.1) (9.2) (9.3)

Thus, xy =

Q2 .

2M E

On the other hand, s = (p+k)2 = (E+M, k)2 = (E+M )2 −k2 = E 2 +2EM +M 2 −k2 = m2 +M 2 +2M E ≈ 2M E , neglecting the masses of the electron (m) and proton (M ). Then xy = i.e. Q2 ≈ s · x · y.

Q2 Q2 ≈ , 2M E s

Hadron Structure via Lepton-Nucleon Scattering

123

t

Fig. 9.1

DIS kinematics for Exercise 9.4(a). (c) The relation 2M ν + M 2 − 4 > Q2 is the condition for DIS. We write ν using the definition ν = p·q M , 2M ν = 2(p · q) = 2p · (k − k 0 ) ≈ 4Ep Ee − 4Ep Ee0 sin2 (

θe ), 2

when ignoring the mass of the proton and electron, which is reasonable for high2Ep Ee energy collider kinematics. The maximum ν is obtained for θe = 0◦ , with ν = M .

Hence, case(ii) :Ep = 940GeV, Ee = 27GeV, ν = 54100GeV

(9.4)

case(iii) :Ep = 250GeV, Ee = 10GeV, ν = 5329GeV

(9.5) (9.6)

Fig. 9.2 is a plot of the kinematics for the different cases.

9.5 Electron-Ion Collider Neglecting electron and proton masses, a) Show that the center-of-mass energy for a lepton beam of energy E0 on a proton √ is ECM = 2M E0 .

124

Hadron Structure via Lepton-Nucleon Scattering

t

Fig. 9.2

DIS kinematics for Exercise 9.4(c). b) Show that the center-of-mass energy for a lepton beam pof energy E0 colliding head-on with a proton beam of energy Ep is ECM = 4E0 · Ep . c) Calculate the energy of a lepton on a proton fixed target necessary to produce the same center-of mass energy as the electron-proton collisions.

Exercise 9.5 Solution: Electron-Ion Collider Neglecting electron and proton masses M and m, we have a) If in laboratory frame have lepton beam with energy E0 then s = M 2 + m2 + √ 2 2M E0 ' 2M E0 . Then in CM frame have s = ECM so EC M = 2M E0 b) In head on collision between lepton beam with energy E0 and proton beam with 2 energy Ep we have s =pM 2 + m2 + 2(E0 Ep + p0 pp ) ' 4E0 Ep . Setting s = ECM we have then ECM = 4E0 Ep c) To produce the same center of mass energy, we must have a lepton beam with energy Ẽ0 with

q p 2Ep 2M Ẽ0 = 4E0 Ep so Ẽ0 = E0 M

10

High-Energy QCD 10.1 Relativistic Collider Kinematics Consider some basic quantities used to analyze high energy collider experiments. a) Determine the electron beam energy in a fixed target ep mode configuration that would be required to reach a center-of-mass energy of 300 GeV? b) How does this electron beam energy compare to the electron beam energy in a ep collider mode configuration for a proton beam energy of 800 GeV to reach a center-of-mass energy of 300 GeV? c) Show that for a ep collider such as HERA we have s = (p + k)2 ' 4Ee Ep . d) Defining x, y variables via x=

Q2 , 2p · q

y=

p·q , p·k

0 ≤ x, y ≤ 1 ,

show that we have q 2 ' sxy, where we have defined Q2 = −q 2 = −(ke − ke0 )2 . e) Derive the relations used in Eq, 9.15-9.19 θ0

Ee0 cos2 2e Q2 xe = e = E0 θ0 sye Ep (1 − e sin2 e ) Ee

ye = 1 −

2

Ee0 E0 θ0 (1 − cos θe0 ) = 1 − e sin2 e 2Ee Ee 2

Q2e = 2Ee Ee0 (1 + cos θe0 ) = 4Ee Ee0 cos2 and Q2e (x, θe0 ) =

p2T,e θe0 = 2 1 − ye

sx θ0 E 1 + (x Epe ) tan2 ( 2e ) E0

Q2e (x, Ee0 ) =

sx(1 − Eee ) E

1 − (x Epe )

.

f) Demonstrate that the relative errors of xe , ye , Qe can be written as

0

1 δEe0 xe θe δxe = ⊕ − 1 tan δθe0 0 xe ye Ee Ee /EP 2

0

δye 1 δEe0 θ 1 = 1− − 1 cot e δθe0 ⊕ ye ye Ee0 ye 2 125

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126

δQ2e Q2e

=

δEe0 ⊕ tan Ee0

0 θe δθe0 .

2

.

Exercise 10.1 Solution: Relativistic Collider Kinematics Define M as the proton mass and m the electron mass. a) For fixed target we have 2 ECM = (pp + pe )2 = M 2 + m2 + 2M Ee ' 2M Ee

Then Ee '

(332M )2 ∼ 55, 000M 2M

b) For ep collider have 2 ECM = (pp + pe )2 = M 2 + m2 + Ep Ee + pp pe ' 4Ep Ee

Then Ee '

(332M )2 ∼ 30M 4Ep

c) See first line of b). d) We have xy =

p·q Q2 Q2 Q2 Q2 · = ' = 2p · q p · k 2p · k (p + k)2 s

e) We have, using Eqs. (10.20) and (10.21) ye =

p · (k − k 0 ) Ee − Ee0 + Ee + Ee0 cos θe0 E0 1 E0 θ0 = = 1− e (1−cos θe0 ) = 1− e sin2 e p·k 2Ee Ee 2 Ee 2 Q2e = −(k − k 0 )2 = 2k · k 0 = 2Ee Ee0 (1 + cos θe0 ) = 4Ee Ee0 cos2 θ0

θ0

4Ee Ee0 cos2 2e cos2 2e Q2 Ee0

xe = e = = 0 0 E0 θ0 sye Ep 1 − Ee sin2 θe 4E E 1 − e sin2 e p

e

Ee

Ee

2

2

Also θ0

cos2 2e Ep E0 xe = e 0 0 Ee Ee 1 − Ee sin2 θe Ee

2

so E0

2 θ

0

e sin 2e Ep θ0 1 1 + xe tan2 e = 1 + EeE 0 0 = 0 θ θ0 E 2 e e e Ee 2 1− sin 1− sin2 e

Ee

2

Ee

2

θe0 2

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127

and sxe

= 4Ee Ee0 cos2 θ0

E

1 + xe Epe tan2 2e

θe0 = Q2e 2

Alternatively E0

θ0

1 − Eee cos2 2e Ep E0 = 1 − xe =1− e 0 0 θ0 E0 Ee Ee 1 − Ee sin2 θe 1 − e sin2 e Ee

2

Ee

2

Then

E0 sxe 1 − Eee E

1 − xe Epe

= 4Ee Ee0 cos2

θe0 = Q2e 2

f) Then

0 0 δQ2e θe0 θe0 0 2 θe 2 θe 0 0 0 = 4E sin δE ⊕ −4E E cos sin δθ e e e e e /4Ee Ee sin 2 Qe 2 2 2 2 =

δEe0 θe0 0 ⊕ tan δθ Ee0 2 e

Also 

θ0

θ0

θ0

cos2 2e Ee0 δEe0 cos2 2e sin2 2e δxe  δE 0 + = e

2 ⊕ 0 0 xe Ep 1 − Ee sin2 θe Ep Ee E0 θ0 Ee 2 1 − Eee sin2 2e    0 0   E0 θ0 θ θ  E 0 cos θe0 sin θe0  3 e e 0 0 1 − Eee sin2 2e Ee Ee cos 2 sin 2 e 0  Ep 2 2 − + 2 δθe  0

E0 θ0 θ0   Ep Ee Ee E0 θ0 cos2 2e  Ep 1 − Eee sin2 2e 1 − Eee sin2 2e  =

Ep 1 δEe0 θe0 0 ⊕ −1 + δθ x e tan 0 ye Ee Ee 2 e

and

δye δE 0 θ0 E0 θ0 θ0 1 = − e sin2 e ⊕ − e cos e sin e δθe0 0 E θ0 ye Ee 2 Ee 2 2 1 − e sin2 e Ee

ye − 1 δEe0 θe0 0 ⊕ ctn δθe = ye Ee0 2

10.2 Collider Kinematic Resolutions

2

High-Energy QCD

128

a) Using the results of the previous problem and assuming a lepton momentum resolution of δEe0 /Ee0 = 1% and an angular resolution δθe0 = 5 mrad for Ee = 10 GeV, Ep = 250 GeV determine δx/x, δQ2 /Q2 , δy/y for lepton energies Ee0 of 2, 5, and 8 GeV and scattering angles θe0 of 30, 90, and 150 degrees, b) For a possible future electron-ion collider where a 10 GeV electron is incident head-on with a 250 GeV proton and assuming that the direction of the proton (electron) is the forward (rear) direction, sketch on a plot of Q2 versus x the directions of the final-state electron and current jet (along the direction of 3-momentum transfer) for (i) low x and low Q2 ; (ii) high x and low Q2 ; and high x and high Q2 .

Exercise 10.2 Solution: Collider Kinematic Resolutions

(a)

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129

(b) From Exercise 10-1(e), we have Q2e (x, θe ) =

sx E 1 + (x Epe ) tan2 ( θ2e )

=

4Ee Ep x E

1 + (x Epe ) tan2 ( θ2e )

.

Hence, we can plot Q2e against x for different electron scattering angles θe . (a)

As shown in this plot, we can discuss three regions: a) low x, low Q2 This corresponds to the kinematic region of large θe and large y. Thus, the outgoing electron is largely along the direction of the incoming electron. It also means that Ee0 is small, hence the 3-momentum transfer is large, and roughly against the direction of the incoming electron.

2 b) high x, low Q This corresponds to the kinematic region of large θe and small y. Thus, the outgoing electron is largely along the direction of the incoming electron. It also means that Ee0 is large, hence the 3-momentum transfer is small, and roughly along the direction of the incoming electron. c) high x, high Q2 This corresponds to the kinematic region of small θe and large y. Hence the outgoing electron is agains the direction of the incoming electron. It also means that

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130

Ee0 is large, hence the 3-momentum transfer is large and against the direction of the incoming electron.

10.3 Collider Physics Now let us consider a number of physics topics studied in collider experiments. a) Explain why the the muon cross section is used to normalize the hadron cross section in equation (9.28) and not the electron cross section. b) Explain the logarithmic slope in F2 . c) Explain the scaling phenomena in Drell-Yan production. d) Using the definition of rapidity

1 E + pz y= , 2 E − pz identify which partonic cross section is largest at mid-rapidity.

Exercise 10.3 Solution: Collider Physics (a) The electron cross section would be the Bhabha process e+ e− → e+ e− which involves both annihilation and scattering contributions. For e+ e− → µ+ µ− , we have only scattering contributions which is what is needed to form the ratio to e+ e− → hadrons. (b) In the small x limit of the DGLAP equations we have 2αs ∂F2 (x, Q2 ) = xG(x, Q2 ) 2 2 ∂ ln(Q /Λ ) 9π where the expected rise of the F2 slope is associated with the logarithmic Q2 growth of the gluon distribution xG(x, Q2 ) implied by the DGLAP equations in the small x limit. (c) The expression for the Drell-Yan cross section contains a factor M −4 that originates from the propagator of the intermediate photon. It is this factor that is responsible for the rapid fall of cross section with increasing mass. Rearranging the cross section, d2 σ M3 ≡ FBT (xF , τ ) , dM dxF where τ = M 2 /s = x1 x2 and FBT is a function that depends on the quark content 2 σ of the beam and target. Thus, M 3 dMd dx for a given combination of beam and F target is predicted to be independent of beam energy, provided measurements are made at identical values of xF and τ . (d) The figure shows the pseudorapidity distributions of produced charged hadrons

High-Energy QCD

131

in pp collisions at the LHC CM energy of 7 TeV. The underlying partial distribution functions including the Jacobian are shown as dash-dotted curves for the fragmentation sources arising from quark-quark and quark-gluon interactions, and as a dashed curve for the central source arising from gluon-gluon interactions. The dotted curve does not include the Jacobian.

10.4 W-Boson Production The production of a W-boson in a collider can be identified by its characteristic weak decay pattern. a) Determine the Bjorken-x scaling variables for initial-state quarks / anti-quarks in hadron-hadron collisions for W boson production in terms of the rapidity y, the W mass and center-of-mass energy. b) What is the threshold energy for W boson production in a e+ e− collider mode configuration? c) Provide an estimate of the threshold energy for W boson production in hadronhadron collisions.

132

High-Energy QCD

d) Determine the ratio of the W decay-width into quarks versus leptons.

Exercise 10.4 Solution: W-Boson Production (a) The simplest process by which to produce a W boson is via the Drell-Yan process: q + q̄ → W → l+ + l− .

The kinematics of this process are very simple. In the hadronic center of mass frame, each quark carries a fraction x1 and x2 of the proton momentum. Then p1 = (x1 Ebeam /2, 0, 0, x1 Ebeam /2)

(10.1)

p2 = (x2 Ebeam /2, 0, 0, −x2 Ebeam /2)

(10.2)

q = ((x1 + x2 )Ebeam /2, 0, 0, (x1 − x2 )Ebeam /2)

(10.3) (10.4)

For on-shell W production, the kinematics are fixed once the gauge boson rapidity is specified. Thus, x1 1 log 2 x2 2 x 1 x 2 s = MW , y=

(10.5) (10.6) (10.7)

and the values of Bjorken x that are relevant in this process are MW x1 = √ ey s

MW x2 = √ e−y . s

LHC Drell-Yan data cover a range in x from 10−4 to 1. (b) The threshold for e+ e− → W + W − is 161 GeV. (c) The classic paper here is Producing Massive Neutral Intermediate Bosons with Existing Accelerators by C. Rubbia, P. McIntyre and D. Cline from March 1976.

They write “A more realistic production process is the one initiated by proton-antiproton collisions: p + p̄ → W + hadrons , which, according to the quark (parton) picture, proceeds by a reaction analog to e+ + e− → W → l + + l − , except that now incoming e+ and e− are replaced with q and q̄. Strong support to the idea that W’s are directly coupled to spin 1/2 point-like constituents comes from neutrino experiments and from semi-leptonic hadron decays. Furthermore, neutrino experiments provide the necessary structure functions and have set limits

High-Energy QCD

133

(> 20 GeV) on any nonlocality in the parton form factor. The main difference with respect to e+ e− is that now the kinematics is largely smeared out by the internal motion of q’s and q̄’s. The average center of mass energy squared of the q− q̄ collision is roughly < sqq̄ >∼ s < xq >p < xq̄ >p̄ , where s is the center of mass energy squared of the pp̄ system and < xq >p ( < xq̄ >p̄ ) is the mean fractional momentum of qs (q̄s) in the proton (antiproton).

From neutrino measurements and < xq >p =< xq̄ >p̄ , we find < sqq̄ >∼ 0.04s. For √ M = 100 GeV, this suggests s ≥ 2 × 105 GeV2 or s ≥ 450 GeV. ” Rubbia’s UA1 experiment discovered the W and Z 0 with masses around 80 GeV using the CERN SPS with a center of mass energy of 540 GeV in 1983. (d) There are three decay final-states: a) leptonic channel W → lν¯l ¯lνl , BR = 45.6% b) hadronic channel W → q q̄q q̄, BR = 10.5% c) semi-leptonic channel W → lν¯l q q̄, BR = 43.9%

11

The Nucleon-Nucleon Interaction 11.1 One-Pion-Exchange Potential (OPEP) The nucleon-nucleon interaction can be, to a reasonable degree of approximation, be described in terms of a sum over exchanges of mesons π, σ, ρ, ω, ... between the nucleons. Of these exchanges, the most important at large distances comes from the exchange of the pion, since it has the lightest mass. The goal of this exercise is to find the form of this long-range interaction.

From isospin and parity invariance, the pion-nucleon interaction must have the form Lint = gπN N N̄ γ5 τ · πN , where gπN N is the pion-nucleon coupling constant and has the experimental value 2 gπN N ' 13.5 .

16π

a) Evaluate the pion-nucleon vertex in the nonrelativistic approximation p/mN << 1, and show that igπN N ū(pf )γ5 u(pi )ei(pf −pi )·x

p/mN <<1

= −gπN N

−→

igπN N

1 χ† σ · (pf − pi )χi ei(pf −pi ) 2mN f

1 χ† σ · ∇χi ei(pf −pi ) .

2mN f

b) Calculate the N N scattering amplitude in the nonrelativistic limit and show that 2 0 M(p1 +p2 → p01 +p02 ) = gπN N ū(p1 )γ5 τ 1 u(p1 )·

p/mN <<1

−→

1 ū(p02 )γ5 τ 2 u(p2 ) 0 (p1 − p1 )2 − m2π

1 g2 − πN2N χ†1f τ 1 σ 1 · ∇1 χ1i 0 χ† τ 2 σ 2 · ∇2 χ2i .

4mN (p1 − p1 )2 + m2π 2f

c) Calculate the OPEP by taking the Fourier transform and show that Z dq −iq·(r2 −r1 ) † † χ2f χ1f VOP EP (r 2 − r 1 )χ1i χ2i = e Amp(q) (2π)3 =− 134

2 gπN 1 N τ 1 · τ 2 σ 1 · ∇1 σ 2 · ∇2 exp −mπ |r 2 − r 1 | .

16πm2N |r 2 − r 1 |

The Nucleon-Nucleon Interaction

135

d) Perform the differentiations and show that VOP EP has the form

2 g2 1 1 1 1 mπ −mπ r VOP EP (r) = − πN N2 τ 1 ·τ 2 + + Ŝ + σ · σ e 12 1 2 2 16πmN (mπ r) mπ r 3 3 r

4π 3 − σ 1 · σ 2 δ (r) , 3 where Ŝ12 = 3σ 1 · r̂σ 2 · r̂ − σ 1 · σ 2 is the tensor interaction.

The OPEP potential thus consists of two pieces, a tensor interaction having a range r ∼ 1/mπ together with a local (zero-range) spin-spin term.

Exercise 11.1 Solution: One-Pion-Exchange Potential a) We have iū(pf )γ5 u(pi )ei(pf −pi )·x = iχ†f

p/mN <<1

σ · (pi − pf 2mN

−→

i(χ†f −χ†f

σ · pf ) 2mN

χi ei(pf −pi )·x = −

0 1

1 0

χi

σ·pi 2mN χi

ei(pf −pi )·x

1 χ† σ · ∇χi ei(pf −pi )·x 2mN f

b) Then 2 0 gπN N ū(p1 )γ5 τ 1 u(p1 )

p/mN <<1

−→

1 ū(p02 )γ5 τ 2 u(p2 ) (p1 − p01 )2 − m2π

g2 1 − πN2N χ†1f τ 1 σ 1 · ∇1 χi1 0 χ† τ 2 σ 2 · ∇2 χ2i 4mN (p1 − p1 )2 + m2π 2f

c) Note Z

1 1 d3 q iq·r e = exp −mπ |r| (2π)3 q 2 + m2π 4π|r|

so VOP EP (r 1 − r 2 ) = −

2 gπN 1 N τ 1 · τ 2 σ 1 · ∇1 σ 2 · ∇2 exp −mπ |r 1 − r 2 | 16πm2N |r 1 − r 2 |

d) Defining R = r 1 − r 2 , we have ∇1i ∇2j = −∇1i ∇1j . Then

1 r1j d 1 −mπ R r1j 1 mπ ∇1j exp −mπ R = e = − 2− e−mπ R R R dR R R R R and

1 r1j 1 mπ exp −mπ R = −∇1i + e−mπ R R R R2 R

1 1 mπ ri rj d mπ −mπ R = −δij + 2 e − + 2 e−mπ R R3 R R dR R3 R ∇1i ∇1j

The Nucleon-Nucleon Interaction

136

3 mπ mπ m2π 1 mπ + 2 + + − δ + e−mπ R ij R5 R4 R4 R3 R3 R2

3mπ m2π 1 mπ 3 + 4 + 3 − δij + 2 e−mπ R = ri rj R5 R R R3 R

= ri rj

But need also to include the feature that 1 ∇21 = −4πδ(r) R so 1 σ 1 · ∇1 σ 2 · ∇2 e−mπ R R

2

1 1 1 mπ −mπ R 1 + + 3σ 1 · R̂σ 2 · R̂ − σ 1 · σ 2 + σ 1 · σ 2 e = (mπ R)2 mπ R 3 3 R −

4π σ 1 · σ 2 δ(r) 3

and VOP EP (R) = −

2 gπN N τ 1 ·τ 2 16πm2N

1 1 1 + + 2 (mπ R) mπ R 3

4π − σ 1 · σ 2 δ(r) 3

2 1 mπ −mπ R Ŝ12 + σ 1 · σ 2 e 3 R

where Ŝ12 = 3σ 1 · R̂σ 2 · R̂ − σ 1 · σ 2 11.2 The Deuteron Wavefunction Fermi was fond of asking the question “Where’s the hydrogen atom for this problem?”, meaning what is the simple model that elucidates the basic physics of a given system? In the case of nuclear structure, the answer is clearly the deuteron, and it is essential to have a good understanding of this simplest of nuclear systems at both the qualitative and quantitative levels.

The deuteron is a J P = 1+ neutron-proton bound state, and we wish to construct the most general ground state wavefunction. For definiteness, we shall construct the state |ψ11 >, but other states may be obtained by use of the angular momentum lowering operator. We begin by combining the two spins to produce a state having total spin S either 0 or 1 r 1 |0, 0i = (|↑1 i |↓2 i − |↓1 i |↑2 i) 2 |1, 1i = |↑1 i |↑2 i r 1 |1, 0i = (|↑1 i |↓2 i + |↓1 i |↑2 i) 2

137

The Nucleon-Nucleon Interaction

|1, −1i = |↓1 i |↓2 i .

The total angular momentum is given by the sum of this total spin with the orbital angular momentum J = L + S. There are then only two possibilities which satisfy the requirements of even parity and unit total angular momentum, L = 0, S = 1 and L = 2, S = 1. a) Use Clebsch-Gordan coefficients to construct these states and show that S ψ11 (r) = RS (r)Y00 (θ, φ) |↑1 i |↑2 i D ψ11 (r) = RD (r) r −

r

3 2 Y (θ, φ) |↓1 i |↓2 i 5 2

r 3 1 1 Y (θ, φ) (|↑1 i |↓2 i + |↓1 i |↑2 i) 10 2 2 ! r 1 0 Y (θ, φ) |↑1 i |↑2 i . + 10 2

b) The observed magnetic moment of the deuteron provides information about the ground-state wavefunction. The total magnetic moment of the deuteron is due to the intrinsic magnetic moments of the neutron and proton, as well as the orbital angular momentum of the (charged) proton but not the (neutral) neutron. The magnetic moment operator along the 3-axis can then be written as 1 µ3 = µN (L3 + gp σ3p + gn σ3n ) , 2 where µN = e/2mN is the nucleon magneton while gp = 5.58 and gn = −3.82 are the experimental gyromagnetic ratios for the proton and neutron respectively. (In the center-of-mass system the angular momentum of the proton is half that of the total, which is the reason for the factor of 12 multiplying L.) Taking the overall deuteron wavefunction to be tot S D ψ11 (r) = aS ψ11 (r) + aD ψ11 (r) with |aS |2 + |aD |2 = 1, calculate the deuteron magnetic moment and show that

E gp + gn 3 − gp − gn 1 D 1 tot 1 tot 2 2 ψ1 |Mz | ψ1 = |aS | + |aD | . µN 2 4 c) Comparing the wavefunction calculation with the experimental value µexp = 0.857µN , show that the two can be made consistent by using |aD |2 ' 0.04, i.e., a ∼4% D-state component in the deuteron wavefunction.

Exercise 11.2 Solution: The Deuteron Wavefunction

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138

a) From table of Clebsch-Gordan coefficients, find r < 01; 11|00; 11 >= 1; r < 21; 11|21; 10 >= −

< 21; 11|22; 1 − 1 >= 3 10

3 5 r

< 21; 11|20; 11 >=

1 10

Then wavefunctions are (ψ11 )S (r) = RS (r)Y00 (θ, φ)| ↑1 > | ↑2 > r 3 2 1 D (ψ1 ) (r) = RD (r) Y (θ, φ)| ↓1 > | ↓2 > 5 2 r −

r 3 1 1 Y2 (θ, φ) (| ↑1 > | ↓2 > +| ↓1 > | ↑2 >) 10 2 ! r 1 0 Y (θ, φ)| ↑1 > | ↑2 > + 10 2

b) Find

1 1 2 3 gn 2 3 3 gp aD 2 · + 1 · + aS + a2D − + 2 5 10 2 2 10 5

3 − gp − gn gp + gn = µB a2S + a2D 2 4

< ψ11 |Mz |ψ11 >= µB

c) Using a2S + a2D = 1, write b) as

1 3 2 1 1 < ψ1 |Mz |ψ1 >= µB (gp + gn ) + aD (1 − gp − gn ) = 0.857µB 2 4 Since gp = 5.58 and gn = −3.82 this becomes 0.857 =

1 3 · 1.76 + (1 − 1.76)a2D 2 4

or 0.880 − 0.857 ' 0.04 0.57 11.3 The Deuteron Quadrupole Moment a2D =

As discussed in Exercise 11.2, understanding the deuteron magnetic dipole moment requires a small D-state admixture. Thus we anticipate that the deuteron is approximately, but not quite, spherical. We can examine this admixture from another perspective by looking at the deuteron quadrupole moment tot E e D 1 tot eQ = ψ1 |(3z 2 − r2 )| ψ11 , 4 which provides a measure of the nonsphericity. (Here the factor of 4 is due to the fact that only the proton contributes to the charge distribution.)

The Nucleon-Nucleon Interaction

139

a) Writing the quadrupole moment as Z tot 1 (r)|2 , drr2 (3 cos2 θ − 1)| ψ11 Q= 4 use the deuteron wavefunction from the previous exercise to show that 1 1 Q = √ Re(a∗S aD ) r2 SD − |aD |2 r2 DD , 20 5 2 where r2 SD = RS |r2 |RD

and

r2 DD = RD |r2 |RD

.

b) Using the experimental result Qexp = 0.286 fm2 , to extract the magnitude and sign of aD and compare with size found in the previous exercise.

Exercise 11.3 Solution: The Deuteron Quadrupole Moment a) Using Z

s 3∗ 2 1 dΩY`m (θ, φ)Y`m (θ, φ)Y`m (θ, φ) = 3 2 1

(2`1 + 1)(2`2 + 1) 4π(2`3 + 1)

× < `1 0; `2 0|`1 `2 `3 0 >< `1 m1 ; `2 m2 |`1 `2 `3 , m3 < calculate quadrupole moment via r r r Z 1 16π 2 1 1 Q= d3 rr2 Y20 (θ, φ)|ψ11 (r|2 = · · aD aS < RD |r2 |RS > 4 5 4 10 5 2 − 4

r

2 7

3 · 5

r

2 3 − · 7 10

r

1 1 − · 14 10

r !

2 a2D < RD |r2 |RD > 7

1 1 = √ aS aD < RS |r2 |RD > − a2D < RD |r2 |RD > 20 5 2 b) Since, from problem 12.2 a2D ' 0.04, we have |aD | ' 0.2 and |aS | ' 0.98. Since 1 1 2 then 5√ |a a | >> 20 aD the relative sign between aS , aD must be the same 2 S D as the sign of Q, which is positive.

To go further, we must choose a specific form for wavefunctions RS (r), RD (r).

For example, if we take the n-p potential to vanish beyond some range, then the √ asymptotic form of the wavefunction is ∼ exp(−κr)/r where κ = 2mr EB ' 45 MeV. We have then 1 < RS |r2 |RD >∼ 2 ∼ 18 fm2 κ

The Nucleon-Nucleon Interaction

140

so √ 1 aS aD ∼ aD ∼ 5 2κ2 Q = √ = 0.12 6 2 which is the expected order of magnitude.

11.4 The Tensor-Force Operator Ŝ12 One requires matrix elements of the tensor operator Ŝ12 = 3σ 1 · r̂σ 2 · r̂ − σ 1 · σ 2 in order to evaluate the influence of pion exchange on the nucleon-nucleon interaction. With this in mind, demonstrate the following properties of Ŝ12 : a) operating on any spin-singlet state, Ŝ12 yields 0; b) from parity considerations, Ŝ12 applied to a spin-triplet function with L = J ± 1 gives a linear combination of the triplet functions for L = J + 1 and J−1 and, applied to a triplet function with L = J, reproduces the same function multiplied by a constant, i.e., M M M Ŝ12 YL1J = aL,J+1 YJ+1,1J + aL,J−1 YJ−1,1J

if

L=J ±1

and M M Ŝ12 YJ1J = aJ,J YJ1J .

c) The operator Ŝ12 commutes with J3 , and hence the coefficients aL,J+1 , aL,J−1 , and aJ,J are independent of M . d) aJ+1,J−1 = aJ−1,J+1 . e) By considering the particular case inpwhich r̂ is along the z-axis and M = 1, and using the fact that Ylm (θ = 0) = (2l + 1)/4πδl0 , show that D E M M aJ,J = YJ1J |Ŝ12 |YJ1J =2. f) Similarly, taking the special case r̂ = k, M =0 and M =1, use the result obtained in b) above to find two equations between aJ−1,J+1 and aJ−1,J−1 . Thus, show that −2(J − 1) 2J + 1 D E 6pJ(J + 1) M M YJ+1,1J |Ŝ12 |YJ−1,1J = 2J + 1 D E −2(J + 2) M M YJ+1,1J |Ŝ12 |YJ+1,1J = .

2J + 1

D

M M YJ−1,1J |Ŝ12 |YJ−1,1J

E

=

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141

Exercise 11.4 Solution: The Tensor-Force Operator (a) Spin singlet means a state |S = 0, MS = 0 >. S12 will give zero when acting on a spin singlet state. h i 2 2 (b) S12 ≡ 2 3 (S·r) − S Now [S 2 , Si ] = 0 so [S12 , S 2 ] = 0 . Thus, S12 cannot 2 r change S. Now note that if we rotate both S and r about some axis by an angle θ, then S12 is invariant. The generator for this rotation is L + S. (L generates the space rotation; S generates the spin rotation.) The invariance of S12 under this rotation implies [S12 , J] = 0 .

Thus, [S12 , J 2 ] = [S12 , J3 ] = 0 .

Hence, S12 cannot change J or M either. However, L can change. For S=1, the possible orbital angular momentum quantum numbers are L = J+1, J, and J-1.

Thus, M M M M S12 YL1J = aYJ−1,1J + bYJ1J + cYJ+1,M J .

Next note that S12 is invariant under a parity transformation (r → −r). This implies that S12 preserves parity. Accordingly, for L = J + 1, the only possible contributions are from terms with J + 1 and J − 1. For L = J, the only state having the same parity is J because the parity eigenvalue is (−)L . Thus, we can write M M M S12 YL1J = aL,J+! YJ+1,1J + aL,J−1 YJ−1,1J (L = J ± 1) M M S12 YL1J = aJ,J YJ1J (L = J) .

(c) We saw that [S12 , J] = 0. Since [S12 , J± ] = 0, J± acting on the equations in (b) will yield the same results with M ± 1. Thus, the coefficients are independent of M . (d) S12 is Hermitian so M M M M < YJ−1,1J |S12 |YJ+1,1J >=< YJ+1,1J |S12 |YJ−1,1J >∗

so aJ+1,J−1 = aJ−1,J+1 . (e) We have M M 1 1 aJJ = < YJ1J |S12 |YJ1J >=< YJ1J |S12 |YJ1J >

M YJ1J = ΣL ML =−L < LSML MS |JM > YLML (θ, φ)|SMS > . q With Ylm (θ = 0) = 2l+1 4π δl0 and M = 1, we have r 2l + 1 1 YJ1J =< L101|J1 > |11 > .

4π

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142

Further, for θ = 0, we have S12 = 2(3Sz2 − S 2 ) so

1 1 1 S12 YJ1J = 2 3(1)2 − 1(2) YJ1J = 2YJ1J so aJ,J = 2. (f) Taking L = J ± 1, θ = 0, and M = 0,1, we have: r 2(J − 1) + 1 1 YJ−1,1J = < (J − 1)101|J1 > |11 > 4π s r (J − 1 + 1)(J − 1 + 1 + 1) (2(J − 1) + 1) |11 > = (2(J − 1) + 1)(2(J − 1) + 2) 4π r (J + 1) = |11 > ; 8π r

2(J + 1) + 1 |11 > 4π s r (J + 1 − 1)(J + 1 − 1 + 1) (2(J + 1) + 1) = |11 > 2(J + 1)(2(J + 1) + 1) 4π r J |11 > ; = 8π

1 YJ+1,1J = < (J + 1)101|J1 >

r

2(J + 1) + 1 |10 > 4π s r (J − 1 − 0 + 1)(J − 1 + 0 + 1) (2(J − 1) + 1) = |10 > (2(J − 1) + 1)(J − 1 + 1) 4π r J = |10 > ; 4π

0 YJ−1,1J = < (J − 1)100|J0 >

r

2(J + 1) + 1 |10 > 4π s r (J + 1 − 0)(J + 1 + 0) (2(J + 1) + 1) =− |10 > (J + 1)(2(J + 1) + 1) 4π r J +1 =− |10 > .

4π

0 YJ+1,1J = < (J + 1)100|J0 >

Note that for θ = 0, we have S12 = 2(3Sz2 − S 2 ), so that S12 |10 > = 2(0 − 2)|10 >= −4|10 > S12 |11 > = 2|11 > .

To determine aJ−1,J−1 and aJ−1,J+1 , consider M M M S12 YJ−1,1J = aJ−1,J+1 YJ+1,1J + aJ−1,J−1 YJ−1,1J .

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143

For M = 0, 1, we have √ √ √ −4 J = aJ−1,J+1 (− J + 1) + aJ−1,J−1 J r r r J +1 J J +1 2 = aJ−1,J+1 + aJ−1,J−1 2 2 2

(M = 0) (M = 1) .

Thus, # √ √ −4 J + aJ−1,J+1 J + 1 √ √ 2 J + 1 = aJ−1,J+1 J + J +1 J

√ √ J +1 6 J + 1 = aJ−1,J+1 J+ √ J p 6 J(J + 1 aJ−1,J+1 = .

2J + 1 √

√

"

Further, r

p −4(2J + 1) + 6(J + 1) J + 1 6 J(J + 1) · = aJ−1,J−1 = −4 + J 2J + 1 2J + 1 2(J − 1) aJ−1,J−1 = − .

2J + 1 Finally, consider 0 0 0 S12 YJ+1,1J = aJ+1,J+1 YJ+1,1J + aJ+1,J−1 YJ−1,1J p √ 6 J(J + 1) √ √ 4 J + 1 = aJ+1,J+1 − J + 1 + J , 2J + 1

so 6J −4 2J + 1 2(J + 2) =− .

2J + 1

aJ+1,J+1 =

11.5 Energy-Dependence of N N Phase Shifts Experimentally, the phase shifts in nucleon-nucleon scattering are determined as a function of kinetic energy for the isovector (T = 1) and isoscalar (T = 0) channels.

They are shown in detail in [Pia15]. a) Note that both the 3 S1 and 1 S0 phase shifts have zeros at approximately the same energy. What does this mean? Estimate the spatial separation between the nucleons at which this vanishing occurs. b) Consider the effect of the tensor force in the P states. Assume V (r) = Vc (r) + VT (r)Sˆ12 . Using the results of Exercise 11.4, show that for two nucleons in the

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144

3

PJ state D

E P0 |Ŝ12 |3 P0 = −4 D E 3 P1 |Ŝ12 |3 P1 = +2 D E 2 3 P2 |Ŝ12 |3 P2 = − .

5 Now determine the sign of VT (r) in the P states. c) Now consider the spin-orbit force VLS (r) L · S, which will play a significant role at higher energies. Show that the effect of this additional term is to change the slope of the 3 P0 phase shift if VLS < 0. d) Similarly, show that for two nucleons in the 3 DJ state, D E 3 D1 |Ŝ12 |3 D1 = −2 D E 3 D2 |Ŝ12 |3 D2 = +2 D E 4 3 D3 |Ŝ12 |3 D3 = − .

7 3

Consequently, assuming central and tensor forces dominate, show that δ(3 D2 ) > δ(3 D3 ) > δ(3 D1 ), which is consistent with the experimental determinations of [Pia15].

Exercise 11.5 Solution: Energy Dependence of the N N Phase Shifts (a) Both the 3 S1 and 1 S0 phase shifts change from positive to negative at a nucleon kinetic energy in the laboratory of about 250 MeV. This corresponds to an incident 3-momentum of about 685 MeV/c. From the Heisenberg uncertainty principle, this corresponds to a nucleon-nucleon distance r of 200 fm ∼ 0.3 fm .

685 This shows that the N N interaction changes from being attractive (positive phase shift) to being repulsive (negative phase shift) at this inter-nucleon distance. It supports the conclusion that the N N potential has a repulsive core at r ∼ 0.3 fm. r∼

(b) Note that we designate the states via 2S+1 LJ where L = 0 is the S-state; L = 1 is the P -state, L = 2 is the D-state, etc.. Using the results of the previous Exercise 11.4, we have < 3 P0 |S12 |3 P0 > = aJ+1,J+1 |J=0 = −2 < 3 P1 |S12 |3 P1 > = aJ,J |J=1 = +2

< 3 P2 |S12 |3 P2 > = aJ−1,J−1 |J=2 = −

2 5

(c) Consider a potential formed from a central potential and the tensor force V (r) = Vc (r) + VT (r)S12 .

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The Nucleon-Nucleon Interaction

If we apply the results of (b) then we expect that the phase shifts would be δ(3 P0 ) = δc − 4δT

δ(3 P1 ) = δc + 2δT 2 δ(3 P2 ) = δc − δT .

5

This will be consistent with the experimental phase shift data if 2 δc − 4δT > δc − δT > δc + 2δT , 5 which implies that δT < 0, i.e. the tensor force is repulsive in odd states. Positive signs for both Vc and VT are predicted by OPEP in 3 P states.

At higher energies the spin-orbit term will become important. Recall

1 2 J − L2 − S 2 2 1 = [J(J + 1) − 4] 2 = −2 (J = 0)

L·S=

= −1 (J = 1)

= +1 (J = 2) .

This effect of the spin-orbit force causes the phase shift for 3 P0 to turn over if VLS (r) < 0. Thus for V (r) = Vc (r) + VT (r)S12 + VLS (r)L · S we expect the resulting phase shifts to be δ(3 P0 ) = δc − 4δT − 2δLS

δ(3 P1 ) = δc + 2δT − δLS 2 δ(3 P2 ) = δc − δT + δLS .

5

Comparison with phase shift data indicates δLS > 0, i.e. the spin-orbit force is attractive. (d) For the D-states, we have < 3 D1 |S12 |3 D1 > = aJ+1,J+1 |J=1 = −2 < 3 D2 |S12 |3 D2 > = aJ,J |J=2 = +2

< 3 D3 |S12 |3 D3 > = aJ−1,J−1 |J=1 = −

4 .

7

For a potential with central, tensor, and spin-orbit terms, we have V (r) = Vc (r) + VT (r)S12 + VLS (r)L · S .

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The Nucleon-Nucleon Interaction

We expect the resulting phase shifts to be δ(3 D1 ) = δc − 2δT − 3δLS

δ(3 D2 ) = δc + 2δT − δLS 4 δ(3 D3 ) = δc − δT + 2δLS .

7

Comparison with experimental phase shift data indicates (i) δ(3 D2 ) > δ(3 D3 ) > δ(3 D1 ); (ii) that the central and tensor forces are attractive in the D-states and (iii) that δLS < 0, i.e. the spin orbit force is repulsive in the D-states.

12

The Structure and Properties of Few-Body Nuclei 12.1. Pre-existing ∆-Components in the Ground State of 3 He Consider an experiment to measure pre-existing ∆ components in the ground state of the 3 He nucleus. These components can arise in interactions among three nucleons as shown. A pion exchanged between nucleons 1 and 2 excites nucleon 2 to a ∆(1232) isobar state. The ∆ decays by emitting a pion which is absorbed by nucleon 3.

∆

An electron beam is incident on the 3 He nucleus and, in the final state, the scattered electron is detected in coincidence with the pion resulting from the decay of the ∆.

The reaction is e + ∆(N N ) → e0 + ∆(N N ) → e0 + πN (N N ). a) Write down the different charge-state possibilities of the (spin 23 and isospin 32 ) ∆(1232). b) Given that the e−∆ coupling is proportional to the square of the charge of the ∆, find the relative probabilities (PQ ) for knockout of the ∆ in its various charge states. c) Consider next the coupling of the pre-existing ∆ with the remaining nucleons in 3 He (spin = 12 and isospin= 12 ). Use the Table of Clebsch-Gordan coefficients to find the relative probabilities (P∆ ) for the different ∆ components: P (∆++ nn) : P (∆+ np) : P (∆0 pp) where n is a neutron and p is a proton. d) Use the Table of Clebsch-Gordan coefficients to obtain the relative probabilities (Pdecay ) of the decays of the different ∆ charge-states into πN states. e) From b), c,) and d), determine the probabilities of detecting π + , π 0 , and π − 147

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The Structure and Properties of Few-Body Nuclei

Table 12.1 Decay probabilities for the different ∆ states ∆ state

Decay channel

Pdecay

∆++

→ p + π+

1

resulting from ∆ knockout from 3 He. Exercise 12.1 Solution: Pre-existing ∆-Components in the Ground State of 3 He (a) The different states are: ∆++ (Tz = 23 ), ∆+ (Tz = 12 ), ∆0 (Tz = − 12 ), and ∆− (Tz = − 32 ). (b) Since the e − ∆ coupling is proportional to the square of the charge, the relative probabilities PC are: PC (∆++ ) : PC (∆+ ) : PC (∆0 ) : PC (∆− ) = 4 : 1 : 0 : 1 . (c) The pre-existing ∆ must couple to the other two nucleons to form the spin and isospin quantum numbers of the 3 He ground state. Thus, there can be no ∆− component. The two nucleons must couple to T = 1 to produce total T = 1/2 for 3 He. Using Clebsch-Gordon coefficients, we have P (∆++ nn) : P (∆+ np) : P (∆0 pp) =

1 1 1 : : .

2 3 6

(d) The decay probabilities in the different channels are listed in Table 12.1: (e) The total probability of detecting pions of different charge states is then given by Pk = PC · P · Pdecay .

Writing out the possible terms, we have Pk (π + ) = PC++ · P ++ · Pdecay (+) + PC+ · P + · Pdecay (+) Pk (π 0 ) = PC+ · P + · Pdecay (0) + PC0 · P 0 · Pdecay (0)

Pk (π − ) = PC0 · P 0 · Pdecay (−) + PC− · P − · Pdecay (−) With PC0 = 0 and P = 0, we have Pk (π + ) : Pk (π 0 ) : Pk (π − ) = (2 + 1/9) : 2/9 : 0 = 19 : 2 : 0 .

