Chapter 4 Symmetries And Conservation Laws
Chapter 4 4.1 Why Conservation And Symmetry 4.2 Angular Momentum 4.3 Flavour Symmetry : Isospin 4.4 Parity and Charge Conjugation 4.5 Interactive Exercise
4.3 Flavour Symmetry: Isospin
Symmetry and Conservation Laws
Particle Physics
Flavor Symmetries: Isospin Due to the observed property of the neutron, proton mass p=938.28 MeV/c2, mn=939.57 MeV/c2, Heisenberg proposed that we regard them as two 'states' of a single particle, the nucleon. In other words, the proton and neutron would be identical, or the strong forces experienced by protons and neutrons are identical. Express the nucleon as a 2component column matrix
→
→
To appreciate the analogy with spin S, define isospin I , which is a vector in an abstract 'isospin space', with components I1, I2 and I3. Thus the nucleon with isospin ½, I3 component has eigenvalue ½(the proton) and (½) for the neutron
the proton is, 'isospin up'; neutron is 'isospin down'.
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Just as electrical forces are invariant under rotation is ordinary configuration space, the strong interactions are invariant under rotation in isospin space. A rotation through 180○ about axis I, in isospin space converts protons into neutrons and viceverse. Noether's theorem states that isospin is conserved in all strong interactions, just as angular momentum is conserved in ordinary space.
Isospin Classification The
analog
of
isospin
invariance
under
strong
interactions, could be seen in the eightfold diagrams. So to each of these multiplets, assign an isospin I, and each member of the multiplet has an I3 value. For example (a) pions have I = 1: (b) For the Λ, I = 0 gives Λ= (c)
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Method: a) To find the isospin of a multiplet, count the number of particles it contains, which should be (2 I+1). b) I3 is related to charge Q of the particle. Assign the maximum value I3 = I to the member of the multiplet with highest charge and complete the rest in order of decreasing Q.
Dynamical Application Of Isospin For two nucleons, the addition of angular momenta gives a total isospin of 1 or 0. i.e. a symmetric isotriplet:
and an antisymmetric isosinglet:
Experimentally, the neutron and proton form a single bound state, the deuteron (d); whereas there is no bound state of two protons or two neutrons. Hence the deuteron must be an isosinglet. If the deuteron were a triplet, all three states would have to occur, to since they differ only by a rotation in isospin space. This implies that there is a strong interaction in the I = 0 state, but not in the I = 1 state. Dayalbagh Educational Institute
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Symmetries: Spin 1/2 systems Some generalities Spin1/2 systems are often studied Example: electron and its spin, isospin, ...
in
physics.
Spinstatistics theorem suggests that such systems are fermionic in nature. Basic building blocks of matter (quarks & leptons) are spin1/2. Simple representation: ↑⟩ = s=1/2, sz= 1/2 ⟩, ↓⟩ = s=1/2, sz= 1/2 ⟩,
Adding two spin 1/2 Often, it is important to add spins Examples: bound states of spin1/2 fermions, spinorbit coupling, etc.. If two spin1/2 systems are added, the following objects can emerge: ↑↑⟩ , ↑↓⟩ , ↓↑⟩ , ↓↓⟩ Naively, they have spin 1, 0, or 1, respectively. But: Need to distinguish total spin s and its projection onto the “measurement axis” sz (here, z has been chosen for simplicity) Then the truly relevant states are for s=1 (triplet) ↑↑⟩ = s = 1, sz= 1⟩ , 1/√2 (↑↓ + ↓↑)⟩ = s = 1, sz= 0⟩ ↓↓⟩ = s = 1, sz= 1⟩ Dayalbagh Educational Institute
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and for s = 0 (singlet) 1/√2 (↑↓  ↓↑)⟩ = s = 0, sz= 0⟩ Note: the triplets are symmetric, the singlet is antisymmetric. Catchy: 2 ⨂ 2 = 3 ⊕ 1
ClebschGordan Coefficients
The coefficients in front of the new s , sz⟩ states can be calculated (or looked up). They go under the name of ClebschGordan coefficients. Formally speaking, they are defined as follows:
⟨s1, s1z; s2, s2zs1, s2; s, sz⟩ indicating that two spin systems s1 and s2 are added to form a new spin system with total spin s. Obviously, it is not only the total spin of each system that counts here, but also its orientation. This is typically indicated through a “magnetic” quantum number, m, replacing sz in the literature. Special cases: For s = 0:
s1 m1
⟨s1, m1; s2, m2s1, s2; 0, 0⟩ = δs1 , s2δm1,m2 ClebschGordan tables for two cases: (a squareroot over each coefficient is implied)
Table 1 Dayalbagh Educational Institute
Notation: m1
m2
m1
m2
· · ·
· · ·
(1)
√2s2 +1 J
J
···
M
M
···
Coefficients
8
Symmetry and Conservation Laws
Table 2
Particle Physics
Table 3
From spin to isospin Who carries isospin? Remember Heisenberg's proposal: p and n are just two manifestations of the same particle, the nucleon. Identify them with the isospinup and isospindown states of the nucleon:  p⟩ =  1/2 , 1/2⟩,  n⟩ =  1/2 , 1/2⟩ Catch: Isospin conserved in strong interactions! Will dwell on that a bit: Play with pions, nucleons and Delta's. (Note: Multiplicity in each multiplet: 2I+1). Dynamical implications: Bound states (deuteron) Add two nucleons: can have isosinglet and isotriplet. 0, 0⟩ = 1/√2 pn – np⟩ 1, 1⟩ = pp⟩ , 1, 0⟩ = 1/√2 pn + np⟩, 1, 1⟩ = nn⟩ No pp, nnbound states ⇨ deuteron = isosinglet !!! Consider processes (+ their isospin amplitudes, below) p + p → d + π+
1 1, 1⟩ → 1, 1⟩
p + n → d + π0
n + n → d + π
1/√2 1/√2 (1,0⟩ + 0,0⟩) → 1,0⟩
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1 1, 1⟩ → 1, 1⟩ 9
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Who carries isospin? Nucleons in isospin notation:  p⟩ =  1/2 , 1/2⟩,  n⟩ =  1/2 , 1/2⟩ Pions in isospin notation:  π+⟩ = 1, 1⟩,  π0⟩ = 1, 0⟩,  π⟩ = 1, 1⟩ Delta's in isospin notation: Δ++⟩ = 3/2, 3/2⟩, Δ+⟩ = 3/2, 1/2⟩, Δ0⟩ = 3/2, 1/2⟩, Δ⟩ = 3/2, 3/2⟩
Isospin and scattering amplitudes Use isospin for pionnucleon scattering amplitudes Elastic processes a) π+ + p → π+ + p b) π0 + p → π0 + p c) π + p → π + p d) π+ + n → π+ + n e) π0 + n → π0 + n f) π + n → π + n Charge exchange processes g) π+ + n → π0 + p h) π0 + p → π+ + n i) π0 + n → π + p j) π + p → π0 + n
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Isospin and scattering amplitudes Use isospin for pionnucleon scattering amplitudes Remember: pions and nucleons are isopin1 and 1/2. The total isospin is either 1/2 or 3/2 and thus, there are only two independent amplitudes: M3 and M1 Use ClebschGordan coefficients: π+ + p = 3/2, 3/2⟩ π0 + p = √2/3 3/2, 1/2⟩ 1/√3 1/2, 1/2⟩ π + p = 1/√3 3/2, 1/2⟩  √2/3 1/2, 1/2⟩ π+ + n = 1/√3 3/2, 1/2⟩ + √2/3 1/2, 1/2⟩ π0 + n = √2/3 3/2, 1/2⟩ +1/√3 1/2, 1/2⟩ π + n = 3/2, 3/2⟩ Then: Reactions (a) and (f) are pure 3/2: a) π+ + p → π+ + p f) π + n → π + n Other reactions are mixtures (coefficients given by the ClebschGordans), e.g. c) π + p → π + p
Mc = 1/3M3 + 2/3M1
j) π + p → π0 + n
Mj = √2/3M3  √2/3M1
Therefore, the cross sections behave like σa : σc : σj = 9M32 : M3 + 2M12 : 2M3  M12 At a c.m. energy of 1232 MeV there is a in the pionnucleon scattering cross discovered by Fermi in 1951. There, nucleon form a shortlived resonance which we know to carry I = 3/2. Dayalbagh Educational Institute
dramatic bump section, first the pion and state, the Δ 11
Symmetry and Conservation Laws
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At c.m. energies around the Δmass, one can expect that M3 ≪ M1, and therefore, there σa : σc : σj ≈ 9 : 1 : 2 Experimentally, it is simpler to combine (c) and ( j), leading to σtot (π+ + p) σtot (π  p) = 3
Fig 1 Dayalbagh Educational Institute
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Isospin and Gparity How does this work for the mesons (the pions)? Pions = bound states of a quark and an antiquark, so naively: “Just add the isospins like the spins”. But: Rules of spin addition not sufficient. How to “bar” a spin? Problem: want to preserve some symmetries like charge conjugation under “barring”. Gparity (a grouptheory construct) demands:
ˆ π⟩ = π⟩, G
ˆ nπ⟩ = (1)nnπ⟩ G
conserved quantum number in strong interactions. Altogether: The pion (isospin=1) multiplet reads π+⟩ = ud⟩, π0⟩ = 1/√2uu  dd⟩, π⟩ = du⟩. The unexpected minussign in the neutral pion (compare with spin) is due to the Gparity acting on the quarks and antiquarks (the former have positive, the latter negative Gparity).
