Quantum Mechanics An Experimentalist's Approach 1st edition Eugene D. Commins Instructor Resources w

SOLUTION MANUAL COMPLETE VERSION

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

Chapter 2 2.1a) Let A u = λ u . Then u A u = λ u u . Thus,

u A u = u A† u = λ * u u If A† = −A , − u A u = λ* u u = −λ u u

Thus:

λ* = −λ so λ is imaginary or zero.

2.1b) If A† = A, B† = B , then

[ A, B ]† = ( AB − BA )† = B† A† − A† B†

= BA − AB = − [ A, B ]

Thus [ A, B ] is skew-Hermitian or zero.

2.1c) Suppose that A, B, I are n × n matrices, where n is a finite positive integer. Then if: AB − BA = iqI we have: trace(AB − BA) = 0 but: trace ( iqI ) = iqn which is a contradiction.

2.2. Straightforward computation using the method described in Sec. 2.5 of the text yields the following results:

n=0:

φ0 (x) = 1

n = 1:

φ1 (x) = 3x

n=2:

1⎞ ⎛3 φ2 (x) = 5 ⎜ x 2 − ⎟ ⎝2 2⎠

3 ⎞ ⎛5 φ 3 (x) = 7 ⎜ x 3 − x ⎟ ⎝2 2 ⎠ Apart from normalization, these are the Legendre polynomials: P0 ( x ) = 1 P1 (x) = x 3 1 P2 (x) = x 2 − 2 2 5 3 P3 ( x ) = x 3 − x 2 2 n = 3:

1

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

2.3. Consider the double integral: x2

I = ∫ dx ∫ x1

2 f (x)g(y) − f (y)g(x)] dy [ x x2

(1)

1

Obviously, I ≥ 0. Expanding the integrand in (1), we have: x2

x2

x2

x2

x2

x2

x1

x1

∫ f (x)dx ∫ g (y)dy + ∫ g (x)dx ∫ f (y)dy ≥ 2 ∫ f (x)g(x)dx ∫ f (y)g(y)dy 2

x1

2

x1

2

x1

x1

which immediately yields: x2

x2

2

(

x2

∫ f (x)dx ∫ g (x)dx ≥ ∫ f (x)g(x)dx 2

x1

2

x1

x1

)

2

(2)

2.4a) Let the (orthonormal) eigenvectors of G be ui and the corresponding (positive) eigenvalues be gi = γ i2 where the γ i are real numbers. Similarly, let the eigenvectors of H be w j and the corresponding eigenvalues be h j = η 2j where the η j are real. We have:

G = ∑ γ i2 ui ui ,

H = ∑ η 2j w j w j

i

Thus,

j

GH = ∑ γ i2η 2j ui ui w j

(1)

wj

i, j

Now tr(GH) can be evaluated in any basis, for example the ui basis. Thus,

tr(GH ) = ∑ γ i2η 2j um ui ui w j

w j um

i, j, m

= ∑ γ η ui w j 2 i

2 j

i, j

w j ui = ∑ γ η 2 i

2 j

ui w j

2

(2)

i, j

since um ui = δ mi . Each term in the double sum (2) is non-negative. Hence: tr(GH ) ≥ 0 (3) To have tr(GH)=0, each term in the double sum must vanish. Since the eigenvalues of G and H are all positive, this implies that ui w j = 0 for all i, j ; hence from (1) that

GH = 0. However, if we assume that the eigenvectors of G and H each form a complete set, it is impossible to have ui w j = 0 for all i, j .

2 2.4b) As in 2.4a) we have: H = ∑ η j w j w j . Let A = ∑ η j w j w j . Then: j

j

A = ∑ η jη k w j w j wk wk 2

j, k

= ∑ η jηk w j δ jk wk = ∑ η 2j w j w j = H j, k

j

Also, A is Hermitian since the η j are all real. Then:

2

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

u H u = u AA u = u A† A u = x x where

x = A u . Similarly,

inequality, x x y y ≥ x y

2

vH v = yy

y =Av .

where

From Schwarz’

. Thus:

u H u v H v ≥ u A† A v

2

= uH v

2

Also tr(H ) = ∑ ηi2 > 0 , since otherwise all ηi = 0 in which case H = 0.