12.2 Electroproduction of ∆s from 3 He

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149

Consider the process e + N (N N ) → [∆(N N ) → πN (N N )] + e0 where the ∆ is produced in interaction with the virtual photon. a) Using the same analysis as in Exercise 12.1, calculate the relative probabilities of detecting π + , π 0 , and π − resulting from ∆ production on a single nucleon in 3 He. b) Compare your results from 12.1 e) and 12.2 a) and suggest an experimental method for distinguishing between them. What particles in the final state should be detected for a definitive measurement?

Exercise 12.2 Solution: Electroproduction of ∆s from 3 He Consider the electroproduction process e + N (N N ) → [∆(N N )] + e0 → πN (N N ) + e0 (a) The virtual photon γ ∗ can produce ∆ states on a single nucleon (p or n) as follows γ ∗ + p → ∆+

γ ∗ + n → ∆0

The states ∆++ nor ∆− cannot be produced. Next we can assume that the ratio of ∆+ to ∆0 production will be 2:1, the proton-neutron ratio in 3 He. This is valid because the dominant N → ∆ transition is M 1, i.e. the photon couples to the nucleon magnetic moment. The proton and neutron magnetic moments are roughly equivalent in magnitude (though opposite in sign).

Using the Clebsch-Gordon coefficients from Exercise 12.1 c), we have Pγ (π + ) = 2 · 1/3

Pγ (π 0 ) = 2 · 2/3 + 1 · 2/3

Pγ (π − ) = 1/3 .

In summary,

Pk (π + ) : Pk (π 0 ) : Pk (π − ) = 2 : 6 : 1 . (b) Using pion detection, one could measure the relative numbers by charge state.

These are very different for knockout and production as summarized in Table 12.2.

In particular, the ∆ knockout mechanism produces a large enhancement of π + s and no π − s. Note that detecting π 0 s requires a photon detector. A magnetic spectrometer would be used to detect the charged pions. Milner and Donnelly have proposed an extension of this idea using polarized electrons on a polarized 3 He target [Mil88].

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150

Table 12.2 Decay probabilities for the different ∆ states

Knockout

π+ 19

π0 2

π− 0

Production

2

6

1

12.3. Neutron and Proton Polarization in Polarized 3 He The polarized 3 He nucleus is widely used as an effective polarized neutron target for scattering experiments. Here we estimate the polarization of the neutron and proton in a polarized 3 He nucleus following [?]. We consider (±) (±) 3 Pn,p =<3 He m = +1/2|Pbn,p | He m = +1/2 > ,

where Pbn(±) =

X 1 − τ3 (i) 1 ± σ3 (i) i

Pbp(±) =

2

2

X 1 + τ3 (i) 1 ± σ3 (i) i

2

2

.

The isospin factor in Pbn (Pbp ) counts the number of neutrons (protons) which are aligned (+) or anti-aligned (−) with the 3 He spin. a) Show that Pn(+) + Pn(−) = 1 Pp(+) + Pp(−) = 2 . b) We assume that the ground-state wavefunction is dominated by the S, S 0 , and D states, i.e., we neglect the small P -state components. Assuming that * + 3 X 1 1 m = 1/2| σ3 (i)τ3 (i)|m = 1/2 = −[P (S) − P (S 0 ) + P (D)] .

3 3 i=1 show that hm = 1/2|σ3 |m = 1/2i = P (S) + P (S 0 ) − P (D)

hm = 1/2|τ3 |m = 1/2i = 1 = P (S) + P (S 0 ) + P (D) .

c) Thus, show that Pn(+) = 1 − ∆

Pn(−) = ∆ ,

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151

where ∆ = [P (S 0 ) + 2P (D)]/3. d) In addition, show that Pp(+) = 1 − 2∆0

Pp(−) = 1 + 2∆0 , where ∆0 = [P (D) − P (S 0 )]/6. e) Using the probabilities of Table 12.1, estimate the neutron and proton polarizations.

Exercise 12.3 Solution: Neutron and Proton Polarization in Polarized 3 He We have (±) Pn,p =

3

He m = +

1 1 b(±) 3 He m = + P 2 n,p 2

where X 1 − τ3 (i) 1 ± σ3 (i) · 2 2 i X 1 + τ3 (i) 1 ± σ3 (i) Pbp(±) = · .

2 2 i Pbn(±) =

(a) We have X 1 − τ3 (i) 1 + σ3 (i) 1 − τ3 (i) 1 − σ3 (i) · + · 2 2 2 2 i 1X = [1 − τ3 (i) + σ3 (i) − τ3 (i)σ3 (i) + 1 − τ3 (i) − σ3 (i) + τ3 (i)σ3 (i)] 4 i X 1 − τ3 (i) = .

(12.1) 2 i

Pbn+ + Pbn− =

The 3 He nucleus has 2 protons and 1 neutron. In terms of isospin operators we have τ3 |pi = |pi

τ3 |ni = − |ni 1 − τ3 |pi = 0 2 1 − τ3 |ni = |ni 2 1 + τ3 |pi = |pi 2 1 + τ3 |ni = 0 .

2

The Structure and Properties of Few-Body Nuclei

152

Hence, we have * Pbn+ + Pbn− =

3

He

X 1 − τ3 (i)

+ 3

2

i

He

= 1 · hn |n i + 2 · 0 · hp |p i = 1 .

Similarly, we have X 1 + τ3 (i) 1 + σ3 (i) 1 + τ3 (i) 1 − σ3 (i) · + · 2 2 2 2 i X 1 + τ3 (i) = , 2 i

Pbp+ + Pbp− =

(12.2)

which yields Pbp+ + Pbp− = 2. (b) We have three states: S, S 0 , and D states. From normalization we must have P (S) + P (S 0 ) + P (D) = 1 .

For both the S and S 0 states, we have L = 0 and S = 1/2 and hS = 1/2, mS = 1/2 |σ3 | S = 1/2, mS = 1/2i = 1.

The same is true for the S 0 -state. For the D-state, we have L=2 and S = 3/2 and it follows that r r

2 3 1 1 3 1 = mL = 2, mS = − − mL = 1, mS = − + J = , mJ = 2 2 5 2 10 2 r

1 1 mL = 0, mS = 5 2

r

−

3 1 mL = −1, mS = 10 2

.

This yields hD − state |σ3 | D − statei =

3 1 1 2 · (−3) + (−1) + (+1) + (+3) = −1 .

5 10 5 10

Similarly, one can also show that h|τ3 |i = 1 , for each of the S, S 0 and D-states. Putting this together, we have hm = +1/2 |σ3 | m = +1/2i = P (S) < σ3 >S +P (S 0 ) < σ3 >S 0 +P (D) < σ3 >D = P (S) · 1 + P (S 0 ) · 1 + P (D) · (−1) = P (S) + P (S 0 ) − P (D) .

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153

(c) We have X 1 − τ3 (i) 1 + σ3 (i) · 2 2 i X 1 = [1 − τ3 (i) + σ3 (i) − τ3 (i)σ3 (i)] 4 i " # X X X X 1 < 1>−< τ3 (i) > + < σ3 (i) > − < σ3 (i)τ3 (i) = 4 i i i i

1 1 1 = 3 − 1 + P (S) + P (S 0 ) − P (D) + P (S) − P (S 0 ) + P (D) .(12.3) 4 3 3

Pbn(+) =

Using the normalization condition P (S) + P (S 0 ) + P (D) = 1, we have

1 4 8 (+) 0 b Pn = 4 − P (S ) − P (D) 4 3 3 1 = 1 − [P (S 0 ) + 2P (D)] ≡ 1 − ∆ 3 where ∆ ≡ 13 [P (S 0 ) + 2P (D)]. Similarly, we have X 1 − τ3 (i) 1 − σ3 (i) · 2 2 i X 1 [1 − τ3 (i) − σ3 (i) + τ3 (i)σ3 (i)] = 4 i " # X X X X 1 = < 1>−< τ3 (i) > + < σ3 (i) > + < σ3 (i)τ3 (i) 4 i i i i

1 4 8 0 = P (S ) + P (D) 4 3 3 1 = [P (S 0 ) + 2P (D)] = ∆ .

3

Pbn(−) =

(d) From above, we have X 1 + τ3 (i) 1 + σ3 (i) · 2 2 i 1X = [1 + τ3 (i) + σ3 (i) + τ3 (i)σ3 (i)] .

4 i

Pbp(+) =

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154

Thus, Pp(+) = < Pbp(+) > " # X X X X 1X = < 1>+< τ3 (i) > + σ3 (i) > + < σ3 (i)τ3 (i) > 4 i i i i i

1 1 1 0 0 = 3 + 1 + P (S) + P (S ) − P (D) − P (S) + P (S ) − P ((D) 4 3 3

1 4 4 = 4 + P (S 0 ) − P (D) 4 3 3 0 = 1 − 2∆ , where ∆0 ≡ 16 [P (D) − P (S 0 )].

Following the same procedure it is straightforward to show that X 1 + τ3 (i) 1 − σ3 (i) Pp(−) = < · > 2 2 i 1 X = < [1 + τ3 (i) − σ3 (i) − τ3 (i)σ3 (i)] > 4 i = 1 + 2∆0 .

(12.4)

0 (e) From Table 12.1, P (S) i ≈ 8%, and P (S ) ≈ 1.2%. Thus, the proton h ≈ 90%, P (D) (+)

polarization in 3 He is 21 Pp (+)

in 3 He is Pn

(−)

− Pn

(−)

− Pp

= −2∆0 ≈ −0.027. The neutron polarization

= 1 − 2∆ ≈ 0.885.

12.4. Polarized Fusion At energies below several hundred keV, the D+3 He reaction is dominated by a channel which involves a single intermediate 5 Li state. This nearly resonant state lies 407 keV above the rest energy of D and 3 He and has an angular momentum and + parity of J π = 23 . The spins of D and 3 He are 1 and 12 , respectively, so that the total spin of the reactants is S = 32 or 12 . At these low energies, the average orbital angular momentum is nearly zero that only the S = 32 states contribute significantly to the D + 3 He reaction. The statistical weights of the S = 32 and S = 12 states are four and two, respectively. Therefore, if the nuclei are unpolarized, only two-thirds of the collisions will contribute to the fusion reactions.

Consider now the effect of polarizing both the initial D and 3 He nuclei. We assume a magnetic field so that the fractions of D nuclei polarized parallel, transverse, and antiparallel to the magnetic field are d+ , d0 , and d− , respectively. Correspondingly, the fractions for the polarized 3 He nuclei are h+ and h− . Then the total cross section can be written as

2 1 2 4 σ = a + b + c f σ0 + b + c (1 − f )σ0 , (12.5) 3 3 3 3 where a = d+ h+ + d− h− , b = d0 , c = d+ h− + d− h+ , σ0 is the unpolarized fusion

The Structure and Properties of Few-Body Nuclei

155

cross section, and f is the fraction of the cross section occurring due to the 32 state. The magnitude of f is assumed to be close to unity.

+

a) Show that for unpolarized D and 3 He, σ = 32 σ0 . b) Show that if the nuclei are all polarized along a magnetic field, the enhancement of the fusion cross section is 32 f ∼ = 1.5.

Exercise 12.4 Solution: Polarized Fusion We consider D + 3 He fusion where we define the population states as: (a) d+ , d0 , d− for the M = +1, 0, −1 states of the deuteron; and h+ and h− for the M = + 21 , − 12 states of the 3 He nucleus. The total fusion spin-dependent cross section can be written as

2 1 2 4 σ = a + b + c f σ0 + b + c (1 − f )σ0 .

3 3 3 3 where a = d+ h+ + d− h− , b = d0 , and c = d+ h− + d− h+ , σ0 is the unpolarized fusion cross section and f is the fraction of the cross section occurring due to the 3+ state.

2 (a) For unpolarized D and 3 He, there should not be any preferred polarization states so all the polarization states are evenly occupied. Hence, d+ = d0 = d− = 13 and h+ = h− = 21 . It follows that a = d+ h+ + d− h− = b = d0 =

1 1 1 1 1 · + · = 3 2 3 2 3

1 3

c = d+ h− + d− h+ =

1 1 1 1 1 · + · = .

3 2 3 2 3

Thus,

1 2 1 1 1 2 1 4 1 + · + · f σ0 + · + · (1 − f )σ0 3 3 3 3 3 3 3 3 3 2 2 = f σ0 + (1 − f )σ0 3 3 2 = σ0 .

3

σ=

(b) Now we assume that the nuclei are all in the positive M substate. Thus, d+ = 1 and d0 = d− = 0 and h+ = 1 and h− = 0. We then have a=1·1+0·0=1 b=0

c=1·0+0·1=0 .

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The Structure and Properties of Few-Body Nuclei

The cross section then becomes

2 1 2 4 σ = 1 + · 0 + · 0 f σ0 + · 0 + · 0 (1 − f )σ0 = f σ0 .

3 3 3 3 The enhancement due to polarization then is 3 f σ0 = f .

2 2 3 σ0 The enhancement factor can be of order 50%, if f ≈ 1.

13

Overview of Many-Body Nuclei 13.1 Matrix Elements of One- and Two-Particle Operators One- and two-particle operators acting in first-quantization (in contrast to oneand two-body operators acting in second-quantization, i.e., occupation number representation) have been discussed in the text. In this exercise some specific matrix elements of such operators are considered. a) For the antisymmetrized state given as a Slater determinant

1 Φα1 α2 ···αA (1, 2, · · · A) = √ A!

··· ···

ϕα1 (A) ϕα2 (A) .. .

ϕαA (1) ϕαA (2) · · ·

ϕαA (A)

ϕα1 (1) ϕα2 (1) .. .

ϕα1 (2) ϕα2 (2) .. .

,

following the developments in Section 13.1, show that the expressions D E XZ [1] F O F = dx1 ϕ†α (1)O[1] (1)ϕα (1) , α≤F

and Z D E 1 X Z dx1 dx2 ϕ†α (1)ϕ†β (2)O[2] (1, 2) F O[2] F = 2 α,β≤F

× (ϕα (1)ϕβ (2) − ϕβ (1)ϕα (2)) , emerge. Specifically, for plane waves as used in descriptions of symmetric nuclear matter, namely for the Fermi gas model, verify D

F T [1] F

E

=K=4

X |k|2 3 → EF · A , 2mN 5

k≤kF

and D E F V [2] F ≡ v =

X

X

k≤kF ms mt k0 ≤kF m0s m0t

[hkms mt , k 0 m0s m0t |V | kms mt , k 0 m0s m0t i

− hkms mt , k 0 m0s m0t |V | k 0 m0s m0t , kms mt i] /2.

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158

b) Making use of these results, prove the basic identity

Z F T [1] + V [2] F B1 (kF ) 3 kF3 (4aW − aM ) dzV (z) ≡ = EF + A A 5 12π 2

2 ) Z 3j1 (kF z) + (4aM − aW ) dzV (z) . kF z for the binding energy per nucleon using the simple toy model of Section 13.2.

Exercise 13.1 Solution: Matrix Elements of One- and Two-Particle Operators a) To simplify the notation, let us denote the A lowest single-particle states, namely, the ones that are filled to form the degenerate ground state |F i, with the running quantum number index α = 1 · · · A. The Slater determinant above is then 1 (A) Φ1,2,··· ,A (1, 2, · · · A) = √ A!

ϕ1 (1) ϕ2 (1) .. .

ϕ1 (2) ϕ2 (2) .. .

··· ···

ϕ1 (A) ϕ2 (A) .. .

ϕA (1)

ϕA (2) · · ·

ϕA (A)

,

where we have added the label A as a superscript to remind us that this Slater determinant is for the A-body system. Expanding along the first column, the one referring to particle 1, we find that 1 X (A) (A−1) Φ1,2,··· ,A (1, 2, · · · A) = √ (−)α+1 ϕα (1)Φ1,2,··· ,A;α (2, · · · A), A α (A−1)

where Φ1,2,··· ,A;α (2, · · · A) is the normalized Slater determinant for A − 1 particles having particle coordinates 2, 3, · · · , A and with the row labeled by single-particle quantum number α missing, namely, 1

(A−1)

Φ1,2,··· ,A;α (2, · · · A) = p (A − 1)!

ϕ1 (2) ϕ2 (2) .. .

ϕ1 (3) ϕ2 (3) .. .

··· ···

ϕ1 (A) ϕ2 (A) .. .

ϕA (2)

ϕA (3) · · ·

ϕA (A)

row α missing

The same procedure can be applied to isolate single-particle wavefunctions with coordinates 1 and 2 to write X 1 (A) (A−2) Φ1,2,··· ,A (1, 2, · · · A) = p (−)α+β ϕα (1)ϕβ (2)Φ1,2,··· ,A;αβ (3, · · · A), A(A − 1) α6=β (A−2)

where as above Φ1,2,··· ,A;αβ (3, · · · A) is the normalized Slater determinant for A − 2 particles having particle coordinates 3, 4, · · · , A and with the rows labeled by single-particle quantum numbers α and β with α 6= β missing.

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For the ground-state matrix element (specifically the degenerate state |F i) of a single-particle operator (one-body operator in second quantization) we then have Z Z E Z D (A)† [1] F O F = dx1 dx2 · · · dxA Φ1,2,··· ,A (1, 2, · · · A) X (A) × O[1] (i)Φ1,2,··· ,A (1, 2, · · · A) Zi

=A

Z

dx2 · · ·

dx1

Z

(A)†

dxA Φ1,2,··· ,A (1, 2, · · · A)

(A)

×O[1] (1)Φ1,2,··· ,A (1, 2, · · · A) using the antisymmetry properties of the many-body wavefunctions to permute the particle coordinate of the operator to particle number 1. Inserting the first expansion derived above to write the A-particle Slater determinant in terms of Slater determinants with A − 1 particles, we then have Z D E X [1] α0 +α F O F = (−) dx1 ϕ†α0 (1)O[1] (1)ϕα (1) α0 α

×

Z

dx2 · · ·

Z

(A−1)†

(A−1)

dxA Φ1,2,··· ,A;α0 (2, · · · A)Φ1,2,··· ,A;α (2, · · · A).

The integrals over coordinates for particles, together with the orthogonality of spin and isospin parts of the single-particle wavefunctions yield simply Z Z (A−1)† (A−1) dx2 · · · dxA Φ1,2,··· ,A;α0 (2, · · · A)Φ1,2,··· ,A;α (2, · · · A) = δα0 α and we obtain the result desired: D E XZ F O[1] F = dx1 ϕ†α (1)O[1] (1)ϕα (1). α≤F

The derivation of the ground-state matrix element of a two-particle operator (two-body operator in second quantization) proceeds in the same way: first one uses the antisymmetry of the A-body Slater determinants to bring the two particle coordinates to the first two positions, namely, for particles 1 and 2. Next one uses the expansion of that Slater determinant into Slater determinants for A − 2 particles derived above. Finally, following similar steps to those for the single-particle operator, canceling the factors A(A − 1) and employing the orthogonality of the Slater determinants for A − 2 particles one arives at the desired result, Z D E 1 X Z [2] dx1 dx2 ϕ†α (1)ϕ†β (2)O[2] (1, 2) F O F = 2 α,β≤F

× (ϕα (1)ϕβ (2) − ϕβ (1)ϕα (2)) . b) The kinetic energy operator for the ith particle is T [1] (i) = −∇2i /2mN and, using the plane-wave single-particle wavefunctions introduced in the book (Eq.

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(13.39)) 1 ms mt ϕα (i) = √ eik·x η1/2 (i)ξ1/2 (i), Ω where now α stands for k, ms , mt , one has T [1] (i)ϕα (i) =

k2 ϕα (i).

2mN

It is then straightforward to compute the ground-state matrix elements of the number operator E D X 1=A F N [1] F = 4 k≤kF

and the kinetic energy operator, using the above, D E X F T [1] F = 4 k≤kF

k2 .

2mN

Converting these to integrals in the so-called thermodynamic limit where the volume and A go to infinity, and noting that R kF 2 k dk · k 2 3 0 = kF2 R kF 5 k 2 dk 0

one finds that D E 3 F T [1] F = EF · A. 5 Finally, the potential energy ground-state matrix element is simply given by the above result for a two-particle operator.

Starting with the direct term and the part of the potential not involving the space-exchange operator (i.e., the term involving aW ), one has Z Z D E

† 0 ms mt ms mt F V [2] F = dx1 dx2 eik·x1 η1/2 (1)ξ1/2 (1)eik ·x2 η1/2 (2)ξ1/2 (2) D,W

1 0 ms mt ms mt ×V (x) eik·x1 η1/2 (1)ξ1/2 (1)eik ·x2 η1/2 (2)ξ1/2 (2) Ω2 Z 1 = dxV (x). Ω For the other part of the direct term (involving the space-exchange operator and going as aM ) the exponential factors enter differently and yield an additional factor 0 of ei(k −k)·x : Z D E 0 1 [2] F V dxei(k −k)·x V (x) F = Ω D,M For the exchange contributions things are reversed and, in addition, there is now a

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factor δms m0s δmt m0t coming from the orthogonality of the spin and isospin spinors.

0 For the two terms that have the extra factor ei(k −k)·x one needs the identity Z kF j1 (kz) k 2 dkj0 (kz) = kF3 kz 0 which is simple to verify upon substituting yj0 (y) = sin y and the fact that j1 (y) = (sin y − y cos y) /y 2 . Putting all of this together yields the desired result for the ground-state energy per nucleon using the simple toy model.

13.2 Schmidt Lines for Magnetic Dipole Moments of One-Particle and One-Hole Nuclei The Schmidt lines discussed in the text are given by

1 g` L + gs σ n0 `0 j0 , hn0 `0 j0 kµ1 k n0 `0 j0 i = n0 `0 j0 2 where one assumes that the nuclear ground state of an odd nucleus is to be approximated simply by a closed shell plus or minus one particle; that is, the dipole moment for the ground state of such a system will arise from the last unpaired nucleon. The single-particle state is assumed to have total angular momentum j0 and to have orbital angular momentum `0 where the orbital and spin angular momenta are coupled (J = L + S), namely, one has configuration (`0 1/2)j0 mj0 . Since the magnetic dipole operator ( 1 j0 = `0 + 1/2 ` 0 g` + 2 gs

µ/µN = 2`0 −1 1 (`0 + 1) g` − 2 gs j0 = `0 − 1/2 2`0 +1 has both orbital and spin contributions, one must compute the matrix elements of operators acting in subspaces. a) Perform the necessary analysis to prove the expressions for the Schmidt lines given above. b) Calculate the ground-state magnetic moments of the following pairs of nuclei, 17 17 209 209 8 O, 9 F, and 82 Pb, and 83 Bi. Assume that they are well represented as having single particles or single holes (proton or neutrons) with respect to closed shells and compare the results with the experimental values.

Exercise 13.2 Solution: Schmidt Lines for Magnetic Dipole Moments of One-Particle and One-Hole Nuclei a) Start by considering a typical vector V, which transforms as a tensor of rank 1. The Wigner-Eckart theorem (see [Edm74] and Chapter 2) tells us that

J 1 J 0 J−M hJM |V1m | JM i = (−) hJ kV k Ji , −M m M 0

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whereas the total angular momentum operator also satisfies the same form of equation:

J 1 J 0 J−M hJM |J1m | JM i = (−) hJ kJk Ji . −M m M 0 Since the reduced matrix elements are independent of the choice of coordinate system and all other factors (the phases and the 3j-symbols) are the same, the two matrix elements must therefore be proportional: hJ kV k Ji = κ, hJ kJk Ji where κ is some constant, to be determined. Next consider the scalar product P J · V = m (−)m J1m V1,−m . One has X X hJM |J · V| JM i = (−)m hJM |J1m | JM 0 i hJM 0 |V1m | JM i m

M0

X

m

=κ

(−)

X M0

m

hJM |J1m | JM 0 i hJM 0 |J1m | JM i

using the above proportionality. This implies that hJM |J · V| JM i = κ JM J2 JM = κJ(J + 1) or, in other words, that κ=

hJM |J · V| JM i J(J + 1)

and hence that hJM |V1m | JM 0 i = hJM |J · V| JM i hJM |J1m | JM 0 i which is known as the Landé formula [Lan21]. We now apply this to the magnetic dipole operator discussed in the text. We wish to evaluate the single-particle matrix elements of the form µ/µN ≡ hn0 `0 j0 , mj0 = j0 |[g` L10 + gs S10 ]| n0 `0 j0 , mj0 = j0 i , where µN is the nucleon magneton and g` = 1(0) and gs = 2µp (2µs ) are the orbital and spin g-factors, respectively, for the proton and neutron. Using the Landé formula we then have that µ/µN ≡

1 hn0 `0 j0 , mj0 = j0 |([g` L + gs S] · J) J10 | n0 `0 j0 , mj0 = j0 i . j0 (j0 + 1)

Using the facts that L · J = L2 + L · S S · J = S2 + L · S

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one has that L2 → `0 (`0 + 1)

S2 → 3/4 1 L · S → [j0 (j0 + 1) − (`0 (`0 + 1) + 3/4)] 2 and hence 1 [j0 (j0 + 1) + `0 (`0 + 1) − 3/4] 2 1 S · J → [j0 (j0 + 1) − `0 (`0 + 1) + 3/4] .

2 It is then straightforward to substitute into the expression for the magnetic dipole moment in the two situations where j0 = `0 ± 1/2 to find the desired results:  `0hg` + 21 gs i     `  j0 = `0 + 1/2 = j0 g` + g2`s0−g  +1

µ/µN ≡ 2`0 −1 1  (` + 1) g` − 2 gs   2`0 +1 h 0 i   `  j0 = `0 − 1/2 = j0 g` − g2`s0−g +1 L·J→

+

b) Both 17 O and 17 F have spin-parity 25 ground states and are reasonably approximated as having `0 = 2, j0 = 25 , namely, 1d5/2 , shell model single-particle states above 16 O. Thus we have 1 µ/µN = 2g` + gs 2 1 n 17 O 2 gs = −1.913 , = 1 p 2 + 2 gs = 4.793 17 F to be compared with -1.89371 and 4.7224, respectively, namely good to 1.3% and 1.5%, respectively.

Both 209 Pb and 209 Bi have spin- 92 ground states, but they are quite different.

209 Pb has positive parity and one might suspect the magnetic moment to arise from a 2g9/2 neutron added to the closed N = 126 shell in 208 Pb (see Fig. 13.4), i.e., `0 = 4, j0 = 29 . On the other hand, 209 Bi has negative parity and there one presumes that this is from a 1h9/2 proton added to the closed Z = 82 shell in 208 Pb, i.e., `0 = 5, j0 = 92 . Substituting these into the formulas above, remembering that the bismuth case is one where j0 = `0 − 1/2, we then find that

1 n 209 −1.913 Pb 2 gs =

µ/µN = , 9 1 p 209 6 − g = 2.624 Bi s 11 2 in contrast to the experimental values of -1.473 and 4.1106, respectively.

13.3 Isospin Multiplets

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Isobars have equal mass number, A, but different atomic number Z. Consider a set of isobars that are members of the same isospin multiplet. Their mass differences are determined by the Coulomb energies and the neutron-proton mass difference.

Assume that the Coulomb energy of the nucleus of atomic number Z and radius R is 3 Z 2α · , 5 R where R = R0 A1/3 . a) Show that the energy difference between two adjacent nuclei of an isospin multiplet is 3α (2Z − 1) − 0.78 MeV . ∆E = 5R0 A1/3

1+ 15.11

12 7N

12 5B

β+

β-­‐

12 6C

b) We consider the T = 1, J π = 1+ isospin triplet consisting of the ground state of 125 B, the 15.1 MeV excited state of 126 C, and the ground state of 127 N. Using the equation derived in a), and R0 = 1.4 fm, show that M (12 C, T = 1) − M (12 B, T = 1) = 2.2 MeV . c) Knowing that M (12 B, T = 1) − M (12 C, T = 0) = 12.9 MeV, show that M (12 C, T = 1) − M (12 C, T = 0) = 15.1 MeV . d) The same mass difference in the mass-14 isobar states is only 2.2 MeV. Can you explain why the ground state of 12 C has such a low energy?

Exercise 13.3 Solution: Isospin Multiplets

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The Coulomb energy is EC =

3 Z 2α 5 R

(13.1)

where R = R0 A1/3 . a) The nucleus transitions from Z to Z − 1, effectively we swap one proton with one neutron. We also lose an electron so

3α Z 2 − (Z − 1)2 + (mp − mn + me ) ∆E = 5R 3α (2Z − 1) + (938.28 − 939.57 + 0.511) MeV = 5R0 A1/3 3α = (2Z − 1) − 0.78 MeV .

(13.2) 5R0 A1/3 1 , and R0 = 1.4 fm. Thus, b) Here A =12, Z =6, α = 4π 0

M (12 C, T = 1) − M (12 B, T = 1) = 2.2 MeV .

(13.3)

c)

M (12 C, T = 1) − M (12 C, T = 0) = 2.2 MeV + M (12 B, T = 1) − M (12 C, T = 0) = 2.2M eV + M (12 B, T = 1) − M (12 C, T = 0)

= 15.1 MeV .

(13.4)

d) 12 C has 6 protons and 6 neutrons. In addition, it can be viewed as the bound system of 3 alpha particles, which are tightly bound. This gives rise to a lower ground state energy.

13.4 Symmetry Energy Consider the members of the A = 135 isobar and look up their masses using a table of mass excess. Then plot the results as a function of Z. Hence, determine the value of the symmetry energy parameter in the Weizsäcker mass formula. Perform the same calculation for members of the A = 136 isobar and estimate the value of the pairing parameter.

Exercise 13.4 Solution: Symmetry Energy Recall that in atomic physics, we use carbon to define a mass scale in terms of the so-called atomic mass unit mu = 931.5 MeV/c2 . Thus, M (Z, N ) = (Z + N )mu .

Since the mass of the atom is dominated by the nucleus a better guess of the mass is simply the sum of the constituent masses: M (Z, N ) = ZMH + N Mn = Z(Mp + Me ) + N Mn .

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Once can then define the mass excess ∆(Z, N ) as the difference between its actual mass and its mass number in atomic mass units: ∆(Z, N ) ≡ M (Z, N ) − (Z + N )mu .

The nuclear binding energy EB is the difference between the actual mass of an atom and the total mass of its constituent electrons, protons and neutrons EB = [Z · 1.007825 + (A − Z) · 1.008665 − (A + ∆)] · 931.5 MeV ,

(13.5)

where 1.007825 amu is the mass of the hydrogen atom and 1.008665 amu is the mass of the neutron. For example, for 135 Ba, Z = 56, A =135, and ∆ = −0.0943.

Thus, the binding energy is EB = [56 · 1.007825 + 79 · 1.008665 − (135 − 0.0943)] · 931.5 MeV = 1133.7 MeV .

(13.6)

In this way, from a table of mass excesses (see www.nndc.bnl.gov) one calculates the binding energy of a given nucleus.

Use the semi-empirical mass formula for the binding energy EB = Evol + Esurf ace + ECoulomb + Esymmetry + Epairing = a1 A − a2 A2/3 − as

Z(Z − 1) (A − 2Z)2 − a5 δA−4/3 . − a4 1/3 A A

(13.7)

Then input data from the mass excess tables to make a fit and extract a4 and a5 .

From the A =135 data, obtain a4 = 19.7. From the A = 136 data, obtain a5 = 29.4, in reasonable agreement with values in book.

13.5 Barrier Penetration Consider tunneling in α-decay from a heavy nucleus. The barrier penetration factor for the α-particle to pass through the Coulomb field of the residual nucleus may be estimated as follows. Consider a triangular-shaped, one-dimensional barrier of height V0 at the radius R which decreases to the value of the kinetic energy Eα of the α-particle at the distance R1 . For a heavy nucleus with Z=72, we can use R = 9 fm, and the height of the Coulomb barrier may be taken as the Coulomb energy at distance R. At R1 the barrier height equals the kinetic energy of the α-particle, which we will assume to be 5 MeV. What is the probability for the α-particle to escape from the nucleus?

Exercise 13.5 Solution: Barrier Penetration The triangular-shaped barrier can be written as V (x) =

Eα − V0 V0 R1 − Eα R x+ , R1 − R R1 − R

(13.8)

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as shown with Eα = 5 MeV and R = 9 fm. For a heavy nucleus with Z = 72, we have V0 =

(Ze)(2e) = 23.04 fm .

R

(13.9)

(Ze)(2e) R1

(13.10)

Similarly, we have Eα =

so R1 = 41.5 fm.

V(r) V0

Eα R1

R

r

Next, the probability for a particle to penetrate the barrier is P ≈ e−2γ where γ=

where p(x) =

p

1 ~

Z dx|p(x)| ,

(13.11)

2m(V (x) − E). Thus,

Pα = e −

2 ~

Z R1

p dx 2m(V (x) − Eα ) R Z R1 √ s 2 Eα − V0 (V0 − Eα )R1 = e− dx 2m x+ . ~ R R1 − R R1 − R

(13.12) (13.13)

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Let x0 = x − R1 so that −2

Pα = e

~

Z R1 −R

0

dx

√

s 2m

0

s

(Eα − V0 ) √ 0 x R − R1

2m(V0 − Eα ) 2 0 3/2 R1 −R · x |0 R1 − R 3 s 2m(V0 − Eα ) − 4 =e (R1 − R)3/2 3~ R1 − R p 4 = e− (R1 − R) 2m(V0 − Eα ) .

3~ 2 = e− ~

(13.14) (13.15) (13.16) (13.17)

Substituting the numerical values yields Pα = e−80.5 = 1 × 10−35 .

13.6 Brueckner Theory and the Bethe-Goldstone Equation For the assumptions made in the discussions of nuclear matter in the text, show that the solution for the Bethe-Goldstone wavefunction inside the repulsive core is found by solving d2 u + K 2 u = Γ [δ(r − c) − χ(r, c)] dr2

Z c + w(r)θ(c − r) − dr0 χ(r, r0 )w(r0 ) .

0

Exercise 13.6 Solution: Brueckner Theory and the Bethe-Goldstone Equation First, one knows that u(r) must vanish inside the core, the product ν(r)u(r) must vanish outside the core and, by integrating over a small region from below to above the core radius, one can show that the product ν(r)u(r) must contain a δ-function at the core radius, so that ν(r)u(r) = Γδ(r − c) + w(r)θ(c − r), that is, the product vanishes outside the core, but is finite inside. Substituting into the s-wave Bethe-Goldstone equation yields

2 d 2 + K u(r) = (Γδ(r − c) + w(r)θ(c − r)) dr2 Z c − dr0 χ(r, r0 ) (Γδ(r0 − c) + w(r0 )θ(c − r0 )) 0

= (Γδ(r − c) + w(r)θ(c − r)) Z c −Γχ(r, c) − dr0 χ(r, r0 )w(r0 ), 0

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the desired result.

13.7 Dirac Equation for Baryons in Uniform Nuclear and Neutron Matter Once the mean-field approximations have been made in Section 13.3, one obtains a Lagrangian from which a Dirac equation can be constructed. a) Perform the necessary steps to find this equation.

Since the classical scalar and vector fields enter simply as constants, it is straightforward to solve this equation. b) Construct this solution and, as in standard treatments of quantum field theory, proceed to write the Dirac field as an expansion involving the spinor solutions, plane waves, and creation and destruction operators. c) Using these results, prove 1 1 HRM F T = − m2V V02 + m2S φ20 + gV V0 ρbB 2 2 q

1X † † k 2 + m∗2 + N Ak,λ Ak,λ + Bk,λ Bk,λ V kλ

with baryon density ρbB =

1 X † † Ak,λ Ak,λ − Bk,λ Bk,λ .

V kλ

and 1X δH = − V k,λ

q

q ∗2 2 2 2 k + mN − k + mN .

d) Explain how the baryon density should be interpreted in terms of the Dirac field operators.

Exercise 13.7 Solution: Dirac Equation for Baryons in Uniform Nuclear and Neutron Matter a) Since the scalar and vector fields are to be replaced by their constant average (mean) values φ → hφi ≡ φ0

Vµ → hVµ i ≡ δµ0 V0 , one has that Fµν = 0, m2V δµ0 V0 = gV Bµ from the first of the field equations in the text plus the fact that on the average no direction in space can be chosen,

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which implies that Bµ = ρB δµ0 with only a time component, and thus that V0 = gV ρB /m2V . From this one finds that L0RM F T =

1 2 2 1 2 2 m V − mS φ0 .

2 V 0 2

Substituting Dµ = ∂µ + igV Vµ into the third of the field equations yields L1RM F T = ψ [i ∂6 − γ0 V0 δµ0 − m∗N ] ψ, where m∗N ≡ mN − gS φ0 , namely, the two parts of the Lagrangian discussed in the text. Assembling all of these parts yields the required result. b) The Dirac equation for the baryons in the problem arises in the usual way from the Lagrangian L1RM F T , [i ∂6 − gV γ0 V0 − m∗N ] ψ = 0, where we see that this takes on the usual form but with the energy shifted by the ∗ offset gV V0 and with mN replaced p by mN = mN − gS φ0 . Accordingly, the energy eigenvalues are E± (k) = gV V0 ± k 2 + m∗2 N . The rest proceeds just as for the free Dirac equation: one can expand in normal modes, involving plane-wave factors, spinors, and creation and destruction operators for baryons and antibaryons,

1 X † b U (k, λ)Ak,λ eik·x + V (k, λ)B−k,λ ψ(x) = e−ik·x , V kλ

where the first (second) contribution involves baryons (antibaryons). The creation and destruction operators satisfy the usual anticommutation relations n o Ak,λ , A†k,λ = δkk0 δλ,λ0 n o † Bk,λ , Bk,λ = δkk0 δλ,λ0 rest = 0.

Everything else proceeds just as it does for the free-particle solutions of the Dirac equation, except for the energy shift and use of the effective mass. c) The energy eigenvalues are given in part b) and, when these are employed together with the baryon creation and destruction operators to obtain the total energy contribution of the baryons in the problem, two terms emerge: one is the baryon density that counts the number of baryons minus the number of antibaryons with respect to the vacuum,

1 X † † Ak,λ Ak,λ − Bk,λ Bk,λ , ρbB = V kλ

while the other term involving creation and destruction operators, q

1X † † H0 ≡ k 2 + m∗2 A A + B B k,λ k,λ N k,λ k,λ V kλ

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together with the term not involving quantum operators, q

q 1X ∗2 2 2 2 δH = − k + mN − k + mN .

V k,λ

contain the remainder. d) The operator ρbB is, in fact, the normal-ordered product of the field operators, obtained by subtracting their vacuum expectation value, b ρbB = : ψb† (x)ψ(x) : E D b b 0 = ψb† (x)ψ(x) − 0 ψb† (x)ψ(x) 1X b = ψb† (x)ψ(x) − 1, V kλ

where the last term counts the number of filled states in the negative-energy Dirac sea. The interested reader is directed to [Wal95] for detailed discussions of extensions to these ideas and the formalism of so-called quantum hadrodynamics (QHD).

13.8 The Equations-of-State for Symmetric Nuclear Matter and for Neutron Matter Using the developments of relativistic nuclear and neutron matter in the text, prove the results Z kF γ 3 γ dk = k . ρB = 3 (2π) 0 6π 2 F ε(ρB , φ0 ) =

gV2 2 m2 γ 2 ρB + S2 (mN − m∗N ) + 2 2mV 2gS (2π)3

γ m2 g2 2 p(ρB , φ0 ) = V2 ρ2B − S2 (mN − m∗N ) + 2mV 2gS 3(2π)3

Z kF 0

Z kF 0

dk k 2 + m∗2 N

1/2

dkk 2 k 2 + m∗2 N

−1/2

.

for the baryon density, energy density, and pressure, and hence for the equations of state.

Exercise 13.8 Solution: The Equations-of-State for Symmetric Nuclear Matter and for Neutron Matter Converting from sums over momenta to integrals, as usual, the baryon density is given by Z kF 1 k3 B =γ 4π k 2 dk = γ F2 , ρB = 3 Ω 6π (2π) 0 where the quantization volume is Ω and where the degeneracy is γ, as in the text.

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The energy density ε = E/Ω, where from the previous developments (see Exercise 13.7) we have upon substituting V0 = gV ρB /m2V and gS φ0 = mN − m∗N that 2

1 2 gV ρB 1 2 1 gV ρB ∗ E = − mV + mS [mN − mN ] + gV 2 m2V 2 gS m2V Z kF q γ k 2 dk k 2 + m∗2 + N 3 4π (2π) 0 Z kF q g2 m2 γ 2 = V2 ρ2B + S2 [mN − m∗N ] + 2 k 2 dk k 2 + m∗2 N. 2mV 2gS 2π 0 Since the pressure is given through the familiar thermodynamic equation dE = −pdΩ, one can show via the identities ∂ε ∂ (E/Ω) /∂Ω = ∂ρB ∂ρB /∂Ω 1 ∂E E ∂ (E/Ω) /∂Ω = − Ω ∂Ω Ω2 at fixed baryon number that ∂ε 1 = (p + ε) ∂ρB ρB implying that p = ρB

∂ ∂ε − ε = ρ2B ∂ρB ∂ρB

ε . ρB

Finally, using the fact that we want the solution where the energy is minimized with respect to variations in φ0 , namely where

∂E = 0, ∂φ0 Ω,B implying that there is no derivative of the expression mN − m∗N , and substituting for ε/ρB using the equation for E obtained above, ε g2 1 m2S 2 = V2 ρB + · [mN − m∗N ] ρB 2mV ρB 2gS2 Z kF q 1 γ + · 2 k 2 dk k 2 + m∗2 N, ρB 2π 0 one can proceed to evaluate the derivative need to obtain the pressure. The derivatives of the first and second terms are simple: the first yields gV2 /2m2V , while the 2 second yields −m2S [mN − m∗N ] /2gS2 ρ2B . The third term yields " Z # q q kF γ ∂k F 2 − k 2 dk k 2 + m∗2 kF2 + m∗2 , N + ρB · k F N 2π 2 ρ2B ∂ρB 0 where the last derivative can easily be obtained using the fact that ρB = γkF3 /6π 2 .

Assembling all of the terms and compactifying the resulting expression a bit yields

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the desired result for the pressure. See [Wal95] for further discussion.

13.9 Euler-Lagrange Equations for the σω-Model Write down the general form of the Euler-Lagrange equations and, using the Lagrangians

1 1 1 ∂µ φ∂ µ φ − m2S φ2 L0 = − Fµν F µν + m2V Vµ V µ + 4 2 2 and L1 = ψ [i D 6 − (M − gS φ)] ψ , show that the field equations ∂ ν Fµν + m2V Vµ = gV Bµ

+ m2S φ = gS ρ [i D 6 − (mN − gS φ)] ψ = 0 emerge.