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The eightfold way: Some SU(3) relations Why SU(3)? In isospin, there are two quarks related by symmetry, u⟩ = 1/2, 1/2⟩ and d ⟩ = 1/2, 1/2⟩ The group related to this is the spin group, or SU(2). Its generators are the Pauli matrices, 0 1
σ1, 2, 3 =
1 0
, 0 i , i 0
1 0
0 1
The pions can be identified with σ3 and the two linear combinations (of definite charge) σ± = 1/√2 (σ1 ± iσ2) For three states u⟩ , d⟩ , s⟩, similarly related through a symmetry, one could think about the group SU(3). Its generators are the GellMann matrices. In SU(3), the mesons can be connected to suitable linear combinations of the GellMann matrices (see next slide) Note: QCD's gauge group is also SU(3). differentiate between SU(3) of flavour (up, down, strange) and SU(3) of colour (red, green, blue), although group theory is the same! The GellMann matrices λ1, 2, 3 =
0 1 0 1 0 0 0 0 0
,
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0 i 0 i 0 0 0 0 0
,
1 0 0 0 1 0 0 0 0
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0 0 i 0 0 0 i 0 0
λ4, 5, 6 =
0 0 1 0 0 0 1 0 0
λ7, 8 =
0 0 0 0 0 i , 1/√3 0 i 0
,
,
0 0 0 0 0 1 0 1 0
1 0 0 0 1 0 0 0 2
Singletoctet mixing Note: In the meson sector, also a “singlet meson” bit contributes, with a wave function of the form ψ1⟩ = 1/√3 (uu⟩ + dd⟩ + ss⟩) It could be realized through a unit matrix. Typically there is a mixing with octet wave functions, most notably examples are the η – η' and the ω – Φ mixing in the pseudo scalar and vector multiplet. So, typically, there are nine mesons per SU(3)multiplet.
The Pseudo Scalar Mesons Here, the spins antialign 1/√2 ↑↓  ↓↑⟩ K0⟩ = ds⟩
K +⟩ = us⟩
π⟩ = du⟩
π+⟩ = ud⟩
K⟩ = su⟩
K0⟩ = sd⟩
π0⟩ = 1/√2 dd – uu⟩
K0 η π0 η'

π
KQ=1
S=+1
K+
K0 Q=0
η⟩ = cosθ /√3 uu + dd + ss⟩ + sinθ /√3 uu  dd  2ss⟩
π+
S=0 S=1
Q=+1 Fig 2
η'⟩ = sinθ /√3 uu + dd + ss⟩ + cosθ /√3 uu  dd  2ss⟩
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The vector mesons Here, the spins align ↑↑⟩, 1/√2 ↑↓ + ↓↑⟩, ↑↑⟩ , ↓↓⟩ K*0⟩ = ds⟩ 
K* +⟩ = us⟩
K*0
+
ρ ⟩ = du⟩
ρ ⟩ = ud⟩
K*⟩ = su⟩
K*0⟩ = sd⟩
ρ
ρ0⟩ = 1/√2 uu – dd⟩ ω⟩ = 1/√2 uu + dd⟩

ω ρ0 φ
ρ
+
S=0 S=1
K*0
K*Q=1
Φ⟩ = ss⟩
S=+1
K*+
Q=+1
Q=0
Fig 3
An alternative look Example for colour: Basic equation: 3 ⨂ 3 = 8 ⊕ 1
This should explain the eight gluons and the absence of the ninth (the singlet) one. Similar for the mesons, just replace colour with flavour. B
R ⨂ R
R
B
G
G
GB
R
≡
GR
Fig 4
g3 g8 BR
B Fig 5 Dayalbagh Educational Institute
RB
RG BG
⊕
gsing 16
Symmetry and Conservation Laws
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Eightfold Way The symmetry group is SU(3): the octets constitute 8dimensional representation of SU(3), decuplet or 10dimensional representation. In this case, no naturally occurring particles fall in the fundamental (3dimension) representation of SU(3), as the nucleons fall for SU(2). In
quarks,
the
u,
d,
s
form
a
3dimensional
representation of SU(3) which breaks down into an isodoublet (u, d) and isosinglet (s) under SU(2). This picture shifted to SU(6) for all six quarks. Isospin, SU(2) is a very 'good symmetry', since the members of isospin multiplet differ in mass by almost 2 or 3%. But the eightfold way, SU(3) is a 'broken' symmetry: mass splittings within the baryon are around 40%. This symmetry breaking increases with inclusions of charm, bottom and top quarks.