2.5. In general, A u = α u + β u⊥ where α , β are complex numbers and u u⊥ = 0 .

Thus, u A u = α u u + β u u⊥ = α u u Hence,

α=

u Au

uu which is real if A is Hermitian. Now, AA u = α A u + β A u⊥ Thus, u AA u = α u A u + β u A u⊥

= α2 u u + β

2

u⊥ u⊥

Hence: 2

u Au 2 u AA u − = β u⊥ u⊥ uu Therefore 2 u A2 u u Au 2 u⊥ u⊥ β = − = A2 − A 2 2 uu uu uu We can choose the norm of u⊥ such that u⊥ u⊥ = u u , and we can choose the phase of u⊥ so that β is real and positive. Then,

β = A2 − A 2 2.6. Let A n = λn n . a) Since [ A, B ] = 0 , AB n = BA n = λn B n . Thus B n is an eigenvector of A with eigenvalue λn . Therefore either B n = consti n = bn n ; or B n and n are linearly independent, in which case λn is a degenerate eigenvalue. Similar remarks hold for C: C n = cn n where cn is a constant, or λn is a degenerate eigenvalue. However, since

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

[ B,C ] ≠ 0 we cannot have B n = bn n and C n = cn n for all n; hence at least one

λn must be degenerate. b) If BC-CB=iq where q is a real constant, there is no state n for which B n = bn n and C n = cn n ; hence each state n is degenerate. To prove that the degeneracy is of infinite degree, let us assume the contrary: that there exists a state n which is mn-fold degenerate, where mn is finite. Then the mn eigenvectors of A form an mn dimensional subspace, and the vectors B n and C n lie in this subspace. Writing BC-CB=iqI we now take the trace of both sides of this equation within the subspace only. As in problem 2.1c we obtain a contradiction if mn is assumed to be finite.

2.7. Suppose A = A† and AA = I . Then A = A −1 = A† .

Suppose A −1 = A† and AA = I. Then A = A −1 = A† .

Suppose A† = A and A† = A −1 . Then A = A −1 so A 2 = I. 2.8. A matrix M has an inverse (is non-singular) if and only if det M ≠ 0 . If det M ≠ 0 and M2 = M (1) −1 then multiplying (1) on the left by M we obtain: M −1 MM = M −1 M which implies M = I . Also (1) implies that det M det M = det M which only has the solutions det M = 1, det M = 0 . Therefore, if M ≠ I , det M = 0 .

2.9. Let A be a matrix that can be diagonalized by a non-singular transformation: A' = UAU −1 (1) where A' is diagonal. Then: 1 exp(A') = I + A'+ A' A'+ ...

2!

1 = UIU −1 + UAU −1 + UAU −1UAU −1 + ...

2! −1 = U exp(A)U Hence det(e A ' ) = det(U )det(e A )det(U −1 ) = det(e A ) = ∏ e Ann '

n

⎛ ' ⎞ = exp ⎜ ∑ Ann ⎟⎠ = exp ( trA') = exp(trA) ⎝ n

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

2.10. Let ⎛ a11 A=⎜ ⎝ a21

a12 ⎞ a22 ⎟⎠

Then, ⎛ a112 + a12 a21 A =⎜ ⎝a a + a a 2

21 11

22 21

a11a12 + a12 a22 ⎞ 2 ⎟⎠ a21a12 + a22

(1)

Now, ⎛ a11 tr(A)iA = (a11 + a22 ) ⎜ ⎝ a21

a12 ⎞ ⎛ a112 + a11a22 = a22 ⎟⎠ ⎜⎝ a11a21 + a21a22

a12 a11 + a12 a22 ⎞ 2 ⎟⎠ a11a22 + a22

(2)

and ⎛ a12 a21 − a11a22 − det(A)iI = ⎜ 0 ⎝ Thus we see that: A 2 = tr(A)iA − det(A)iI

0 ⎞ a12 a21 − a11a22 ⎟⎠

(3) (4)

2.11. a) Given that A 2 = 0, AA† + A† A = I, B = A† A , we have:

(

)

B 2 = A† AA† A = A† I − A† A A = B (1) b) Assuming B to be diagonal and non-degenerate we try a 2x2 matrix: ⎛b 0⎞ B=⎜ ⎝ 0 b '⎟⎠ Then (1) implies: b 2 = b, b '2 = b ' Since b,b ' are distinct, one of them must be equal to unity, the other to zero. Choose b = 1,b ' = 0 : ⎛ 1 0⎞ (2) B=⎜ ⎝ 0 0 ⎟⎠ The given conditions also imply that AB = A (3) BA = 0 Now writing ⎛ a11 a12 ⎞ A=⎜ ⎝ a21 a22 ⎟⎠ and employing (3) with (2) we obtain: ⎛ a11 0 ⎞ AB = ⎜ =A ⎝ a21 0 ⎟⎠ whereas (4) with (2) yields:

(4)

(5)

5

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

⎛ a11 a12 ⎞ BA = ⎜ =0 0 ⎟⎠ ⎝ 0 Equations (5,6) imply that: ⎛ 0 0⎞ A=⎜ ⎝ a21 0 ⎟⎠ which cannot be diagonalized in any representation. Finally, the condition: AA† + A† A = I implies that a21 = eiθ where θ is a real number.

2.12. We start with: Ai A j + A j Ai = 2δ ij I

(6)

(7)

(1)

If i ≠ j we have: Ai A j = −A j Ai = (−I )A j Ai

Thus,

( )

( )

det(Ai )det A j = det(−I )det A j det(Ai )

(2)

Since the matrices A are non-singular, their determinants are all non-zero so (2) implies: det(−I ) = (−1)N = +1 Hence N must be an even number. Can we have N =2? No, because the set of all 2x2 matrices forms a 4 dimensional vector space, so the number of linearly independent 2x2 matrices, including the identity matrix, is 4. Since there are 4 A-matrices, none of which is equal to the identity, N must be greater than 2. Hence, N ≥ 4 .

2.13a) Suppose there is a 2x2 matrix A such that: Aσ x = −σ x A, Aσ y = −σ y A, Aσ z = −σ z A Then we would have: Aσ xσ y = iAσ z

= −iσ z A = −σ xσ y A = +σ x Aσ y = −Aσ xσ y which implies that A = −A = 0 . b) This follows immediately from the relation: σ xσ y = iσ z c) Making use of the repeated index summation convention, we have:

6

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

σ i Aσ iB = σ i Aiσ j B j

= Ai B jσ iσ j = Ai B j ⎡⎣δ ij I + iε ijkσ k ⎤⎦ d)

= AiB + iσ i A × B

i2 i3 i4 (σ i n̂ )2 θ 2 + (σ i n̂ )3 θ 3 + (σ i n̂ )4 θ 4 + ...

2!

3!

4!

Now we use the result of c) to obtain: (σ i n̂ )2 = n̂i n̂ = 1 Hence (1) becomes: 1 i 1 exp [ iσ i n̂θ ] = I + iσ i n̂θ − θ 2 I − (σ i n̂ )θ 3 + θ 4 I + ...

2!

3!

4! = I cosθ + iσ i n̂sin θ exp [ iσ i n̂θ ] = I + iσ i n̂θ +

(1)

∂ −λA . Then: e ∂x ∂g ⎛ ∂ ⎞ ⎛ ∂ ⎞ = eλ A A ⎜ e− λ A ⎟ − eλ A ⎜ Ae− λ A ⎟ ⎝ ∂x ⎠ ⎝ ∂x ⎠ ∂λ

2.14. Let g(λ ) = eλ A

(

)

∂A − λ A ⎛ ∂ ⎞ ⎛ ∂ ⎞ = eλ A A ⎜ e− λ A ⎟ − eλ A e − eλ A A ⎜ e− λ A ⎟ ⎝ ∂x ⎠ ⎝ ∂x ⎠ ∂x ∂A − λ A e ∂x ∂g ∂A ∴ =− ∂λ λ = 0 ∂x

(1)