Exercise 13.9 Solution: Euler-Lagrange Equations for the σω-Model Using the Euler Lagrange equation ∂µ

δL δL − =0 δ∂µ φ δφ

we find a) For the field φ we have δL = gS ψ̄ψ − m2S φ δφ

δL = ∂µφ δ∂µ φ

and

so EOM is ∂µ ∂ µ φ = φ = gS ψ̄ψ − m2S φ or

+ m2S φ = gS ψ̄ψ b) For field ψ we have the covariant derivative iDµ = i∂µ − gV Vµ δL = −ψ̄(M − gS φ + gV 6 V ) δψ

and

δL = iψ̄γ µ δ∂µ ψ

so EOM is (∂µ ψ̄)iγ µ ψ = −ψ̄(M − gS φ + gV 6 V )ψ or, integrating by parts to transfer the derivative from ψ̄ to ψ, ψ̄ [i D 6 − (M − gS φ)] ψ = 0

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c) For field Vµ we have δL = m2V V ν − gV ψ̄γ ν ψ δVν

and

δL = −∂ µ V ν + η µν ∂ · V δ∂µ Vν

so EOM is ∂µ (−∂ µ V ν + η µν ∂ · V ) = m2V V ν − gV ψ̄γ ν ψ or ∂µ F µν + m2V V ν = gV ψ̄γ ν ψ

14

Models of Many-Body Nuclei 14.1 Quantum Numbers of the Nucleus From the level scheme given by the independent-particle shell model, if 25 12 Mg is a spherical nucleus, what are the most likely spin, parity, and isospin quantum numbers of its ground state? If, instead, it is a deformed nucleus with prolate deformation, what are the most likely spin and parity?

Exercise 14.1 Solution: Quantum Numbers of the Nucleus 25 12 Mg

has 12 protons and 13 neutrons. Thus, there is one unpaired neutron.

Assuming a spherical shape, i.e. β = 0, and filling up the orbitals from the bottom, we find that the unpaired neutron will be in the 1d5/2 orbit. Thus, the spin is 5/2, the parity is (−)2 = +1. The isospin is −1 since it is a neutron.

If it is deformed with a prolate shape, we have β > 0. We consider two cases: (i) Small deformation 0 < β < 0.2: We have the 13th neutron in the 1d5/2 orbital so the spin is 5/2 and the parity is positive. (ii) Large deformation β > 0.2: Now the 13th neutron will be in the 2s1/2 orbital so the spin is 1/2 and the parity is positive.

14.2 Quadrupole Moment and Shape Consider the quadrupole moment of a non-spherical nucleus in a rotational model.

This is discussed in detail in [Won98]. Assuming axial symmetry, the quadrupole moment is characterized by the parameter δ, the difference between R3 and R⊥ , defined by δ=

2 3 R32 − R⊥ · 2 2 , 2 R3 + 2R⊥

where R3 is the radius of the nucleus along the body-fixed symmetric 3-axis and R⊥ is the radius in the direction perpendicular to it. Consider now the ground state − of 152 63 Eu which is known to be 3 with an electric (Coulomb) quadrupole moment 2 2 of +3.16 × 10 e·fm . Determine δ and find the intrinsic quadrupole moment of the nucleus in its the ground state. What is the shape of this nucleus?

Exercise 14.2 Solution: Quadrupole Moment and Shape 175

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From Wong Eq. 6-19, the measured spectroscopic quadrupole moment of a rotating nucleus is related to the intrinsic quadrupole moment by Qlab =

3k 2 − J(J + 1) Qint .

(J + 1)(2J + 3)

− We consider 152 63 Eu in the ground state 3 . This means that J = 3 and k = 0. Then

Qlab = −

J Qint = 3.16 × 102 e · fm2 , 2J + 3

yielding Qint = −9.48 × 102 e · fm2 for the intrinsic quadrupole moment.

We have 3 Qint = √ R2 Zβ(1 + 0.36β) , 5π where Z = 63, A =152, and R = R0 A1/3 with R0 = 1.4 fm. Solving for β yields β = −0.419. β is related to δ by r 4 π β= δ, 3 5 so δ = −0.396. The nuclear shape is oblate.

14.3 α-Particle Model of the Nucleus Consider the α-particle model of the nuclei 12 C (three α-particles) and 16 O (four α-particles). a) Evaluate the difference between the total binding energy of 12 C and the total binding energy of the three α-particles. b) Evaluate the difference between the total binding energy of 16 O and the total binding energy of the four α-particles. c) By counting the number of αα bonds in each of the two nuclei, estimate the bond energy between two α-particles in this model.

Exercise 14.3 Solution: α-Particle Model of the Nucleus (a) The mass excess of an α-particle is 2.4249 MeV. The mass excess of 12 C is 0. A 12 C nucleus has 3 × 2.4249 = 7.2747 MeV of additional binding energy relative to three α-particles. (b) The mass excess of a 16 O nucleus is −4.7370 MeV. An 16 O nucleus has 4 × 2.4249 + 7.737 = 14.437 MeV of additional binding energy relative to four α particles. (c) A nucleus composed of Nα α-particles will have Nα (Nα − 1)/2 bonds. 16 O has 6 bonds, while 12 C has 3. Thus, we would expect the additional binding energy for 16 O (≈14.4 MeV) to be twice of 12 C (≈ 7.2 MeV). The additional binding energy

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per bond is ≈ 2.4 MeV. 14.4 Rotational Model of the Nucleus For an axially-symmetric nucleus of mass M with density given by

ρ0 , for r ≤ R0 (1 + βY2,0 (θ, φ)) ρ(r) = 0 otherwise , show a) that the intrinsic (charge) quadrupole moment up to second-order in deformation parameter β is given by 1 Q0 = √ ZR02 β(1 + 0.36β) 5π and b) the moment of inertia about the laboratory z-axis is given by I=

2 M R02 (1 + 0.31β) .

5

Exercise 14.4 Solution: Rotational Model of the Nucleus From Wong Eq. 6-19, the measured spectroscopic quadrupole moment of a rotating nucleus is related to its intrinsic quadrupole moment via Qlab =

3K 2 − J(J + 1) Qintrinsic .

(J + 1)(2J + 3)

P We consider 152 = 3− which implies K = 0 and 63 Eu in the ground state with J J = 3. Set J Qlab = − Qintrinsic = 3.16 × 102 e fm2 .

2J + 3

Therefore, the intrinsic Coulomb quadrupole moment is Qintrinsic = −3Qlab = −9.48 × 102 e fm2 .

From the lecture notes 3 Qintrinsic = √ R2 Zβ(1 + 0.36β) 5π 1

where Z = 63e and R = R0 A 3 with R0 = 1.4 fm and A = 152. Solving for β we find then β = −0.419.

Since β is related to δ by 4 β= 3

r

π δ 5

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we have δ = −0.396 and the negative sign implies an oblate shape.

14.5 Hartree and Hartree-Fock Equations with the Coulomb Interaction While our main focus in this book is placed on issues of nuclear and particle physics where the strong interaction plays a dominant role, it is still informative to see how other systems, such as those relevant in atomic and condensed matter physics, also employ the same many-body methods discussed in this chapter. For example, in the case of atomic physics, start with the Hamiltonian X 1 Z XZ † dr∇ψs (r) · ∇ψs (r) + drψs† (r)Vc (r)ψs (r) H= 2m s s +

1X 2 0 ss

Z

Z dr

dr0 ψs† (r)ψs†0 (r0 )V (r − r0 )ψs0 (r0 )ψs (r) ,

where Vc (r) = −Ze2 /r and V (r0 −r) = e2 /|r0 −r| and use this to calculate hψ|H|ψi. a) If |ψi is taken to be the state hr1 r2 ...rN |ψi = φ1 (r1 )φ2 (r2 )....φN (rN ) , minimize hψ|H|ψi and show that what results are the Hartree equations

Z Ze2 e2 X ∇2 − |φj (r0 )|2 φi (r) = i φi (r) . φi (r) + dr0 − 2m r |r − r0 | j6=i

b) If |ψi is taken to be the fully antisymmetrized state 1 |ψi = √ N!

φ1 (r1 ) . . . φN (r1 )

··· ··· ··· ··· ···

φ1 (rN ) . . . φN (rN )

minimize < ψ|H|ψ > and show that what results are the HF equations,

Z ∇2 Ze2 e2 X ∗ 0 − φi (r) + dr0 φj (r )[φj (r0 )φi (r) − δsi ,sj φj (r)φi (r0 )] − 2m r |r − r0 | j6=i

= i φi (r) .

Note that in both parts a) and b) the states φi (r) are assumed to be orthogonal normalized states: Z drφi (r)φj (r) = δij .

Exercise 14.5 Solution: Hartree and Hartree-Fock Equations with the

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Coulomb Interaction Note that ap |φ1 φ2 . . . φN >= |φ1 φ2 . . . φN −1 > −|φ1 φ2 . . . φN −2 φN > + . . . + (−)

N −1

Z

Z

d3 p0 ap |p0 >< p0 |φN > (2π)3

d3 p 0 ap |p0 >< p0 |φN −1 > (2π)3

|φ2 φ3 . . . φN >

Z

d3 p 0 ap |p0 >< p0 |φ1 > (2π)3

= ΦN (p)|φ1 φ2 . . . φN −1 > −ΦN −1 (p)|φ1 φ2 . . . φN −2 φN > + . . . + (−)N −1 Φ1 (p)|φ2 φ3 . . . φN −1 > Then ψ(r) = |φ1 φ2 . . . φN >=

Z

d3 p ip·r e ap |φ1 φ2 . . . φN > (2π)3

= φN (r)|φ1 φ2 . . . φN −1 > + . . . + (−)N −1 φ1 (r)|φ2 φ3 . . . φN > so that < φ1 φ2 . . . φN |

XZ

=

d3 rψs† (rVc (r)ψs (r)|φ1 φ2 . . . φN >

s N Z X

d3 rφ∗i (r)Vc (r)φi (r)

i=1

and < φ1 φ2 . . . φN |

X 1 Z d3 r∇ψs† (r · ∇ψs (r)|φ1 φ2 . . . φN > 2m s

=

N Z X i=1

d3 r

1 ∇φ∗i (r) · ∇φi (r) 2m

Also d3 p ip·r0 e {φN (r) [φN −1 (p)|φ1 φ2 . . . φN −2 > (2π)3

+ . . . + (−)N −2 φ1 (p)|φ2 φ3 . . . φN −1 > + . . . + (−)N −1 φ1 (r) [φN (p)|φ2 φ3 . . . φN −1 > ψ(r0 )ψ(r)|ψ1 φ2 . . . φN >=

Z

+ . . . + (−)N −2 φ2 (p)|φ3 φ4 . . . φN >

= {φN (r) [φN −1 (r0 )|φ1 φ2 . . . φN −2 >

+ . . . + (−)N −2 φ1 (r0 )|φ2 φ3 . . . φN −1 > + . . . + (−)N −1 φ1 (r) [φN (r0 )|φ2 φ3 . . . φN −1 >

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+ . . . + (−)N −2 φ2 (r0 )|φ3 φ4 . . . φN > Then 1X 2

s,s0

d3 r

Z

d3 r0 ψs† (r)ψs†0 (r0 )V (r − r0 )ψs0 (r0 )ψs (r)|φ1 φ2 . . . φN >

X 1 ·2 2

Z

d3 r

Z

d3 r0 V (r − r0 )δsi ,sj φ∗i (r)φj (r)φi (r0 )φ∗j (r0 )

< φ1 φ2 . . . φN | =

Z

Z

pairs

X 1 − ·2 2

Z

d3 r

pairs

d3 r0 V (r − r0 )|φi (r)|2 |φj (r0 )|2

so that < ψ|Ĥ|ψ >=

N Z X i=1

N

+

N

1 XX 2 i=1 j=1

Z

d3 r

Z

d3 r

1 ∇φ∗i (r) · ∇φi (r) + φ∗i (r)Vc (r)φi (r) 2m

d3 r0 V (r − r0 ) |φi (r|2 |φj (r0 )|2 − δsi ,sj φ∗i (r)φj (r)φ∗j (r0 )φi (r0 )

Now vary φ∗i (r) → φi (r) + δφi (r), whereby

N Z X 1 3 ∗ ∗ < ψ|Ĥ|ψ >−→< ψ|Ĥ|ψ > + d r ∇δφi (r) · ∇φi (r) + δφi (r)Vc (r)φi (r) 2m i=1 N

+

1 X 2 i,j=1

Z

d3 r

Z

d3 r0 V (r − r0 ) δφ∗i (r)φi (r|φj (r0 )|2 + δφ∗j (r)φj (r|φi (r0 )|2

−δsi ,sj δφ∗i (r)φj (r)φ∗j (r0 )φi (r0 ) + δφ∗j (r)φi (r)φ∗i (r0 )φj (r0 ) Set

δ < ψ|Ĥ|ψ > − < ψ|ψ > = 0 whereby

XZ

1 2 − ∇i + Vc (r) φi (r) + d3 r0 V (r − r0 ) φi (r|φj (r0 )|2 2m j6=i

−δsi ,sj φj (r)φ∗j (r0 )φi (r0 ) − φi (r) = 0 where is a Lagrange multiplier inserted in order to maintain the constraint < ψ|ψ >= 1. a) For Hartree equations omit the δsi ,sj term so

XZ 1 2 ∇ + Vc (r) φi (r) + d3 r0 V (r − r0 )φi (r|φj (r0 )|2 − φi (r) = 0 − 2m i j6=i

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b) For Hartree-Fock retain the anti-symmetrizing term so

XZ

1 2 − d3 r0 V (r − r0 ) φi (r|φj (r0 )|2 ∇i + Vc (r) φi (r) + 2m j6=i

−δsi ,sj φj (r)φ∗j (r0 )φi (r0 ) − φi (r) = 0

14.6 Example of an Infinite System: The Electron Gas Continuing the theme of the previous exercise, we again use the Coulomb interaction. Now, however, we consider an infinite system, a “gas” of electrons, as is appropriate in many studies of condensed matter physics [Fet71]. Indeed, the electron gas is the beginning paradigm for electrons in condensed matter systems. Note that the developments here are very closely related to those in Chapter 13 where nuclear matter was the focus. As noted previously, the treatment of nuclear (and neutron) matter presented there can be extended to a full HF analysis.

The Hamiltonian for an electron gas may be written as H = H0 + H1 + H2 , where N

H0 = − H1 =

1 X 2 ∇ 2m i=1 i

N X

U (ri )

i=1

H2 =

e2 1X exp −κ|ri − rj | , 2 |ri − rj | i6=j

where U (ri ) is a uniform potential of magnitude −U0 over a box of volume Ω. U (ri ) may be thought of as the average effect of the positive ions if the electron gas is in a solid. The parameter κ is the inverse of a scattering radius, which may be permitted to become infinite later (κ → 0).

Begin by choosing plane waves with periodic boudary conditions as a basic orthonormal set of single-particle states (inside the box) conditions: 1 φk (r) = √ eik·r . Ω Ignore the spin degrees of freedom. a) Show that in terms of creation and destruction operators H0 =

X k2 k

a† ak 2m k

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H1 = −U0 H2 =

X † ak ak k

1 2πe2 X † † a`0 ak0 ak a` 2 δk0 +`0 ,k+` , Ω Q + κ2 0 0 kk ``

where Q = k − k0 = `0 − ` and δ is a Kronecker delta. b) Neglecting the interaction H2 , the ground state |g.s.i is clearly given by a degenerate Fermi gas; i.e., the occupation numbers are

1 |k| ≤ kF nk = 0 |k| > kF , where kF is the Fermi momentum. Show that in this state < H2 >=< g.s.|H2 |g.s. >= Edir + Eexc , where Edir is the direct Coulomb interaction energy of all the electrons Edir 2πe2 N = 2 Ω κ Ω and Eexc is the exchange energy Eexc e2 Ω =− Ω (2π)5 N

Z Z

dkd` , |k − `|2

where N , the total number of electrons in the gas, is very large and the two integrals are both over the Fermi sphere. The term Edir can be canceled by a suitable choice of U0 . Note that in obtaining the last formula we let κ → 0 and made the replacement Z X dk . →Ω (2π)3 k

c) Show that Eexc 3e2 =− kF , N 8π and estimate Eexc /N for a typical metal.

Exercise 14.6 Solution: Example of an Infinite System: The Electron Gas a) We have < k|H0 |` >=

k2 δk,` 2m

so Ĥ0 =

X † X k2 † ak < kk|H0 |` > a` = a ak 2m k k,`

k

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Also < k|H1 |` >= −U0 δk,` so Ĥ1 =

X † X † ak < kk|H1 |` > a` = −U0 ak ak k,`

k

and < k0 , `0 |H2 |k, ` >=

1 2πe2 δk+`,k0 +`0 2 Ω q + κ2

so Ĥ2 =

2πe2 Ω

X

a†k0 a†`0 ak a`

k,`,k0 ,`0

1

δk+`,k0 +`0 q2 + κ2

where q = k − m = n − `. Here we have used Z 1 e2 < k, `|H2 |m, n >= 2 d3 r1 d3 r2 ei(k−m)·r1 ei(`−n)·r2 exp(−κ|r1 − r2 |) Ω |r1 − r2 | Z Z r1 +r2 1 3 r1 + r2 = 2 d d3 (r1 − r2 )ei(k−m+`−n)·( 2 ) Ω 2 r1 −r2 e2 ei(k−m−`+n)·( 2 ) e−κ|r1 −r2 | |r1 − r2 | Z ∞ Z 1 1 e2 sin qs −κs e2 = δk−m,n−` d3 s ei(k−m)·s−κ|s| = δk−m,n−` · 4π dss2 e Ω |s| Ω s qs 0 Z 1 4πe2 ∞ 4πe2 1 δk−m,n−` · ds sin qse−κs = δk−m,n−` 2 Ω q Ω q + κ2 0

×

b) We have then < H2 >=

2πe2 1 X 2πe2 X 1 n n − nk n` 2 k ` 2 Ω κ Ω q + κ2 k,`

=

so

2πe2 N 2 − 2πe2 Ω Ω κ2

k,`

Z

d3 k (2π)3

Z

d3 ` 1 3 (2π) |k − `|2

c) We show below that Z Z d3 ` 1 N d3 k = 4π 2 kF4 = 12π 4 kF 3 (2π) (2π)3 |k − `|2 Ω Eexch = −

3 2 e kF 8π

We have used Z Z Z Z Z d3 k d3 ` 1 d3 k d3 ` 1 i(k−`)·r e = d3 r 3 3 2 3 3 (2π) (2π) |k − `| (2π) (2π) 4πr

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1 = 4π Z = 4π

Z

1 d r r 3

1 d r 5 r 3

"Z

#2

kF

3

d ke

ik·r

3

= 4π

0

#2

"Z kF r dzz sin z 0

= (4π)2 kF4 =

1 d r 3 r

Z

Z ∞ 0

Z = 4π

"Z

#2

kF

2

dkk j0 (kr) 0

1 2 d3 r 5 (sin kF r − kF r cos kF r) r

dy π j1 (y)2 = (4πkF2 )2 y 2

Z ∞ 0

dy J 3 (y)2 y2 2

π Γ(1)Γ(1) 1 (4πkF2 )2 2 3 = (4πkF2 )2 3 2 4 2 Γ( 2 )Γ( 2 )Γ(3)

14.7 Zero Temperature Interacting Bose Gas Now consider another condensed matter system, a Bose gas. Consider the binding energy per particle of a zero temperature dilute Bose gas with a repulsive interaction. Dilute means that the scattering length a is much less than the average 1 particle spacing n− 3 , where n = N/Ω is the number density. In this limit S-wave scattering dominates and the relevant low momentum matrix elements of the potential are independent of the momentum, so that the interaction can be represented by a simple contact term, X p2 U0 X H= a†p ap + a†p4 a†p3 ap2 ap1 .

2m 2Ω p p p p p 1 2 3 4

If there were no interaction the system at zero temperature would be represented by a simple wavefunction with all N particles in the ground state, D E g.s. N̂ g.s. = N , P where N̂ = p a†p ap is the number operator.

However, with the interaction turned on this will change. Still, provided that U0 is not too large we expect that the occupation number of any particular excited state will be O(1/N ) campared to the ground state and that interactions between pairs of excited states will be O(1/N 2 ) relative to the ground state and can be ignored. We expect then that a†0 a0 |g.s.i = N0 |g.s.i ∼ O(N ) |g.s.i a0 a†0 |g.s.i = (a†0 a0 + 1) |g.s.i = (N0 + 1) |g.s.i .

But, since N0 1, the 1 can be neglected and we can pretend that both a0 and a†0 are c-numbers 1

a0 ' a†0 = N02 .

Now rewrite the interaction by separating out excited and ground state terms.

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a) Show that: N0 = N −

X

a†p ap

p6=0

a†0 a†0 a0 a0 = N 2 − 2N

X

a†p ap .

p6=0

b) Show in addition that X (a†p a†−p a0 a0 + a†0 a†0 ap a−p + a†0 a†p ap a0 + a†p a†0 a0 ap + a†p a†0 ap a0 ) p6=0

≈N

X p6=0

(a†p a†−p + ap a−p + 4a†p ap )

and that this leads to the result X p2 N 2 U0 N U0 X † † H' + (ap a−p + ap a−p + 2a†p ap ) + a† ap .

2Ω 2Ω 2m p p6=0

c) Now perform a Bogliubov transformation, by defining ap + αp a†−p bp = q 1 − αp2

a†p + αp a−p b†p = q 1 − αp2

and show that [bp , b†p0 ] = δpp0 b†p bp =

1 (a† ap + αp a†p a†−p + αp ap a−p + αp2 (1 + a†−p a−p )) .

1 − αp2 p

d) Pick αp so that the Hamiltonian can be written as X H = E0 + (p)b†p bp p6=0

and find both (p) and E0 . Show that E0 =

p2 N U0 N 2 U0 X + ( (p) − − ) 2Ω 2m 2Ω p6=0

r (p) =

(

p2 2 N U0 p2 ) + .

2m Ωm

Note here then that E0 is the new ground state energy while (p) is the energy of excitations (phonons) in the system. e) Show that E0 can be expressed by merely adding to the lowest order energy the vacuum energy of the Bogliubov oscillators (with appropriate terms included to subtract the divergences).

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f) Now convert from the potential U0 (which is not an observable) to the S-wave scattering length (a). Use second-order perturbation theory to show that 4π U2 X 1 a = U0 − 0 p2 m Ω

p6=0 m

and rewrite your expressions for (p) and E0 in terms of a. g) Show that as p → 0 the phonon excitation energy is of the form (p) = pv , where v is the phonon velocity and thus also the velocity of sound in the system.

R d3 p P h) By integrating over the excitation spectrum using p = Ω (2π) 3 show that the ground state energy is of the form

Eg.s. hg.s. |H| g.s.i 2πan2 128 3 21 = = 1 + √ (na ) + . . . . Ω Ω m 15 π i) Find the number of particles in the lowest energy state, N0 = g.s. a† a0 g.s. , and show that 1 8 N0 = 1 − √ (na3 ) 2 + . . .

N 3 π j) Find the speed of sound via the usual thermodynamic identity r Ω2 ∂ 2 E vs = mN ∂Ω2 and show that this yields the same result as found using the phonon velocity. k) Show that an explicit expression for the ground state is h i Yq 1 − αp2 exp −αp a†p a†−p |0i , |g.s.i = p

i.e., verify that the state is normalized to unity and that bp |g.s.i = 0 .

Exercise 14.7 Solution: Zero Temperature Interacting Bose Gas a) We have N=

X p

a†p ap = a†0 a0 +

X

a†p ap = N0 +

p6=0

X p6=0

so N0 = N −

X

a†p ap

p6=0

and a†0 a0 a†0 a0 = a†0 (1 + a†0 a0 )a0 ' a†0 a†0 a0 a0

a†p ap

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= (N − b) Look at X

X p6=0

a†p4 a†p3 ap2 ap1 =

a†p ap )2 ' N 2 − 2N

X

a†p ap

p6=0

X † (a0 + ap4 )(a†0 + ap3 )(a0 + ap2 )(a0 + ap1 )

p1 ,p2 ,p3 ,p4

= a†0 a†0 a0 a0 +

X † † a0 a0 ap a−p + a†p a†−p a0 a0 + a†p a†0 a0 a−p + a†p a†0 a−p a0 p

+a†0 a†p ap a0 + a†0 a†p a0 ap + . . . = N02 + N0

X (ap a−p + a†p a†−p + 4a†p ap ) + . . . p6=0

= (N −

X p6=0

a†p ap )2 + N (ap a−p + a†p a†−p + 4a†p ap ) + . . .

= N2 + N

X p6=0

(ap a−p + a†p a†−p + 2a†p ap ) + . . .

Hence H'



 X U0  2 a† ap + N +N (ap a−p + a†p a†−p + 2a†p ap ) 2m p 2Ω

X p2 p6=0

p6=0

c) We have [bp , b†p ] =

1 [ap , a†p ] + αp [ap , a−p ] + αp [a†−p , a†p ] + αp2 [a†−p , a−p 2 1 − αp = b†p bp =

=

1 − αp2 =1 1 − αp2

1 a†p + αp a−p ap + αp a†−p 2 1 − αp

1 † † † † 2 a a + α a a + α a a + α (1 + a )a −p p p p p p p p p −p −p 1 − alpha2p

d) Pick (p)

(p)

αp N U0 = 1 − αp2 2Ω

1 + αp2 p2 N U0 = + 2 1 − αp 2m Ω

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Then H'

N 2 U0 X N U0 p2 † N U0 + (ap a−p + a†p a†−p ) + ( + ap ap ) 2Ω 2Ω Ω 2m p6=0

" # 1 + αp2 † αp N 2 U0 X † † (p) (ap a−p + ap a−p ) + (p) a ap + = Ω 1 − α + p2 1 − αp2 p p6=0

αp2 N 2 U0 X ' + (p) b†p bp − 2Ω 1 − αp2

!

p6=0

Solve: 1 + αp2 = 2 + x with αp

x=

p2 Ω mN U0

or αp2 − (2 + x)αp + 1 = 0 so p 1 1 αp = (2 + x − (2 + x)2 − 4) = 1 + x − 2 2

r x+

x2 4

For (p) use (p)

1 + αp2 αp (1 − αp )2 1 − αp − 2 (p) = (p) = (p) 2 2 2 1 − αp 1 − αp 1 − αp 1 + αp =

N U0 N U0 p2 p2 + −2 = 2m Ω 2Ω 2m

Then q q 2 1 x2 1 2 + x − x + x + x + x4 p 1 + αp p 2 4 2 q q (p) = = 2m 1 − αp 2m − 1 x + x + x2 1 x + x + x2 2 4 2 4 2

2

q r 2 2 x2 x + + 2 x + x4 − x − x4 p p2 1 1 4 = = + 2 2 2m m 4 x x + x4 − x4 s 2 p2 p2 N U0 = + 2m m Ω 2

Ground state energy is then Eg.s. =

αp2 N U0 X N 2 U0 X αp − (p) = − (p) αp 2Ω 1 − αp2 2Ω 1 − αp2 p6=0

p6=0

N 2 U )u X N U0 = − 2Ω 2Ω p6=0

x 1+ − 2

r

x2 x+ 4

!

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N 2 U0 1X = + 2Ω 2

r

p6=0

p2 N U0 p4 N U0 p2 + − − 4m2 m Ω Ω 2m

!

Note that r

= so

p2 2m

p4 p2 N U0 p2 + = 2 4m m Ω 2m

1+

2m N U0 + ... p2 Ω

=

s 1+

4m N U0 p2 Ω

p2 N U0 1 + + O( 2 ) 2m Ω p

P

p6=0 is convergent. e) Note that

X

(p)b†p bp

p6=0

is simoly a harmonic oscillator Hamiltonian with ground state energy 1X (p) 2 p6=0

removed. Here zero point energy is r p2 N U0 1 X p2 N U0 1 1X p4 + = + + O( 2 ) E0 pt = 2 4m2 m Ω 2 2m Ω p p6=0

p6=0

and the term involving 1/p2 is convergent while the others are divergent. Hence include terms which subtract off the divergence leaving ! r 1X p2 N U0 p2 N U0 p4 E0 pt = + − − 2 4m2 m Ω 2m Ω p6=0

f) Now rewrite in terms of the scattering length T =V +V so T (E = 0) = V − V

1 V + ...

E − H0

1 U2 X 1 4π V + . . . = U0 − 0 a 2 = p H0 Ω p m m

We have N 2 U0 1X Eg.s. = + 2Ω 2 p

r

p4 p2 N U0 p2 N U0 − − + 2 4m m Ω 2m Ω

Now use ( 4π a)2 X 1 4π a+ m = U0 p2 m Ω p m

!

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190

so N2 Eg.s. = 2Ω

4πa 16π 2 a2 X 1 + m m2 Ω p p2

!

m

 −

1 X p2 2

v u u  t 1 − 1 −

4πaN/mΩ

+

4πaN mΩ

=

N 2 2πa 8π 2 N 2 a2 X 1 + p2 Ω m m2 Ω2 p

2m

p

!2

p2 4πaN 2m + mΩ

  

m

−

p4 p2 4πa + 2 2 4m m Ω

!

p4 p2 N 4πa + ' pvS 4m2 mΩ m

1 vS = m

r

p

g) Look at phonons r (p) =

r

p2 2πaN 1 + − 4m mΩ 2

X

so

4πa

N Ω

h) Now perform the sum over p, using cutoff regularization ! r Z p4 d3 p p2 2πaN 1 p2 4πaN + − + (2π)3 4m mΩ 2 4m2 m2 Ω 4π = (2π)3

Z Λ dp 0

p4 p2 2πaN p3 + − 4m mΩ 2

r

p2 4πaN + 2 2 4m m Ω

!

  3  4πaN 2 Λ2 2  Λ + 5 3 4m2 m2 Ω 1 1 Λ 2πaN Λ 1 = + −  1 2 2  4m 5 2π  mΩ 3 2 5 · 4m2 2 4πaN m2 Ω − 1 2 15 ( 4m 2) 1 = 2π 2

(

Λ2 4πaN + 2 4m2 m Ω

−

2πaN Λ3 2m2 2 Λ3 Λ5 + − Λ 20m 3mΩ 5 8m3

1 2 4πaN + · 16m4 2 2 15 m Ω 1 = 2π 2

23

Λ3 − 8m3

4πaN m2 Ω

32 !#)

16πaN 1+ Λ2 Ω

4πaN m2 Ω

23

23 !)

6πaN 2 3 1 4πaN 2 1 2πaN Λ3 − Λ3 · · · + Λ3 · ·2· 3mΩ mΩ 5 2 8 mΩ 15 2

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191

1 2 − · · 16m4 2 15 1 = 2π 2

(

πaN Λ3 mΩ

2 6 8 − + 3 5 15

52 )

4πaN m2 Ω

1 (32)2 − 2 15m

πaN Ω

52 )

12

=−

(16)2 1 π2 15

aN Ω

52

Hence full result is Eg.s.

N 2 2πa = 2 Ω Ω m

1+

128 1

15π 2

a

3 2

N Ω

! + ...

i) Note bp + αp b†−p ap = q 1 − αp2

b†p + αp b−p and a†p = q 1 − αp2

α2

p Then a + p† ap has VEV Np = 1−α 2 so N0 = N −

P

p Np and

p

X 1 αp2 N0 =1− N N 1 − αp2 p Here   p2 N U0 αp2 + N U0 1 1 Ω  = αp = −1 + q 2m 1 − αp2 2Ω (p) 2 p4 p2 N U 0 4m2 + m Ω so 1 X N

p



=

αp2 Ω 1 = 1 − αp2 N 2π 2 3

Ω 1  Λ 1 − + N 4π 2 3 2m

Z Λ 0

Z Λ 0

dpp3 q

r

1 p2 N U0 4m2 + mΩ

r

4πaN Ω

32

+

N U0 Ω

r

0

dpp q r

32 +Λ

1

N 4πa Ω m2

8πaN Ω

32πaN 8πaN −Λ +Λ 3Ω 3Ω

!



p2 N U0 4m2 + mΩ

Λ2 N 4πa + − 4m2 Ω m2

4πaN Ω



Z Λ

Λ2 N 4πa + − 2 4m Ω m2

Λ3 1 Λ3 + − 4m − 3 3 m

1 1 + 16m 3 m

2

p N U0 4m2 + m Ω

4πa Λ2 N 4πa 1 2 N Ω m2 + − 1 2 4m2 Ω m2 2m 3 ( 4m 2)

Ω 1 = N 4π 2



p2 N U0 2m + Ω

1 dpp2 (−1 + q 2 p4

" Ω 1 Λ3 N 4πa 1 = − + 1 2 N 4π 3 Ω m 4m 2 1 Λ2 + 3 2m 4m 2



r



!

N 4πa Ω m2

!#

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192

Ω 1 4 3 = 8π 2 N 4π 2 3

32 12 8 3N 3N a a = 1 Ω Ω 3π 2

Then 8 N0 =1− 1 N 3π 2

a3

N Ω

21 + ...

j) Determine speed of sound via ∂2E ∂ω 2

ρvS2 = Ω Here Ω

∂ 2 N 2 2πa 2N 2 2πa 4πa N 2 = = Ω = ∂ω 2 Ω m Ω3 m m Ω2

m

N Ω

4πa N m Ω

so 1 vS = m

r 4πa

N Ω

k) Claim |g.s. >= =

Y 1 (1 − αp2 ) 2 exp(−αp a†p a†−p )|0 > p

Y X (−)n 1 (1 − αp2 ) 2 αpn (a†p a†−p )n |0 > n! p n

Check normalization: (1 − αp2 ) < 0|

X (−)n

= (1 − αp2 )

n!

n

X n

= (1 − αp2 ) Check also bp |g.s. >= 0:

αpn (a−p ap )n

X (−)` `

`!

1 α2n < 0|an−p anp (a†p )n (a†−p )n |0 > (n!)2 p

X

αp2n = (1 − αp2 )

1 =1 1 − α + p2

(ap + αp a†−p ) exp(−αp a†p a†−p )|0 >= (ap + αp a†−p ) =

αp` (a†p a†−p )` |0 >

X (−)n n

n!

αpn (a†p a†−p )n |0 >

X (−)n X (−)n αpn (a†p a†−p )n−1 a†−p |0 > + αpn+1 (a†p a†−p )n a†−p |0 > (n − 1)! n! n n = −αp a†−p exp(−αp a†p a†−p )|0 > +αp a†−p exp(−αp a†p a†−p )|0 >= 0

14.8 Diagrammatic Expansion of the HF Green’s Function In the text the basics of expressing the HF approximation in terms of single-particle

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193

Green’s functions were summarized. Taking the form of the Dyson’s equation involved for GHF and the diagrammatic representation of the HF self-energy insertion (Figs. 14.5 and 14.4, respectively), start making expansions counting diagrams with specific numbers of interaction lines. Do this up to three interaction lines and detail the number of topologically distinct diagrams found in each order.

Exercise 14.8 Solution: Diagrammatic Expansion of the HF Green’s Function We begin with the two first-order diagrams in the text (Fig. 14.1) and add one more self-energy insertion (S) or tadpole insertion (T) to obtain the eight topologically distinct second-order diagrams shown in the first figure. There are two labeled A and B with two S insertions and no T insertions (2S0T), four labeled C, D, E and F of type (1S1T), and two labeled G and H of type (0S2T). Any two diagrams that can be manipulated topologically into the same shape are counted only once, otherwise one might think that there are 10 diagrams here.

Next, for each of the eight cases we add another S insertion or another T insertion — to make clear what is being inserted, and to make the figures a bit simpler than their final forms, we indicate an S insertion with an X and a T insertion with a dot. First we simply show all 80 diagrams without paying attention to which constitute the topologically distinct subset; these are illustrated in the following figures based on the original eight. Now examining each of the parent sets (A, B, C, etc.,) we can immediately drop A3, A5; B4; C3, C9; D7; E8; F4, F9; G7, G8; and H4, H9, as these are not topologically distinct. What is more work is to examine the remaining cases and observe that B1=A4, B5=A2; C1=A10, C2=B10, C4=A8; D1=A8, D2=A6, D3=B6, D4=A6, D9=C6; E1=A9, E2=B9, E3=B7, E4=A7, E6=D8; F1=C5, F2=D5; G1=C8, G2=C6, G3=D6, G4=F7, G5=F6; and H1=C10, H2=D10, H6=G10, H7=G9. This leaves 41 topologically distinct diagrams.

14.9 The Vibrational Model of Low-Lying Nuclear Excitations Following the steps outlined in the text, fill in the following details: a) Obtain the Lagrangian L=T −V =

X 2 1 1 5X 1 · 2 ρR0 αL,M − σR02 (L − 1)(L + 2) |αL,M | .

2 L 2 L,M

L,M

and Hamiltonian

X 1 1 2 2 H= |pL,M | + CL |αL,M | , 2BL 2 L,M L≥2

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194

A C

D

B

G

E

H

t

Fig. 14.1

F Topologically distinct second-order diagrams.

respectively, where

BL ≡ ρR05 /L

CL ≡ σR02 (L − 1)(L + 2) 2 ωL ≡ CL /BL

·

pL,M = BL (−)L αL,−M .

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195

A X X X X

t

Fig. 14.2

X 1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram A shown above. b) Show that the quantized Hamiltonian

X1 † † H= ωL αL,M αL,M + αL,M αL,M 2 L,M L≥2

emerges. c) Prove that the matrix elements of the creation and destruction operators involved are D E p † nL,M + 1 αL,M nL,M = nL,M + 1

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B

X X X X

t

Fig. 14.3

X 1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram B shown above. hnL,M − 1 |αL,M | nL,M i =

√

nL,M ,

and obtain the oscillation frequency 2 ωL ≡

1 L(L − 1)(L + 2)ω22 .

8

d) Confirm that the parity of these surfon states is (−)L .

Exercise 14.9 Solution: The Vibrational Model of Low-Lying Nuclear

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197

C X

X

X X

t

Fig. 14.4

X 1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram C shown above.

Excitations a) Let us begin by defining A≡ B≡

X

αL,M YLM (θ, φ)

L,M

1 X 2 |αL,M | 4π L,M

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D

X X X X

X

t

Fig. 14.5

1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram D shown above. for two of the terms in Eq. (14.35), so that the radius becomes R (θ, φ) = R0 {1 + A − B} .

The surface then becomes Z S=

Z dS =

dΩ R2 ,

surf ace

where for R2 we have

R2 = 1 + (A + A∗ ) − (B + B ∗ ) + A∗ A + O α3 ,

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199

E

X X X X

t

Fig. 14.6

X 1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram E shown above. retaining terms only up to second order, since we are making a small-oscillation approximation. We have for the contributions above X A + A∗ = 2 αL,M YLM (Ω) L,M

1 X 2 |αL,M | 4π L,M X X 0 ∗ ∗ A∗ A = αL,M YLM (Ω) αL0 ,M 0 YLM0 (Ω) ∗

B+B =2·

L,M

L0 ,M 0

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200

F X

X

t

Fig. 14.7

X

X

X 1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram F shown above.

using the symmetry properties given in the text. Upon performing the integral over solid angle and using the orthogonality relation for the spherical harmonics we obtain the expression for ∆S given in the text. The rest of the developments needed to find the Lagrangian and Hamiltonian are then those in the text. b) Equation (14.42) expresses the expansion coefficients in the usual form for treating small oscillations, i.e., in terms of normal modes (a similar strategy is adopted when quantizing the harmonic oscillator: see standard texts on quantum

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201

G X X X X X

t

Fig. 14.8

1

2

3

4

5

6

7

8

9

10

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram G shown above. mechanics if these procedures are not familiar; and see also Exercise 13.7) h i 1 M ∗ αL,M e−iωL t + (−) αL,M αL,M → p √ eiωL t 2 BL C L which yields h i 1 · M ∗ αL,M → −iωL p √ αL,M e−iωL t − (−) αL,M eiωL t 2 BL CL From this one goes to quantization by treating the expansion coefficients as surfon

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202

H X

X X

X 1

2

3

4

5

6

7

8

9

10

X

t

Fig. 14.9

Results from adding one S insertion (X) or one T insertion (dot) to second-order diagram H shown above.

† creation and destruction operators αL,M and αL,M , respectively. Substituting in the classical Hamiltonian immediately yields the quantized Hamiltonian, where the relationships between the factors BL and CL and the oscillation frequency ωL given just below Eq. (14.41) have been employed.

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203

c) Consider the creation operator acting on the state with nL,M − 1 surfons:

nL,M −1 1 † † † p αL,M |nL,M − 1i = αL,M αL,M |0i (nL,M − 1)!

nL,M 1 † αL,M |0i =p (nL,M − 1)! √ = nL,M |nL,M i and thus hnL,M − 1 |αL,M | nL,M i =

√

nL,M .

The other matrix element follows by taking the complex conjugate of this identity and changing nL,M to nL,M + 1. To obtain the oscillation frequency we use the fact that

CL σ 2 ωL = = L (L − 1) (L + 2) , BL ρR03 and since then ω22 =

σ ρR03

×8

we have 2 ωL =

1 L (L − 1) (L + 2) ω22 , 8

as required. d) Recall that under spatial inversion the spherical harmonics transform as L

YLM (θ, φ) → YLM (π − θ, φ + π) = (−) YLM (θ, φ) , (normal parity; see also Exercise 2.5). We require that R (θ, φ) behave the same way, L

R (θ, φ) → R (π − θ, φ + π) = (−) R (θ, φ) and hence that the creation and destruction operators also have normal parity.

15

Electron Scattering from Discrete States 15.1 Elastic Scattering: Parity and Time Reversal Symmetries have been discussed in Chapter 2 and, in Chapter 15, we have seen that parity and time-reversal invariance, together with conservation of angular momentum, limit the multipoles allowed for elastic scattering; this exercise is focused on proving this statement. a) Using the properties of the elastic multipoles discussed in Chapter 7, show that when spatial inversion is assumed to be a good symmetry, one has only even-J Coulomb and electric multipoles together with only odd-J magnetic multipoles. b) If, additionally, time reversal symmetry is invoked, show that the even-J electric multipoles also cannot occur and that one is left only with even-J Coulomb and odd-J magnetic multipoles. c) Supposing the ground state of a given nucleus to have angular momentum J0 = 7/2 (for example, the ground state of 41 Ca) what multipoles occur for elastic electron scattering? Which are coherent and which are not?

Exercise 15.1 Solution: Elastic Scattering: Parity and Time Reversal a) The parity properties of the EM multipoles have been discussed in the text.

There we found that the Coulomb (CJ) and electric (EJ) multipoles have normal (natural) parity, while the magnetic multipoles (M J) have abnormal (non-natural) parity, meaning that under spatial inversion one has cJM → (−)J M cJM M J el el TbJM → (−) TbJM

J+1 bmag mag TbJM → (−) TJM .