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Question 1 The deuteron, symbol d, is a bound state of a proton and neutron with total strong isospin quantum number I = 0. Using the appropriate ClebschGordan coefficients, calculate the relative rates for the following reactions which occur via the strong force. p + p → d + π+ n + p → d + π0
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Answer
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Symmetry and Conservation Laws
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Answer 1 The isospin assignments of the various particles in the question are as follows: d
p
n
π+
π0
π
I
0
I
½
½
I
1
1
1
I3
0
I3
+½
½
I3
+1
0
1
From the ClebschGordan tables we get the following wave functions (in form (I, I3)): p + p ≡ (1, 1) n + p ≡ √½ (1, 0)  √½ (0, 0) d + π+ ≡ (1, 1) d + π0 ≡ (1, 0) (since the deuteron has zero isospin the result of adding it to another particle is to just get the isospin of that particle). For the relative overlaps (matrix elements) and weights we get: Matrix Element
Rate
p + p : d + π+
1
1
n + p : d + π0
√½
½
Thus the first reaction has twice the rate of the second.
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Question 2 A 4He nucleus is a bound state of two protons and two neutrons. What possible strong isospin states (I and I3) could in principle this nucleus have? Explain.
Answer
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Answer 2 Each proton has I3 = +1/2 and each neutron has I3 = 1/2 so a system consisting of two of each will have I3 = 1/2 + 1/2 1/2  1/2 = 0. The two protons are in a (1, 1) state and the two neutrons a (1, 1) state. So the possibilities are given by combining (1, 1) with (1, 1). We use the 1 x 1 table:
So the 4He nucleus could be in a (2, 0), (1, 0) or (0, 0) state. Remember: One combines isospin vectorially. This is why there are more possibilities that just (2, 0). Note that one could have combined two p + n pairs instead. This is a little more involved but leads to the same result.
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Question 3 If you are told that experimentally the following reaction d + d â†’ 4He + Ď€0 is forbidden on grounds of strong isospin conservation, what additional information does this give you about the strong isospin state of the 4He nucleus? You can find a table of ClebschGordan coefficients here.
Answer
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Answer 3 Since the deuteron has isospin zero, a system of two deuterons also has isospin zero. So the question then becomes, if the given reaction is forbidden, what isospin states can be combined with that of a Ď€0 (1, 0) to give zero total isospin? Those isospin states are then ruled out. Of the three possibilities found in part (b), only the (1, 0) state can combine with a (1, 0) state to give a (0, 0) state. So this is ruled out, and we then conclude that 4He can only be in a (2, 0) or (0, 0) state.
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Question 4 a) The deuteron, symbol d, is a bound state of a proton and neutron with total strong isospin quantum number I = 0. Using the appropriate ClebschGordan coefficients, calculate the relative rates for the following reactions which occur via the strong force. p + p → d + π+ n + p → d + π0 b) A 4He nucleus is a bound state of two protons and two neutrons. What possible strong isospin states (I and I3) could in principle this nucleus have? Explain. c) If you are told that experimentally the following reaction d + d → 4He + π0 is forbidden on grounds of strong isospin conservation, what additional information does this give you about the strong isospin state of the 4He nucleus? You can find a table of ClebschGordan coefficients here.
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Solution 4 The isospin assignments of the various particles in the question are as follows: d
p
n
π+
π0
π
I
0
I
½
½
I
1
1
1
I3
0
I3
+½
½
I3
+1
0
1
a) From the ClebschGordan tables we get the following wave functions (in form (I, I3)): p + p ≡ (1, 1) n + p ≡ √½ (1, 0)  √½ (0, 0) d + π+ ≡ (1, 1) d + π0 ≡ (1, 0) (since the deuteron has zero isospin the result of adding it to another particle is to just get the isospin of that particle). For the relative overlaps (matrix elements) and weights we get: Matrix Element
Rate
p + p : d + π+
1
1
n + p : d + π0
√½
½
Thus the first reaction has twice the rate of the second.
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b) Each proton has I3 = +1/2 and each neutron has I3 = 1/2 so a system consisting of two of each will have I3 = 1/2 + 1/2  1/2  1/2 = 0. The two protons are in a (1, 1) state and the two neutrons a (1, 1) state. So the possibilities are given by combining (1, 1) with (1, 1). We use the 1 x 1 table:
So the 4He nucleus could be in a (2, 0), (1, 0) or (0, 0) state. Remember: One combines isospin vectorially. This is why there are more possibilities that just (2, 0). Note that one could have combined two p + n pairs instead. This is a little more involved but leads to the same result. c) Since the deuteron has isospin zero, a system of two deuterons also has isospin zero. So the question then becomes, if the given reaction is forbidden, what isospin states can be combined with that of a Ď€0 (1, 0) to give zero total isospin? Those isospin states are then ruled out. Of the three possibilities found in part (b), only the (1, 0) state can combine with a (1, 0) state to give a (0, 0) state. So this is ruled out, and we then conclude that 4He can only be in a (2, 0) or (0, 0) state.
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