= −eλ A

Also, from (1): ∂2 g ∂A ∂A − λ A = −eλ A A e− λ A + eλ A Ae 2 ∂λ ∂x ∂x ⎡ ∂A ⎤ = −eλ A ⎢ A, ⎥ e− λ A ⎣ ∂x ⎦ 2 ∂ g ⎡ ∂A ⎤ ∴ 2 = − ⎢ A, ⎥ ∂λ λ = 0 ⎣ ∂x ⎦ Similarly, ∂ 3g ⎡ ⎡ ∂A ⎤ ⎤ = − ⎢ A, ⎢ A, ⎥ ⎥ 3 ∂λ λ = 0 ⎣ ⎣ ∂x ⎦ ⎦ and so forth. Now

(2)

(3)

(4)

(5)

7

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

g ( λ ) = g(0) + λ

∂g λ 2 ∂2 g λ 3 ∂ 3g + + + ... ∂λ λ = 0 2! ∂λ 2 λ = 0 3! ∂λ 3 λ = 0

Thus,

∂ −λA ∂A λ 2 ⎡ ∂A ⎤ λ 3 ⎡ ⎡ ∂A ⎤ ⎤ e = 0−λ − A, − A, A, − ... ∂x ∂x 2! ⎢⎣ ∂x ⎥⎦ 3! ⎢⎣ ⎢⎣ ∂x ⎥⎦ ⎥⎦ Let λ = −i. Then: 2 3 ∂A ( −i ) ⎡ ∂A ⎤ ( −i ) ⎡ ⎡ ∂A ⎤ ⎤ − iA ∂ iA e e = −(−i) − A, − A, A, − ... ∂x ∂x 2! ⎢⎣ ∂x ⎥⎦ 3! ⎢⎣ ⎢⎣ ∂x ⎥⎦ ⎥⎦ eλ A

(

)

( )

( −i )n A( n) ⎧ ∂A ⎫ = i∑ ⎨ ⎬ ⎩ ∂x ⎭ 0 ( n + 1)! ( 0) (1) where A { B} = B, A { B} = [ A, B ] , A( 2 ) { B} = ⎡⎣ A, [ A, B ] ⎤⎦ , ... ∞

(6)

2.15.a) Given Ap = ⎡⎣ H , Ap −1 ⎤⎦ with A0 = A we may write the identity

eiHt Ae−iHt = A + it [ H , A ] + as: e Ae iHt

−iHt

=∑

p=0

n∑

p=0

(it ) p A . Therefore:

p!

p

p!

(it ) p A† A n = p

1 (it )2 ⎡⎣ H , [ H , A ]⎤⎦ + ...

2!

∑ nAe

† iHt

k k Ae−iHt n

k

= ∑ ei(Ek − En )t n A† k k A n k

(it ) p E − E p n A† k

kAn ( k n) p!

Since this last relation is valid for any value of t, we must have: S p = n A† Ap n =∑ p, k

which is the desired result. b) We have:

n A† Ap n = n A† HAp −1 − A† Ap −1 H n = n A† HAp −1 n − En n A† Ap −1 n = n A† HAp −1 n − n HA† Ap −1 n = n ⎡⎣ A† , H ⎤⎦ Ap −1 n = n A1† Ap −1 n This procedure can be repeated m times to yield: 8

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

n A† Ap n = n Am† Ap − m n

(1)

If we choose p = 2m , S2 m = n Am† Am n . The procedure can also be repeated m+1 times to yield: n A† Ap n = n Am† +1 Ap − m −1 n

(2)

Then, using p = 2m + 1 in (1) or (2) we obtain:

S2 m +1 = n Am† Am +1 n = n Am† +1 Am n 2.16. According to the Fourier integral theorem, if:

f (x) =

1 ∞ g(t)eixt dt 2π ∫−∞

(1)

then:

1 ∞ f (z)e−izt dz ∫ −∞ 2π Substituting (2) in (1) we obtain: g(t) =

(2)

⎡ 1 ∞ i( x − z )t ⎤ f (z) ⎢ (3) ∫ e dt ⎥⎦dz −∞ ⎣ 2π −∞ 1 ∞ i( x − z )t Thus we have δ ( x − z ) = e dt . Note from (3) that if x ≠ z , the integral in 2π ∫−∞ brackets must vanish because f is an arbitrary function, so f (x) cannot depend on f (z) . Now consider: f (x) = ∫

∞

1 sin 2 ω T T sin 2 ω T U (ω ) = lim = lim T →∞ π T →∞ π ω 2T 2 ω 2T T It is easy to see from (2.42) that U ( 0 ) = lim = δ ( 0 ) , and that U (ω ) = 0 for all ω ≠ 0 .