These are essentially built in to the operators by definition. Let us fill in some of the details that led to these properties. First we know that under spatial inversion, r → −r, that the spherical harmonics transform as follows J

YJM → (−) YJM (see also Exercises 2.5 and 14.9) and hence for the Coulomb multipoles the above result is immediate. For the transverse multipoles one needs to show that under spatial inversion one has the following for the vector spherical harmonics L+1

M YJL1 → (−)

204

M YJL1 ,

205

Electron Scattering from Discrete States

which is proven by writing out the vector spherical harmonics in terms of ordinary spherical harmonics and unit vectors (see the text), and then using the above parity properties of the former. Upon examining the definitions of the electric and magnetic multipoles in terms of how they involve specific vector spherical harmonic projections of the three-vector EM current one finds the above results.

These results are general and can be used when taking matrix elements of the operators for any type of transition. The case of elastic scattering is special, however, since the initial and final states involved when taking matrix elements are the same, namely, the ground state. It follows that only even-J Coulomb and electric elastic multipole matrix elements and odd-J magnetic elastic multipole matrix elements are nonzero; the others are forbidden by having a quantity equal to minus itself.

b) When discussing the properties of the multipoles under time reversal it is more convenient to make new definitions that incorporate factors of i (recall that time reversal involves an antiunitary operator and complex conjugation). Following, for instance, Edmonds [Edm74] or De Forest and Walecka [Def66], one defines primed operators via c0 ≡ iJ M cJM M JM 0el el TbJM ≡ iJ TbJM

0mag mag TbJM ≡ iJ TbJM .

Since all observables involve bilinear products of matrix elements of the multipole operators and their complex conjugates this redefinition leaves all observables unchanged. For the states themselves we choose the phase convention under time reversal Tb J+M Tb |JM i = (−) |J, −M i .

Again, as in part a) we start with properties of the spherical harmonics: using the fact that ∗ M YJM = (−) YJ−M and the fact that the charge density operator is hermitian one has immediately that

† J+M c0 0 cJM MJ,−M .

M = (−) And again, by writing out the vector spherical harmonics in terms of ordinary spherical harmonics and unit vectors, one can easily prove that ∗ M L+1−J −M M YJL1 = (−) (−) YJL1 which upon examining the definitions of the transverse multipoles leads to the

Electron Scattering from Discrete States

206

following:

0el TbJM

0mag TbJM

†

= (−)

†

= (−)

J+M +1 b0el TJ,−M J+M +1 b0mag TJ,−M .

Now, as stated above, under time reversal one has the following for c-numbers and matrix elements ∗ b b−1 E∗ D T cT E = cD a Tb−1 b = Tba | b .

Under time reversal the charge density operator, convection current operator and magnetization current operator have the following properties: TbρbTb−1 = ρb TbbjTb−1 = −bj

Tbµ bTb−1 = −b µ and hence, using the above, one immediately has that c0 Tb−1 = (−)J+M M c0 TbM JM J,−M J+M b0el 0el b−1 TJ,−M TbTbJM T = (−) J+M b0mag 0mag b−1 T = (−) TJ,−M , TbTbJM J+M

i.e., all with the phase (−) . The final step in the proof is to take matrix elements: for the Coulomb case one has D E D

E c0 c0 b−1 Tb Ji Mi Jf Mf M Tb−1 Tb M JM Ji Mi = Jf Mf JM T D

E c0 Tb−1 Tb Ji Mi = Jf Mf Tb−1 TbM JM D E Jf −Ji +J c0 = (−) Jf , −Mf M J,−M Ji , −Mi and then, going to reduced matrix elements (to take the known magnetic quantum number dependences into account), one can show that D E D E∗ c0 Ji = (−)Jf −Ji +J Ji M c0 Jf .

Jf M J J The transverse multipole matrix elements are obtained analogously, differing in that they have the extra minus sign above when a three-vector current is time-reversed: D D E E∗ J −J +J+1 0el 0el Jf TbJM Ji = (−) f i Ji TbJM Jf D E D E∗ J −J +J+1 0mag 0mag Jf TbJM Ji = (−) f i Ji TbJM Jf .

Finally, as in part a) for elastic scattering where the initial and final states are the same ground state one has that only even-J Coulomb multipole matrix elements and odd-J electric and magnetic multipole matrix elements are nonzero. For the

207

Electron Scattering from Discrete States

Table 15.1 Allowed elastic scattering multipoles when J0 = 7/2 J

CJ

EJ

MJ

0 1 2 3 4 5 6 7

C0 – C2 – C4 – C6 –

– – – – – – – –

– M1 – M3 – M5 – M7

Coulomb and magnetic cases this is the same constraint as came from the parity discussions in part a); however, for the electric multipole this means that under both spatial and temporal inversion no elastic electric multipoles are allowed. c) For ground-state angular momentum J0 = 7/2 angular momentum conservation means that only multipoles from J = 0 to J = 7 are potentially nonzero. Using the symmetry constraints of parts a) and b) we then have the allowed elastic multipoles given in the table. For all multipoles except for the Coulomb multipole C0 when the momentum transfer goes to zero one is left with some nontrivial power of r coming from the low-q limits of the spherical Bessel functions involved (see text for details). For instance, the Coulomb quadrupole goes as r2 . Hence, when computing the matrix elements the so-called one-body density matrix elements involved are of order unity. Said in terms of Slater determinants (see Chapter 13), typically only single rows and columns of the Slater determinants enter, with the rest of the rows and columns simply giving unity through the orthogonality of the single-particle wavefunctions involved. In contrast, the C0 multipole is special: it involves the spherical Bessel function j0 (qr) which, when q → 0 (the long wavelength limit, LWL), goes to unity. In this case all rows and columns of the Slater determinants contribute and the total goes as the particle number A (or Z or N if one is projecting protons or neutrons) and is therefore coherent. The most familiar case of this type is for elastic electron scattering where the Coulomb monopole is coherent and leads to the fact that the cross section goes as Z 2 in the LWL. All other multipoles (7 others in the present example) are incoherent. Indeed, the concepts of elastic scattering and coherence of multipoles are not the same and are often confused.

15.2 Elastic Charge Form Factor of a Spin-0 Nucleus Using the formalism presented in the present chapter, this exercise involves a specific

Electron Scattering from Discrete States

208

example of how to implement a model for the ground-state charge distribution and to find the related elastic C0 form factor. a) Using the phenomenological model for the ground-state charge distribution of 28 Si discussed in the text ρ(x)00 =

ρ0 , 1 + exp x−R a

compute the elastic charge form factor C0 using F0 (q) =

1 Z

Z dx

sin qx ρ(x)00 . qx

b) Compare this result with that obtained using harmonic oscillator wavefunctions in a closed-shell 1s41/2 1p83/2 1p141/2 1d12 5/2 model (the tables of Donnelly and Haxton [Don79a] may be used for the latter; as an approximation, consider only protons and neglect the contributions from neutrons). c) Plot the results for momentum transfers up to 4 fm−1 . What value of the oscillator parameter yields the best agreement between the two models? What are the RMS charge radii obtained?

Exercise 15.2 Solution: Elastic Charge Form Factor of a Spin-0 Nucleus a) The numerical result for the 2PF distribution is shown in the figure. b) Using the tables of Donnelly and Haxton [Don79a] one has that the harmonic oscillator (HO) reduced matrix elements of the C0 operator can be written in the form 1 n(`j hn(`1/2)j kM0 k n(`1/2)ji = √ y −1 e−y pM (y)J=0 4π 2

with y = (bq/2) and b the HO parameter. The polynomials required here (Tables I, V and XI in [Don79a]) are given in the table, which leads to the required Coulomb single-particle matrix elements:  √ 

 2  √ 2√ 1 − 23 y −y 4π hn(`1/2)j kM0 k n(`1/2)ji = e ×  2 1 − 23 y 

 √ 4 2 6 1 − 43 y + 15 y

1s1/2 1p3/2 .

1p1/2 1d5/2

These should be multiplied by the center-of-mass correction fcm (y) = ey/A , by the charge form factor of the proton (F1p (Q2 ) in [Don79a], although GpE (Q2 ) is √ used here), and by a factor [j] = 2j + 1 to account for the degeneracy of each

Electron Scattering from Discrete States

209

10 10 10

10

-1

-2

HO

-3

-4

2

F0 (q)

10

0

2PF 10 10 10 10 10 10

t

Fig. 15.1

-5

-6

-7

-8

-9

-10

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm ) 28

Elastic C0 form factors for Si normalized to unity at q = 0 obtained with the phenomenological two-parameter Fermi charge distribution (2PF) and harmonic oscillator model (HO) discussed in the text. single-particle level in the many-body model being used for 28 Si, which yields

2 2 FC0 = e−y ey/A GpE (Q2 ) 2 + 4 1 − y + 2 1 − y 3 3

4 4 +6 1 − y + y 2 3 15

6 4 = e−y ey/A GpE (Q2 )14 1 − y + y 2 7 35 = ZF0 .

Note that FC0 goes to Z = 14 in the LWL when q → 0, as it should. c) The results are plotted in the figure from part a). By varying the oscillator parameter it is found that a reasonable compromise fit to the phenomenological form factor is found for b = 877 fm, although this is somewhat subjective, since typically one requires the agreement to be optimized in some specific range of momentum transfer. This particular value forces the first diffraction minimum to occur at the same place. Using the fact that in the LWL one has F0 → 1 −

1 2 (qRch ) + · · · 6

Electron Scattering from Discrete States

210

Table 15.2 HO polynomials for Exercise 15.2 a) n(`j

n(`j

pM (y)J=0

1s1/2 1p3/2 1p1/2 1d5/2

√ 2y

2y 1 − 32 y √

2y 1 − 23 y √

4 2 y 6y 1 − 43 y + 15

and therefore that r Rch = lim

q→0

6 (1 − F0 ) q2

one finds that the RMS charge radii obtained are 3.350 fm and 3.207 fm for the phenomenological model of part a) and the HO model, respectively, given that the above value is chosen for the oscillator parameter. If, on the other hand, one chooses to make the charge radii agree, both at the value 3.350 fm, then the HO model must have b = 1.966 fm. Note that in obtaining these numbers both the cm correction ey/A and the proton charge form factor GpE (Q2 ) have been included, and that both contribute to the coefficient of q 2 when taking the above limit.

15.3 Elastic Form Factors of a Non-Spin-0 Nucleus This exercise proceeds from Exercise 15.2 to consideration of a non-spin-0 nucleus.

4 7 39 Assume a 1s41/2 1p83/2 1p141/2 1d12 K. 5/2 2s1/2 1d3/2 model for the ground state of a) Using harmonic oscillator wavefunctions with oscillator parameter b = 2 fm and the Donnelly-Haxton tables [Don79a], obtain expressions for all of the elastic form factors that enter, assuming angular momentum and parity are good quantum numbers and that one has time reversal invariance. b) What are the RMS charge radius, magnetic dipole moment and quadrupole moment that result?

Exercise 15.3 Solution: Elastic Form Factors of a Non-Spin-0 Nucleus +

a) The ground state of 39 K has J π = 23 , in the simple model being used arising from a 1d73/2 proton hole in a closed shell description of 40 Ca. The allowed elastic multipoles (see Exercise 15.1) are C0, M 1, C2 and M 3. For the first, one can employ the results found in Exercise 15.2 and adding the remaining 2s1/2 and 1d5/2 C0

211

Electron Scattering from Discrete States

matrix elements from Table XI in [Don79a]:  √  2 

   2 1 − 32 y  √   2 √ √2 1 − 43 y

4π hn(`1/2)j kM0 k n(`1/2)ji = e−y × 4 2  y 6 1 − 3 y + 15  √   2 1 − 4 y + 2 y2   3 3   4 2 y 2 1 − 34 y + 15

1s1/2 1p3/2 1p1/2 , 1d5/2 2s1/2 1d3/2

which yields the C0 form factor FC0 = ZF0, protons + N F0, neutrons , where the contributions of protons and neutrons are obtained just as in Exercise 15.2, yielding

2 2 2+4 1− y +2 1− y ZF0, protons = e e 3 3

4 4 2 4 2 2 +6 1 − y + y + 2 1 − y + y 3 15 3 3

4 4 +3 1 − y + y 2 3 15

56 56 2 2 −y y/A p y+ y = e e GE (Q )19 1 − 3 × 19 15 × 19

2 2 −y y/A n 2 N F0, neutrons = e e GE (Q ) 2 + 4 1 − y + 2 1 − y 3 3

4 4 4 2 +6 1 − y + y 2 + 2 1 − y + y 2 3 15 3 3

4 4 2 +4 1 − y + y 3 15

1 2 −y y/A n 2 = e e GE (Q )20 1 − y + y , 5 −y y/A

GpE (Q2 )

using the normalization conventions introduced in the text. Note that for completeness here we have kept the contributions from neutrons, even though they are a small correction due to the smallness of the neutron charge form factor GnE (Q2 ) (see Chapter 8), and note the weighting of the 1d3/2 protons and neutrons, namely, via the factors of 3 and 4, respectively, since that shell is filled for neutrons, but has one hole for protons.

The C2 form factor is obtained in a similar fashion using Table XI in [Don79a],

Electron Scattering from Discrete States

212

although now only the 1d3/2 proton hole contributes: √ 1d 4π 1d3/2 kM2 k 1d3/2 = e−y pM 3/2 (y)J=2

28 2 1d3/2 pM (y)J=2 = − y 1 − y 15 7 FC2 = −e

−y y/A

e

28 GpE (Q2 ) y 15

2 1− y .

7

As in the text writing 4π 2 F ≡ F02 + F22 + · · · Z2 L E2 D X 4π 1 cJ J0 = 2 , J0 M Z 2J0 + 1

2 Fch =

J≥0,even

we see that F2 =

√

E 4π D c2 J0 .

J0 M Z[J0 ]

Now, the handling of isospin is a bit involved (all necessary details may be found in [Don79a]). In this solution we present the details that will prove useful in performing similar steps in other cases discussed below, including for inelastic electron scattering. First one can form doubly-reduced matrix elements of a general operator bJM ;T M by using the Wigner-Eckart theorem in isospin space, O J T

D E T0 T T0 T0 −MT0 b J0 ; T0 MT0 OJ;T 0 J0 ; T0 MT0 = (−) −MT0 0 MT0 D E bJ;T 0 J0 ; T0 , × J0 ; T0 O where here the ground state is assumed to have isospin T0 = 1/2 with projection MT0 = (Z −N )/2 = −1/2 and ||| indicates matrix elements reduced both in angular momentum and isospin. From Edmonds [Edm74] one has

( 1 T =0 [T0 ] T0 T T0 T0 −MT0 MT0 . (−) = √ T =1 −MT0 0 MT0 [T0 ]

T0 (T0 +1)

Upon using the relationship between many-body matrix elements and the results written in terms of doubly-reduced matrix elements of single-particle operators (we are assuming here that the EM current operators are one-body operators; two-body meson-exchange currents can also play a role, although we do not consider them here): D E X (00) bJ;T 0 J0 ; T0 = J0 ; T0 O ΨJ;T (a0 a) ha0 |||OJ;T ||| ai aa0

with one-body density matrix elements (see [Don79a]) (00)

T

ΨJ;T (a0 a) = (−) δa0 1d3/2 δa1d3/2

Electron Scattering from Discrete States

213

in the present simple situation of a single hole in a filled 1d3/2 shell, i.e., one hole in 40 Ca viewed as a simple nucleus with shells 1s − 1p − 2s/1d all completely filled.

Finally, one needs the single-particle reduced matrix elements in isospin space (see Eqs. (20) and (21b) in [Don79a]): √ h1/2 kIT k 1/2i = [T ] / 2, where I00 ≡ 1/2 and I10 ≡ τ3 /2. Assembling these factors we have

1 1 b OJ;T 0 J0 ; T0 = −MT0 = J0 ; T0 = −MT0 = 2 2 1 = 1d3/2 kOJ k 1d3/2 T =0 + 1d3/2 kOJ k 1d3/2 T =1 , 2 and accordingly, for the Coulomb quadrupole operator the result is proportional to (0) (1) GE + GE /2 = GpE as expected. Thus we have

7 2 −y y/A p 2 F2 = −e e GE (Q ) y 1− y .

13 × 15 7

The abnormal parity M 1 and M 3 multipoles are handled likewise. In these cases the above implies that one has convection current contributions with the factor GpE as above, and magnetization current contribution contributions with the factor

(0)

(1)

GM + GM

/2 = GpM . One has FT2 =

1 2J0 + 1

X

D

J0 iTbJmag J0

E2

J≥1,odd

≡ F12 + F32 + · · · in general for elastic magnetic scattering (here only dipole and octupole, of course).

Table XII in [Don79a] yields for the M 1 case √ 1d 4π 1d3/2 k∆1 k 1d3/2 = e−y p∆ 3/2 (y)J=1 √ 1d 4π 1d3/2 kΣ01 k 1d3/2 = e−y pΣ03/2 (y)J=1 , where 3√ 1d p∆ 3/2 (y)J=1 = − 10

2 1− y 5 5

2√ 34 8 2 1d3/2 pΣ0 (y)J=1 = − 10 1 − y + y 5 15 15

and √

=

√

4π 1d3/2 kiT1mag k 1d3/2

q 1 1d 1d = e−y ey/A GpE (Q2 )p∆ 3/2 (y)J=1 − GpM (Q2 )pΣ03/2 (y)J=1 mN 2 4πF1 .

Electron Scattering from Discrete States

214

10

10

10

2

39

K

1

C0 0

2

F (q)

10

3

10

-1

M3

M1 10

10

-2

-3

C2 10

10

t

Fig. 15.2

-4

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

-1

q (fm ) 39

Elastic form factors squared for K in a simple model having one proton hole with respect to a closed shell model of 40 Ca. The M 3 case is completely analogous: √ 1d 4π 1d3/2 k∆3 k 1d3/2 = e−y p∆ 3/2 (y)J=3 √ 1d 4π 1d3/2 kΣ03 k 1d3/2 = e−y pΣ03/2 (y)J=3 , where 8√ 15y 75 8√ 1d pΣ03/2 (y)J=3 = 15y (1 − 2y) 75 1d

p∆ 3/2 (y)J=3 =

and √

=

√

4π 1d3/2 kiT3mag k 1d3/2

q 1 1d 1d = e−y ey/A GpE (Q2 )p∆ 3/2 (y)J=3 − GpM (Q2 )pΣ03/2 (y)J=3 mN 2 4πF3 .

The results are shown in the accompanying figure.

Electron Scattering from Discrete States

215

b) The relevant expression for the RMS charge radius was given in Exercise 15.2: r 6 Rch = lim (1 − F0 ). q→0 q2 As we saw above the monopole form factor is given by F0 = F0, protons + F0, neutrons , and hence the corresponding RMS charge radius is 3.521 fm when only protons are taken into account or 3.502 fm when adding in the contribution from neutrons. For the magnetic dipole form factor we have the limiting expression (see also [Don84]) 1 q→0 6π

FT2 →

J0 + 1 J0

q 2MN

2

µ2

or (s µ = −lim

6π

q→0

J0 J0 + 1

2mN q

)

FT

,

which yields µ=

3 (3 − µp ) = 0.124 5

in units of nucleon magnetons, namely, the value obtained from the expression for the jack-knifed proton Schmidt line discussed in Chapter 13 of the text. This differs by only 0.26 n.m. from the experimental value, which should be viewed as good agreement when measured against the characteristic scale of several n.m. Likewise the quadrupole moment arises from the limit (see also [Don84]) F22 →

1

q→0

2 4

180(2J0 + 1)

J0 −J0

2 0

J0 J0

J0 −J0

2 0

2 (Q/Z) q

or p Q = −lim 6 5(2J0 + 1) q→0

6Z = −lim F2 (q) q→0 q2

J0 J0

Z F2 (q) q2

yielding Q = 4.09 fm2 , quite close to the experimental value of 4.9 fm2 . The facts that the magnetic dipole moment and quadrupole moment are quite well reproduced in this simple model indicates that the one-proton-hole description of the ground state of 39 K is a good starting point for more sophisticated treatments.

15.4 Parity-Violating Elastic Electron Scattering I

216

Electron Scattering from Discrete States

Building on the results of the preceding two exercises, here PV elastic electron scat4 7 tering is considered. Assume that 40 Ca is a closed 1s41/2 1p83/2 1p141/2 1d12 5/2 2s1/2 1d3/2 shell nucleus and that 48 Ca is built from that closed shell plus eight neutrons in the 1f7/2 -shell.

a) Obtain the parity-conserving elastic scattering cross section when the singleparticle wavefunctions are taken to be harmonic oscillators with oscillator parameter b = 2.2 fm (again the Donnelly-Haxton tables [Don79a] may be used). b) Take into account both protons and neutrons in computing the elastic form factor.

How important are the neutrons? c) Obtain the parity-violating elastic electron scattering asymmetry (assume that strangeness can be neglected). d) Suppose that one has a facility with an 85% polarized electron beam of luminosity L = 1039 cm−2 sec−1 and a detector of 100 msr, how long will it take to measure the PV asymmetry to δA/A ∼ 5%?

Exercise 15.4 Solution: Parity-Violating Elastic Electron Scattering a) Now in addition to the results obtained in Exercise 15.3 we need contributions from the 1f7/2 -shell: we have from the tables in [Don79a]  √  2 

   2√ 1 − 23 y    2    √2 1 − 3 y

√ 4 2 6 1 − 43 y + 15 y 4π hn(`1/2)j kM0 k n(`1/2)ji = e−y × √   2 1 − 43 y + 32 y 2    4 2  y 2 1 − 43 y + 15  

 √  4  8 1 − 2y + y 2 − 8 y 3 5 3×35

1s1/2 1p3/2 1p1/2 1d5/2 2s1/2 1d3/2 1f7/2

where the 1f7/2 result comes from Table XVII. For the 48 Ca case we can write the total form factor as in Exercise 15.3

F0 = F0, protons + F0, neutrons

Electron Scattering from Discrete States

217

with

2 2 ZF0, protons = e e 2+4 1− y +2 1− y 3 3

4 4 2 4 2 2 +6 1 − y + y + 2 1 − y + y 3 15 3 3

4 4 +4 1 − y + y 2 3 15

1 2 −y y/A p 2 = e e GE (Q )20 1 − y + y 5

2 2 −y y/A n 2 N F0, neutrons = e e GE (Q ) 2 + 4 1 − y + 2 1 − y 3 3

4 4 4 2 +6 1 − y + y 2 + 2 1 − y + y 2 3 15 3 3

4 4 2 4 2 8 3 +4 1 − y + y + 8 1 − 2y + y − y 3 15 5 3 × 35

9 13 2 16 −y y/A n 2 3 = e e GE (Q )28 1 − y + y − y .

7 5×7 7 × 105 −y y/A

GpE (Q2 )

For the 40 Ca case similar expressions obtain, except that for neutrons we have the result above for protons but with GpE → GnE . Using b = 2.2 fm the parity-conserving form factors are shown in the first figure. Clearly the form factors for the two nuclei are very similar, as long as the same oscillator is assumed for both. b) The ratios of the contributions from neutrons to that from protons are shown in the second figure for the two nuclei. For the 40 Ca case one sees a smooth curve that rises as the momentum transfer increases, reaching about 17% at q = 4 fm−1 .

This behavior stems from the fact that the momentum transfer dependence in the polynomials involved has been taken to be the same, and accordingly the diffraction zeros fall in the same places. In contrast, for the 48 Ca case those polynomials are difference, since the 1f7/2 -shell now enters for neutrons and forces the diffraction zeros for protons and neutrons to fall in different places. Except where the proton contributions go through their zeros, the results are not too different. In more sophisticated treatments it is important to treat the incoming and scattered electrons as distorted waves, in which case no zeros are enountered but only diffraction minima. c) As discussed in the text the PV asymmetry is given by A=

GF Q2 W P V √ · , 2πα 2 W P C

where for elastic scattering from spin-0 nuclei the hadronic ratio is given by WPV FeC0 . = aA P C W FC0 Here FC0 is the parity-conserving elastic C0 form factor found in part a), while

Electron Scattering from Discrete States

218

10 10 10 10 10

2

Fel (q)

10 10 10 10 10 10 10 10 10 10

3 2 1 0

-1 -2 -3 -4 -5

40

-7 -8 -9

-10 -11

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

t

Fig. 15.3

Ca

-6

q (fm )

3.5 48

4.0

Ca

Elastic form factors squared for 40,48 Ca.

FeC0 arises from the interference product of the EM monopole form factor with its WNC counterpart; aA is the leptonic axial-vector coupling, as in the text. We have found in part a) that FC0 = FC0,protons + FC0,neutrons = ZF0,protons + N F0,neutrons , where the two contributions can be re-written in terms of isoscalar and isovector contributions i 1 h (0) (1) F0 + F0 2 i 1 h (0) (1) F0 − F0 .

F0,neutrons = 2 F0,protons =

Thus one has FC0 =

i 1h (0) (1) (Z + N ) F0 + (Z − N ) F0 .

2

The WNC analog to this is obtained simply by weighting the isoscalar and isovector contributions, not by unity as (by definition) occurs for the EM form factor, but

Electron Scattering from Discrete States

219

0.20

(Fel(q))n/(Fel(q))p

0.15 0.10 0.05 0.00 -0.05 -0.10 -0.15 -0.20 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.4

Ratio of the contributions from neutrons and protons to the elastic form factors of 40 Ca (red) and 48 Ca (blue).

(0)

(1)

with factors of βV and βV , respectively i 1 h (0) (0) (1) (1) FeC0 = βV (Z + N ) F0 + βV (Z − N ) F0 2 i 1h (0) (1) = (Z + N ) βV + (Z − N ) βV F0,protons 2 i 1h (0) (1) + (Z + N ) βV − (Z − N ) βV F0,neutrons 2 = [ZβVp + N βVn ] F0,protons + [ZβVn + N βVp ] F0,neutrons , where we have employed the hadronic WNC couplings from Chapters 8 and 15 in the text. The resulting PV asymmetries for the two nuclei are shown in the figure.

As above the diffraction zeros in the 48 Ca case arise from the different proton and neutron distributions as reflected in their polynomials being different. The 40 Ca case is much simpler in the present model: there the polynomials cancel in forming the hadronic ratio (one finds that W P V /W P C is a constant, 0.92). d) The parity-conserving elastic cross section is given by P C dσ −1 2 = σM ott frec FC0 dΩ with σM ott being the Mott cross section and frec the recoil factor discussed in the

Electron Scattering from Discrete States

220

48

PV Asymmetry

10

10

10

10

Ca

-4

40

-5

Ca

-6

-7

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.5

Parity-violating asymmetries for 40,48 Ca. text. One can form the so-called figure-of-merit as P C dσ F= A2 dΩ and show that the fractional uncertainty in the PV asymmetry A is obtained from 1 δA =√ , A FX0 where X0 = p2 · L · ∆Ω · T with p the polarization of the electron beam, L the luminosity, ∆Ω the solid angle of the detector and T the length of time over which the measurement is made (this formula is derived in detail in [Mus94], pages 66–70).

The figure-of-merit for 40 Ca and 48 Ca are shown in the accompanying figures, together with the corresponding times required. Three scattering angles have been assumed, 100 , 200 and 300 ; of course, to obtain the values of momentum transfer in the figures, the incoming electron energies need to be computed (high energies for the forward angle, typically conditions at JLab, and low energies at more backward angles, typically conditions at Mainz). While the minimum times, roughly 250 s (1100 s, 2500 s) for 40 Ca at 100 (200 , 300 ) may seem quite reasonable, the

Electron Scattering from Discrete States

221

10

-11

40

2 -1

FOM (fm s )

10 10

0

-12

20 10

10

0

-13

30 10

Ca

0

-14

-15

0.0

0.5

1.0

1.5

2.0

2.5

-1

q (fm )

t

Fig. 15.6

Figure-of-merit for 40 Ca.

conditions stated are very challenging and the 5% precision not the desired state-ofthe-art. For instance, in the high-energy 100 case (JLab at 2.5 GeV) the luminosity is typically an order of magnitude lower and the solid angle is an order of magnitude smaller, which increases the time by two orders of magnitude; also, something like 1% for δA/A is desirable, which incurs an additional factor of 25 in the time, yielding a total time of roughly 6.2x105 s under optimal conditions. Similar considerations hold for 48 Ca with times being somewhat shorter due to the larger PV asymmetries in that case.

15.5 Parity-Violating Elastic Electron Scattering II In this exercise, a different approach to PV elastic electron scattering is considered. a) Prove

N n WPV p ≡ aA βV + βV [1 + Γnp (q)] , WPC Z and Γnp (q) = −βV00

R

dxj0 (qx)ρn (x)/N R 1− , dxj0 (qx)ρp (x)/Z

Electron Scattering from Discrete States

222

10

-11

2 -1

FOM (fm s )

10 10

0

48

20 10

10

0

-13

30 10

Ca

-12

0

-14

-15

0.0

0.5

1.0

1.5

2.0

2.5

-1

q (fm )

t

Fig. 15.7

Figure-of-merit for 48 Ca. where βV00 ≡

N βVn p ZβV + N βVn

.

. b) Consider the 2PF function ρ(x)00 =

ρ0

1 + exp x−R a

discussed in the text as an approximation for the proton and neutron distributions of 208 Pb. Assume that these have slightly different radius parameters.

Compute the transforms R

dxj0 (qx)ρn (x)/N 00 Γnp (q) = −βV 1 − R , dxj0 (qx)ρp (x)/Z where βV00 ≡

N βVn p ZβV + N βVn

and hence the deviation function Γnp (q) as a function of the difference in radius parameter. (Assume that the average radius parameter is 7 fm and the surface diffuseness parameter is 0.5 fm; of course the distributions must be normalized to the correct numbers of protons and neutrons).

Exercise 15.5 Solution: Parity-Violating Elastic Electron Scattering II

Electron Scattering from Discrete States

223

4000 3500

30

Time (s)

3000

0

2500

20

0

2000 1500

10

0

1000 40 500

Ca

0 0.2

0.4

0.6

0.8

1.0

-1

q (fm )

t

Fig. 15.8

Time required to perform the PV measurement for 40 Ca assuming the conditions posed in the exercise.

a) For these developments we follow [Don89] (see also [Mor09] and references therein). Let us define the Fourier transforms of the proton and neutron distributions to be Z Sp,n (q) ≡ dx j0 (qx)ρp,n (x), then one has in the long wavelength limit Sp (q) → Z q→0

Sn (q) → N. q→0

For simplicity we set GnE to zero, although it can be retained if desired; also strangeness is ignore here. Note that Sn is the Fourier transform of the distribution of neutrons in the nucleus, not the neutrons’ contribution to the charge density that was considered in Exercise 15.4. The isoscalar and isovector Fourier transforms are then S (0) (q) = Sp (q) + Sn (q) S (1) (q) = Sp (q) − Sn (q)

Electron Scattering from Discrete States

224

4000 3500

Time (s)

3000 2500 2000

30

0

1500

20

0

1000

10

500

0 48

Ca

0 0.2

0.4

0.6

0.8

1.0

-1

q (fm )

t

Fig. 15.9

Time required to perform the PV measurement for 48 Ca assuming the conditions posed in the exercise. with inverses i 1 h (0) S (q) + S (1) (q) 2 i 1 h (0) Sn (q) = S (q) − S (1) (q) .

2 Sp (q) =

For the usual parity-conserving elastic scattering form factor we have FC0 = Sp (q) =

i 1 h (0) S (q) + S (1) (q) 2

and to obtain the WNC analog as in the previous exercise one weights the isoscalar (0) (1) and isovector contributions with βV and βV , respectively, i 1 h (0) (0) (1) βV S (q) + βV S (1) (q) FeC0 = 2 = βVp Sp (q) + βVn Sn (q).

If the shapes of the distributions of protons and neutrons in the nuclear ground state are the same (indicated by a subscript 0 below) one has Sn (q)/N = Sp (q)/Z

Electron Scattering from Discrete States

225

and then the above WNC result would be

h i N n p e FC0 = βV + βV Sp (q).

Z 0 the hadronic ratio would then become proportional to h i FeC0 N 0 = βVp + βVn .

FC0 Z The strategy here is to remove this result as an overall factor and only look at deviations of the general result from this answer:

FeC0 N ≡ βVp + βVn [1 + Γnp (q)] , FC0 Z where, assembling the contributions above, one has for the deviation

Sn (q)/N Γnp (q) = −βV00 1 − Sp (q)/Z with βV00 ≡

N βVn p ZβV + N βVn

,

as desired. b) Employing the 2PF function one obtains results such as those shown in the figure.

15.6 Inelastic Electron Scattering in 39 K In Exercise 15.3 a model for the ground state of 39 K was introduced. Suppose that an excited state of the same nucleus is obtained by promoting a proton from the 1d3/2 shell in the ground state to the 1f7/2 shell. a) What multipoles are allowed? b) Obtain the form factor for the lowest multipolarity allowed, remembering that a proton transition is assumed. c) Obtain the form factors for the highest multipolarity allowed.

The tables cited above may be used; take b = 2 fm.

Exercise 15.6 Solution: Inelastic Electron Scattering in 39 K +

π

−

a) The transition here goes from Jiπi = 23 to Jf f = 72 and thus J = 2, 3, 4 and 5, and the parity changes; hence the allowed multipoles are the following: M 2, C3/E3, M 4 and C5/E5. b) The lowest multipolarity allowed is M 2 and hence we need from Table XVI

Electron Scattering from Discrete States

226

1.0

0.5

0.95

0.95 Γnp (q)

0.99 1.01 0.0

0.99 1.01 1.05

1.05

-0.5

-1.0 0.2

t

Fig. 15.10

0.4

0.6

0.8

1.0

-1

q (fm )

The deviation function Γnp (q) obtained as discussed assuming that the radius parameter R in the 2PF distribution of neutrons is 0.95, 0.99, 1.01 or 1.05 that for the proton distribution. Of course, when the radius parameters are equal the deviation is zero, by definition.

in [Don79a] √

1f

1d

4π 1f7/2 k∆2 k 1d3/2 = y 1/2 e−y p∆ 7/2 3/2 (y)J=2 √ 1f 1d 4π 1f7/2 kΣ02 k 1d3/2 = y 1/2 e−y pΣ07/2 3/2 (y)J=3 ,

where 4√ 1f 1d 6 p∆ 7/2 3/2 (y)J=2 = −

2 1− y 15 7

8√ 5 2 2 1f7/2 1d3/2 pΣ0 (y)J=2 = − 6 1− y+ y 5 7 21

Electron Scattering from Discrete States

227

10

-1

2

2

F (q)

10

10

(FT )M2

-2

-3

2

(FT )E5

2

(FL )C5 10

10

10

-4

-5

-6

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.11

Squares of the three form factors for transition in 39 K discussed in the exercise.

and √

4π

1f7/2 kiT2mag k 1d3/2

×

2

FT M 2 =

1 4

=

q mN

y 1/2 e−y ey/A

1 1f 1d 1f 1d GpE (Q2 )p∆ 7/2 3/2 (y)J=2 − GpM (Q2 )pΣ07/2 3/2 (y)J=2

1f7/2 kiT2mag k 1d3/2

2

2

Here we have assumed that the excited state also has isospin 1/2, as does the ground state, and thus the isospin arguments are as in Exercise 15.3. The result is shown in the figure. c) The highest multipolarity allowed is 5, having C5 and E5 multipoles from Table XV: √

1f

1d

4π 1f7/2 kM5 k 1d3/2 = y 3/2 e−y pM7/2 3/2 (y)J=5 16 1f 1d pM7/2 3/2 (y)J=5 = √ y 3 105

1 2 FL2 C5 = 1f7/2 kM5 k 1d3/2 4

Electron Scattering from Discrete States

228

for the C5 multipole, and √

1f

1d

4π 1f7/2 k∆05 k 1d3/2 = y 3/2 e−y p∆07/2 3/2 (y)J=5 √ 1f 1d 4π 1f7/2 kΣ5 k 1d3/2 = y 3/2 e−y pΣ 7/2 3/2 (y)J=5 ,

where 4 √ 14 105 16 √ 1f 1d pΣ 7/2 3/2 (y)J=5 = 14y 105 1f

p∆07/2

1d3/2

(y)J=5 = −

and √

1 FT2 E5 = 4

q 4π 1d3/2 T5el 1d3/2 = y 3/2 e−y ey/A mN

1 1f 1d 1f 1d × GpE (Q2 )p∆07/2 3/2 (y)J=5 + GpM (Q2 )pΣ 7/2 3/2 (y)J=5 2 1f7/2 T5el 1d3/2

2

The results are shown in the figure for part b).

15.7 Inelastic Electron Scattering in 16 O In the text the case of pure 1p1h excitations in 16 O was considered, with focus placed on two specific configurations, namely, the T = 1 states of the type 1d5/2 1p−1 3/2 and 2s1/2 1p−1 .

1/2 a) Compute form factors for the analogous T = 0 states and compare the results with those already found. b) Extend the analysis to all T = 1 1p1h states involving particles in the 2s − 1d shell and holes in the 1p shell.

The tables cited above may be used; take b = 1.5 fm.

Exercise 15.7 Solution: Inelastic Electron Scattering in 16 O [Note to instructors: This exercise may easily be made shorter by assigning only a subset of cases or varied by choosing different shells.] a) For the 1d5/2 1p−1 3/2 case we have six types of multipoles, C1/E1, M 2, C3/E3

Electron Scattering from Discrete States

229

and M 4 with

√

(T ) 1d

1p

4π 1d5/2 kM1 k 1p3/2 = y −1/2 e−y ey/A GE pM 5/2 3/2 (y)J=1

√ q el 4π 1d5/2 T1 1p3/2 = y −1/2 e−y ey/A mN

1 (T ) 2 1d5/2 1p3/2 (T ) 2 1d5/2 1p3/2 × GE (Q )p∆0 (y)J=1 + GM (Q )pΣ (y)J=1 2

√ q 4π 1d5/2 kiT2mag k 1p3/2 = y 1/2 e−y ey/A mN

1 (T ) 1d5/2 1p3/2 (T ) 1d5/2 1p3/2 × GE p∆ (y)J=2 − GM pΣ0 (y)J=2 2 √ 1p (T ) 1d 4π 1d5/2 kM3 k 1p3/2 = y 1/2 e−y ey/A GE pM 5/2 3/2 (y)J=3

√ q 4π 1d5/2 T3el 1p3/2 = y 1/2 e−y ey/A mN

1 (T ) 1p 1d 1p (T ) 1d × GE p∆05/2 3/2 (y)J=3 + GM (Q2 )pΣ 5/2 3/2 (y)J=3 2

√ q 4π 1d5/2 kiT4mag k 1p3/2 = y 3/2 e−y ey/A mN

1 (T ) 1d 1p 1p (T ) 1d × GE p∆ 5/2 3/2 (y)J=4 − GM (Q2 )pΣ05/2 3/2 (y)J=4 , 2

(T )

with isoscalar (T = 0) and isovector (T = 1) single-nucleon form factors GE,M , depending on the type of transition being considered, namely, T = 1 in the text and T = 0 in the present exercise. For the 2s1/2 1p−1 1/2 case only C1/E1 multipoles occur and they have

√

(T ) 2s

1p

4π 2s1/2 kM1 k 1p1/2 = y −1/2 e−y ey/A GE pM1/2 1/2 (y)J=1

√ q el 4π 2s1/2 T1 1p1/2 = y −1/2 e−y ey/A mN

1 (T ) 2 2s1/2 1p1/2 (T ) 2s1/2 1p1/2 × GE p∆0 (y)J=1 + GM (Q )pΣ (y)J=1 .

2

Electron Scattering from Discrete States

230

The corresponding polynomials from Tables IX and X in [Don79a] are

√ 2 (y)J=1 = 2 2y 1 − y 5

4 1d 1p p∆05/2 3/2 (y)J=1 = − 1 − y 5

2 1d 1p pΣ 5/2 3/2 (y)J=1 = 2y 1 − y 5 √ 2 1d 1p 35 p∆ 5/2 3/2 (y)J=2 = − 15

10 2√ 1d 1p pΣ05/2 3/2 (y)J=2 = 35 1 − y 5 21 √ 8 1d 1p 3y pM 5/2 3/2 (y)J=3 = − 15 4 1d 1p p∆05/2 3/2 (y)J=3 = 15 4 1d5/2 1p3/2 (y)J=3 = − y pΣ 15 1d 1p p∆ 5/2 3/2 (y)J=4 = 0 4√ 1d 1p pΣ05/2 3/2 (y)J=4 = − 7 7 1d

pM 5/2

1p3/2

and from Table IX 2√ 2y (1 − y) 3 1 1p 2s p∆01/2 1/2 (y)J=1 = − (1 + y) 3 4 2s1/2 1p1/2 pΣ (y)J=1 = y (1 − y) .

3 2s

pM1/2

1p1/2

(y)J=1 =

The complete form factors are then simply obtained as the squares of the above matrix elements divided by 2 from handling the isospin content; see the accompanying figures. By comparing with the T = 1 results shown in the text (Figs. 15.10–12) we see that the T = 0 results found here are quite similar, the main trend being that the latter for electric and magnetic multipoles are somewhat lower at medium to high values of momentum transfer due to the smaller isoscalar magnetization current contributions. b) The remaining cases involving particles in the 2s − 1d shell and holes in −1 the 1p shell are 1d5/2 1p−1 1/2 with M 2 and C3/E3; 2s1/2 1p3/2 with C1/E1 and M 2; −1 −1 1d3/2 1p3/2 with C1/E1, M 2 and C3/E3; and 1d3/2 1p1/2 with C1/E1 and M 2. The expressions in part a) for reduced matrix elements of the five types of multipole operators may be used again with the replacements only of the polynomials for these remaining cases. For the latter one has from Tables IX and X of [Don79a] for

Electron Scattering from Discrete States

231

1d5/21p3/2 10

-1

0

C1 10

-1

C3

10

-2

2

F (q)

E1

M2 10

10

10

-3

-4

E3

M4

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.12

Squares of form factors for 1d5/2 1p−1 3/2 contributions to T = 0 to T = 0 transitions in 16 O.

the 1d5/2 1p−1 1/2 case

2√ 10 15

4√ 1 1d5/2 1p1/2 pΣ0 (y)J=2 = − 10 1 − y 5 3 4√ 1d5/2 1p1/2 pM (y)J=3 = − 15y 15 2√ 1d 1p p∆05/2 1/2 (y)J=3 = 5 15 8√ 1d 1p pΣ 5/2 1/2 (y)J=3 = − 5y; 15 1d

p∆ 5/2

1p1/2

(y)J=2 = −

Electron Scattering from Discrete States

232

2s1/21p1/2 10

-1

0

C1

2

F (q)

10

10

-1

-2

E1 10

10

10

-3

-4

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.13

Squares of form factors for 2s1/2 1p−1 1/2 contributions to T = 0 to T = 0 transitions in 16 O.

for the 2s1/2 1p−1 3/2 case

4 y (1 − y) 3 1√ 2s 1p p∆01/2 3/2 (y)J=1 = − 2 (1 + y) 3 2√ 2s 1p pΣ 1/2 3/2 (y)J=1 = − 2y (1 − y) 3 2s 1p p∆ 1/2 3/2 (y)J=2 = 0 2√ 2s 1p p∆ 1/2 3/2 (y)J=2 = 6 (1 − y) ; 3 2s

pM1/2

1p3/2

(y)J=1 =

Electron Scattering from Discrete States

233

for the 1d3/2 1p−1 3/2 case 2√ 1d 1p pM 3/2 3/2 (y)J=1 = − 2y

2 1− y 3 5

1 4 1d3/2 1p3/2 p∆0 (y)J=1 = 1− y 3 5

8 2 1d3/2 1p3/2 pΣ (y)J=1 = y 1 − y 3 5 2√ 1d3/2 1p3/2 15 p∆ (y)J=2 = 15 √ 4 1d 1p pΣ03/2 3/2 (y)J=2 = 15 15 √ 4 1d 1p pM 3/2 3/2 (y)J=3 = 2y 5 2√ 1d 1p p∆03/2 3/2 (y)J=3 = − 6 15 √ 8 1d 1p pΣ 3/2 3/2 (y)J=3 = − 6y 15 and for the 1d3/2 1p−1 1/2 case 2√ 1d 1p pM 3/2 1/2 (y)J=1 = 10y

2 1− y 3 5

1√ 4 1d3/2 1p1/2 p∆0 (y)J=1 = − 5 1− y 3 5

2√ 2 1d3/2 1p1/2 (y)J=1 = − 5y 1 − y pΣ 3 5 √ 2 1d 1p p∆ 3/2 1/2 (y)J=2 = − 15 15 2√ 1d 1p pΣ03/2 1/2 (y)J=2 = − 15 (1 − 2y) .

15

These yield the following for the remaining cases involving these two major shells: for the 1d5/2 1p−1 1/2 case √

√

q 4π 1d5/2 kiT2mag k 1p1/2 = y 1/2 e−y ey/A mN

1 (T ) 1d 1p 1p (T ) 1d × GE p∆ 5/2 1/2 (y)J=2 − GM pΣ05/2 1/2 (y)J=2 2 (T ) 1d

1p

4π 1d5/2 kM3 k 1p1/2 = y 1/2 e−y ey/A GE pM 5/2 1/2 (y)J=3

√ q el 4π 1d5/2 T3 1p1/2 = y 1/2 e−y ey/A mN

1 (T ) 2 1d5/2 1p1/2 (T ) 1d5/2 1p1/2 × G E p ∆0 (y)J=3 + GM (Q )pΣ (y)J=3 2

Electron Scattering from Discrete States

234

for the 2s1/2 1p−1 3/2 case √

(T ) 2s

1p

(T ) 1d

1p

(T ) 1d

1p

4π 2s1/2 kM1 k 1p3/2 = y −1/2 e−y ey/A GE pM1/2 3/2 (y)J=1

√ q el 4π 2s1/2 T1 1p3/2 = y −1/2 e−y ey/A mN

1 (T ) 2s 1p 2s 1p (T ) × GE (Q2 )p∆01/2 3/2 (y)J=1 + GM (Q2 )pΣ 1/2 3/2 (y)J=1 2

√ q 4π 2s1/2 kiT2mag k 1p3/2 = y 1/2 e−y ey/A mN

1 (T ) 2s 1p 1p (T ) 2s × GE p∆ 1/2 3/2 (y)J=2 − GM pΣ01/2 3/2 (y)J=2 2 for the 1d3/2 1p−1 3/2 case √

4π 1d3/2 kM1 k 1p3/2 = y −1/2 e−y ey/A GE pM 3/2 3/2 (y)J=1

√ q el 4π 1d3/2 T1 1p3/2 = y −1/2 e−y ey/A mN

1 (T ) 2 1d3/2 1p3/2 (T ) 2 1d3/2 1p3/2 (y)J=1 × GE (Q )p∆0 (y)J=1 + GM (Q )pΣ 2

√ q 4π 1d3/2 kiT2mag k 1p3/2 = y 1/2 e−y ey/A mN

1 (T ) 1d3/2 1p3/2 (T ) 1d3/2 1p3/2 (y)J=2 × GE p∆ (y)J=2 − GM pΣ0 2 √ 1p (T ) 1d 4π 1d3/2 kM3 k 1p3/2 = y 1/2 e−y ey/A GE pM 3/2 3/2 (y)J=3

√ q 4π 1d3/2 T3el 1p3/2 = y 1/2 e−y ey/A mN

1 (T ) 1d 1p 1p (T ) 1d × GE p∆03/2 3/2 (y)J=3 + GM (Q2 )pΣ 3/2 3/2 (y)J=3 2 and for the 1d3/2 1p−1 1/2 case √

4π 1d3/2 kM1 k 1p1/2 = y −1/2 e−y ey/A GE pM 3/2 1/2 (y)J=1

√ q el 4π 1d3/2 T1 1p1/2 = y −1/2 e−y ey/A mN

1 (T ) 2 1d3/2 1p1/2 (T ) 2 1d3/2 1p1/2 × GE (Q )p∆0 (y)J=1 + GM (Q )pΣ (y)J=1 2

√ q 4π 1d3/2 kiT2mag k 1p1/2 = y 1/2 e−y ey/A mN

1 (T ) 1d3/2 1p1/2 (T ) 1d3/2 1p1/2 × GE p∆ (y)J=2 − GM pΣ0 (y)J=2 .

2

The results for the remaining configurations are shown in the accompanying figures.

Electron Scattering from Discrete States

235

10

1d5/21p1/2

-1

-1

C3 -2

2

F (q)

10

10

-3

E3

M2 10

10

-4

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.14

Squares of form factors for 1d5/2 1p−1 1/2 contributions to T = 0 to T = 0 transitions in 16 O. [Discuss results] 15.8 Parity-Violating Inelastic Electron Scattering π

Consider an inelastic transition Jiπi Ti = 0+ 0 → Jf f Tf = 1+ 0 in a nucleus such as 12 C where a single nucleon is excited from a 1p3/2 level to a 1p1/2 level. a) Compute the inelastic PC electron scattering cross section for this transition and extend this calculation to compute the PV asymmetry (the tables of [Don79a] may be used). b) Assuming that the same “typical” experimental conditions as in Exercise 15.4 are available, and that PV elastic scattering from 12 C yields a significant determination of the elastic asymmetry in 1000 hours, by computing both the elastic and inelastic figures-of-merit provide an estimate of how much beam time would be required to determine the inelastic asymmetry to the same precision as the elastic asymmetry.

Exercise 15.8 Solution: Parity-Violating Inelastic Electron Scattering

Electron Scattering from Discrete States

236

2s1/21p3/2 10

-1

-1

C1 -2

F (q)

10

2

E1 10

10

10

M2

-3

-4

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.15

Squares of form factors for 2s1/2 1p−1 3/2 contributions to T = 0 to T = 0 transitions in 16 O. a) For an M 1 transition with this pure ph matrix element one needs √ 4π 1p1/2 kiT1mag k 1p3/2

q 1 (T ) 1p1/2 1p3/2 (T ) 1p1/2 1p3/2 −y y/A = (y)J=1 , e e GE p∆ (y)J=1 − GM pΣ0 mN 2 where T = 0 for the present situation, with 1√ 2 3

4√ 1 1p 1p pΣ01/2 3/2 (y)J=1 = − 2 1− y 3 2 1p

p∆ 1/2

1p3/2

(y)J=1 = −

from Table VI in [Don79a]. The total EM form factor squared forming the denominator of the PV asymmetry is given by W P C = F 2 = vT FT2 having only a transverse contribution 1 √ (FT )M 1 ≡ FM 1 = √ 4π 1p1/2 kiT1mag k 1p3/2 , 2 √ where the factor 1/ 2 arises from isospin as in the other exercises in this chap-

Electron Scattering from Discrete States

237

1d3/21p3/2 10

-1

-1

C1 C3

10

-2

F (q)

E1

2

E3 10

10

-3

-4

M2 10

-5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.16

Squares of form factors for 1d3/2 1p−1 3/2 contributions to T = 0 to T = 0 transitions in 16 O. ter. For the numerator in the PV asymmetry one in principle has two transverse contributions 0

T W P V = vT WAV + vT 0 WVT A W T = aA FM 1 FeM 1 AV

0 WVT A = −aV FM 1 FeE15

involving the weak neutral current form factors (0) FeM 1 = βV FM 1 FeE1 = 0 5

upon making two assumptions, (1) that there are no strangeness contributions and (2) that the hadronic WNC couplings take on their tree-level Standard Model (0) values, implying that βA = 0 and hence that the axial-vector electric multipole above is zero. This means that for this simple case one has WPV (0) = aA βV , WPC yielding for the PV asymmetry A = A0

WPV , WPC

Electron Scattering from Discrete States

238

10

1d3/21p1/2

0

-1

C1 -1

2

F (q)

10

10

-2

E1 M2 10

10

-3

-4

0.0

0.5

1.0

1.5

2.0

2.5

3.0

-1

q (fm )

t

Fig. 15.17

Squares of form factors for 1d3/2 1p−1 1/2 contributions to T = 0 to T = 0 transitions in 16 O. where A0 is given by Eq. (7.149) in the text. Note that the result does not depend on the specific nature of the 0+ 0 → 1+ 0 transition and is completely general, given only the good quantum numbers and the two assumptions above. The PC (purely EM) cross section is given by P C dσ −1 2 = 4πσM ott frec F dΩ with σM ott and frec given in the text, and thus one has the results shown in the figures. b) From the results in part a) one can form the figure of merit (see also Exercise 15.4 d) P C dσ A2 F= dΩ and show that the fractional uncertainty in the PV asymmetry A is obtained from

where

δA 1 =√ , A FX0

X0 = p2 · L · ∆Ω · T

Electron Scattering from Discrete States

239

10

-5

-6

2

-1

cm sr )

10

dσ/dΩ (10

-30

10 10

10

10

0

-7

20 30

-8

0

0

-9

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.18

Inelastic PC cross section for three choices of scattering angle (see Exercise 15.4 for discussion).

with p the polarization of the electron beam, L the luminosity, ∆Ω the solid angle of the detector and T the length of time over which the measurement is made (this formula is derived in detail in [Mus94], pages 66–70). The square of the C0 elastic form factor may be written

F 2 el = vL FL2 el = vL Z 2 F02 , where

2 F0 = e−y ey/A GpE (Q2 ) 1 − y 3 from Eq. (15.32) in the text may be used for F0 . The square of the inelastic M 1 form factor is that obtained in part a),

2 F 2 inel = vT FM 1

Electron Scattering from Discrete States

240

PV Asymmetry A

10

10

10

10

-4

-5

-6

-7

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.19

PV asymmetry for the isoscalar transition discussed in the exercise. and so

vL FL2 el F 2 el = R≡ 2 (F 2 )inel v T FM 1 2 

ZGpE (Q2 ) 1 − 32 y vL 

h = i  (0) (0) q 1 1 vT G − 2G 1− y 3

mN

E

M

2

using the polynomials derived in part a). We are to assume the same fractional uncertainty in the elastic and inelastic asymmetries,

δA δA = , A inel A el and so (FX0 )inel = (FX0 )el .

Since all conditions except the runtime are assumed to be the same, this means that (FT )inel = (FT )el , and, since F equals the PC cross section times the square of the PV asymmetry,

Electron Scattering from Discrete States

241

0

2 -1

cm s )

10

-1

FOM (x10

-38

10

10 10

-2

30 10

0

20

0

0

-3

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.20

Elastic scattering figure-of-merit (FOM) discussed in the solution for the three selected scattering angles. that

F 2 A2 T inel = F 2 A2 T el .

Above we found that the inelastic PV asymmetry for the chosen isoscalar transition is the same as the isoscalar elastic PV asymmetry since PV PV W W (0) = = aA βV W P C inel W P C el and therefore one has Tinel = RTel .

The elastic and inelastic scattering figures-of-merit are shown in the accompanying figures, and from the developments given above, one obtains the time required for the stated conditions. One finds that the optimal situation for the 100 case where both elastic and inelastic FOMs have peaks (roughly 0.7–0.8 fm−1 ) incurs a time of approximately 7x107 s or approximately 2 y of beamtime. And this is for atypical conditions: as discussed in the solution to Exercise 15-4, more typically the luminosity is lower and the solid angle smaller, making such a measurement more like 200 y — clearly completely unfeasible.

Electron Scattering from Discrete States

242

-4

2 -1

cm s )

10

20

10

0

FOM (x10

-38

10

0

-5

30 10

10

0

-6

-7

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.21

Inelastic scattering figure-of-merit (FOM) discussed in the solution for the three selected scattering angles.

Electron Scattering from Discrete States

243

10

10

Time, T (s)

10

9

30 10

0

8

20 10

10

0

10

7

0

6

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

-1

q (fm )

t

Fig. 15.22

Time required to perform a measurement of the PV asymmetry to the same precision as required for elastic scattering.

Electroexcitation of High-Lying Excitations of the Nucleus

16

16.1 The Relativistic Fermi Gas The basic definition of the RFG model is given by Eq. (16.12), Z 3m2N dp µν WRF (q, ω) = δ[ω − (E(p + q) − E(p))] G 4πkF3 E(p)E(p + q) f µν ×θ(kF − |p|)θ(|p + q| − kF )W nucleon (p + q, p) , with the on-shell single-nucleon EM responses given by

qµ qν µν µν + W2 (τ )V µ V ν , Wnucleon (p + q, p) = −W1 (τ ) g + Q2 where Vµ ≡

p·q 1 pµ + qµ . mN Q2

a) Substituting the latter into the former, show that 3 ( F − Γ) θ ( F − Γ) 4mN κηF3 i ( 2h κ f2 (τ )∆ for L e 2 (τ ) + W G E τ × f2 (τ )∆ for T e 2 (τ ) + W 2τ G

RL,T (q, ω) =

M

results both when one has Pauli blocking and when Pauli blocking is absent. b) Perform the steps outlined in Section 16.3 to prove that the universal scaling functions 3 fLRF G (ψ) = fTRF G (ψ) ≡ f RF G (ψ) = (1 − ψ 2 )θ(1 − ψ 2 ) 4 emerge when Pauli blocking is absent.

Exercise 16.1 Solution: The Relativistic Fermi Gas a) We begin by developing the appropriate parts of the hadronic tensor W µν (q, ω).

For inclusive electron scattering we have shown in the text that only two terms enter, namely, RL (q, ω) = W 00 (q, ω) RT (q, ω) = W 11 (q, ω) + W 22 (q, ω), 244

245

Electroexcitation of High-Lying Excitations of the Nucleus

and as shown in the text (and repeated above) this means that we require the corresponding components of the single-nucleon tensor: 2 2 q 00 Wnucleon (p + q, p) = −W1 (τ ) + W2 (τ ) V 0 Q2 2 κ 2 = −W1 (τ ) + W2 (τ ) ( + λ) τ h 2 2 i 11+22 Wnucleon (p + q, p) = 2W1 (τ ) + W2 (τ ) V 1 + V 2 2

= 2W1 (τ ) + W2 (τ ) (η sin θ) , where the standard dimensionless variables introduced in the text have been used: κ ≡ q/2mN

λ ≡ ω/2mN

τ ≡ Q2 /4m2N = κ2 − λ2 η ≡ p/mN

≡ E(p)/mN =

p

1 + η2

and where θ is the angle between q and p. Naturally, when used in the expression for the RFG model we will require the combination of single-nucleon tensors for protons and neutrons weighted by the proton and neutron numbers, Z and N , respectively µν µν f µν W nucleon (p + q, p) ≡ ZWp (p + q, p) + N Wn (p + q, p).

The on-shell energy conservation condition q q ω = m2N + (p + q)2 − m2N + p2 yields the relationship

q 1 2 2 2 cos θ ≡ cos θ0 = 2ω mN + p − Q 2pq which implies that ηκ cos θ0 = λ − τ and hence that r η sin θ0 ≡ χ =

τ 2 ( + λ) − (1 + τ ). κ2

Accordingly the single-nucleon tensors above become 2 κ 00 Wnucleon (p + q, p) = −W1 (τ ) + W2 (τ )χ2 τ 11+22 Wnucleon (p + q, p) = 2W1 (τ ) + W2 (τ )χ2 .

Electroexcitation of High-Lying Excitations of the Nucleus

246

Next let us change variables in the integral above that defines the RFG model.

Let q

q 2 2 2 2 Ω ≡ ω − [E(p + q) − E(p)] = ω − mN + (p + q) − mN + p ; clearly we want to have Ω = 0. Using the fact that

∂E(p + q) pq = ∂ cos θ E(p + q) p (both q and ω are assumed to be constant here), we have for the δ-function in the RFG integral E(p + q) δ(cos θ − cos θ0 ) δ(Ω) = pq and hence the integral over cos θ yields

Z +1 E(p + q) d cos θδ(Ω)X = θ(1 − cos θ0 )θ(1 + cos θ0 )X , pq −1 Ω=0 where X contains the rest of the integrand and where the θ-functions guarantee that Ω is equal to zero occurs in the physical range. The last implies that −2pq ≤ 2ωE(p) − Q2 ≤ 2pq, that is, that −ηκ ≤ λ − τ ≤ ηκ.

Upon squaring one finds that 2

( + λ) ≥ κ2 (1 + 1/τ ) and therefore that the minimum occurs at p min = κ 1 + 1/τ − λ ≡ γ− , namely that ≥ γ− . Substituting into the expression for the RFG and converting from an integral over p to an integral over energy E(p), one now has Z ∞ h i 3 µν µν f d θ( − )θ(( + λ) −

) W (p + q, p) , WRF (q, ω) = F F G nucleon 4mN ηF3 κ γ− Ω=0 where we have used dimensionless variables everywhere, defining ηF ≡ kF /mN F ≡ EF /mN =

q

1 + ηF2

ξF ≡ F − 1;

the last will be used below. The first θ-function here requires that the upper limit of the integral is F , while the second θ-function imposes a second constraint on the lower limit, namely, the lower limit must be Γ ≡ max {γ− , F − 2λ}

Electroexcitation of High-Lying Excitations of the Nucleus

247

leaving µν WRF G (q, ω) =

3 4mN ηF3 κ

Z F Γ

h i f µν d W (p + q, p) nucleon

Ω=0

.

If κ > ηF , that is, q > 2kF , the first condition (Γ = γ− ) pertains and one has no Pauli blocking; when κ ≤ ηF , that is, q ≤ 2kF , the second condition (Γ = F − 2λ) pertains and one has Pauli blocking. The final steps are to evaluate the two types of integral that occur: Z F d · 1 = ( F − Γ) θ ( F − Γ) Γ Z F d · χ2 = ( F − Γ) θ ( F − Γ) ∆, Γ

using the expression above for χ2 in terms of . The quantity ∆ is the result of doing the simple integration:

τ 1 2 2 2 ∆ = 2 λ + λ ( F − Γ) + + F Γ + Γ − (1 + τ ) . κ 3 F Substituting these into the expression gives the required result. b) Scaling is discussed in the context where Pauli blocking is absent and, accordingly, where Γ = γ− . We have that F = 1 + ξ F and, defining a scaling variable ψ via γ− = 1 + ξF ψ 2 with ψ > 0 when above the QE peak (λ > τ ) and ψ < 0 when below the QE peak (λ < τ ), where ψ = 0 corresponds to λ = τ or ω = Q2 2mN , we have

F − γ − = ξ F 1 − ψ 2 , which immediately yields the desired result for the universal RFG scaling function discussed in Chapter 16; see also [Alb88].

16.2 The RFG Model for Excitations involving Baryon Resonances In Section 16-4 electroexcitation of the ∆ in the RFG model of nuclei was summarized. Generalize this discussion for any baryon resonance X of given mass MX . a) Show that the inelastic scaling variable, the extension of v " r # u u 1 1 +1 λ ≥ λ∆ 0 2 t κ ρ∆ + − (1 + ρ∆ λ) × ψ∆ (κ, λ) = −1 λ < λ∆ ξF τ 0 naturally emerges from considerations of the kinematics.

,

Electroexcitation of High-Lying Excitations of the Nucleus

248

b) Prove RL,T (q, ω) = ( ×

κ2 τ

h

3ξF 2 2 1 − ψ∆ θ 1 − ψ∆ 3 4mN κηF

i f ∆ (τ ) − W f ∆ (τ ) + W f ∆ (τ )D(κ, λ) for L (1 + ρ2∆ τ )W 2 1 2 , f ∆ (τ ) + W f ∆ (τ )D(κ, λ) 2W for T 1 2

the RFG model of quasielastic baryon production.

Exercise 16.2 Solution: The RFG Model for Excitations involving Baryon Resonances a) First we must generalize the kinematics from the QE case considered in Exercise 16.1. There we introduced the quantity Ω = ω − [E(p + q) − E(p)] and, given a final-state energy for a specific rest mass MX we now have

q q 2 + (p + q)2 − MX m2N + p2 Ω=ω− which yields q

1 2 2 2 2 2 cos θ ≡ cos θ0 = 2ω mN + p − Q − MX − mN 2pq

1 1 2 = λ − τ − µ −1 ηκ 4 X 1 = [λ − ρX τ ] ηκ with µX ≡ MX /mN µ2 − 1 . ρX ≡ 1 + X 4τ Clearly these expressions revert to the QE case when MX → mN , for then µX = ρX = 1. The inequalities obtained in Exercise 16.1 now become −ηκ ≤ λ − ρX τ ≤ ηκ

κ2 2 ( + ρX λ) ≥ 1 + ρ2X τ τ and yield the extensions of the quantities γ− and ψ obtained for the QE case: q X 2 X min = κ ρX + 1/τ − ρX λ ≡ γ− 2 ≡ 1 + ξF ψX

Electroexcitation of High-Lying Excitations of the Nucleus

249

and lead to the required expression for the inelastic scaling variable ψX , the generalization of ψ∆ given above and in the text, v " r # u u 1 1 +1 λ ≥ λX 0 2 t κ ρX + − (1 + ρX λ) × ψX (κ, λ) = , −1 λ < λX ξF τ 0 1 where λX 0 = 2

hp i µ2X + 4κ2 − 1 .

b) The quantity χ for the QE case generalizes directly from the results in part a); from the expression obtained above for cos θ0 we have r τ 2 ( + ρX λ) − (1 + ρ2X τ ) η sin θ0 ≡ χX = κ2 and thus the single-nucleon inelastic tensors become 2 κ 00 N →X + W2N →X (τ )χ2X WN →X (p + q, p) = −W1 (τ ) τ N →X WN11+22 (τ ) + W2N →X (τ )χ2X →X (p + q, p) = 2W1 N →X which involve the inelastic structure functions W1,2 (τ ) evaluated at invariant mass MX . In general, these should be continuous functions of both τ and a continuous invariant mass W and the result obtained in this exercise for a specific mass MX should be further generalized to have an integral over W from the inelastic threshold to an upper limit determined by the kinematics of the reaction; see [Bar04]. Naturally the proton and neutron contributions (p → X + and n → X 0 ) f µν (p + q, p) as in the QE should be weighted by Z and N and added to form W N →X situation. The rest proceeds as before: one has a lower limit for the integral over of X ΓX ≡ max γ− , F − 2λ

and, building on the QE results and performing the similar integrals, one finds the desired result,

3ξF N →X 2 2 1 − ψX θ 1 − ψX RL,T (q, ω) = 4mN κηF3 i ( 2h κ 2 f N →X (τ ) − W f N →X (τ ) + W f N →X (τ )DX (κ, λ) for L 1 2 τ (1 + ρX τ )W2 × , f N →X (τ ) + W f N →X (τ )DX (κ, λ) for T 2W 1

2

where DX (κ, λ) ≡

2

τ 1 2 2 4 2 1 + ψ + ψ ξ + (1 + ρ λ) 1 + ψ ξ + (1 + ρ λ) X F X X X F X κ2 3 −(1 + ρ2X τ ) .

Finally, note that the results obtained here for the RFG model of electron scattering, both quasielastic (in Exercise 16.1) and inelastic (Exercise 16.2) are straightforwardly generalized to charge-changing neutrino reactions (see Chapter 18 and,

Electroexcitation of High-Lying Excitations of the Nucleus

250

in particular, Exercise 18.4).

16.3 Quasielastic Scattering and the Width of the QE Peak Consider inclusive quasielastic electron scattering from a nucleus in the Fermi gas model. The Fermi momentum of the nucleons, each of mass mN , is kF and the energy and three-momentum of the virtual photon are ν and q, respectively. Show that the width of the quasielastic peak is given by ∆E 0 ≈

2|q| · |kF | , mN

assuming the proton is nonrelativistic, i.e., that the nonrelativistic Fermi gas model (NRFG) is being employed. Compare with what happens when the RFG model discussed in the text is used. Note that experimental data agree with the RFG at large values of q, but not with the NRFG. Typically at what momentum transfer do the two models differ significantly, say by about 20%?

Exercise 16.3 Solution: Quasielastic Scattering and the Width of the QE Peak We begin with (Eq. 16.1) d2 σ = σM (νL RL ((q, ω) + νT RT (q, ω)) dΩe d 0 which is the cross section for quasi-elastic scattering, where σM is the Mott cross section and νL , νT are Rosenbluth kinematic factors. Here RL,T (q, ω) = ( ×

L

2τ G̃2M (τ ) + +W̃2 (τ )∆

T

κ2 τ

3 ( F − Γ)θ( F − Γ) 4mN κη 3 G̃2E (τ ) + W̃2 (τ )∆

where r Γ(κ, λ) = max F − 2λ, κ

!

1 1+ −λ τ

and

τ 1 2 2 2 ∆= 2 + F Γ + Γ + λ ( F + Γ) + λ − 1 − τ . κ 3 F Here we have defined κ = |q|/2mN , λ = ω/2mN , τ = κ2 − λ2 , ξF = F − 1, and η = |p|/mN .

In the Pauli non-blocking region q > kF (κ > ηF ) and F − Γ becomes ξF (1 − Ψ2 )

Electroexcitation of High-Lying Excitations of the Nucleus

251

where Ψ= √

1 ξF

λ−τ √ (1+λ)τ +κ τ (1+τ )

.

Hence the peak value for d2 σ d2 σ → 0 dΩe d dΩe dω is maximized when ξF (1 − Ψ2 ) is maximized, i.e, when Ψ = 0 so require λ = τ . We have then τ = κ2 − λ2 = λ

λ2 + λ − κ2 = 0

so

which has the solution r λ = −1 ±

1 + 4κ2 .

2

Thus, ω = 2mN

−1 ±

q

|q| 1 + 4 4m 2

2

N

=

2

p −mN ± m2N + q 2 2mN

so ω=

q

q 2 + m2N − mN .

The uncertainty is then δω = p

|q|

q 2 + m2N

δ|q| .

Here δ|q| is the range of momentum transfer required to knock out a nucleon so |q + p| − kF ≥ 0 where p is the nucleon momentum. Since 0 ≤ |p| ≤ kF the range of |q| is 0 ≤ |q| ≤ 2kF . We have then δω = ∆E 0 = p

|q|

q 2 + m2N

× 2kF .

In the nonrelativistic limit, we have kF << mN so ∆E 0 '

p q 2 + m2N ' mN so

2|q|kF . mN

Then for the relativistic FG and nonrelativistic Fermi gas to differ by 20% we have q p q 2 + m2N = 1.2mN or |q| = (1.2)2 − 1mN = 0.66mN .

16.4 Semi-Inclusive Electron Scattering Consider the kinematic boundaries for (e, e0 N ) reactions discussed in Section 16.5. a) Verify the boundaries for energy transfers below the QE peak.

Electroexcitation of High-Lying Excitations of the Nucleus

252

b) Verify the boundaries for energies above the peak. c) Determine where the peak occurs and obtain a relationship between this result and the peak position found in the RFG model.

Exercise 16.4 Solution: Semi-Inclusive Electron Scattering Let us begin by summarizing some of the kinematic relations to be used. From four-momentum conservation one has pµA−1 = q µ + pµA − pµN with q µ = (ω, q) pµA = (MA0 , 0) pµN = (EN , pN ) pµA−1 = (EA−1 , pA−1 ), where the laboratory system is assumed in that pA = 0. and one has that EN = p m2N + p2N . Defining p ≡ −pA−1 0 E(p) ≡ EA−1 − EA−1 =

q q 2 2 0 2 (MA−1 ) + p − MA−1 + p2 ,

where MA−1 is the invariant mass of the undetected A − 1 daughter system and 0 0 ) is the energy (invariant mass) of the daughter system when in its (MA−1 EA−1 ground state, one has p = pN − q E(p, θ) = MA0 + ω −

q

0 MA−1

2

+ p2 −

q m2N + q 2 + p2 + 2qp cos θ,

where θ is the angle between q and p. By construction E ≥ 0 and p ≡ |p| ≥ 0, and one must have −1 ≤ cos θ ≤ +1. Hence limiting curves are formed by q q 2 2 0 0 2 EM ≡ E(p, θ = π) =MA + ω − MA−1 + p − m2N + (q − p) q q 2 2 0 Em ≡ E(p, θ =0) =MA0 + ω − MA−1 + p2 − m2N + (q + p) . a) For kinematics where ω is low, namely, below the QE peak one can show by examining these limiting curves that in this case Em < 0 and thus the allowed region lies below the EM curve and above E = 0. Where these two curves intersect one has, by definition, the minimum and maximum values of p, namely, −y and Y .

Setting EM (p0 ) ≡ 0, one must solve the equation q q 2 0 MA0 + ω = MA−1 + p20 + m2N + q 2 + p20 + 2qp0 cos θ

Electroexcitation of High-Lying Excitations of the Nucleus

253

for p0 : after some work one finds that q 1 n 0 0 p0 = 2qΛ ± M + ω W 2 − (MA−1 + mN )2 A 2W 2 q o 0 × W 2 − (MA−1 − mN )2 , where q 2 0 (MA0 + ω) − q 2 ≥ WT = MA−1 + mN h i 2 1 0 W 2 + MA−1 − m2N Λ≡ 2

W ≡

with WT the threshold value of W . From this one immediately has the values of y and Y given in the text. b) For ω > ωQE (i.e., where y > 0) the boundaries are a bit more complicated.

In the region +y ≤ p ≤ Y the same boundaries as in part a) occur; however, when 0 ≤ p ≤ +y the limit curve Em now plays a role, since in this region one has Em ≥ 0, and one finds that Em ≤ E ≤ E M . c) Since one expects the largest contributions to semi-inclusive scattering to arise from kinematics where both p and E are small (see the discussions in the text), one equivalently expects the QE peak to occur when y is small (either positive or negative). Setting EM = p0 = 0 in the above equation, which defines y = 0, one finds that this yields an estimate of the QE peak position q ωQE = m2N + q 2 − mN + ES with 0 ES ≡ MA−1 + mN − MA0

defining the separation energy. From the discussion of the RFG model in the text one sees that the spectral function in that case lies along the line q q E(p) = ERF G (p) = m2N + kF2 − m2N + p2 p

TF = m2N + kF2 − mN p = 0 = 0 p = kF with 0 ≤ p ≤ kF . The QE peak in the RFG model therefore occurs when the maximum line integral along this trajectory occurs, namely, when one has p = 0 and E =TF . From the expression for Em given above one then has the standard answer q ω = m2N + q 2 − mN + (ES + TF ) q = m2N + q 2 − mN = Q2 /2mN , since for the RFG one has ES = −TF

Electroexcitation of High-Lying Excitations of the Nucleus

254

16.5 Inclusive Quasielastic Scattering: the Scaling Function and the Spectral Function In Section 16.5 the scaling function was written in terms of the spectral function assuming the validity of the PWIA, namely, as Z Y

f (q, y) = kF × 2π

kdk −y

Z EM 0

dES(k, E) .

Consider the large-q limit on that expression, the regime where scaling is supposed to be applicable. a) By applying the Leibniz theorem take the derivative with respect to y for fixed q. b) Assume that q becomes very large compared with y and provide expressions for Y and EM . Obtain the q → ∞ limit for the latter and show that this tends to a finite limit. What does this imply for the derivative obtained in part a)?

Exercise 16.5 Solution: Inclusive Quasielastic Scattering: the Scaling Function and the Spectral Function a) Leibniz’ theorem states that for some function Z b(x) F (x) =

dyG(x, y) a(x)

one has ∂F = ∂x

∂b ∂x

G(x, y = b) −

∂a ∂x

Z b(x)

G(x, y = a) +

dy a(x)

∂G(x, y) ∂x

with partial derivatives throughout, since the functions involved may depend on other variables, as they do below. In the case of interest (we limit the discussion to the y < 0 situation as the y > 0 situation proceeds similarly) we have Z Y f (q, y) = 2πkF −y

kdk n e(q, y; k),

where n e(q, y; k) =

Z EM 0

dE S(k, E).

Electroexcitation of High-Lying Excitations of the Nucleus

255

Thus,

(

∂f (q, y) ∂Y = 2πkF [ke n(q, y; k)]k=Y + [ke n(q, y; k)]k=−y ∂y ∂y q q

) Z Y ∂e n(q, y; k) + kdk ∂y −y q,k (

∂Y = 2πkF Yn e(q, y; k = Y ) − yke n(q, y; k = −y) ∂y q

Z Y ∂EM kdk + S(k, EM ), ∂y q,k −y

where the last line is obtained by applying Leibniz’ theorem to the inner integral defining n e in terms of S. The two remaining partial derivatives may be evaluated by employing the expressions given in the text for y, Y and EM , although we do not do so here. b) One has in the text or in Exercise 16.4 that q q 2 2 0 + k 2 − m2N + (q − k) MA−1 EM (k)=MA0 + ω − which implies that EM (−y)=0=MA0 + ω −

q

0 MA−1

2

+ y2 −

q

2

m2N + (q + y)

and hence that EM (k)=X1 + X2 , where q

q 2 2 m2N + (q + y) − m2N + (q − k) q q 2 2 0 0 X2 = MA−1 + y2 − MA−1 + k2 .

X1 =

Expanding for large q one has the following:   2q + y − k  → y+k q X1 = (y + k)  q q→∞ 2 2 m2N + (q + y) + m2N + (q − k) X2 = q

y2 − k2 q → 0, 2 2 q→∞ 0 0 2 MA−1 + y + MA−1 + k2

0 the latter since we typically assume that k, y MA−1 , although the second term can be retained if desired. Thus in this limit we expect that

EM (k) → y + k q→∞

Electroexcitation of High-Lying Excitations of the Nucleus

256

and that for the partial integral in part a) we have

∂EM ' 1. ∂y q,k For the other variables we can proceed as follows: define q 2 0 M≡ MA−1 + y2 X ≡ MA0 + ω,

where then W =

p

X 2 − q2 .

The above expression for EM (k) and, in particular, the fact that it is zero when k = −y (or Y ), allows us to write q 2 X = m2N + (q + y) + M s 2

mN +M = (q + y) 1 + q+y → q + (M + y) +

q→∞

m2N + ··· q+y

and thus W 2 = X 2 − q2

→ 2 (M + y) q [1 + a + · · · ]

q→∞

2

a≡

m2N + (M + y) .

2 (M + y) q

Since

2 0 Λ = W 2 + MA−1 − m2N /2

2

0 = X 2 + MA−1 − q 2 + m2N /2, we have Λ → (M + y) q [1 + b+ + · · · ] q→∞ 2 2 0 (M + y) + MA−1 b+ ≡ 2 (M + y) q q 2 0 Λ2 − MA−1 W 2 → (M + y) q [1 + b− + · · · ] q→∞ 2 2 0 (M + y) − MA−1 b− ≡ 2 (M + y) q

257

Electroexcitation of High-Lying Excitations of the Nucleus

and thus, given that

q 2 1 0 2 2 Y = 2 X Λ − MA−1 W + qΛ , W it is straightforward to show that Y → q [1 + d + · · · ] q→∞

2

d≡

(M + y) − m2N .

2 (M + y) q

The derivative required in part a) is then

2 ∂Y (M + y) + m2N = . ∂y q 2 (M + y) M Some final comments are in order. One can write the spectral function in the following way S(k, E) ≡ S0 (k, E) + S1 (k, E), where the first term is proportional to a δ-function and the second vanishes when E =0: S0 (k, E) ≡ n0 (k)δ (E) S1 (k, 0) = 0,

implying that n e(q, y; k) = n0 (k) + n e1 (q, y; k) Z EM n e1 (q, y; k) ≡ dE S1 (k, E) 0+

and therefore that f (q, y) = f0 (q, y) + f1 (q, y) Z Y f0 (q, y) = 2πkF kdk n0 (k) −y

Z Y f1 (q, y) = 2πkF

−y

kdk n e1 (q, y; k).

Upon inserting these into the expression in part a) for the partial derivative of f (q, y) with respect to y one then obtains

∂f0 (q, y) ∂f1 (q, y) ∂f (q, y) = + ∂y ∂y ∂y q q q ( )

∂Y ∂f0 (q, y) = 2πkF −yn0 (k = −y) + n0 (k = Y ) ∂y ∂y q q

Z Y ∂f1 (q, y) ∂EM = kdk S1 (k, EM ), ∂y ∂y q,k −y q

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Electroexcitation of High-Lying Excitations of the Nucleus

where we have used the facts that n e1 (q, y; k = −y) = n e1 (q, y; k = Y ) = 0.

Since at high momentum transfers we have seen that Y is also large and since n0 (k = Y ) at large Y is very small, one has approximately that

1 ∂f0 (q, y) n0 (k = −y) ' − 2πkF y ∂y q while the other term involves a line integral of S1 (k, E) with (∂EM /∂y)q,k ' 1 at high momentum transfers, as shown above. A special case where these ideas have been applied is to electrodisintegration of the deuteron at kinematics below pion production. There one has only the type-0 term and thus the above relationship provides a way to use inclusive electron scattering to obtain the ground-state momentum distribution.

16.6 General Characteristics of (e, e0 x) Reactions In Chapter 7, the general form for inclusive electron scattering was obtained by building the hadronic tensor in terms of the available four-momenta. Extend that analysis to semi-inclusive (e, e0 x) reactions, where now another four-momentum comes into play, namely, pµx . a) Build the most general symmetric and antisymmetric hadron tensors. b) Invoke the properties of the polar-vector EM current to eliminate some of the contributions found in Part a). c) Impose the continuity equation to further limit the possibilities. How many terms survive? d) Contract the result above with the general (polarized as well as unpolarized) leptonic tensor obtained in Chapter 7. In particular, show that Rf i = vL RfLi + vT RfTi + vT L RfTiL cos φx + vT T RfTiT cos 2φx results when the electrons are unpolarized. What happens when the electron beam is linearly polarized?

Exercise 16.6 Solution: General Characteristics of (e, e0 x) Reactions a) In addition to the four-momenta q µ and pµi encountered in the text when treating inclusive electron scattering one now has pµx . Equivalently, since they span the same space, one may use q µ together with

pi · q 1 µ µ p − qµ vi ≡ Mi i q2

px · q µ µ vx ≡ p x − qµ , q2

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Electroexcitation of High-Lying Excitations of the Nucleus

which have the useful properties that q· vi = q· vi = 0. The hadronic tensor can always be decomposed into pieces that are symmetric (s) and antisymmetric (a) under interchange µ ↔ ν: Wsµν = X1 g µν + X2 q µ q ν + X3 viµ viν + X4 vxµ vxν +X5 (q µ viν + viµ q ν ) + X6 (q µ vxν + vxµ q ν ) +X7 (viµ vxν + vxµ viν ) Waµν = i {Y1 (q µ viν − viµ q ν ) + Y2 (q µ vxν − vxµ q ν ) +Y3 (viµ vxν − vxµ viν )

+ µναβ (Z1 qα viβ + Z2 qα vxβ + Z3 viα vxβ ) ,

where all of the invariant response functions, the Xs, Y s and Zs, depend on four dynamical scalars, namely, q 2 , q·pi , q·px and pi ·px , since the remaining combinations p2i = Mi2 and p2x = Mx2 are fixed by their masses. Also, the factor of i in the antisymmetric case is conventional and related to the properties under time reversal.

Given the basis four-vectors involved these are the most general forms for the two types of hadronic second-rank Lorentz tensors. b) The EM case involves tensors that must go as bilinear products of polarvector current matrix elements, whereas the terms proportional to µναβ behave as products of polar and axial vectors and thus cannot enter in the EM case; they do, however, enter for the weak interaction where both types of current play roles.

Thus, in the EM case one has Z1 = Z2 = Z3 = 0. c) The continuity equation for the EM current matrix elements plus the fact that the tensors are bilinear productions of these matrix elements means that qµ Wsµν = qµ Waµν = 0.

This yields the following

X1 + X2 q 2 q ν + X5 q 2 viν + X6 q 2 vxν = 0

Y1 q 2 viν + Y2 q 2 vxν = 0, and, since the four-vectors are linearly independent, this immediately means that the terms above in parentheses are each separately zero, X1 + X2 q 2 = X5 = X6 = 0 Y1 = Y2 = 0, yielding the final forms for the EM hadron tensors

qµ qν µν µν + X3 viµ viν + X4 vxµ vxν Ws = X1 g − 2 q +X7 (viµ vxν + vxµ viν ) Waµν = iY3 (viµ vxν − vxµ viν ) for a total of 4 independent symmetric contributions and 1 antisymmetric combination, the last being the so-called 5th response.

Electroexcitation of High-Lying Excitations of the Nucleus

260

d) We want to contract the symmetric results in part c) with the unpolarized (symmetric) leptonic tensor discussed in the text, Lunpol = µν

1 kµ kν0 + kµ0 kν − gµν k · k 0 − m2e .

2 me

As in the text we use qµ = kµ − kµ0

1 kµ + kµ0 , Rµ = 2 where then 1 kµ kν0 + kµ0 kν = 2Rµ Rν − qµ qν 2 and, as in part c), contractions with q µ become simplified, since q · R = 0. One can also prove that 1 k · k 0 − m2e = Q2 2 with Q2 = −q 2 , as usual, and thus write the leptonic tensor as

1 1 2 Lunpol = q q + Q g ; R R − µ ν µν µ ν µν m2e 4 = 0, as required. The contractions of symmetric leptonic clearly one has q µ Lunpol µν and hadronic tensors can then be performed: 1 1 m2e Lunpol Wsµν = A − B − Q2 C µν 4 4 with A = Rµ Rν Wsµν B = qµ qν Wsµν C = gµν Wsµν , upon inserting the results from part c) yielding 2

2

A = X1 R2 + X3 (R · vi ) + X4 (R · vx ) + X7 2 (R · vi ) (R · vx )

B=0

C = 3X1 + X3 vi2 + X4 vx2 + X7 2 (vi · vx ) .