T →∞ π Hence U (ω ) = δ (ω ) . From now on we dispense with the limit signs, which are understood. Then:

δ 2 (ω ) =

1 sin 2 ω T ⎛ 1 sin 2 ω T ⎞ T =⎜ = δ (ω ) i δ ( 0 ) π2 ω2 ⎝ π ω 2T ⎟⎠ π

9

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

3.1. The initial state is: ⎛ a⎞ ⎛ 1⎞ ψ (0) = ⎜ ⎟ ⎜ ⎟ ⎝ b⎠ S ⎝ 0⎠ A The time development operator is: ⎡ i T ⎤ U(T , 0) = exp ⎢ − ∫ H (t)dt ⎥ ⎣  0 ⎦ S A ⎡ i σzσy T ⎤ = exp ⎢ − g(t)dt ⎥ ∫0 ⎣  4 ⎦ ⎡ π ⎤ = exp ⎢ −i σ zSσ yA ⎥ ⎣ 4 ⎦ π π = I S I A cos − iσ zSσ yA sin 4 4 1 ⎡⎣ I S I A − iσ zSσ yA ⎤⎦ = 2 Thus, ⎛ a ⎞ ⎛ 0⎞ ⎤ 1 ⎡⎛ a ⎞ ⎛ 1 ⎞ ψ (T ) = U(T , 0) ψ (0) = ⎢⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ ⎥ 2 ⎣⎝ b ⎠ S ⎝ 0 ⎠ A ⎝ −b ⎠ S ⎝ 1 ⎠ A ⎦

3.2. The density matrix is: M=

1 ( I + σ i p̂ ) 2

Thus,

dM 1 dp̂ 1 1 ⎡ gµ 1 ⎤ = σi = [ H , M ] = ⎢− S i B, σ i p̂ ⎥ dt 2 dt i i ⎣  2 ⎦ gµ =− [σ i B, σ i p̂ ] 4i Hence

dp̂ gµ gµ (1) =− σ i B, σ i p̂ ] = σ i p̂ × B = γσ i p̂ × B [ dt 2i  where γ ≡ gµ / . Using the repeated index summation convention, equation (1) can be written: dp̂ σ i i = γσ i p̂ j Bk ε ijk dt Multiplying on the left by σ m we have:

σi

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

dp̂i = γ (δ im I + iε miσ  ) p̂ j Bk ε ijk dt Taking the trace of both sides and dividing the resulting equation by 2 we obtain: dp̂ (2) = γ p̂ × B dt The unit vector p̂ is in the direction of the expectation value of the spin vector. Equation (2) reveals that the expectation value of the spin follows the classical equation of motion: rate of change of angular momentum = torque (on the magnetic moment due to the magnetic field).

(δ im I + iε miσ  )

3.3. For 1 S0 positronium, J = 0, and the parity is odd. Thus the two-photon state is: 1 ψ = (1) ( +1 ) ( + 2 ) − ( −1 ) ( − 2 ) 2 This can be expressed in terms of the linear polarization states, using: 1 1 ±1 = x1 ± iy1 , ± 2 = x2  iy2 2 2 Thus: 1 ψ = ( x1 + iy1 )( x2 − iy2 ) − ( x1 − iy1 )( x2 + iy2 ) 2 2 (2) i = y1 x2 − x1 y2 2 Suppose we arrange the polarizers for γ 1 , γ 2 to be oriented along the x1 ' and x2 axes respectively, where: x1 ' = x1 cosθ + y1 sin θ

y1 ' = −x1 sin θ + y1 cosθ Then

Px2 Px1 ' ψ = ( x2 x2 ) ( x1 ' x1 ' )

i y1 x2 - x1 y2 2 i i = x1 ' ( x2 x2 ) x2 sin θ = sin θ x1 ' x2 2 2 1 Hence the polarization correlation function is sin 2 θ .