The various four-vectors may be worked out in the laboratory system where pµi = (Mi , 0) and where then pi · q = Mi ω and thus viµ =

q2 Q2

ω 1, 0, 0, q

.

Electroexcitation of High-Lying Excitations of the Nucleus

261

With pµx = (Ex , px ) one has pi · px = Mi Ex

q · px = ωEx − qpx cos θx ,

where θx is the angle between q and px . From this one can obtain the four-vector vxµ :

vxµ = vx0 , vx1 , vx2 , vx3

q2 ω vx0 = 2 Ex − px cos θx Q q vx1 = px sin θx cos φx vx2 = px sin θx sin φx ω vx3 = vx0 q (see Fig. 16.6 in the text for the coordinate system being used, where here a general particle x is assumed). The leptonic variables are obtained by noting that k0 = k 1 = 1q kk 0 sin θe ≡ ξ k2 = 0 k 3 = 1q k (k − k 0 cos θe )

k 00 = 0 k 01 = ξ k 02 = 0 k 03 = − 1q k 0 (k 0 − k cos θe )

and then 1 ( + 0 ) ≡ E 2 R1 = ξ R0 =

R2 = 0 ω R3 = E. q From these the various required scalar products may be obtained in the laboratory frame, Q2 2 E − ξ2 q2 q2 vi2 = 2 Q

2 q2 ω 2 vx2 = 2 Ex − px cos θx − (px sin θx ) Q q R · vi = E

ω R · vx = E Ex − px cos θx − ξpx sin θx cos φx q 2

2 q ω vi · vx = 2 Ex − px cos θx , Q q R2 =

262

Electroexcitation of High-Lying Excitations of the Nucleus

from which the quantities A and C above may be obtained as kinematic factors times the invariant response functions.

To get the general form for the unpolarized semi-inclusive response we need the individual Lorentz components conventionally defined in Eqs. (7.64–7.67) of the text, namely, q 2 WfLi = Ws(00) = Ws00 ω WfTi = Ws(+1+1) + Ws(−1−1) = Ws11 + Ws22 o n q √

Re Ws(0+1) − Ws(0−1) = 2 2Re Ws01 WfTiL = −2 o nω WfTiT = 2Re Ws(+1−1) = Ws22 − Ws11 .

Note: to avoid confusion here we have put the spherical vector labels in parentheses: there “0” refers to the m = 0 component, whereas in the usual Cartesian coordinates (the Lorentz indices µ = 0, 1, 2, 3), this is the 3-component and the 0-component refers to Lorentz index 0. Remember that the µ = 0 and µ = 3 components are related by the continuity equation which then accounts for the extra factors involving q/ω. Substituting from the expressions obtained above we have 2 2 "

2 2 Q q ω 00 Ws = 2 X1 + X3 + X4 Ex − px cos θx q Q2 q

ω +X7 2 Ex − px cos θx q 2

Ws11 = −X1 + X4 (px sin θx ) cos2 φx 2

Ws22 = −X1 + X4 (px sin θx ) sin2 φx

ω q2 Ws01 = 2 (px sin θx ) X4 Ex − px cos θx + X7 cos φx Q q from which we get RfLi = Ws00 RfTi = Ws11 + Ws22 2

= −2X1 + X4 (px sin θx ) √

TL Rf i cos φx = 2 2Re Ws01 2

√ q ω =2 2 (px sin θx ) X4 Ex − px cos θx + X7 cos φx Q2 q RfTiT cos 2φx = Ws22 − Ws11

2

= X4 (px sin θx ) cos 2φx , hence proving that Rf i = vL RfLi + vT RfTi + vT L RfTiL cos φx + vT T RfTiT cos 2φx ,

263

Electroexcitation of High-Lying Excitations of the Nucleus

as required. Note the explicit behavior in terms of the azimuthal angle φx , namely, no dependence in the L and T terms, cos φx in the T L term, and cos 2φx in the T T term. And also observe that the T L term has an explicit factor of sin θx , while the T T terms has an explicit factor of sin2 θx . A note for the interested reader: the general situation with two or more particles detected in coincidence with the scattered electron, namely, (e, e0 x1 x2 ), (e, e0 x1 x2 x3 ), etc., was presented in [Don85].

The final question was mis-stated: it should have read “What happens when the exchanged virtual photon is linearly polarized?” The answer is that it is always linearly polarized, which corresponds to a combination of the 11 and 22 components above, i.e., to a linear combination of Ws11 and Ws22 or equivalently of RfTi and RfTiT .

16.7 The Real-Photon Limit of Semi-Inclusive Electron Scattering: (~γ , N ) Reactions In Chapter 7 the basics for studies of unpolarized inclusive electron scattering from unpolarized targets were introduced, while in this chapter some extensions to unpolarized, but semi-inclusive electron scattering were considered. In this exercise, the real-photon limit for (γ, x) reactions, those where some outgoing particle x is presumed to be detected, is taken as the limit of semi-inclusive (e, e0 x) reactions where the electron scattering angle goes to zero and hence where Q2 → 0. While this could be done directly using a Feynman diagram with one external (on-shell) photon line, it can also be obtained as the limit of electron scattering. As seen in Eq. (16.40) the general form for (unpolarized) semi-inclusive electron scattering has a response that goes as Rf i = vL RfLi + vT RfTi + vT L RfTiL cos φx + vT T RfTiT cos 2φx .

For unpolarized real-γ reactions we have already seen that only the transverse response RT survives (see Eqs. (7.102) and (7.103)); i.e., real-γ reactions have purely transverse photons, while in electron scattering, where virtual photons enter, all four terms (L, T , T L and T T ) occur. For polarized real-γ reactions the L and T L terms can be shown to be absent, although now both transverse contributions, T and T T , play a role. The point of this exercise is to explore what happens in this more general situation.

We begin by focusing on the two combinations of the matrix elements of the transverse projections of the current that can occur (see Eqs. (7.65) and (7.67)), 2

2

WfTi = |Jf i (+1)| + |Jf i (−1)| ≡ RfTi

WfTiT = 2Re (Jfi (+1)∗ Jfi (−1)) ≡ RTT fi cos φx , where in the T T case the explicit dependence on φx has been extracted (the T case has no φx dependence), as in the text. Noting that from Eqs. (7.69) and (7.71) one

Electroexcitation of High-Lying Excitations of the Nucleus

264

has 1 ρ + tan2 θe /2 2 1 vT T = − ρ , 2 vT =

where ρ ≡ Q2 /q 2 , one then has for the transverse parts of the response Rtrans = vT RfTi + vT T RfTiT cos φx fi ≡ vT Rreal photons ,

and thus for the effective response for real-γ kinematics, vT T T T photons Rfreal = RfTi (q = ω) + R (q = ω) cos φx , i vT f i showing that for real photons in situations where both T and T T terms contribute one must generalize RT to Rreal photons . a) Assuming the ERL and taking the Q2 → 0 limit show that

2 0 vT T . →− 2 vT + 0 2 For ω, as required when keeping ω fixed while taking the limit as the scattering angle goes to zero (see Eq. (7.5) and Exercise 7.7), this becomes −1, and hence one has photons Rfreal = RfTi − RfTiT cos φx . i

b) Recalling the definitions of the helicity projections of the transverse current matrix elements (see the developments in Section 7.3), obtain expressions for the Cartesian projections (x and y, corresponding to 1 and 2, respectively, using the conventional coordinate system), and hence rewrite RT and RT T using the latter basis. c) In Eq. (7.92) the general multipole expansion involving transverse projections of the current were given. Assuming that only E1 and M 1 multipoles are important, derive expressions for Jfxi and Jfyi . d) Finally, consider transitions from a 0+ ground state to excited states with spinparity 1− (E1) or 1+ (M 1). Write the matrix elements that occur above in terms of reduced matrix elements of the E1 and M 1 multipole operators. For linearly polarized real photons use the expressions derived above to determine which multipoles enter and what φx -dependence occurs.

We note the relevance of these developments: for real-γ facilities such as the HIγS at Duke University where an intense source of polarized real photons exists many very interesting measurements can be undertaken; here the ability to separate contributions of E1 from M 1 contributions in photoexcitation provides the prime motivation.

Electroexcitation of High-Lying Excitations of the Nucleus

265

Exercise 16.7 Solution: The Real-Photon Limit of Semi-Inclusive Electron Scattering: (~γ , N ) Reactions a) For the ratio of the electron kinematic factors we have 1 vT T 2ρ =−1 2 vT 2 ρ + tan θe /2

=−

Q2 , Q2 + 2q 2 tan2 θe /2

and, using the fact that Q2 = 2 0 sin2 θe /2 in the ERL, this yields 2 0 vT T =− 0 vT 2 + q 2 / cos2 θe /2

2 0 →− 2 , + 0 2 since q → ω = − 0 in the real-γ limit. b) We have that 2

WfTi = |Jf i (+1)| + |Jf i (−1)|

2

WfTiT = 2Re (Jfi (+1)∗ Jfi (−1)) with

1 Jf i (±1) = ∓ √ Jf1i ± iJf2i 2 and so

1 1∗ Jf i − iJf2∗i Jf1i + iJf2i 2 o 1n 1 2 2 = Jf i + Jf2i + i Jf1∗i Jf2i − Jf2∗i Jf1i 2

1 1∗ 2 |Jf i (−1)| = Jf i + iJf2∗i Jf1i − iJf2i 2 o 1n 1 2 2 = Jf i + Jf2i − i Jf1∗i Jf2i − Jf2∗i Jf1i 2 2

|Jf i (+1)| =

which yields WfTi = Jf1i

2

2

+ Jf2i ,

again unpolarized, but now written in terms of Cartesian coordinates, i.e., linear polarizations of the virtual photon. The other case proceeds similarly:

1 1∗ Jf i − iJf2∗i Jf1i − iJf2i 2 o 1n 1 2 2 =− Jf i − Jf2i − i Jf1∗i Jf2i + Jf2∗i Jf1i 2

Jf i (+1)∗ Jf i (−1) = −

Electroexcitation of High-Lying Excitations of the Nucleus

266

and so WfTiT = − Jf1i

2

+ Jf2i

2

which give RfTi and RfTiT . c) Equation (7.90) gives the following when J = 1: Eo E D √ nD mag el i i + m f Tb1m Jf i (m) = −i 6π f Tb1m and hence using the inverses of the relationships between Cartesian and spherical vector notation given above in part b), 1 Jf1i = − √ [Jf i (+1) − Jf i (−1)] 2 i 2 Jf i = √ [Jf i (+1) + Jf i (−1)] 2 it is easy to show that E D E √ n D el el Jfxi = Jf1i = i 3π f Tb1,+1 i − f Tb1,−1 i D E D E o mag mag + f Tb1,+1 i + f Tb1,−1 i E D E √ n D el el f Tb1,+1 i + f Tb1,−1 i Jfyi = Jf2i = 3π D E D E o mag mag + f Tb1,+1 i − f Tb1,−1 i . d) If the final states have good total angular momentum and parity, then, as stated in the text, there cannot be any interferences between electric and magnetic multipoles with the same value of J (here J = 1). Accordingly, one gets the usual answer with only one of the two terms, namely, E2 D 1− Tb1el 0+ or D E2 .

1+ Tb1mag 0+ However, in (γ, x) reactions (also (e, e0 x) reactions) the final state contains a particle x which is presumed to be detected together with some residual system. Since all partial waves are involved there can be coherent interferences between contributions with various values of angular momentum and parity, thereby making the developments in this exercise relevant.

17

Beta Decay 17.1 Beta-Decay Classification Classify the following β-decays as allowed, forbidden, Fermi, or Gamow-Teller transitions: a) 3 H →3 He+e− + ν̄e , b) 8 Li(2+ ) → 8 Be∗ (2+ )+e− + ν̄e , c) 12 N(1+ ) → 12 C(0+ ) → e+ + νe , d) 16 N∗ (0− ) → 16 O(0+ ) + e+ + νe , +

+

e) 21 Na∗ ( 32 ) → 21 Ne( 32 ) + e+ + νe .

Exercise 17.1 Solution: Beta Decay Classification +

a) 3 He and 3 H both have J P = 21 and are isotopic analogs so decay is allowed and mixed Fermi and Gamow-Teller. b) 8 Li has J P = 2+ and 8 Be∗ also has J P = 2+ but the are not isotopic analogs so decay is allowed and is pure Gamow-Teller.

12 c) N has J P = 1+ and 12 C has J P = 0+ so decay is allowed and is pure GamowTeller. d) 16 N∗ has J P = 0− and 16 O also has J P = 0+ so decay is first forbidden. + + e) 21 Na∗ has J P = 23 and 21 Ne also has J P = 32 and they are isotopic analogs so decay is allowed and is mixed Fermi and Gamow-Teller.

17.2 Second-Class Currents In this problem we explore the second-class currents discussed in Section 17.5. Using the general definition of weak form factors a, b, c, ...j3 given in Eq. (17.64), a) Suppose that the parent and daughter states of a nuclear beta decay are members of a common isotopic multiplet, i.e., |i >= |T, T3 > and |f >= |T, T3 ± 1 >.

Examples are the decay of 19 Ne to 19 F or of 3 H to 3 He. Suppose that the weak currents are purely first-class. That is, suppose that the G-parity of the polar (axial) vector current is positive (negative), respectively. Show that the form 267

Beta Decay

268

factors d(q 2 ), e(q 2 ), f (q 2 ), j2 (q 2 ) must vanish, while there is no constraint on the remaining form factors. b) Consider an SU (3) generalization of this result. That is, suppose we consider a strangeness-changing beta decay such as Ξ− → Λ + e− + ν̄e . Show that in the SU (3) symmetry limit the induced-tensor and induced-scalar form factors g2 (q 2 ) and f3 (q 2 ) form factors are required to vanish. c) Of course, G-parity and its SU (3) generalization are only approximate symmetries. Estimate the size of the induced tensor for a realistic beta decay, both ∆S = 0 and ∆S = 1. d) Consider the case of mirror beta decays, that is, electron and positron decays which occur from initial states which are members of the same isotopic multiplet to final states which are members of a common isotopic multiplet. Examples are the decay of 12 B and 12 N to the 12 C ground state or of 8 B and 8 Li to the J P = 2+ first-excited state of 8 Be. Suppose that the weak currents are purely first-class. Demonstrate that each of the electron-decay form factors is required to have the same size as its positron-decay counterpart.

Exercise 17.2 Solution: Second Class Currents a) Suppose that parent is |T, T3 > while daughter is |T, T3 ± 1 > and use G-parity.

Note that for first class currents GVµ± G−1 = Vµ±

−1 and GA± = −A± µG µ

We have CVµ± C −1 = −Vµ∓ −1 CA± = A∓ µC µ

so

e−iπT2 Vµ∓ eiπT2 = −Vµ±

iπT2 so e−iπT2 A∓ = −A± µe µ

Then if the initial and final states are members of a common iso-multiplet < T, T3 ±1; p0 JM |Jµ± |T, T3 ; pJM >= − < T, T3 ±1; p0 JM |e−iπT2 Jµ∓ eiπT2 |T, T3 ; pJM > = − < T, −T3 ∓1; p0 JM |Jµ∓ |T, −T3 ; pJM >= − < T, −T3 ; pJM |Jµ± |T, −T3 ∓1; p0 JM >∗ 0

0

0

−k;−M k;M Note that under p ↔ p0 and using the identity CJ−M = (−)J +I−J CJM0 I;J , 0 I;J we have for vector current

a(q 2 ) → a(q 2 ) e(q 2 ) → −e(q 2 )

since q

b(q 2 ) → b(q 2 )

since q and I=1

f (q 2 ) → −f (q 2 )

since q and I=2

g(q 2 ) → g(q 2 )

since I=2

Beta Decay

269

while for axial current c(q 2 ) → c(q 2 ) d(q 2 ) → −d(q 2 )

since q

h(q 2 ) → h(q 2 )

since q·q

j2 (q 2 ) → −j2 (q 2 )

since I=2 and (−)1+1−2

j3 (q 2 ) → j3 (q 2 )

since I=2 and (−)1+1−3

so e(q 2 ), f (q 2 ), d(q 2 ), j2 (q 2 ) must vanish if there are no second class currents. b) Use V-spin and define GV ≡ e−iπV2 CeiπV2 . Then analysis for ∆S = 1 semileptonic decay is parallel to that for ordinary G-parity and ∆S = 0 semileptonic decay, which implies that the weak electricity term d must vanish in the SU(3) limit where V-spin conservation obtains. m −m c) Since isospin (SU(2)) breaking is O( mnn +mpp ) ' 1% while SU(3) breaking is Σ −mN O( m mΣ +mN ) ' 20% a reasonable estimate is

|d| 0.01 SU(2) ∼ 0.20 SU(3) b d) If first class, then for both axial and vector currents we have e−iπT2 Jµ∓ eiπT2 = −Jµ± . Then, we require < T 0 , T30 ; p0 J 0 M 0 |Jµ± |T, T3 ; pJM >=< T 0 , T30 ; p0 J 0 M 0 |e−iπT2 Jµ∓ eiπT2 |T, T3 ; pJM > =< T 0 , −T30 ; p0 J 0 M 0 |Jµ∓ |T, −T3 ; pJM > so that electron and positron form factors of mirror decays must be identical.

17.3 Testing CVC Even before the development of the Standard Model, it was theorized that the strangeness-conserving component of the weak polar-vector current was related to the electromagnetic current by a simple isospin rotation [T± , Vµem ] = Vµ± , which is called the conserved vector current (CVC) hypothesis. Various tests of this suggestion were proposed and verified. One of the first was a proposal by Gell-Mann to measure the electron energy dependence of the unpolarized 12

B →12 C + e− + ν̄e

beta-decay spectra.

and

12

N →12 C + e+ + νe

Beta Decay

270

a) Since both 12 B and 12 N have J π = 1+ , while 12 C is an even-even nucleus and has J π = 0+ ,show that, neglecting the electron mass compared to the ∼ 12M eV decay energy, the only weak form factors which contribute to these decays are the Gamow-Teller axial-vector term c, the weak magnetism term b, and the weak electricity term d and that the size of these terms should be identical for β + and β − transitions. b) Evaluate the decay spectrum and show that it has the expected form 4 E dΓ± ∼ |c|2 ± Re(c∗ b) . dEe 3 M c) Calculate the Gamow-Teller amplitudes c+ and c− from the measured 12 B and 12 N lifetimes and show that they have the same magnitude. d) Show that the size of the weak magnetism term b is predicted by CVC in terms of the M1 electromagnetic width of the decay of the 15.11 MeV J π = 1+ excited state of 12 C to the 12 C ground state via ΓM1 =

1 αem |b|2 ∆3 , 6M 2

where ∆ = 15.11 MeV is the excitation energy.

This prediction was tested in the experiment of Lee et al., and was one of the first experimental verifications of the CVC hypothesis [Lee63].

Exercise 17.3 Solution: Testing CVC a) Since 1+ stae decays to 0+ state, only operators carrying J = 1 can contribute– b, c, d, and h. However, q µ ū(pe )γµ (1 − γ5 )u(pν ) = ū(pe )(6 pe + 6 pν )(1 − γ)5u(pν ) = me ū(pe )(1 − γ5 )u(pν ) which may be dropped in the massless limit. b) We have

1 ib M 0 k;M λ η λ η (q × `)k + ijk ijλη ( c` P − d` q C01;1 2M 4M 0

M k;M ' C01;1

q0 qk ib (q × `)k + (c − d )`k + c`0 ) 2M 2M 2M

Then for electron decay `η `ξ∗ →

1 Tr(6 pe + me )γ η (1 − γ5 ) 6 pν γ ξ (1 − γ5 ) 8 '

1 Tr 6 pe γ η 6 pν γ ξ (1 − γ5 ) 4

= pηe pξν + pξe pην − η ηξ pe · pν − i αηβξ peα pνβ

Beta Decay

271

while for positron decay we have u ↔ v. Thus `η `ξ∗ → pηe pξν + pξe pην − η ηξ (Ee Eν − pe · pν ) ∓ i αηβξ peα pνβ Since 0

0

M k;M M j;M C01;1 C01;1 = δjk

X M,M 0

we have X

0

0

M k;M M j;M C01;1 C01;1

M,M 0

(c − d

q0 qk b )`k + d`0 +i (q × `)k 2M 2M 2M

b∗ q0 ∗ qk )`j + d∗ `∗0 −i (q × `∗ )j × (c∗ − d∗ 2M 2M 2M = |c − d

1 2Rec∗ b ∗ q0 2 | ` · `∗ + (cd∗ `∗0 ` · q + c∗ d`0 `∗ · q) + i (` × `) · q 2M 2M 2M

' (|c|2 − 2Rec∗ d

E0 2Rec∗ d )(3Ee Eν − 3pe · pν + 2pe · pν ) + (Ee pν · q + Eν pe · q) 2M 2M ±

= (|c|2 −2Rec∗ d

2Rec∗ b · 2(Ee pν · q − Eν pe · q) 2M

E0 2Rec∗ d )(3Ee Eν −pe ·pν )+ (Ee (p2ν +pe ·pν )+Eν (p2e +pe ·pν )) 2M 2M

4Rec∗ b (Ee (p2ν + pe · pν ) − Eν (p2e + pe · pν )) 2M which, integrating over neutrino solid angle, becomes ±

E0 2Rec∗ d )3Ee Eν + (Ee Eν2 + Eν (Ee2 − m2e )) 2M 2M

4Rec∗ b 4Rec∗ b 2 2 2 2 ∗ E0 ± (Ee Eν −Eν (Ee −me )) ' 3Ee Eν |c| − 4Rec d ± (E0 − 2Ee ) 2M 6M 6M

4Ee 2 ∗ ' 3Ee Eν |c| ± Rec b 3M R

dΩν

−→ (|c|2 − 2Rec∗ d

c) The Q-value for the two decays are d) According to CVC, the weak matrix element and electromagnetic matrix elements are related by r 1 12 < C|V + |12 B >=<12 C|V em |12 C > 2 Then the electromagnetic decay amplitude is eb M 0 0;M √ Amp(C ∗ →12 C + γ) = C01;1 (ˆ × q)k 2 2M

Beta Decay

272

Then Γ(C ∗ →12 C + γ) =

× 1 = 3

Z

1 X 3 0

Z

M,M

2 X λ=1

d3 k 2πδ(E12 C ∗ − E12 C − k 0 ) (2π)3 2k 0

k0 3π

eb √ 2 2M

2

0

0

M u;M M v;M λ C01;1 (ˆ × q)u (ˆ ∗λ × q)v C01;1

d3 k 2πδ(E12 C ∗ − E12 C − k 0 ) (2π)3 2k 0 =

eb √ 2 2M

2

q2 =

eb √ 2 2M

2 X 2 λ=1

(ˆ λ × q) · (ˆ ∗λ × q)v

αem b2 (E12 C ∗ − E12 C )3 6M 2

17.4 Muon Capture Besides beta decay and neutrino scattering, another manifestation of the semileptonic weak interaction is muon capture, in which a negatively-charged muon is captured by an atom. Typically, the muon is captured in a high angular momentum state, but then cascades down to the 1S state, which has a significant overlap with the nucleus. Because of this overlap, at some point the muon interacts with the nucleus of charge Z and is captured, changing the nucleus to one with charge Z − 1 and releasing a muon neutrino in the process.

For simplicity, suppose we are dealing with capture by a proton, which after capture becomes a neutron. Using the hadronic current matrix element

i < n(p2 )|Vµ + Aµ |p(p1 ) >= ū(p2 ) gV (q 2 )γµ − gM (q 2 ) σµν q ν + gA (q 2 )γµ γ5 2mN −gP (q 2 )qµ γ5 + gII (q 2 )

i σµν q ν γ5 u(p1 ) , 2mN

where q = p1 − p2 is the momentum transfer and mN = 21 (mp + mn ) is the mean mass, a) Show that the muon capture on the proton takes place at

mn − mp mn − mp − mµ mn − mp q 2 = −m2µ 1 − 2 + 1 − = −0.88m2µ . m2µ M1 + mµ mµ b) In the approximation that the muon is at rest, show that the T -matrix can be written in the form G T = √ Vud χ†2 χ†ν (1 − σ · k̂) 2

Beta Decay

273

h i × GV 1 · 10 − GA σ · σ 0 − GP σ · k̂σ 0 · k̂ χµ χ1 , where the unprimed operators act on the lepton spinors and the primed operators act in the hadron spinor space and

k0 mµ k0 2 GV = gV (q ) 1 + − gM (q 2 ) E2 + M2 2M (E2 + M2 )

k0 k0 − mµ 2 2 2 k0 GA = gA (q ) + gV (q ) + gM (q ) 1+ E 2 + M2 2M E 2 + M2

mµ − k0 k02 2 −gII (q ) − 2M 2M (E2 + M2 )

k0 gA (q 2 ) − gV (q 2 ) − mµ gP (q 2 ) E 2 + M2

k0 mµ k0 2 2 − gII (q ) 1 + + gM (q ) 1 + , 2M E2 + M2 E2 + M2 GP =

with M12 − M22 + m2µ 1+ 2mµ M1

m µ M1 k0 = M1 + mµ

!

being the neutrino energy. c) In gaseous hydrogen most of the capture takes place in the hyperfine singlet state.

In this case, by taking the muon to be in a simple Coulomb field, show that

3 2 G2 Vud m3µ α3 E2 + M2 k02 M1 |GV + 3GA + GP |2 . ΓH−singlet = k0 2π 2 2E2 1 + E M1 + mµ 2

d) Calculate the triplet capture rate and show that 2 m3µ α3 E2 + M2 k02 G2 Vud ΓH−triplet = k0 2π 2 2E2 1 + E

2

M1 M1 + mµ

3

∗

×Re(GV − GA ) (3GV − 3GA − 2GP ) . e) Assuming the axial-tensor induced-pseudoscalar form factor to be dominated by the pion pole show that gP (q 2 ) =

2M gA (q 2 ) . m2π − q 2

f) Show that CVC requires that the weak magnetic form factor have the value gM (0) = κp − κn = 3.70 , where κp = 1.79 and κn = −1.91 are the anomalous magnetic moments of the proton and neutron respectively in units of the nucleon magneton. g) Show that G-parity requires the vanishing of gII (q 2 ).

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274

h) Calculate the expected singlet and triplet capture rates and show that ΓH−singlet ' 710 s−1 ΓH−triplet ' 15 s−1 .

Exercise 17.4 Solution: Muon Capture a) Energy conservation yields mµ + M = k0 +

q

k02 + M 2

so

k02 + M 2 = (mµ + M − k0 )2

= k02 + M 2 − 2(M + mµ )k0 + 2M mµ + m2µ or

k0 = mµ

M + 21 mµ M + mµ

= mµ 1 −

1 2 mµ

M + mµ

Momentum transfer is then 2

2

q = (pµ − kν )

= m2µ − 2mµ k0 = −m2µ

mµ 1− M + mµ

= −0.9m2µ

b) Contract lepton and hadron currents L : H = L0 H 0 − L · H where 0

0

L = ūν (kν )γ (1 − γ5 )uµ (pµ ) =

χ†ν

σ · kν − χ†ν Eν

1 1

−1 −1

χµ 0

−σ −σ

σ σ

χµ 0

= χ†ν (1 − σ · k̂ν )χµ

L = ūν (pν )γ(1 − γ5 )uµ (pµ ) =

χ†ν − χ†ν

σ · kν Eν

= −χ†ν (1 − σ · k̂ν )σχµ HV0

1 i 2 0i 2 2 0 gM (q )σ qi + gS (q )q0 u1 (p1 ) = ū2 (p2 ) gV (q )γ − 2M 2M

σ 0 · kν = χ†2 χ†2 E2 + M = χ†2

m −k

0

·kν µ 0 gV (q 2 ) + 2M gS (q 2 ) −gM (q 2 ) σ2M 0 mµ −k0 ·kν −gM (q 2 ) σ2M −gV (q 2 ) + 2M gS (q 2 )

mµ − k0 Eν2 2 2 2 gV (q ) + gS (q ) − gM (q ) χ1 2M 2M (E2 + M )

!

χ1 0

Beta Decay

275

i 1 2 2 ij 2 a · H V = ū2 (p2 ) gV (q )a · γ − gM (q )ai σ qj − a · kν gS (q ) u1 (p1 ) 2M 2M

σ 0 · kν = χ†2 χ†2 E2 + M

1 i a × kν · σ 0 gM (q 2 ) − 2M a · kν gS (q 2 ) gV (q 2 )σ 0 · a − 2M i 1 2 0 −gV (q )σ · a − 2M a × kν · σ 0 gM (q 2 ) − 2M a · kν gS (q 2 )

χ1 0

= χ†2

0 i 1 0 2 2 2 σ · kν 0 − a × kν · σ gM (q ) − a · kν gP (q ) − gV (q ) σ · a χ1 2M 2M E2 + M

0 HA = ū2 (p2 )

i 1 2 0 2 0i 2 gA (q )γ − gII (q )σ qi + gP (q )q0 γ5 u1 (p1 ) 2M 2M

σ 0 · kν = χ†2 χ†2 E2 + M = χ†2

m −k

0

·kν µ 0 gP (q 2 ) −gII (q 2 ) σ2M gA (q 2 ) + 2M mµ −k0 2 σ 0 ·kν 2 −gII (q ) 2M −gA (q ) + 2M gP (q 2 )

!

0 χ1

0

0 σ · kν mµ − k0 2 2 σ · kν 2 gP (q ) − gII (q ) χ1 −gA (q ) + 2M E2 + M 2M

i 1 a·H A = ū2 (p2 ) gA (q 2 )a · γ − gII (q 2 )ai σ ij qj − a · kν gP (q 2 ) γ5 u1 (p1 ) 2M 2M

σ 0 · kν = χ†2 χ†2 E2 + M

1 i a × kν · σ 0 gII (q 2 ) − 2M a · kν gP (q 2 ) gA (q 2 )σ 0 · a − 2M i 1 2 0 −gA (q )σ · a − 2M a × kν · σ 0 gII (q 2 ) − 2M a · kν gP (q 2 )

0 χ1

= χ†2

1 i 0 2 2 2 0 a × kν · σ gII (q ) − a · kν gP (q ) − gA (q )σ · a χ1 − 2M 2M

c) If spin singlet then σ · σ 0 = −3 and, quantizing along the neutrino direction, σ · k̂ν σ 0 · k̂ν = −1 so rate must be proportional to |GV + 3GA + GP |2 . Can also be seen from

1 Tr GV 1 · 1 − GA σ · σ 0 − GP σ · k̂ν σ 0 · k̂ν (1 − σ · σ 0 ) 4

× GV 1 · 1 − GA σ · σ 0 − GP σ · k̂ν σ 0 · k̂ν = |GV |2 +|GA |2 (3+2·3)+|GP |2 +2ReG∗V GP +2·3ReG∗V GA +2·(1+2)ReG∗A GP = |GV + 3GA + GP |2

Beta Decay

276

To connect with the capture rate calculate a if plane wave and then multiply by 1S wavefunction at the origin, which is |ψ1S (0)|2 = α3 m3r /π. For phase space we have Z 3 Z q 4π d kν d3 p 2 4 4 2 (2π) δ (p +p −p −k ) = k dk δ(k + k02 + M22 −mµ −M1 ) 1 µ 2 ν 0 0 0 (2π)3 (2π)3 (2π)2 =

k02 k0 ) π(1 + E 2

Including normalization factor from the neutron Dirac wavefunction we have

3 2 2 E 2 + M2 α 3 k02 mµ M1 G Vud Γµ (singlet) = |GV + 3GA + GP |2 k0 π M1 + mµ 2 π(1 + E ) 2E2 2

0

d) If spin triplet then σ · σ = +1 and result is found from

1 (3 + σ · σ 0 ) Tr GV 1 · 1 − GA σ · σ 0 − GP σ · k̂ν σ 0 · k̂ν 4

× GV 1 · 1 − GA σ · σ 0 − GP σ · k̂ν σ 0 · k̂ν = 3|GV |2 +|GA |2 (3+2·3)+|GP |2 −2ReG∗V GP −2·3ReG∗V GA +2·(1+2)ReG∗A GP = Re(G∗V − G∗A )(3GV − 3GA − 2GP ) + 3|GP |2 Using the same arguments from item c) we have

3 2 2 mµ M1 k02 E 2 + M2 α 3 G Vud Γµ (triplet) = k0 π M1 + mµ 2 π(1 + E ) 2E2 2

× Re(G∗V − G∗A )(3GV − 3GA − 2GP ) + 3|GP |2

e) Pion pole contribution is, using the Goldberger-Treiman relation, √ so

√ 2M gA 1 2Fπ qµ 2 ū(p2 )γ5 u(p1 ) 2gπN N = − 2 ū(p2 )γ5 u(p1 ) 2 q − mπ mπ − q 2 gP =

2M gA m2π − q 2

f) According to CVC we have [I − , Vµem ] = −Vµwk so < n|Vµwk |p >= − < n|[I − , Vµem ]|p >=< p|Vµem |p > − < n|Vµem |n > Thus we require gV (0) = Qp − Qn = 1 − 0 = 1 and gM (0) = κp − κn = 1.79 − (−1.91) = 3.70.

Beta Decay

277

g) G-parity invariance requires that GAµ G−1 = Aµ and charge conjugation invariance gives CAµ C −1 = A†µ so we require e−iπI2 Aµ eiπI2 = A†µ Thus we must have < n(p2 )|Aµ |p(p1 ) >=< p(p2 )|A†µ |n(p1 ) >=< n(p1 )|Aµ |p(p2 ) >∗ That is, we must set q → −q. Then ū(p1 )γµ γ5 u(p2 ) → u† (p2 )γ5 㵆 γ0 u(p1 ) = −ū(p)2)γ5 γµ u(p1 ) = ū(p)2)γµ γ5 u(p1 ) and ū(p1 )γ5 u(p2 )qµ → −ū(p2 )γ5 γ0 u(p1 )qµ = ū(p2 )γ5 u(p1 )qµ but ū(p1 )[γµ , 6 q]γ5 u(p2 ) → −u† (p2 )γ5 [6 q † , 㵆 ]γ0 u(p1 ) = ū(p2 )[6 q, γµ ]γ5 u(p1 ) = −ū(p1 )[γµ , 6 q]γ5 u(p2 ) so we must have gII = 0. h) Numerical values are GV (0) = 1 +

mµ k0 k0 mµ k0 − 3.70 =1+ 1 − 3.70 E2 + M2 2M (E2 + M2 ) E2 + M2 2M

' 1 + 0.05 · 0.80 = 1.04

k0 k0 − mµ k0 GA (0) = 1.27+ 1+ +3.70 ' 1.27+0.05(1+3.70) = 1.52 E 2 + M2 2M E2 + M2 GP =

k0 2M mµ ) ' −0.44 (1.27 − 1 − 1.27 2 E 2 + M2 mπ + 0.88m2µ

Then |GV + 3GA + GP |2 = 27 while (GV − GA )(3GV − 3GA − 2GP ) + 3|GP |2 = 0.6 Also, we have at q 2 = 0

3 2 G2 Vud mµ M1 k02 E2 + M2 1 −10 3 α ' 10 ·1.4−3 ·10−6 ·10−4 ·100 MeV k0 2π 2 M 1 + mµ 2E 20 1 + E2 2 = 1.9 × 10−20 MeV/6 × 10−16 eV s = 32 s−1 so would have Γ(singlet) ' 855 s−1

and

Γ(triplet) ' 19 s−1

But muon capture takes place at q 2 ' −0.9m2µ so form factors reduce rate by !2 m2ρ ' 0.95 m2ρ − q 2

18

Neutrino Physics 18.1 Neutrino Propagation Through Matter When neutrinos pass through matter, the propagation can be described in terms of the usual mixing matrix modified by an index of refraction term which describes the interaction of the neutrinos with the medium. This exercise studies how this comes about. a) Show that when a beam travels through a medium the associated index of refraction n is given by 2πN n − 1 = 2 fk (0) , k where fk (0) is the forward scattering amplitude, N is the number density of scatterers, and k is the wavenumber of the incident beam.

Suggestion: Consider a slab of matter extending from z = 0 to z = L and assume an incident beam eikz which travels along the z-direction. Calculate the wavefunction at location z = Z, where z > L, as a superposition of the incident wave and the scattered wave in the slab. Taking the distribution of scatterers to be uniform with number density N and using cylindrical coordinates ρ, φ, z this is given by Z L Z ∞ 1 ψ(Z) = eikZ + N dzeikz dρρ eikr fk (θ) . r 0 0 Introducing a factor which guarantees convergence at ρ = ∞, integrate by parts and show that the result 2πN L 2πiN L fk (0)]eikZ ≈ eikZ ei k fk (0) k implies the relation given above between the index of refraction and the forward scattering amplitude. b) Show that the forward scattering amplitude of muon- or tau-neutrinos from electrons arises from Z 0 -boson exchange and is given by √ 1 fνµ e (0) = fντ e (0) = 2GF (− + 2 sin2 θw ) .

2

ψ(Z) = [1 +

c) Show that the forward scattering amplitude of electron-neutrinos from electrons arises from both Z 0 - and W ± -boson exchange and is given by √ 1 fνe e (0) = 2GF [1 + ( + 2 sin2 θw )] .

2 278

Neutrino Physics

279

Suggestion: Use a Fierz transformation to rewrite GF eνe HW = √ ēγµ (1 − γ5 )eν̄e γ µ (1 − γ5 )νe 2 and then note that one can drop the axial-vector electron current in looking at coherent scattering. d) Neglecting the mixing between νe and ντ (θ13 = 0) so only two-channel mixing is assumed, demonstrate that the resulting mixing matrix can be described via  q 

∆m2 ∆m212 − 4E12 cos 2θ12 + 12 GF Ne sin 2θ 12 d aνe aνe 4E   q = , i ∆m212 ∆m212 1 aνµ aνµ dt 4E sin 2θ12 4E cos 2θ12 − 2 GF Ne where Ne is the electron number density.

Exercise 18.1 Solution: Neutrino Propagation Through Matter a) Note that since r2 = ρ2 + L2 we have ρdρ = rdr. Then Z ∞ Z ∞ 1 ikr dreikr fk (θ) dρρ e fk (θ) = r Z−z 0 1 ikr Writing dreikr = d ik e , integrate by parts, dropping terms at infinity and omitting the small term involving the derivative of the scattering amplitude. Then Z ∞ 1 1 dρρ eikr fk (θ) ' eik(Z−z) fk (0) r ik 0

so total amplitude at location Z becomes Z L i 2πiN L ikZ Ψ(Z) = e + 2πN dz eik(Z−z) fk (0)eikz = eikZ + fk (0)eikZ k k 0 = eikZ

1+

2πiN L i2πN L fk (0) ' eikZ exp( fk (0)) k k

b) For νµ , ντ scattering, only neutral current interaction is relevant. From Eq. 4.40 effective Hamiltonian is √ Hw ' 2GJµZ JZµ† with JµZ =

1 [ν̄e γµ (1 − γ5 )νe + ν̄µ γµ (1 − γ5 )νµ + ν̄τ γµ (1 − γ5 )ντ 2

−ē γµ (1 − 4 sin2 θw ) − γµ γ5 e

Coherent scattering is only possible for the vector current since the leading axial

Neutrino Physics

280

current involves the spin operator. Note that spinors are normalized via the factor q 1 2E u(p) so we have in the case of forward scattering √

µ 1 p2 →p1 2p1 ū(p2 )γ µ u(p1 ) −→ 2E 2E1 2E2

Then we have in the case of either νµ e or ντ e scattering √ 1 Amp ' − 2GF (1 − 4 sin2 θw ) (ν̄γ µ ν ēγµ e + ēγ µ eν̄γµ ν) 4

√ √ 1 4pe · pν 1 forward ≈ − 2GF − 2 sin2 θw −→ − 2GF (1 − 4 sin2 θw ) 2 4Ee Eν 2 c) For νe scattering, there exists both charged and neutral current scattering.

The charged current Hamiltonian is GF ch Hw = √ ν̄e γµ (1 − γ5 )eēγ µ (1 − γ5 )νe 2 which can be Fierz-transformed to the form ch F ierz GF Hw −→ √ ēγµ (1 − γ5 )eν̄e γ µ (1 − γ5 )νe 2 so

GF 1 2 ch neu µ Hw + Hw = √ ν̄e γµ νe ē γ 1 − (1 − 4 sin θw ) e 2 2 where again we have kept only the vector current pieces. Then

√ 4pe · pν 1 1 forward √ + 2 sin2 θw + 2 sin2 θw Amp −→ 2GF ≈ 2GF 2 4Ee Eν 2 d) Because of scattering, the νe νe element picks up a new piece so can write ! √

∆m2 ∆m212 d aνe aνe sin 2θ12 − 4E12 cos 2θ12 + 2GF Ne 4E i = , 2 2 ∆m12 ∆m12 aνµ aνµ dt sin 2θ12 cos 2θ12 4E

4E

However can subtract unit matrix since gives only an overall phase so can write q  

∆m212 ∆m212 1 − cos 2θ + G N sin 2θ 12 F e 12 d aνe aνe 4E 2 4E  q i = , ∆m212 ∆m212 aνµ aνµ dt sin 2θ12 cos 2θ12 − 1 GNe 4E

4E

2

18.2 Neutrino Mixing: Theoretical The weak flavor (e, ν, τ ) and mass (1,2,3) eigenstates of the neutrino generations are related via a unitary mixing matrix U 

  e Ue1  µ  =  Uµ1 τ Uτ 1

Ue2 Uµ2 Uτ 2

  Ue3 1 Uµ3   2  .

Uτ 3 3

Neutrino Physics

281

The matrix U can be written in various forms, but one way is that given in the text   c12 c13 s12 c13 s13 e−iδ U =  −s12 c23 − c12 s23 s13 eiδ c12 c23 − s12 s23 s13 eiδ s23 c13  s12 s23 − c12 c23 s13 eiδ

−c12 s23 − s12 c23 s13 eiδ

c23 c13

a) The meaning of this rather complicated form can be seen by decomposing it into several pieces. Demonstrate that the above form for Uν follows from the matrix multiplication of the simpler forms       1 0 0 c13 0 s13 e−iδ c12 s12 0  ×  −s12 c12 0  Uν =  0 c23 s23  ×  0 1 0 0

−s23

−s13 eiδ

c23

0

c13

0

0

1

b) Suppose that at time t = 0 the neutrino (or antineutrino) is produced in weak flavor eigenstate |νa i (or |ν̄a i). Demonstrate that the probability to be observed in state |νb i (or |ν̄b i) at later time t is given by Pνa →νb (t) = δab − 4 3 X

+2

3 X

Re(U ∗ ai Ubi Uaj U ∗ bj ) sin2 (∆m2ij

i>j=1

Im(U ∗ ai Ubi Uaj U ∗ bj ) sin2 (∆m2ij

i>j=1

L ) 4E

L ) 2E

or, for antineutrinos, Pν̄a →ν̄b (t) = δab − 4

−2

3 X

3 X

Re(U ∗ ai Ubi Uaj U ∗ bj ) sin2 (∆m2ij

i>j=1

Im(U ∗ ai Ubi Uaj U ∗ bj ) sin(∆m2ij

i>j=1

L ) 4E

L ), 2E

where E is the neutrino energy and L is the distance between production and observation. c) Show that in the case of two flavors with mixing angle θ the survival probability of state |νa > in the presence of mixing with state |νb > can be written in the familiar form L ).

Pνa →νa (t) = 1 − sin2 2θab sin2 (∆m2ab 4E d) In the three-flavor case suppose that we measure the probabilities for both νa → νb and ν̄a → ν̄b as functions of time. Show that a nonzero value for the difference Pνa →νb (t) − Pν̄a →ν̄b (t) is a measure of CP violation, and is proportional to s12 s23 s13 , so that it must vanish if any of the mixing angles θ12 , θ23 , θ13 vanishes.

Neutrino Physics

282

Exercise 18.2 Solution: Neutrino Mixing: Theoretical a) Perform matrix multiplications  c13 0 s13 e−iδ c12 s12    −s12 c12 0 1 0 −s13 eiδ 0 c13 0 0 

  0 c12 c13 0 = −s12 −c12 s13 eiδ 1

s12 c13 c12 −s12 s13 eiδ

 s13 e−iδ  0 c13

and 

1  0 0

 0 c12 c13   s23 −s12 c23 −c12 s13 eiδ

0 c23 −s23

 s13 e−iδ  0 c13

s12 c13 c12 −s12 s13 eiδ



c12 c13 s12 c13  −s12 c23 − c12 s13 s23 eiδ c12 c23 − s12 s13 s23 eiδ s12 s23 − c12 c23 s13 eiδ −c12 s23 − c23 s12 s13 eiδ p m2 b) We have, writing Ei = p2 + m2i ' p + 2pi + . . . < να |νβ (t) >=

3 X i=1

m2 i

< να |νi > e−i 2p t < νi |νβ >=

3 X

 s13 e−iδ c13 s23  c13 c23

m2 i

∗ Uiα Uiβ e−i 2p t

i=1

so 3 X

Pνβ →να (t) =

+2Re

2

m2

i ∗ Uiα Uiβ e−i 2p t

i=1

=

3 X i=1

3 X

|Uiα |2 |Uiβ |2

∆m2 ij

∗ ∗ −i 2p t Uiα Uiβ Ujα Ujβ e

i>j=1

=

3 X i=1

|Uiα |2 |Uiβ |2 +2

Now write cos

3 X

∆m2ij ∆m2ij ∗ ∗ ∗ ∗ Re Uiα Uiβ Ujα Ujβ t+Im Uiα t cos Uiβ Ujα Ujβ sin 2p 2p i>j=1

2 ∆m2ij 2 ∆mij 2p t = 1 − 2 sin 4p t so, since t/p ' L/E

Pνβ →να (t) =

3 X i,j=1

∗ Uαi Uβi Uα∗j Uβj − 4

+2Im

= δαβ − 4

i>j=1

"

i>j=1

∗ ∗ Uiα Uiβ Ujα Ujβ

3 X

+2Im

3 X

2 ∆m2ij ∗ ∗ Re Uiα Uiβ Ujα Ujβ sin L 4E

# ∆m2ij sin L 2E

"

2 ∆m2ij ∗ ∗ Re Uiα Uiβ Ujα Ujβ sin L 4E

∗ ∗ Uiα Uiβ Ujα Ujβ

# ∆m2ij sin L 2E

Neutrino Physics

283

For antineutrinos, take complex conjugate so " 3 X 2 ∆m2ij ∗ ∗ Pν̄β →ν̄α (t) = δαβ − 4 L Re Uiα Uiβ Ujα Ujβ sin 4E i>j=1 −2Im

∗ ∗ Uiα Uiβ Ujα Ujβ

∆m2ij sin L 2E

#

c) If only two flavors then mixing matrix is real and Pνα →να (t) = δαα − 4

2 X ∆m2αβ L (Uiα Uiα )2 sin2 4E i=1

= 1 − 4 sin2 θαβ sin2

∆m2αβ L 4E

d) If three flavors, then from b) we have Pνβ →να (t) − Pν̄β →ν̄α (t) =

3 X

∆m2ij ∗ ∗ 4Im Uiα Uiβ Ujα Ujβ sin L 2E i>j=1

Suppose α = 1 and β = 3. Then, taking ∆m2ij to be the same 3 X

∗ ∗ ∗ ∗ ∗ ∗ ∗ Im Uiα Uiβ Ujα Ujβ = Im (U31 U33 (U21 U23 + U11 U13 ) + U21 U23 U11 U13 )

i>j=1 ∗ ∗ ∗ = U23 U33 ImU31 U21 + U11 U33 ImU21 U13 + U11 U23 ImU21 U13 ∗ For first term ImU31 U21 ∝ −s12 s13 − s23 s13 while U23 U33 ∝ s23 . For second term ∗ 2 ∗ ImU13 U21 ∝ −2s13 s23 + s23 s13 while U11 U33 ∝ 1. For third term ImU13 U21 ∝ 2 −2s13 s23 + s23 s13 while U11 U13 ∝ s13 .

18.3 Neutrino Mixing: Experimental In the previous exercise, the formalism to study neutrino oscillations was set up. In the present exercise, we examine various experimental aspects of mixing. a) Atmospheric Neutrinos: At Super-Kamiokande a giant water detector was used to detect electron and muon neutrinos which are emitted when muons are created in cosmic ray collisions at the top of the atmosphere. and travel to the earth. Since the average energy of such neutrinos is ∼1 GeV, their interactions with water can produce electrons or muons, but not taus. By looking at the Cherenkov radiation that the charged particles make as they pass through the water, the electrons can be distinguished from the muons and the direction of the neutrino determined. If there were no mixing, one would expect roughly two muon neutrinos for each electron neutrino from the decay chain π + , K + → µ+ + νµ → e+ + νe + ν̄µ + νµ .

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However, it was found that there was a severe reduction in the population of νµ (ν̄µ ) which were produced on the opposite side of the globe and traveled through the earth before interacting, presumably because they oscillated into ντ (ν̄τ ). Use the two-channel mixing formula to show that this observation implies that θ23 ≈ 45◦ and estimate the mass difference ∆m223 . Compare both numbers with those found experimentally. b) νe −ντ Mixing: A recent experiment at Daya Bay in China involved the use of eight 20-ton ν̄e detectors at distances of about 0.5 km and 1.8 km from a powerful nuclear reactor. The nearby detector was used to normalize the ν̄e flux and then the ν̄e rates measured at the more distant detector provided a measure of possible oscillation. Presuming that both ν̄e − ν̄µ and ν̄e − ν̄τ oscillations are present and that θ13 is small, show that the survival probability at distance L from the reactor is given by Pee (L) ≈ 1 − sin2 2θ13 sin2

∆m212 ∆m213 L − cos4 θ13 sin2 2θ12 sin2 L. 4E 4E

A 6% rate reduction was found at the 1.8 km detector compared with that expected from normalizing to the nearby detector and assuming no oscillation.

Using a typical reactor antineutrino energy of ∼ 4MeV and experimental values of ∆m212 and ∆m223 determine sin2 2θ13 . Compare with the experimental value. c) Solar Neutrinos: In the standard pp-chain, three types of neutrinos are produced: low-energy Eν < 0.42 MeV neutrinos from the reaction p + p → d + e+ + νe , medium-energy 0.86 MeV neutrinos from the electron capture reaction e− + 7 Be → νe + 7 Li , and high-energy E < 14 MeV neutrinos from the beta-decay 8

B → 8 Be + e+ + νe .

In the standard solar model and assuming no mixing, the predicted capture rates for these reactions in gallium radiochemical experiments, which are primarily sensitive to the low-energy pp neutrinos, are ∼130-140 SNU. The two such experiments SAGE and GALLEX, however, yielded measurements about 70 SNU. (Here a SNU — solar neutrino unit — is a capture rate of 10−36 per second per atom.) Similarly Davis’s Homestake experiment, which is primarily sensitive to the high-energy 8 B neutrinos, found a flux only about 1/3 of the standard solar model prediction. Using the experimental values of mixing angles and mass differences, show that theory and experiment can only be resolved with the inclusion of MSW mixing, which is small for low- and medium-energy neutrinos (Eν 2 MeV), but is significant for high-energy neutrinos (Eν 2) MeV.

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Exercise 18.3 Solution: Neutrino Mixing: Experimental a) Suppose consider only νµ − ντ mixing. Then we have Pνµ ,ντ (t) ' 1 − sin2 2θ23 sin2 (∆m223

L ) 4E

In order that probability become very small it is required that sin2 2θ23 ' 1 so θ23 ' 45o . In order that the oscillation is maximum over a distance L ∼ 2REarth ' L ) ' π2 so 8000 Mi for 1 GeV neutrinos, we require (∆m223 4E ∆m223 '

2πE 1 = 6 · 109 eV · 2 × 108 eVfm · ∼ 1 × 10−4 eV2 L 8 · 1.6 × 1021 fm

b) From previous problem

∗ ∗ Pνe ,νe (L) = 1 − 4U13 U13 U11 U11 sin2 (∆m213

= 1 − 4c212 c213 s213 sin2 (∆m213 ' 1 − sin2 2θ13 sin2 (∆m213

L L ∗ ∗ ) − 4U11 U11 U12 U12 sin2 (∆m212 ) 2E 2E

L L ) − 4c212 s212 c413 sin2 (∆m212 ) 4E 4E

L L ) − cos4 θ13 sin2 θ12 sin2 (∆m212 ) 4E 4E

For L = 1.8 km and E = 4 MeV we have ∆m213 and ∆m212

1 L ' 2 × 10−3 eV2 · 1.8 × 1018 fm · ' 1.1 6 4E 4 · 4 × 10 eV · 2 × 108 eV fm

L 1 ' 8 × 10−5 eV2 · 1.8 × 1018 fm · ' 0.045 4E 4 · 4 × 106 eV · 2 × 108 eV fm

Use sin2 θ12 ' 0.3 so sin2 2θ12 ' 0.8 and so last term here is quite small. Keeping L ) ' 0.8 we have sin2 2θ13 ∼ only the second term then, and using sin2 (∆m213 4E .06/.8 ∼ 0.08. c) Note that the Davis experiment is sensitive primarily to the ”high” energy 8 B neutrinos. After traveling the 93 million miles from the sun to the detector, if only vacuum oscillations then the probability that they are still electron neutrinos 2 L ' 12 and we would have then would be found by averaging the factors sin2 ∆m 4E using two state mixing, 1 Pνe ' 1 − sin2 2θ12 2 which would at best offer a 50% reduction. In order to explain a reduction by a factor of 3 we need additional help, which comes from the MSW effect and causes the electron neutrinos to turn into muon neutrinos as they pass through the solar interior. This effect is relevant if E>

∆m212 cos 2θ12 8 × 10−5 eV2 · 0.4 √ = 2 · 0.71 · 10−5 GeV−2 · 6 × 1025 cm−3 2 2GF Ne

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=

0.38 × 1032 eV4 fm3 ∼ 4.5 M eV (2 × 108 eV fm)3

where Ne ∼ 6 × 1025 cm−3 is the density of electrons at the core of the sun and hence only for 8 B neutrinos.

18.4 The RFG Model for CCν Reactions In Chapter 16, the RFG model was developed for electron scattering in the quasielastic and ∆ regions of excitation; the former was extended to discussions of CCν reactions from nuclei in the present chapter. a) Verify the hadronic responses given in Eqs. (18.91). b) Show that the “Rosenbluth” factors VK , K = L, T, . . . in Eqs. (18.92) are obtained when the mass of the outgoing charged lepton, m0 is kept nonzero. c) Generate an extension of the electron scattering RFG ∆-region model of Chapter 16 which is appropriate to CCν reactions.

Exercise 18.4 Solution: The RFG Model for CCν Reactions a) The basic developments of the RFG have already been presented in Chapter 16 (see, in particular, Exercises 16.1 and 16.2) and we will draw on those discussions.

We begin by noting that, for inclusive scattering, as we are considering, only the symmetric terms of the leptonic and hadronic tensors involving Lorentz components µν = 00, 03, 30, 33 and 11 + 22, and the antisymmetric terms having µν = 12 and 21 can occur; the arguments needed to arrive at these statements proceed just as in Section 7.1, with two extensions, namely (1) that the axial-vector contributions require separate 00, 03, 30 and 33 terms, since the axial-vector current is not conserved whereas the vector current is conserved, and (2) that the VA interference term contributes and involves the anti-symmetric tensors (recall the discussions of PV electron scattering in Section 7.5). Forming the contraction of the terms that remain for inclusive scattering we have Lµν W µν → L00 W 00 + 2L03 W 03 + L33 W 33

1 + (L11 + L22 ) W 11 + W 22 + 2ImL12 ImW12 , 2 allowing us to define (by convention) the leptonic kinematical factors and their corresponding hadronic responses: with v0 = ( + 0 )2 − q 2 , as usual. The complete contraction for inclusive scattering then becomes h i 0 Lµν W µν = v0 VCC W CC + VCL W CL + VLL W LL + VT W T + VT 0 W T .

Note that, for semi-inclusive processes (ones where particles in addition to the finalstate charged lepton are detected in coincidence) a much larger set of responses must occur, in fact, 16 types in all of which 6 do not enter for semi-inclusive CCν

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Table 18.1 Leptonic and hadronic responses Leptonic

Hadronic

L00 ≡ v0 VCC L03 ≡ v0 VCL L33 ≡ v0 VLL 1 (L11 + L22 ) ≡ v0 VT 2 ImL12 ≡ v0 VT0

WCC = W 00 WCL = W 03 + W 30 WLL = W 33 WT = W 11 + W 22 WT 0 = − 12 ImW12

processes, as discussed in [Mor14a]. Upon labeling the responses by their vector V V,AA,V A and axial-vector content and renaming them RK with K = CC, CL, etc., we see that four combinations must be computed:

VV XL = VL RL AA AA AA XC/L = VCC RCC + 2VCL RCL + VLL RLL

XT = VT RTV V + RTAA

XT 0 = 2VT 0 RTV 0A .

In the semi-inclusive case upon counting the various V V , AA and V A contributions one finds a total of 18 responses, each a function of 4 variables.

To obtain the required responses we proceed exactly as we did for the RFG model of inclusive electron scattering in Chapter 16. For the VV responses we require the single-nucleon tensor already obtained in Chapter 16, with the one change that only the isovector form factors enter for CC neutrino reactions, rather than both isoscalar and isovector (or proton and neutron) contributions for electron scattering.

We need

1 0 Tr [GM γ µ − F2 T µ ] /p +mN [GM γ ν − F2 T µ ] (p / +mN ) 2 4mN

qµ qν µ ν = 2 −W1 g µν + + W T T , 2 Q2

wVµνV =

where T µ ≡ (p0µ + pµ ) /2mN , just as we found in discussing electron scattering.

The extensions needed here to incorporate the axial-vector current contributions

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proceed in an analogous way:

1 1 0 µ µ Tr GA γ − GP Q γ5 /p +mN 2 4mN 2mN

1 ν ν × GA γ + GP Q γ5 (p / +mN ) 2mN

µ ν qµ qν µ ν 02 q q 2 µν + W2 T T + GA 2 = 2 GA −(1 + τ ) g + Q2 Q

1 0 µν µν µ µ wV A = wAV = Tr [GM γ − F2 T ] /p +mN 4m2N

1 ν ν GP Q γ5 (p / +mN ) × GA γ + 2mN i GM GA µναβ qα Tβ , = mN µν wAA =

where G0A ≡ GA − τ GP . The rest follows the steps already presented in Chapter 16 (see, in particular Exercise 16.1). In fact the VV and AA contributions immediately yield Eqs. (18.88), while the VA contribution requires a bit more discussion. For the inclusive CC neutrino cross section we require the µν = 12 component of the anti-symmetric tensor wVµνA given above. This means that we need 12αβ qα Tβ = ωT 3 − qT 0 τ = −2mN ( + λ) . κ Following the steps discussed for the solution to Exercise 16.1, we get to an integral of the form Z F IV A = Γ

d ( + λ) = ( F − Γ) θ ( F − Γ)

1 ( F + Γ) + λ 2

and thus immediately the last of Eqs. (18.88) together with the quantity ∆0 defined in Eq. (18.89). Note that all of these expressions have been organized to have a leading-order contribution plus terms involving ∆ or ∆0 ; often a good approximation emerges when one neglects the latter, since they are proportional to ηF2 which is typically only a few percent or less. b) To compute the leptonic factors we can begin with the general V-A leptonic current jµ (k; k 0 ) = u(k 0 ) [aV γµ + aA γµ γ5 ] u(k), where to begin we keep the leptonic couplings general through the arbitrary factors aV and aA (see also Section 7.5, for instance), allow for arbitrary initial and final

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lepton masses mi and mf , and form the leptonic tensor Lµν = mi mf

X if

jµ (k; k 0 )∗ jν (k; k 0 )

1 0 Tr [aV γµ + aA γµ γ5 ] /k +mf [aV γν + aA γν γ5 ] (k / +mi ) 4

1 2 0 = aV Tr γµ /k +mf γν (k / +mi ) 4

0 +a2A Tr γµ γ5 /k +mf γν γ5 (k / +mi )

0 +aV aA Tr γµ /k +mf γν γ5 (k / +mi )

0 +aV aA Tr γµ γ5 /k +mf γν (k / +mi )

= a2V kµ kν0 + kµ0 kν − gµν (k · k 0 − mi mf )

+a2A kµ kν0 + kµ0 kν − gµν (k · k 0 + mi mf )

=

−aV aA 2i µναβ k α k 0β .

Note that the purely vector and purely axial-vector contributions differ only in the mass factors mm0 . One can keep everything general and proceed to evaluate the leptonic tensor for use in neutral current reactions (for instance, PV electron scattering, as discussed in Chapter 7) or in BSM studies. The interested reader can consult [Mor14a]) for more detail. Here we specialize to the SM for charge-changing neutrino reactions, in which case aV = −aA = 1; this yields the leptonic tensor

Lµν = 2 kµ kν0 + kµ0 kν − gµν k · k 0 + i µναβ k α k 0β .

From part a) we have the five leptonic kinematic factors given there; as noted above, for the purely vector terms the µν = 00, 03, 30, 33 terms coalesce in the usual way VL = VCC + 2νVCL + ν 2 VLL , where ν ≡ ω/q, and it is convenient to introduce the following dimensionless quantities: Q2 = 1 − ν2 q2 q ρ0 ≡ + 0 mi,f δi,f ≡ p Q2 e Q2 ≡ 4 0 sin2 θ/2 ρ≡

e v0 ≡ 4 0 cos 2 θ/2.

Let us illustrate what results by evaluating one case and then leaving the rest for the reader to fill in. The CC contribution is given by

1 0 2 2 2 v0 VCC = 2 2k0 k0 − mi + mf − Q 2

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using the fact that k · k0 =

1 m2i + m2f − Q2 .

2

Thus we have 1 [2 0 − ( 0 − kk 0 cos θ)] v0

1 = v0 − m2i + m2f v0

e = 1 − δ 2 + δ 2 tan 2 θ/2.

VCC =

i

f

For massless neutrinos and CC neutrino reactions one simply sets δi to zero, but, in general, retains δf ≡ δ in kinematic regions where the final-state charged lepton p mass cannot be assumed to be negligible with respect to Q2 . These results have been organized so that when the charged lepton mass can indeed be neglected (the Extreme Relativistic Limit, ERL) all that is necessary is to set δ to zero. The complete set of kinematics factors is the following: e VCC = 1 − δ 2 tan 2 θ/2 2 δ e VCL = ν + 0 tan 2 θ/2 ρ

2ν 2 2 e VLL = ν + 1 + 0 + ρδ δ 2 tan 2 θ/2 ρ

2ν 2 2 2 e VL = ρ − ρ 1 − 0 − ν δ δ 2 tan 2 θ/2 ρ

1 1 0 2 δ2 2e e VT = ρ + tan θ/2 − 0 ν + ρρ δ tan 2 θ/2 2 ρ 2

1 e VT 0 = 0 1 − νρ0 δ 2 tan 2 θ/2. ρ (Note that in the above-cited paper a different convention was adopted, amounting to various sign changes here.) c) The ∆ analogs of the responses obtained above (see also Exercise 16.1) are straightforward extensions of the purely vector contributions found in Exercise 16.2.

As usual the two four-vectors that characterize the problem are q µ and

1 q·p µ U ≡ p + qµ . mN Q2 µ

Substituting q·p=

1 2 Q2 + M∆ − m2N 2

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one finds that

1 1 µ µ π ≡ p + ρ∆ q = η µ + ρ∆ κµ mN 2  µ=0   + ρ∆ λ  η cos θ + ρ∆ κ µ = 3 =  η sin θ cos φ µ=1   η sin θ sin φ µ=2 µ

Using these extensions together with the developments of the single-nucleon tensor in part a) the rest is straightforward. For the basic hadronic tensors we have

κµ κν + w2V π µ π ν g µν + τ

κµ κν κµ κν µν µ ν ν µ wAA = −w1A g µν + + w2A π µ π ν − uA + uA 1 2 (κ η + κ η ) τ τ

wVµνV = −w1V

wVµνA = 2iw3A µναβ ηα κβ , obtained using our general arguments for the structure that such current×current interaction tensors must have (see, for instance, Chapter 7). The structure functions V A w1,2 , w1,2,3 and uA 1,2 must, of course, be obtained from experiment or from models of the N → ∆ transition (see [Ama05] for more extensive discussions). The results obtained assuming the kinematics are chosen to be where Pauli blocking is absent and using the procedures above and in Chapter 16 for the RFG model are the following:

κ2 −w1V + 1 + τ ρ2∆ w2V + Dw2V τ

λ2 A κ2 ∆ 00 UAA = −w1A + 1 + τ ρ2∆ w2A + Dw2A − uA 1 + λ F + γ− u2 τ τ 2

λ κ2 κ 03 UAA = −w1A + 1 + τ ρ2∆ w2A + Dw2A − uA κ τ τ 1

2

κ + λ2 ∆ F + γ − − ρ∆ τ uA + 2 2λ 2 2

λ κ κ4 33 UAA = −w1A + 1 + τ ρ2∆ w2A + Dw2A − 2 uA κ τ τλ 1 2

κ ∆ + 2 F + γ − − 2ρ∆ τ uA 2 λ

UV00V =

UV11V + UV22V = 2w1V + Dw2V

11 22 UAA + UAA = 2w1A + Dw2A q UV12A = 2i τ (1 + τ ρ2∆ ) [1 + D0 ] w3A ,

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where

τ 1 2 ∆ 2 2 ∆ ∆ 2 + ρ λ

+ γ + ρ λ − 1 + τ ρ2∆

+

γ + γ ∆ F F − − ∆ F − 2 κ 3

r

1 τ 1 ∆ D0 =

+ γ ρ λ + − 1, F ∆ − κ 1 + τ ρ2∆ 2 D=

the latter being natural extensions of the QE CCν developments in part a) and of both the QE and ∆ electron scattering developments in Exercises 16.1 and 16.2, that is, where the VV terms were all that entered and where ρ∆ → 1 to obtain the QE result. Finally, the rest involves simply assembling the above results to obtain the RFG model for N → ∆ CCν reactions.

19

The Physics of Relativistic Heavy-Ions 19.1 Heavy-Ion Hydrodynamics A great deal of the phenomenology of heavy-ion collisions has been shown to follow from hydrodynamic considerations. a) Consider an ultra-relativistic heavy-ion collision in the center-of-mass frame. The nuclei become two Lorentz-contracted pancakes. A short time (say 3 fm/c) after the collision, the pancakes recede from the collision point at the speed of light (γ 1). Assuming an initial thermalization time of about 1 fm/c, show that the energy density is in the range 1-10 GeV/fm3 . b) A basic feature of Bjorken’s description [Bjo83] is that there is a central plateau structure for the rapidity distribution in nucleon-nucleon collisions, where rapidity is defined as 1 t+z , y = ln 2 t−z √ and proper time τ is defined as τ = t2 − z 2 . The system evolves so that throughout the central plateau region, the initial conditions, imposed at a proper time ≈ 1 fm/c after the collision time, are invariant with respect to Lorentz transformations. By applying Landau hydrodynamics to the longitudinal evolution of the plasma, show that

2 2 fm (dNπ /dy)A−A ≈ A2/3 , (dNπ /dy)pp d0 where 1/d20 is the number of independent nucleon-nucleon collisions per unit area.

Exercise 19.1 Solution: Heavy-Ion Hydrodynamics For more detail, see the discussion by J.D. Bjorken, Phys. Rev. D27, 140 (1983).

a) The isotropic component of particle production is approximately the same as in a nucleon-nucleon collision so at SPS energies dNch '3. dy 293

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Assume < E >=400 MeV and Nneu /Nch ∼ 0.5. Then d<E> ∼ 3 × 0.4 × 1.5 ∼ 1.8 GeV . dy Now consider a thin slab of thickness 2d. Then d<E> d<E >1 E'N ∆y = N dy dy 2

2d t

.

The corresponding energy density is '

Nd<E> 1 .

A dy 2t

For ion-ion collisions we replace the number of incident nucleons per unit area by some effective area d20 , i.e., N 1 ∼ 2 .

A d0 Then '

1 1.8 GeV 0.9 GeV ∼ . d20 2t d20 t

Using 1

N A3 A 1 = = 1 2 = d2 .

2 A 3 4.5 fm π(1.2 fmA ) 0 Then for d0 ∼ 0.3 − 1 fm and t = 1 fm/c we have ∼ 1 − 10 GeV/fm3 . b) Apply Landau’s hydrodynamical model. Define (x)=local energy density, p(x)=pressure, T (x) = 1/β(x)=temperature, and uµ (x)=four-velocity the energy-momentum tensor is Tµν = ( + p)uµ uν − pgµν and is conserved, so ∂ µ Tµν = 0. The rapidity is y=

1 t+z ln 2 t−z

√ and the proper time is τ = t2 − z 2 . In our definition of proper time we consider only one dimensional flow. That is, we have translational symmetry in the transverse coordinate, an approximation which is valid for small times compared to the radius of the nucleus, that is 1

t << 1.2 fm A 3 ∼ 7 fm/c for Pb or U .

Under this assumption we consider the variables to be functions only of τ and y = (τ, y), p = p(τ, y), T = T (τ, y)

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and uµ = (u0 (τ, y), 0, 0, uz (τ, y)) .

However, the initial conditions (τ0 , y) ' 0 as shown in a) and uµ (τ0 , y) ' τ10 (t, 0, 0, z) = x̃µ τ0 possess a symmetry with respect to Lorentz transformations which is preserved by the dynamics, so that no y-dependence should show up at later times. That is, = (τ ), p = p(τ ), β = T −1 (τ ) = β(τ ),

and uµ =

x̃µ τ

with 1 1 1 ∂uµ = g̃µν − 3 x̃µ x̃ν = . ∂xν τ τ τ (g̃µν − uµ uν ) From energy-momentum conservation we have ∂ (( + p)g µν − puµ uν ) = 0 , ∂xµ so that ∂ +p =− . ∂τ τ The Landau model assumes that entropy is conserved during the expansion stage.

From thermodynamics we have TS = U + PV which gives the entropy density s=

S = β( + p) V

and the entropy current S µ = β( + p)uµ = Suµ .

Contracting uν with the above expression for ∂x∂ µ T µν we find then that ∂µ S µ = 0 Thus, dS S S(τ ) τ0 =− or = . dτ τ S(τ0 ) τ The bottom line is that the entropy content per unit rapidity is a constant of the motion. To see this in a fluid rest frame we have d3 x = d2 xT τ dy .

Then Z dS =

Z

3

S0 d x = τ S

d2 xT dy

so d dS d = τS dτ dy dτ

Z

d2 x = 0 .

Given that entropy is conserved during the hydrodynamic expansion, the final pion

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(or other hadron) multiplicity per unit rapidity is directly proportional to the entropy density.

We can relate the pp and AA results via

1 dS dS 1 = 2 (π(1.2 fmA 3 )2 ) dy AA d0 dy pp where 1/d20 is the number of effective independent NN collisions (which are the entropy producers) per unit area. Since the pion multiplicity is proportional to entropy we have

dS

2 dy 1 2 1 2 fm 2 AA ' 2 (π(1.2 fmA 3 ) ) = A3 . dS d0 d0 dy

pp

19.2 Scaling and Multiplicity For a variety of nuclear collisions at low energies up through RHIC energies (200 GeV/nucleon N N center-of-mass), it was observed that there was approximate pp Nparticipant scaling [Bus75] of the total charged-particle multiplicity, i.e., Nch /Nch = 1 N , provided one scales heavy-ion data to pp data at twice the center-of-mass part 2 energy, as shown in the figure from [Bac05]. The ratio of the total particle multiplicity data for a wide range of hadron-nucleus and nucleus-nucleus collisions over the multiplicity in proton (antiproton)-proton interactions is plotted as a function of the number of participating nucleons.

pp RA = Nch /Nch

200

d+Au(200 GeV) Au+Au(200 GeV) Au+Au(19.6 GeV)

150

100 RA =

50

0

100

200

300

Npart Make a plot that extends the range in energy also using available data from the LHC program to determine if this scaling is preserved at higher energies. Can one

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understand the origin of any deviations from scaling?

Exercise 19.2 Solution: Scaling and Multiplicity To clarify this scaling relationship, we are testing whether or not the following relationship holds at LHC energies: √ √ 1 pp AA Nch. (2 s) . ( s/A) = Npart. Nch. 2 If we were to plot the fraction: √ AA Nch. ( s/A) √ pp Nch. (2 s) the result, 12 Npart. , is independent of the beam energy, independent of the rapidity bin, and independent of the particular heavy ion projectile used.

Taking the data from Fig. 2 of the 2011 ALICE paper: PRL 106 032301, which shows the scaling law to be roughly upheld, and combining it with data points from the PHOBOS white paper, one can make the following plot

√ √ pp AA Nch. ( s/A)/Nch. (2 s)

1000

R = 0.5Npart ALICE Pb+Pb (2.76 TeV) PHOBOS Au+Au (200 GeV) PHOBOS d+Au (200 GeV)

100

10

1

10

100

1000

Npart.

One can see that the LHC data has a steeper trend than the RHIC data. In LHC data, there are more particles in AA collisions than one would expect from pp data.

The reason for this has to do with the rates of hard scatterings at RHIC and LHC. In pp collisions, the jet cross section increases with energy. With higher energy, it’s possible to access lower and lower values of x, (where the sea-quarks dominate), and the cross section for jets from hard scatterings increases dramatically. At RHIC, the rate of jets is low, but at the LHC, the rate of jets from pp collisions is very high.

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Jets in pp collisions do not produce an excess of particles. The energy is contained in a spray of a few collimated higher energy particles. But jets in AA collisions will be quenched by passing through QGP. The jet energy will be converted into more particles. This is why there is a particle excess in LHC AA collisions.

19.3 Viscosity The concept of viscosity has proven to be useful in describing heavy-ion collisions. a) Provide a definition for viscosity from a classical perspective and find the viscosity for water, oil and mercury. b) Discuss your result for a) from the perspective of the Navier-Stokes equation ∂ρvi = −∇j (ρvj vi + δij P − σij ) , ∂t where ρ, v are the fluid density, velocity respectively. Here

∂vi 2 ∂vj + − δij ∇ · v + ζδij ∇ · v , σij = η ∂xi ∂xj 3 where η is the shear viscosity and ζ is the bulk viscosity.

Exercise 19.3 Solution: Viscosity a) Informally, viscosity describes a fluid’s resistance to flow. More precisely the viscosity η can be described as the ratio of the shear stress F/A to the velocity dv i gradient dx j . That is, F dv i =η j .

A dx In words it is the force per unit area resisting a flow in which parallel layers a unit distance apart have unit velocity with respect to one another. Typical values are 8.9×10−4 for water at 25 degrees Centigrade, 1.53×10−3 for mercury, and 2.9 for oil, all of which are in units of Pascal seconds. b) The Navier Stokes equation is ∂ρvi = −∇j (ρvi vj + δij P − σij ) ∂t where

σij = η

∂vi ∂vj 2 + − δij ∇ · v + ξδij ∇ · v ∂xj ∂xi 3

is the viscosity term. Here the left hand side is of the form m dv ∼ Force density .

V dt The Navier-Stokes equation is derived by applying Newton’s second law to

299

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fluid particles. It applies to both Newtonian and non-Newtonian fluids as well as to compressible an incompressible fluids. As shown above the LHS is the force density, while on the RHS the term −∇j δij P is a pressure term, which captures the finite volume of the fluid, preventing it from shrinking indefinitely. ∂v ∂vi The σij contains two tyres of viscosity. In the η term ∂x + ∂xji has off-diagonal j pieces and measures the shear force due to the relative motion of different layers of fluid, so η is the shear viscosity. On the other hand, the ξ term is purely diagonal and contributes only when ∇ · v 6= 0. This piece contributes only for force along the direction of motion and for compressible fluids. ξ is associated with compression and rarefaction in the longitudinal direction and is called the bulk viscosity.

19.4 Stokes Law for Viscous Flow In order to gain familiarity with the effects of viscosity, it is useful to examine a classic problem: Stokes law for the flow of a viscous fluid.

During the 1850s, G.G. Stokes began investigations of highly viscous flow. One of his results is that, for creeping viscous flow of an incompressible fluid past a sphere, the drag force is given by F drag = 6πηav 0 , where v 0 is the asymptotic flow velocity, a is the radius of the sphere, and η is the viscosity. In this problem we derive this result.

Assume that the most important term for large η is the diffusion term and that the inertial term ρ ∂v ∂t can be neglected. Then, neglecting gravity, we have −∇P + η∇2 v = 0 . a) Show that the pressure satisfies Laplace’s equation ∇2 P = 0 .

Define the stream potential ψ in terms of axial co-ordinates r, θ: 1 ∂ψ , r2 sin θ ∂θ b) Show that for axially-symmetric flow vr =

vθ =

1 ∂ψ . r sin θ ∂r

∇×v =0 , is required, and that ψ obeys the equation 2 2 ∂ 1 ∂2 cotθ ∂ + − ψ = 0. ∂r2 r2 ∂θ2 r2 ∂θ Now insert the boundary conditions

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i) no slip at the sphere surface ∂ψ ∂ψ = =0, ∂r ∂θ ii) asymptotic flow v0 êz vr ∼ v0 cos θ,

vθ ∼ v0 sin θ

as r → ∞ .

c) Show that condition ii) is satisfied by the asymptotic form r→∞

ψ(r, θ) −→

1 , 2v0 r2 sin2 θ

which suggests the use of ψ(r, θ) =

1 v0 f (r) sin2 θ 2

as a trial solution. d) Find the differential equation obeyed by f (r) and solve to show that

3 1 3 a 2 ψ(r, θ) = v0 − ar + r sin2 θ .

2 2r 2 e) Show that the pressure is given by P (r, θ) = P∞ − ηv0

3a cos θ .

2r2

f) Calculate the total drag force on the sphere due to the pressure differential. g) Calculate the total drag force on the sphere due to viscous friction. h) Add your results for f) and g) to yield Stokes Law F drag = 6πηav 0 .

Exercise 19.4 Solution: Stokes Law for Viscous Flow a) We have ∇P = η∇2 v so that ∇2 P = η∇2 ∇ · v = 0 because of incompressibility. b) Note ∇·v = =

1 ∂ 2 1 ∂ r vr + vθ sin θ r2 ∂r r sin θ ∂θ

1 ∂ 2 1 ∂ψ 1 ∂ 1 ∂ψ r · 2 + sin θ · r2 ∂r r sin θ ∂θ r sin θ ∂θ r sin θ ∂r =

1 ∂ 1 ∂ψ 1 ∂ 1 ∂ψ − =0 r2 ∂r sin θ ∂θ r sin θ ∂θ r ∂r

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so ∇ · v = 0 is automatically satisfied. We also have η∇2 v = −η∇ × (∇ × v) = −η∇ × Ω = ∇P so that ∇ × (∇ × Ω) = 0 Note also that 1 r

∂ 1 ∂ψ ∂ 1 ∂ψ ·− − ∂r sin θ ∂r ∂θ r2 sin θ ∂θ 2

1 1 ∂2 ctnθ ∂ ∂ = φ̂ + − ψ ≡ φ̂Õψ r sin θ ∂r2 r2 ∂θ2 r2 ∂θ

Ω = ∇ × v = φ̂

But ∇ × Ω = −r̂

1 ∂ 1 ∂ (sin θÕψ) + θ̂ (rÕψ) r sin θ ∂θ r ∂r

and

1 ∂ ∂ ∂ 1 ∂ · rÕψ + · sin θÕψ r ∂r r ∂θ r sin θ ∂θ

∂ 1 ∂ 1 ∂ 1 ∂ r sin θÕψ + r sin θ Õψ = φ̂ r ∂r sin θ ∂r ∂θ r2 sin θ ∂θ

1 1 ∂2 ∂ 1 ∂ = φ̂ r sin θÕψ + r sin θÕψ r sin θ ∂r2 ∂θ r2 sin θ ∂θ

0 = ∇ × (∇ × Ω) = φ̂

=

φ̂ r sin θ

∂2 1 ∂2 ctnθ ∂ + 2 2− 2 2 ∂r r ∂θ r ∂θ

Õψ =

φ̂ Õ2 ψ r sin θ

c) We have vr =

1 ∂ψ r→∞ 1 ∂ 1 −→ 2 v0 r2 sin2 θ = v0 cos θ r2 sin θ ∂θ r sin θ ∂θ 2

vθ = −

1 ∂ψ r→∞ 1 ∂ 1 −→ − v0 r2 sin2 θ = −v0 sin θ r sin θ ∂r r sin θ ∂r 2

d) Since

∂2 1 ∂2 ctnθ ∂ + 2 2− 2 2 ∂r r ∂θ r ∂θ

1 v0 sin2 θf (r) 2

2

d f 1 ∂ 2 2 sin θ cos θf − ctnθ · 2 sin θ cos θf sin θ + dr2 r2 ∂θ 2

1 d f 1 2 2 2 2 2 = v0 sin θ + 2 (2 cos θ − 2 sin θ)f − 2 cos θf 2 dr2 r r 2

1 d f 2 = v0 sin2 θ − 2f 2 dr2 r

=

1 v0 2

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then

∂2 1 ∂2 ctnθ ∂ + 2 2− 2 2 ∂r r ∂θ r ∂θ

2

1 1 v0 sin2 θf (r) = sin2 θ 2 2

2 2 d f − f =0 dr2 r2

Solve via f (r) =

X n

an r n ,

X df = nan rn−1 , dr n

X d2 f = n(n − 1)an rn−2 2 dr n

so X d2 f 2 − 2f = (n(n − 1)an − 2an ) rn−2 2 dr r n and 2 2 X d f 2 = (n(n − 1) − 2) ((n − 2)(n − 3) − 2) an rn−4 = 0 − f dr2 r2 n Solution is n = 4, 2, 1, −1 so f (r) =

s + br + cr2 + dr4 r

Now require r→∞

f (r) −→ r2 ∂ψ ∂r ∂ψ ∂θ

= r=a

s

1 v0 sin2 θ − 2 + b + 2cr 2 r

= v0 sin θ cos θ r=a

so c = 1, d = 0

s r

+ br + cr2

=0

so

−s + b + 2a = 0 a2

=0

so

s + ba + a2 = 0 a

r=a

r=a

3 We have then b = − 23 a and s = 21 a3 so f (r) = 12 ar − 32 ra + r2 and

1 ψ(r, θ) = v0 sin2 θ 2

1 a3 3 − ra + r2 2 r 2

e) Calculate

1 ∂2 1 ∂ 2 ∂ 1 ∂ ∂ 1 ∂ψ 1 ∂ψ r + sin θ r̂ − θ̂ + r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin θ ∂θ r sin θ ∂r r2 sin2 θ ∂φ2

3

1 ∂2 1 ∂ 2 ∂ 1 ∂ ∂ a 3a = + r + sin θ r̂v cosθ − + 1 0 r2 ∂r ∂r r2 sin θ ∂θ ∂θ 2r3 2r r2 sin2 θ ∂φ2 ∇2 v =

+θ̂v0 sin θ

a3 3a + −1 4r3 4r

" #

ˆr 1 ∂ 2 ∂v ∂vθ 1 ∂ ∂vr ∂vθ = 2 r r + θ̂ + 2 sin θ r̂ + θ̂ r ∂r ∂r ∂r r sin θ ∂θ ∂θ ∂θ

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i 1 ∂ h v sin θ φ̂ + v cos θ θ̂ r θ r2 sin2 θ ∂φ

∂vr ∂vθ 1 ∂ 2 vr ∂ 2 vθ = 2 r̂ 2r + r2 2 + θ̂ 2r + r2 2 r ∂r ∂r ∂r ∂r

1 ∂vr ∂vθ + 2 cos θ r̂ − vθ + θ̂ + vr r sin θ ∂θ ∂θ 2

2

∂vθ ∂vr ∂ vθ ∂ vr + sin θ r̂ − + + θ̂ ∂θ2 ∂θ ∂θ2 ∂θ

∂vθ ∂vr − vθ − r̂ + vr +θ̂ ∂θ ∂θ i ˆθ + +θ̂vr − rv

h i 1 − 2 2 (vr sin θ + vθ cos θ)(sin θr̂) + cos θθ̂ r sin θ

2 ∂vr ∂ 2 vr cos θ 1 ∂vr 2 ∂vθ 1 ∂ 2 vr = r̂ + + − v − 2 − 2v + θ r r ∂r ∂r2 sin θ r2 ∂θ r2 r2 ∂θ2 ∂θ

∂ 2 vθ cos θ 1 ∂vθ 1 ∂ 2 vθ ∂vr 1 2 ∂vθ + + + 0 + + 2 − v +θ̂ θ r ∂r ∂r2 sin θ r2 ∂θ r2 ∂θ2 ∂θ r2 sin2 θ with

3a a3 − +1 vr = v0 cos θ 2r3 2r

∂vr 3a3 3a = v0 cos θ − 4 + 2 ∂r 2r 2r 3

a a ∂ 2 vr = v cos θ 6 − 3 0 ∂r2 r5 r3 3

a 3a ∂vr = −v0 sin θ − + 1 ∂θ 3r3 2r 3

∂ 2 vr a 3a = −v cos θ − + 1 0 ∂θ2 3r3 2r

a3 3a vθ = v0 sin θ + −1 4r3 4r

3a3 3a ∂vθ = v0 sin θ − 4 − 2 ∂r 4r 4r 3

∂ 2 vθ a 3 a = v sin θ 3 − 0 ∂r2 r5 2 r3 3

∂vθ a 3a = v0 cos θ + − 1 ∂θ 4r3 4r 3

∂ 2 vθ a 3a = −v sin θ + − 1 0 ∂θ2 4r3 4r

Hence, we have

3

2 3 a3 3 a a a v0 cos θ − 4 + + v cos θ 6 − 3 0 r 2r 2 r2 r5 r3 3

3

a 3a 1 a 3a 1 −v0 cos θ − 3 + 2 − 2v0 cos θ + 3− 2 2r5 2r r 4r5 4r r 3

3

a 3a 1 a 3a 1 −v0 cos θ − 3 + 2 − 2v0 cos θ − 3+ 2 2r5 2r r 2r5 2r r

∇2 v = r̂

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304

−2v0 cos θ

a3 3a 1 + 3− 2 5 4r 4r r

3

2 3a 3a 3a 3a3 +θ̂ v0 sin θ − 4 − 2 + v0 sin θ + 3 r 4r 4r r5 2r

cos2 θ +v0 sin θ

−2v0 sin θ = r̂v0 cos θ

3a 1 a3 + 3− 2 4r5 4r r

−v0 sin θ

3a a3 + −1 3 4r 4r

a3 3a 1 − 3− 2 5 2r 2r r

v0 − sin θ

3a a3 + −1 4r3 4r

3 1 1 1 1 a 3 3 3 3 a (−3 + 6 − − − − 1 − ) + 3 (3 − 3 + − + + 3 − ) r5 2 2 2 2 r 2 2 2 2 1 + 2 (−1 + 2 − 1 − 2 + 2) r

3 3 1 1 a 3 3 3 3 a +θ̂ 5 (− + 3 − − − 1) + 3 (− + − − + 3) r 2 4 4 r 2 2 4 4 1 + 2 (1 + 1 − 2) r

a a 3 1 = r̂3v0 cos θ 3 + θ̂ v0 sin θ 3 = r 2 r η

∂P 1 ∂P r̂ + θ̂ ∂r r ∂θ

Thus ∂P a = 3ηv0 cos θ 3 ∂r r

and

∂P 3 = ηv0 sin θ ∂θ 2

so 3 a P = P0 − ηv0 cos θ 2 2 r f) Then F =−

Z

2

r̂P (r, θ) r=a

a dΩ = −

Z π

Z 2π dθ sin θ

0

dφa2

0

1 3 ×(cos θêz + sin θ cos φêx + sin θ sin φêy )(P0 − ηv0 2 cos θ) 2 a v0 = 2πa η eˆz 2 a 23

Z 1 −1

dzz 2 = 2πηv0 a

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305

g) We need to calculate n̂ · F = −

Z dSi Tij êj

where

dS = a2 dΩr̂

surf

Here Tij = ρvi vj + δij P − η

∂vj ∂vi + ∂xj ∂xi

Pressure piece was found in f) so we need the rest Z Z 0 n̂ · F = −ρ dS · vv · n̂ + η (n̂ · ∇v · S + dS · ∇v · n̂) Hence F 0 = −ρ

Z

a2 dΩvr v + η

Z

dΩa2

∂ v + ∇vr ∂r

Note

∂vθ ∂vr 1 ∂vr ∂vr ∂ r̂vr + θ̂vθ + ∇vr = r̂ + θ̂ + r̂ + θ̂ ∂r ∂r ∂r ∂r r ∂θ

∂vr ∂vθ 1 ∂vr r = r̂2 + θ̂ + ∂r ∂r r ∂θ

3 a3 3 a 3 a3 3 a = r̂2v0 cos θ − 4 + + θ̂ v sin θ − − 0 2r 2 r2 4 r4 4 r2

−v0 sin θ

a3 3a 1 − 2+ 2r4 2r r

3a3 3a = (cos θêz + sin θ cos φêx + sin θ sin φêy )2v0 cos θ − 4 + 2 2r 2r

3a3 3a a3 3a 1 +(− sin θêz +cos θ cos φêx +cos θ sin φêy )v0 sin θ − 4 − 2 − 4 + 2 − 4r 4r 2r 2r r Then

Z ∂ η dΩa2 v + ∇vr ∂r =

r=a

= −ηa2 2π

Z 1

3 3 1 3 1 dz(1−z 2 )v0 (− − − + −1) êz 4 4 2 2 a −1

3 2 1 2 ηa 2πv0 (2 − )êz = 4πηv0 aêz 2 a 3

Also note that Z dΩa2 vr (vr r̂ + vθ θ̂)

=0 r=a

since vr = vθ = 0

on the surface.