2 Since each photon has energy .511 MeV, the only practical way to measure polarization is by the azimuthal dependence of Compton scattering, and this has limited analyzing power. Thus we have to consider the Compton scatterings γ 1 → γ 1 ', γ 2 → γ 2 ' . It is important to note that γ 1 ' and γ 2 ' are not in a coherent 2-photon final state (there is no definite phase relation between the 2 single- photon state vectors) because each Compton scattering takes place on a random electron in each of two separate macroscopic slabs of matter. Hence there are no interference terms between γ 1 ' and γ 2 ' in the transition probability.

11

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

The transition probability for a single Compton scattering at 90 deg is proportional to: 1 2 (3) + 4 ( ε̂ iε̂ ') 2 Since each detector is insensitive to the polarization of the scattered photon, we must sum over the polarizations of γ 1 ' and of γ 2 ' . The geometry is shown in the following diagrams: k' 2 ϴ k

2

• Ps

! k

1

ϴ

k' 1

Fig.3.1S(=Fig.3.7) k'1

ε'1a

ε1a

ε1b

ε'1b θ

k1

Fig. 3.2S

k'2

ε'2a

ε'2b ε2a

k2 ε2b

Fig.3.3S We label the matrix elements for the various scatterings as follows:

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

γ1 a: b: c: d:

γ2 e : ε 2a → ε '2a f : ε 2a → ε '2b g : ε 2b → ε '2a h : ε 2b → ε '2b

ε1a → ε '1a ε1a → ε '1b ε1b → ε '1a ε1b → ε '1b

In view of equation (2), if γ 1 ' and γ 2 ' formed a coherent 2-photon final state, the coincidence rate would be proportional to: I = ag − ec + ah − ec + bg − ec + bh − ec 2

2

2

2

+ ag − ed + ah − ed + bg − ed + bh − ed

2

+ ag − fc + ah − fc + bg − fc + bh − fc

2

2

2

2

2

2

2

+ ag − fd + ah − fd + bg − fd + bh − fd Expanding this expression and dropping all interference (cross) terms, we obtain: 2

2

2

2

2 2 2 2 2 2 2 2 I = 4 ⎡⎣ a + b ⎤⎦ ⎡⎣ g + h ⎤⎦ + 4 ⎡⎣ e + f ⎤⎦ ⎡⎣ c + d ⎤⎦ Now taking into account Figs. 1,2,3 and (3) we have: 1 2 a = + 4 sin 2 θ 2 1 2 b = 2 1 2 c = + 4 cos 2 θ 2 1 2 d = 2 1 2 e = 2 1 2 f = 2 1 9 2 g = +4= 2 2 1 2 h = 2

(4)

Thus (4) yields:

13

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

{(

) (

I = 4 5 1 + 4 sin 2 θ + 1 1 + 4 cos 2 θ

{ = 4 {26 − 16 cos θ }

= 4 6 + 20 sin 2 θ + 4 cos 2 θ

}

)}

(5)

2

I = const(1 −

8 cos 2 θ ) 13

If the analyzing power were perfect the factor 8/13 would be replaced by unity. Finally, Bell’s inequality for the Wu-Shaknov experiment is:

⎛ 3π ⎞ ⎛π⎞ 1 R⎜ ⎟ − R⎜ ⎟ − ≤ 0 ⎝ 8 ⎠ ⎝ 8⎠ 2 This is not violated by the quantum mechanical result (5):

(6)

8 ⎛π⎞ ⎛π⎞ R ⎜ ⎟ = 1 − cos 2 ⎜ ⎟ = .474736 ⎝ 8⎠ ⎝ 8⎠ 13 8 ⎛ 3π ⎞ ⎛ 3π ⎞ R ⎜ ⎟ = 1 − cos 2 ⎜ ⎟ = .909879 ⎝ 8 ⎠ ⎝ 8 ⎠ 13 .909879 − .474736 − .500 = −.064857

For violation of Bell’s inequality it would be necessary to replace the factor 8/13=.615385 by a number greater than 2-1/2 =.707107.