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306

h) Thus we have F tot = 2πηv0 aêz + 4πηv0 aêz = 6πηv0 aêz where the first piece comes from pressure and the second from friction.

20

Astrophysics 20.1 Solar Fusion Reaction Rate Show that the reaction rate near the most effective temperature T = Emax /kB , 2 given by the value Emax = ( 21 bkB T ) 3 for solar nuclear fusion reactions, may be approximated by a normal distribution centered around Emax with a width σ = p 2Emax kB T /3. The parameter b is defined by the relation √

e−2πZ1 Z2 αc/v = e−b/ E , where Z1 and Z2 are the charge numbers of the projectile and target nucleus, α is the fine structure constant, and v is the relative velocity.

Exercise 20.1 Solution: Solar Fusion Reaction Rate For non-resonant reactions at low energy, σ= where b = 2πZ1 Z2 Λ ∼ σ · v so

p νr 2

1 −2πZ1 Z2 αc 1 − √b v S(E) = e e E S(E) E E

and S9E) is the astrophysical S-factor. The reaction rate Λ =< σ · v >=

Z ∞

σ(E)vp(v)dv ,

0

where p(v) is the Maxwell-Boltzmann velocity distribution for hot gases. Thus, Z ∞ Z ∞ −µv 2 −E S(E) √−b µ S(E) √−b µ 2E dE 3/2 2kB T 2 E Λ= e ( ) e v4πv dv = e E( )3/2 e kB T 4π E 2πk T E 2πk T µ µ B B 0 0 where we have changed integration variables with E = 12 µv 2 so dE = µvdv. Thus, we have r Z 8 1 3/2 ∞ −( E + √b ) S(E)e kB T E dE . Λ= ( ) µπ kB T 0 −E

−b √

Here, S(E) is slowly varying with E, e kB T decreases with E but e E increases with E. To evaluate the integral, we use the method of steepest descent approximation.

In the integral Z ∞ I= f (x)e−g(x) dx 0

if f (x) is assumed to be slowly varying, then the greatest contribution to the integral 307

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308

comes from where g(x) is stationary. Then, for g 0 (x0 ) = 0, values near x0 contribute the most to the integral.

Using a Taylor expansion 1 g(x) = g(x0 ) + g 0 (x0 )(x − x0 ) + g 00 (x0 )(x − x0 )2 + .....

2 we can approximate g(x) ≈ g(x0 ) + 21 g 00 (x0 )(x − x0 )2 . Then the integral becomes Z ∞ −1 00 2 −g(x0 ) dx e 2 g (x0 )(x−x0 ) .

I ≈ f (x0 )e −∞

For our case, f (x) = S(E) and g(x) = kBET + √bE so we set 0=

d E b 1 1 ( +√ )= − bE −3/2 = 0 . dE kB T kB T 2 E

This gives E0 = (

bkB T 2/3 ) .

2

Further, we have d2 3 −5/2 E b 3 ( + √ ) = bE0 = >0. dE 2 kB T 4 2kB T E0 E Thus, g(E0 ) has a minimum and −g(E0 ) a maximum. Renaming E0 as Emax , we have Z ∞ −1 2 2kB T Emax (E−Emax ) −( k ET + √b ) ) E B 3 dE e 2( I ≈ S(Emax )e . −∞

This is an integral over a normal q distribution centered around the mean of Emax = ( bk2B T )3/2 with a width σ =

2kB T Emax .

3

20.2 Building a Star: the Lane-Emden Equation As discussed in the text, stellar dynamics is based on a few simple principles. When given an equation-of-state, which relates the pressure to the density, we can construct a model star, and that is the goal of this problem.

We assume a spherically symmetric star, which is governed by the equations dp(r) Gm(r) =− ρ(r) , dr r2 which is the equation of hydrostatic equilibrium that relates the pressure gradient dp(r)/dr to the density ρ(r), and dm(r) = 4πr2 ρ(r) , dr

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309

which relates the density to the mass gradient dm(r)/dr. Now assume a simple (polytropic) equation-of-state p = ργ , where γ = cP /cV is the ratio of the specific heat at constant pressure to the specific heat at constant volume. For the sun, a reasonable value is γ = 4/3. Now define the function

γ−1 γ−1 ρ(r) p(r) γ Φ(r) = = , ρ0 p0 where ρ0 (p0 ) is the density (pressure) at the star’s center. a) Demonstrate that γ − 1 ρ0 Gm(r) dΦ(r) =− dr γ p0 r2 and 1 d2 Φ(r) 2 dΦ(r) γ − 1 ρ20 + = −4πG Φ(r) γ−1 .

2 dr r dr γ p0

b) Defining the dimensionless quantity x = r/r0 where

p0 γ r0 = 4πGρ20 (γ − 1)

21 ,

show that we obtain the Lane-Emden equation 1 d2 Φ(x) 2 dΦ(x) + + Φ(x) γ−1 = 0 , 2 dx x dx

which must be solved subject to the boundary conditions ρ(0) = ρ0 and ρ0 (0) = 0, i.e., Φ(0) = 1 and Φ0 (0) = 0. c) Solve this equation numerically in the case that γ = 4/3 and show that the first zero is at x0 = 6.897. d) Perform the integral Z x0 1 J= dxx2 Φ(x) γ−1 0

numerically and show that J = 2.018. e) Using the central density ρ0 = 148 g/cm3 , use your solution to determine the solar radius and compare to the known value of Rsun = 6.96 × 105 km. f) Compare the calculated value of p0 with the known solar value of 2.29×1017 erg/cm3 . g) Plot the predicted solar density and pressure profiles.

Exercise 20.2 Solution: Building a Star: the Lane-Emden Equation

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310

a) We have dΦ d = dr dr

γ−1 = −( ) γ

p(r) p0

p(r) p0

γ−1 γ

− γ1

=(

γ−1 ) γ

p(r) p0

− γ1

1 dp(r) p0 dr

γ − 1 ρ0 Gm(r) 1 Gm(r) ρ(r) = − p0 r 2 γ p0 r 2

Also γ − 1 ρ0 d2 Φ =− 2 dr γ p0 so

γ − 1 ρ0 G dm(r) G G m(r) = − m(r) − 2 4πGρ(r) − r2 dr r3 γ p0 r3

1 d2 Φ(r) 2 dΦ(r) γ − 1 ρ0 γ − 1 ρ20 + = −4π Gρ(r) = −4π G(Φ(r)) γ−1 2 dr r dr γ p0 γ p0

b) Multiply both sides by r02 and find Lane-Emden equation 1 d2 Φ(x) 2 dΦ(x) + + Φ(x) γ−1 = 0 2 dx x dx

c) If γ = 43 then r0 =

q

p0 πGρ20

f) We have

4 4 3 3 p0 = ρ03 = 148 g/cm

20.3 White Dwarfs In the text we described the evolution of a star from its birth to death as a supernova. However, what happens to a star that runs out of fuel, but is not massive enough to explode as a supernova? Collapse still occurs, but ends when the electron degeneracy pressure halts the implosion. That is, one has a core with a mass of the order of the solar mass which supports itself from the degeneracy pressure of the electron gas, leading to densities which are of the order of 104 times the solar density. The purpose of this problem is to estimate the degeneracy pressure. a) At low temperature the electrons will occupy the low-lying levels with two electrons in each. Show that this leads to a Fermi momentum kF given by Z kF dk kF3 n=2 = , (2π)3 3π 2 0 where n is the electron number per unit volume. b) Since nearly all the mass of the star is due to nucleons, calculate the Fermi energy in terms of the star density as 2 13 3π ρ kF = , mN r

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311

where r is the ratio of nucleons to electrons. c) For electrons having momentum k show that the pressure on the walls is given by nk 2 , 3E

p=

p where E = k 2 + m2e is the electron energy. d) Integrate over the spectrum and show that Z 2 kF dk k2 p p= 3 3 0 (2π) k 2 + m2e

m4e kF3 kF kF2 1 −1 kF 2 = (2( 3 ) − 3 )(1 + 2 ) + 3sinh , 24π 2 me me me me which represents the equation-of-state, i.e., the pressure in terms of the density. e) If kF << me show that the equation-of-state becomes p=

1 15π 2 me

3π 2 ρ rmN

35 .

Given the equation-of-state, the Lane-Emden equation, derived in Exercise 20.2 above, allows the construction of the star and yields 2.72 M= 2 r

ρ(0) ρc

12 MSun − 61

R=

2.0 × 104 r

ρc =

mN r 3 m ' 1 × 106 g/cm3 3π 2 e

ρ(0) ρc

km ,

where

is the critical density where kF = me . Thus, the white dwarf has a mass of order the solar mass but is confined to a radius of order that of the earth!

Exercise 20.3 Solution: White Dwarfs a) Quantize in a box, with one corner at the origin and sides a distance L apart.

Then solutions are of the form ny πy nz πz nx πx sin sin ψ(x, y, z) = sin L L L with nx , ny , nz integers 0, 1, 2..... The momentum is then πq 2 k= nx + n2y + n2z .

L

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312

If there are two neutrons per state and the total number of neutrons is N, then we have, using kF = πnmax /L Z kF 3 1 N 4πn3max d k 3 2 = kF /3π = 2 n= 3 = ·2· 3 L 8 3L3 (2π) 0 where the factor 1/8 is due to the fact that only positive integers are used so that the volume in number space involves only one octant. b) If the number density of electrons is n and the ratio of nucleons to electrons is r then the mass density is ρ = n · r · mN = kF3 rmN /3π 2 , so 2 13 3π ρ kF = rmN p c) For electron with momentum k and energy E = k 2 + m2e the velocity is v = k/E. The momentum transfer in a perpendicular collision with a wall is 2k and the number of electrons which strike the wall per second per unit area is 1 1 2 · 3 nv, since all electrons in a cylinder with area A and length vt will strike the wall in time t, where the factor of 12 is because half are moving to the right and half to the left and the factor 13 is because only the component of momentum perpendicular to the wall is relevant. Then the pressure, which is the momentum transfer per second per unit area is p = 2k ·

1 1 1 k2 · nv = n 2 3 3 E

d) The total pressure is given by integrating up to the Fermi momentum, so Z kF Z kF 3 k2 1 k 4 dk d k p p p=2 = 3 2 (2π) 3 k 2 + m2e 3π 0 k 2 + m2e 0 m4 = e2 3π

kF Z m

e

0

m4 = e2 3π

kF

m e p p x4 dx 3 3 m4e 1 3 p 2 − x 1 + x2 + 2) √ x ln(x + = 1 + x 1 + x 3π 2 4 8 8 x2 + 1 0

"

1 kF3 4 m3e

s

k2 3 kF 1 + F2 − me 8 me

s

k2 3 kF 1 + F2 + ln( + me 8 me

s

# kF2 1 + 2) me

s " # kF2 m4e kF3 kF −1 kF = (2 3 − 3 ) 1 + 2 + 3sinh 24π 2 me me me me e) Energy density is Z kF ρ=2 0

m4 = 2e π

kF Z m

e

0

dxx2

p

d3 k p 2 1 k + m2e = 2 (2π)3 π

Z kF 0

dkk 2

p k 2 + m2e

kF

m e 4 p 3 1 m 1 1 1 1 + x2 = 2e x(1 + x2 ) 2 − x(1 + x2 ) 2 − ln(x + 1 + x2 ) π 4 8 8

0

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m4 = 2e π

"

s s # kF2 kF2 1 1 kF 1 kF2 kF ( (1 + 2 − ) 1 + 2 − ln( + 1 + 2) me 4 me 8 me 8 me me s " # kF2 m4e kF kF3 −1 kF = (2 3 + ) 1 + 2 − sinh 8π 2 me me me me

f) Now suppose that kF << me . Then

kF3 kF 1 kF4 kF 3 kF5 1 kF2 1 kF3 m4e (2 − 3 − . . .) + 3( + + . . .) )(1 + − p' 24π 2 m3e me 2 m2e 8 m4e me 6 m3e 40 m5e =

kF5 + ...

15π 2 me

and ρ ' rmN ·

kF3 kF 1 kF2 kF 1 kF3 m3e (2 + )(1 + + . . .) − ( − + . . .)

8π 2 m3e me 2 m2e me 6 m3e =

rmN 3 k + ...

3π 2 me F

Thus kF5 =

3π 2 ρ rmN

53

+ . . . = 15π 2 me p + . . .

20.4 Neutron Stars In the previous exercise we looked at what happens when a star collapses and is stabilized by the electron degeneracy pressure. However, Chandrasekhar showed that if the mass of the star is greater than about 1.4 solar masses, the electron degeneracy pressure is insufficient to stop the implosion. What happens if this is the case? The answer is that the collapse continues, but it can be halted by the neutron degeneracy pressure, i.e., we have a neutron star. We can understand the physics here by arguments similar to those given for the white dwarf, but with important differences. a) At low temperature the neutrons will occupy the low-lying levels, with two neutrons in each. Show that this leads to a Fermi momentum kF given by Z kF dk kF3 n=2 = , (2π)3 3π 2 0 where n is the neutron number per unit volume.

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314

b) Since nearly all the mass of the star is due to nucleons, calculate the Fermi energy in terms of the star density as 2 13 3π ρ kF = . mN c) For neutrons having momentum k show that the pressure on the walls is given by nk 2 p= , 3E p where E = k 2 + m2N ' mN is the neutron energy. d) Integrate over the spectrum and show that Z 2 kF dk k2 p p= 3 0 (2π)3 k 2 + m2N =

kF kF3 m4N kF2 1 −1 kF 2 + 3sinh ) − 3 ) (2( )(1 + , 24π 2 m3N mN m2N mN

which represents the equation of state, i.e., the pressure in terms of the density. e) Calculate the neutron star energy density and show that Z kF q dk ρ = 2i k 2 + m2N (2π)3 0

kF m4 k3 kF k2 1 = N2 (2( F3 ) + )(1 + F2 ) 2 − sinh−1 .

8π mN mN mN mN f) Relating p to ρ yields a complicated equation of motion. However, in the nonrelativistic limit, show that we have a simple relation 53 ρ 1 , p = ρc 5 ρc where m4N ' 6 × 1015 g/cm3 3π 2 is the critical density where kF = mN . ρc =

Again, given the equation of state, the Lane-Emden equation allows the construction of the neutron star and yields

1 2.72 ρ(0) 2 M= 2 MSun r ρc

R = 10.9

ρ(0) ρc

− 16

km .

Thus, a neutron star has a mass of order the solar mass, but is now confined to a

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315

radius of order 10 km!

Exercise 20.4 Solution: Neutron Stars a) Quantize in a box, with one corner at the origin and sides a distance L apart.

Then solutions are of the form nx πx ny πy nz πz ψ(x, y, z) = sin sin sin L L L with nx , ny , nz integers 0, 1, 2..... The momentum is then πq 2 k= nx + n2y + n2z .

L If there are two neutrons per state and the total number of neutrons is N, then we have, using kF = πnmax /L Z kF 3 d k 1 4πn3max N 3 2 = k /3π = 2 n= 3 = ·2· F 3 L 8 3L3 (2π) 0 where the factor 1/8 is due to the fact that only positive integers are used so that the volume in number space involves only one octant. b) If the number density is n then the mass density is ρ = nmN = kF3 mN /3π 2 , so

kF =

3π 2 ρ mN

13

p c) For neutron with momentum k and energy E = k 2 + m2N the velocity is v = k/E. The momentum transfer in a perpendicular collision with a wall is 2k and the number of neutrons which strike the wall per second per unit area is 1 1 2 · 3 nv, since all neutrons in a cylinder with area A and length vt will strike the wall in time t, where the factor of 21 is because half are moving to the right and half to the left and the factor 13 is because only the component of momentum perpendicular to the wall is relevant. Then the pressure, which is the momentum transfer per second per unit area is p = 2k ·

1 1 1 k2 · nv = n 2 3 3 E

d) The total pressure is given by integrating up to the Fermi momentum, so Z kF 3 Z kF d k k2 1 k 4 dk p p p=2 = (2π)3 3 k 2 + m2N 3π 2 0 k 2 + m2N 0 m4 = N2 3π

Z mkF

N

0

m4 = N2 3π

"

x4 dx m4 √ = N2 2 3π x +1

1 kF3 4 m3N

s

mkFN p p 1 3p 3 3 x 1 + x2 − x 1 + x2 + ln(x + 1 + x2 ) 4 8 8

k2 3 kF 1 + F2 − mN 8 mN

0

s

k2 3 kF 1 + F2 + ln( + mN 8 mN

s

# kF2 1+ 2 ) mN

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316

s " # kF2 m4N kF3 kF −1 kF = (2 3 − 3 ) 1 + 2 + 3sinh 24π 2 mN mN mN mN e) Energy density is Z kF ρ=2 0

m4 = N π2

Z mkF

=

N

dxx2

p

0

m4N π2

"

d3 k (2π)3

q

k 2 + m2N =

m4 1 + x2 = N π2

1 π2

Z kF

dkk 2

0

q k 2 + m2N

mkFN p 3 1 1 1 1 x(1 + x2 ) 2 − x(1 + x2 ) 2 − ln(x + 1 + x2 ) 4 8 8 0

s

k2 kF 1 k2 1 1 kF ( (1 + F2 − ) 1 + F2 − ln( + mN 4 mN 8 mN 8 mN

s 1+

kF2 ) m2N

#

s " # kF2 kF m4N kF3 −1 kF = (2 3 + ) 1 + 2 − sinh 8π 2 mN mN mN mN f) Now suppose that kF << mN . Then

m4N kF 1 kF4 kF 3 kF5 kF3 1 kF2 1 kF3 p' − . . .) + 3( + + . . .)

(2 3 − 3 )(1 + − 24π 2 mN mN 2 m2N 8 m4N mN 6 m3N 40 m5N =

kF5 + ...

15π 2 mN

and, defining ρc = m4N /3π 2

kF kF m4N kF3 1 kF2 1 kF3 + . . .) − ( + . . .) ρ' (2 3 + )(1 + − 8π 2 mN mN 2 m2N mN 6 m3N =

mN 3 kF3 k + . . . = ρ + ... c 3π 2 F m3N

Thus kF5 = m5N

ρ ρc

53 + ... =

5p + ... ρc

or 1 p = ρc 5

ρ ρc

35 + ...

20.5 The GZK Limit This problem explains why it is expected that no cosmic ray protons with energies greater than 1020 GeV should be observed.

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317

a) The universe is filled with cosmic microwave background (CMB) radiation consisting of photons at an average temperature 2.72 K. Calculate the energy of these photons in eV. b) Consider an ultrarelativistic proton of energy Ep passing through space filled only with CMB radiation. Calculate the threshold value of Ep = EGZK such that the reaction p + γ → ∆+ (1232) → n + π + can occur. One would not expect cosmic rays from deep space having Ep > EGZK because any such cosmic ray particles would scatter off the CMB photons and be lost. This value is known as the Greisen-Zatsepin-Kuzmin (GZK) cutoff [Gre66].

Exercise 20.5 Solution: The GZK Limit a) Energy is 2.72 K or 2.3 × 10−4 eV b) In rest frame of proton, photon energy must be O 300 MeV, so there is a boost of γ ∼ 3 × 108 eV/2.3 × 10−4 eV = 1.3 × 1012 , so proton energy in our frame is EGZK ' γmp = 1.2 × 1021 eV 20.6 Lifetime of the Sun The solar constant — 1.36 kW/m2 — measures the solar flux incident on the Earth’s surface from the sun. a) As discussed in the text, the major source of energy production in the sun is the “pp cycle” which may be summarized as 4p →4 He + 2e+ + 2νe + 25.7 MeV .

Assuming that the sun has been burning fuel at the same rate since its birth 4.6 billion years ago, estimate the fraction of its proton “fuel” which has been used during this time. b) Assuming that the solar burn rate stays the same, how many years does the sun have remaining before running out of fuel? c) Explain why the number calculated in b) is too large and why the sun (and life on the earth) will die long before this time.

Exercise 20.6 Solution: Lifetime of the Sun a) Solid angle subtended by the earth is ∆Ω =

π · (4 × 103 Mi)2 ' 1.85 × 10−9 (93 × 106 ) Mi2

318

Astrophysics

Total power received by earth is PEarth = (1.36 kW/m2 ) · π · (6.4 × 106 m)2 = 5.57 × 1016 w so total power put out by the sun is Ptot = PEarth × 4π/∆Ω = 5.57 × 1016 w × 4π/1.85 × 10−9 = 3.78 × 1026 w Since each proton consumed releases 6.4 MeV, the number of protons consumed per second is dNp = 3.78 × 1026 j/sec j/6.4 MeV = 3.70 × 1038 sec−1 dt Number of protons in the sun is Ntot = Msun /mp = 1.99 × 103 0 kg/1.67 × 10−27 kg = 1.19 × 1057 so fraction of fuel used so far is f = 3.7 × 1038 sec−1 × 4.6 × 109 yr × 3 × 107 sec/yr/1.19 × 1057 = 4.4% b) Time left would be 4.6 billion years × 95.6/4.4' 96 Billion years c) Helium burning will commence, then carbon burning, etc. long before this time.

21

Beyond the Standard Model Physics No exercises in this chapter.

319

22

Useful information

No exercises in this appendix.

320

23

Quantum Theory

B.1 The Optical Theorem One can use the result that, under an integral, 1 1 =P − iπδ(E − E0 ) E − E0 + i E − E0

where P signifies the principle value integral, to prove both Fermi’s golden rule and the optical theorem.

The counter in a scattering experiment can be considered as a device for measur(+) ing the probability that a particle described by wavefunction ψa (r) will be in state |φb > which takes it into the counter. Here |φb i is an eigenstate of the free Hamiltonian Ĥ0 |φb i = Eb |φb i

(+)

while |ψa i is an eigenstate of the full Hamiltonian Ĥ = Ĥ0 + V̂ Ĥ|ψa(+) i = Ea |ψa(+) i . a) Show that the state |ψa(+) i = |φa i +

1 Ea − Ĥ0 + i

V̂ |ψa(+) i

(+)

has this property. The overlap of state |ψa i with |φb i is then found to be

1 (+) (+) < φb |ψa >= hφb | |φa > + V̂ |ψa > Ea − Eb + i = δab +

1 < φb |T̂ |φa > Ea − Eb + i (+)

where < φb |T̂ |φa >=< φb |V̂ |ψa (+)

b) If wba = < φb |ψa

>

2

> is the transition matrix element.

then the transition rate is

dwba = < φ̇b |ψa(+) > + < φb |ψ̇a(+) > < φb |ψa(+) >∗ +c.c. . dt (+)

Use the equations satisfied by |ψa i and |φb i to show h i dwba = −i < φb |V̂ |ψa(+) >< φb |ψa(+) >∗ −c.c. . dt 321

Quantum Theory

322

c) Show that if b 6= a

dwba 2 = 2πδ(Ea − Eb ) |F(Ea )| dt

which is Fermi’s Golden Rule.

P d) Use the fact that b dwdtba = 0 to derive pi Im < φa |V̂ |ψa(+) >∼ σa =− 2m where σa is the total scattering cross section. Since fk (θ) = −

m < φb |V̂ |ψa(+) > 2π

is the scattering amplitude, we have Imfk (0) =

pi σa 4π

which is the standard optical theorem.

Problem B.1 Solution: The Optical Theorem a) We have (Ea − Ĥ0 )|ψa(+) >= (Ea − Ĥ0 )|φa > +(Ea − Ĥ0 ) = V̂ |ψa(+) >

so

1 (Ea − Ĥ0 ) + i

V̂ |ψa(+) >

(Ea − Ĥ0 − V̂ )|ψa(+) >= 0

b) Note that i|φ̇a >= Ĥ0 |φa >

and i|ψ̇a(+) >= (Ĥ0 + V̂ )|ψa(+) >

Then

dwba = < φ̇b |ψa(+) > + < φb |ψ̇a(+) > < φb |ψa(+) > +c.c. dt

= i < φb |Ĥ0 |ψa(+) > −i < φb |Ĥ0 + V̂ |ψa(+) > < φb |ψa(+) > +c.c. h i = −i < φb |V̂ |ψa(+) >< φb |ψa(+) > −c.c. c) Then if b 6= a

dwba 1 (+) (+) ∗ = −i < φb |V̂ |ψa > P + iπδ(Ea − Eb ) < φb |V̂ |ψa > −c.c dt Ea − Eb = 2πδ(Eb − Ea )| < φb |V̂ |ψa(+) > |2

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323

d) Also since

P dwba b

dt

= 0 we have X dwba dwaa =− dt dt b6=a

so h i X −i < φa |V̂ |ψa(+) > −c.c. = − 2πδ(Eb − Ea )| < φb |V̂ |ψa(+) > |2 b

pi 2π · σa while LHS= −2Im m · fk (0) so we have Here RHS= − m a a

pi σa 4π B.2 Effective Lagrangian for a Constant B or E Field Imfk (0) =

Consider a charged scalar field interacting with an external magnetic field B = Bêz for which the corresponding Klein-Gordon equation reads (D2 + m2 )φ(x) = 0 where Dµ = ∂µ + ieAµ is the covariant derivative and Aµ (x) is the vector potential associated with the magnetic field. The effective action is then given by R 4 ∗ R 2 2 [dφ(x)][dφ∗ (x)]ei d xφ (x)(D +m )φ(x) iSeff (B) R e = R 4 ∗ 2 [dφ(x)][dφ∗ (x)]ei d xφ (x)( +m )φ(x) or

D 2 + m2 Seff (B) = iTr log . + m2 Of course, the operation “Tr log” applied to a differential operator is not a trivial one, and the purpose of this problem is to evaluate this quantity for the case at hand. a) Demonstrate that log

a = b

so that Seff (B) = iTr

Z ∞ 0

Z ∞ 0

ds −bs (e − e−as ) s

2 ds −m2 s − s e (e − e−D s ) . s

In order to evaluate the trace we require a complete set of solutions to the equations D2 φn (x, y, z, t) = λn φn (x, y, z, t) φn (x, y, z, t) = κn φn (x, y, z, t) .

Then we may write Seff (B) = i

X Z ∞ ds n

0

s

2

e−m s (e−κn s − e−λn s ) .

Quantum Theory

324

b) Show that when B=0 the eigenstates are given by φ(x, y, z, t) = exp i(kx x + ky y + kz z − kt t) with eigenvalues κn = −kt2 + kx2 + ky2 + kz2 . c) For the gauge choice Aµ = (0, Bxêy ) show that the eigenstates in the presence of the magnetic field become φ(x, y, z, t) = exp i(kz z + ky y − kt t)ψn (x −

ky ), eB

where ψn (x) is an eigenstate of the harmonic oscillator Hamiltonian with frequency ω = eB and that the corresponding eigenvalues are given by λn = −kt2 + kz2 + eB(2n + 1) .

Now rotate to Euclidean space kt → ik0 so that κ2n = k02 + kx2 + ky2 + kz2 λ2n = k02 + kz2 + eB(2n + 1) and evaluate the trace using box quantization. Taking a box with sides L1 , L2 , L3 and a time interval T, we have Z ∞ 4 X d k κ: → L1 L2 L3 T 4 (2π) −∞ n λ:

X n

→ L2 L3 T

Z eBL1 dky 0

∞ dk0 dkz X 2 −∞ (2π) n=0

Z ∞

where the integration on ky is over all values with x0 = x−ky /eB kept positive. d) Now evaluate the effective action Z ∞ Z ds ∞ dk0 dkz −(m2 +k02 +kz2 )s Seff (B) = L1 L2 L3 T e s −∞ (2π)2 0 "

∞

eB X −eB(2n+1)s e − × 2π n=0

Z ∞

dkx dky −(kx2 +ky2 )s e 2 −∞ (2π)

#

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325

and show that Z ∞

1 Seff (B) = L1 L2 L3 T 16π 2

0

ds −m2 s e s3

eBs −1 sinh eBs

.

The “physics” of this result can be seen via an alternative derivation—the effective action is simply the shift in the vacuum energy due to the presence of the magnetic field times the interaction time. Since the zero point energy P is given by n ωn (i.e., 21 ωn associated with the positively charged states and 1 2 ωn with the negatively charged states) we have X Seff (B) = T (ωn (B) − ωn (0)) . n

e) Show that ωn (B) =

p

ωn (0) =

kz2 + eB(2n + 1) + m2

q

kx2 + ky2 + kz2 + m2

so that Seff (B) = L2 L3 T

×

"Z eBL1 0

− L1

Z ∞

dkz −∞ 2π

dkx X p 2 kz + eB(2n + 1) + m2 2π n

Z ∞

dky dkx q 2 kx + ky2 + kz2 + m2 2 −∞ (2π)

.

f) Use the representation √

1 a= √ 2 π

Z ∞ 0

ds −as 3 e s2

to prove the identity of the two expressions for Seff (B).

Now suppose we have a constant electric field E = Eêz . g) Working in the gauge φ = −Ez, show that the Euclidean Klein-Gordon equation becomes [(p̂0 − eφ)2 + p̂2 + m2 ]χn (x, t) = λn χn (x, t) with χn (x, t) = eip0 t+ipx x+ipy y ψn (z −

p0 ) eE

h) Then proceeding as before and making the substitution E → iE to rotate back to Minkowski coordinates, show that

Z ∞ 2 1 ds eEs Seff (E) = −T L1 L2 L3 e−m s .

16π 2 0 s3 sin eEs − 1

Quantum Theory

326

Although the forms of Seff (E) and Seff (B) appear similar, there is an important difference in that sin eEs has a series of zeroes at the points s = nπ/eE along the real axis so that contour integration must be used. i) Show that the effective action becomes complex, with Im(Seff (E)) = T L1 L2 L3

∞

nπm2 e2 E 2 X (−)n+1 exp(− ) 2 2 16π n=1 n eE

indicating that in the presence of sufficiently strong electric fields, the vacuum breaks down and e+ e− pairs are produced. Sometimes this is called “sparking the vacuum” and this vacuum breakdown is the analog of the Hawking effect in gravity.

Problem B.2 Solution: Effective Lagrangian for a Constant B or E Field a) We have Z ∞

dse−as =

0

so dn d(−a)n

Z ∞ 0

−as

dse

=

Z ∞

1 a

dssn e−as =

0

n! Γ(n + 1) = an+1 an+1

Now set n − 1 = . Then, using Γ( ) = 1 + . . . and a− = 1 − ln a + . . . we have Z ∞ 1 ds −as e = − ln a + . . . s

0 so Z ∞ ds −bs a (e − e−as ) = ln s b 0 b) For plane wave with box normalization φ(x) = exp(−ikt t + kx x + ky y + kz z) and φ(x) = (−kt2 + k2 )φ(x) c) Now change → D2 = ∂02 − ∂x2 − (∂y + iexB)2 − ∂z2 and write φ(t, x, y, z) = exp(−ikt t + iky y + ikz z)χ(x) so D2 φ(t, x, y, z) = (−kt2 + kz2 − ∂x2 + (ky − exB)2 ) exp(−ikt t + iky y + ikz z)χ(x)

Quantum Theory

327

k

Defining x̃ = x − eBy we require

1 (−∂x̃2 + e2 B 2 x̃2 )χn (x̃) = En χn (x̃) 2 which is a harmonic oscillator with frequency eB and unit mass. Since the eigenvalues are En = eB(n + 21 ) we have D2 φ(t, x, y, z) = (−kt2 + kz2 + eB(2n + 1))φ(t, x, y, z)

d) Find effective action Sef f (B) = L1 L2 L3 T

Z ∞ 0

"

∞

ds s

Z ∞

eB X × exp(−eB(2n + 1)s) − 2π n=0 Note that

dk0 dkz exp(−(k02 + kz2 + m2 )s) 2 (2π) −∞ Z ∞

dkx −∞ 2π

Z ∞

dky exp(−(kx2 + ky2 )s) −∞ 2π

#

Z ∞

1 dki exp(−ki2 s) = √ 2 πs −∞ 2π

and ∞ X

exp(−eB(2n+1)s) =

n=0

so

exp(−eBs) 1 1 = = 1 − exp(−2eBs) exp(eBs) − exp(−eBs) 2sinheBs

Sef f (B) = L1 L2 L3 T

1 16π 2

Z ∞ 0

ds −m2 s eBs e ( − 1) s3 sinheBs

e) Alternate representation is "Z Z ∞ Z ∞ ∞ eBL1 dkz 1 ds dkx X √ Sef f (B) = L2 L3 T exp(−(m2 + kz2 + eB(2n + 1)) 3 2π n=0 s2 −∞ 2π 2 π 0 0 −L1

Z ∞

dkx −∞ 2π

Z ∞

dky exp(−(m2 + kx2 + ky2 + kz2 )s) −∞ 2π

f) In Euclidean spacetime result for E field is simple permutation of B field result with x, y → 0, z so solution is now φ(t, x, y, z) = exp(ikt t + ikx x + iky y)φn (z −

k0 ) eE

where again φn is a unit mass harmonic oscillator solution. g) Have then Euclidean representation Z ∞ 1 ds −m2 s eEs eu Sef f (E) = L1 L2 L3 T e ( − 1) 2 16π 0 s3 sinheEs However, we need the Minkowski space result, so must rotate t → it. Since E = −∇φ and φ → iφ we require E → iE so, using sinhieEs = sin eBs Z ∞ 1 ds −m2 s eEs min Sef f (E) = L1 L2 L3 T e ( − 1) 2 16π 0 s3 sin eEs

Quantum Theory

328

h) Now when performing integration there are poles along the real axis when eEs = nπ so must deform the contour by using small half circle around each pole nπ having radius . Then write s = eE + eiθ and use ds = i eiθdθ . By Taylor’s nπ + eiθ ) ' sin nπ + cos nπeE eiθ = (−)n eE eiθ theorem sin eE( eE Z nπ Z 0 eE + ds eE i eiθ dθ nπ 2 −m2 s eEs e → exp(− m ) nπ 2 3 n eE ei θ nπ ) s sin eEs ( (−) eE −π eE eE − = −i(−)n+1

nπm2 e2 E 2 exp(− ) 2 n eE

so that effective action now has imaginary part. Now sum over poles to get total ∞ nπm2 e2 E 2 X (−)n+1 exp(− ) Im Sef f (E) = T L1 L2 L3 16π 2 n=1 n2 eE

B.3 Euler-Heisenberg Lagrangian and Photon-photon Scattering Now repeat the previous problem but with parallel electric and magnetic fields. a) Show that the effective action becomes

Z ∞ e2 EB 1 ds −m2 s e −1 Lef f [E, B] = − 16π 2 0 s3 sinh(eBs) sin(eEs) b) Expanding in powers of E, B show that Z ∞ Z ∞ 2 α ds −m2 s α2 2 2 2 2 2 2 2 (E −B ) e + 7(E − B ) + 4E B L= dsse−m s +. . .

24π s 360 0 0 Here the first piece, which is divergent and proportional to Fµν F µν = 2(B 2 − E 2 ), is absorbed into renormalization of the field tensor, while the second piece is finite. c) Write this finite component in terms of the vector potential Aµ and show that it has the general form i α2 h µν 2 µν 2 7(F F ) + 4(F F̃ ) + ...

Lef f = µν µν 1440m4 where F̃ µν =

1 µναβ

Fαβ 2

is the dual tensor. This result was first derived by W. Heisenberg and H. Euler and is called the Euler-Heisenberg Lagrangian.

We can use this result to derive the cross section for low energy (ω << m) photonphoton scattering.

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329

d) Summing over final polarizations and averaging over initial polarizations show that 1X 556α4 2 2 (s t + s2 u2 + t2 u2 ) |Ampγγ |2 = 4 8100m8 pol.

e) Use the result of d) to calculate the total photon-photon scattering cross section and show that 973α4 ω 6 σtot = 10125π m8

Problem B.3 Solution: Euler-Heisenberg Lagrangian and Photon-photon Scattering a) Since effective action must reduce to forms from previous problem if either E or B vanishes and must itself vanish if both fields go to zero, we must have Z ∞ 1 ds −m2 s eEs eBs Sef f (E, B) = L1 L2 L3 T e ( − 1) 16π 2 0 s3 sin eEs sinheBs b) Expand in E,B Z ∞

1 1 2 2 2 1 4 4 4 1 − 3! e E s + 5! e E s + ...

0

1 − 1 1 2 2 2 1 4 4 4 1 + 3! e B s + 5! e B s + ...

Z ∞ ds −m2 s e2 2 1 e 1 + (E − B 2 )s2 = L1 L2 L3 T 16π 2 0 s3 3!

e4 4 e4 4 4 4 4 2 2 4 (E + B − E B )s + . . . − 1 − (E + B )s + 5!

(3!)2 Z ∞ Z ∞ 2 e2 ds −m2 s e4 2 2 4 4 2 2 = (E −B ) e + (E +B −10E B ) dsse−m s +. . .

2 2 96π s 16 · 360π 0 0 Z ∞ Z ∞ 2 α ds −m2 s α2 = (E 2 −B 2 ) e + (7(E 2 −B 2 )2 +4E 2 B 2 ) dsse−m s +. . .

24π s 360 0 0

Sef f (E, B) = L1 L2 L3 T

1 16π 2

ds −m2 s e s3

c) Write in terms of Lorentz invariants. We note Z ∞ 1 dss exp(−m2 s) = 4 m 0 Also Fµν F µν = 2(B 2 − E 2 )

and Fµν F̃ µν = 2E · B

so LEH ef f =

h i α2 7(Fµν F µν )2 + 4(Fµν F̃ µν )2 4 4 · 360m

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330

e) Since s = 4ω 2 , t = −2ω 2 (1 − cos]theta), u = −2ω 2 (1 + cos]theta) we have s2 t2 +s2 u2 +u2 t2 = 16ω 4 ·4ω 4 (1−cos θ)2 +16ω 4 ·4ω 4 (1+cos θ)2 +4ω 4 ·4ω 4 (1−cos2 θ)

= ω 8 64 · 2(1 + cos2 θ) + 16(1 − cos2 θ) = ω 8 (144 + 112 cos2 θ) Then 1 dσ = 2

Z

d3 k4 1X d3 k3 (2π)4 δ 4 (k1 + k2 − k3 − k4 ) |Ampγγ |2 0 0 3 3 (2π) 2k3 (2π) 2k4 4 pol.

=

1 dΩ 556 1 ω 8 (144 + 112 cos2 θ) 8π 2 (2ω)2 2 8100m8

so σ=

4πω 6 556α4 112 (144 + ) 64π 2 m8 8100 3

References

[Alb88] W. M. Alberico, A. Molinari, T. W. Donnelly, L. Kronenberg and J. W. Van Orden, Phys. Rev. C38, 1801 (1988).

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