3.4. This is a well-known example: See N.D. Mermin, Phys. Rev. Lett. 65 3373 (1990); Revs. Mod. Phys. 65, 803 (1993). To verify the assertions in the statement of the problem, we use the fact that any component of σ for one of the particles commutes with any component of σ for either of the other 2 particles. a) Equation (1) is demonstrated by the following argument: σ 1xσ 2 yσ 3yψ = f σ 1xσ 2 yσ 3y ( u1u2u3 − v1v2 v3 ) = f ⎡⎣ v1 ( iv2 ) ( iv3 ) − u1 ( −iu2 ) ( −iu3 ) ⎤⎦ = f ( u1u2u3 − v1v2 v3 ) = ψ

and cyclic permutations.

Equation (2) is demonstrated by: σ 1xσ 2 xσ 3xψ = f σ 1xσ 2 xσ 3x [ u1u2u3 − v1v2 v3 ]

= f [ v1v2 v3 − u1u2u3 ] = −ψ

To demonstrate that the 4 operators

(σ 1xσ 2 xσ 3x ) commute, we have:

(σ σ σ ), (σ σ σ ), (σ σ σ ) and 1x

2y

3y

1y

2x

3y

1y

2y

3x

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

(

)(

)

⎡ σ 1xσ 2 yσ 3y , σ 1yσ 2 xσ 3y ⎤ = σ 1xσ 2 yσ 3yσ 1yσ 2 xσ 3y − σ 1yσ 2 xσ 3yσ 1xσ 2 yσ 3y ⎣ ⎦ = σ 1xσ 1yσ 2 yσ 2 xσ 3yσ 3y − σ 1yσ 1xσ 2 xσ 2 yσ 3yσ 3y = ( iσ 1z ) ⋅ ( −iσ 2 z ) ⋅1

− ( −iσ 1z ) ⋅ ( iσ 2 z ) ⋅1 = 0

and cyclic permutations.

Also,

(

)

⎡ σ 1xσ 2 yσ 3y , (σ 1xσ 2 xσ 3x ) ⎤ = σ 1xσ 2 yσ 3yσ 1xσ 2 xσ 3x − σ 1xσ 2 xσ 3xσ 1xσ 2 yσ 3y ⎣ ⎦ = σ 1xσ 1xσ 2 yσ 2 xσ 3yσ 3x − σ 1xσ 1xσ 2 xσ 2 yσ 3xσ 3y = 1⋅ ( −iσ 2 z ) ⋅ ( −iσ 3z ) −

1⋅ ( iσ 2 z ) ⋅ ( iσ 3z ) = 0

and cyclic permutations. b)

(σ σ σ ) (σ σ σ ) (σ σ σ ) = σ σ σ σ σ σ σ σ σ = (σ σ ) (σ σ ) (σ σ ) σ σ σ 1x

2y

3y

1y

2x

3y

1y

2y

3x

1x

1x

1y

2y

3y

2y

2x

1y

2x

3y

3y

3y

1y

1y

2y

2y

3x 3x

= ( iσ 1z ) ( −iσ 2 z ) ⋅1⋅ σ 1yσ 2 yσ 3x = σ 1zσ 1yσ 2 zσ 2 yσ 3x

= ( −iσ 1x ) ( −iσ 2 x ) σ 3x = −σ 1xσ 2 xσ 3x c) The reasoning presented in the statement of 4 c) neglects the fact that ⎡⎣σ ix , σ iy ⎤⎦ ≠ 0

for i=1,2,3. The point is that while we may measure either σ x or σ y for a given particle without disturbing the other particles, a measurement of σ x( y) for a given particle changes the state and hence the outcome of a subsequent measurement of σ y( x ) for that particle.

3.5.a) We start with:

ψ = α 1R,1G + β 1G,1R + γ 1G,1G

(1)

We require 1G, 2G ψ = 2G,1G ψ = 0 . Then:

α 1G, 2G 1R,1G + β 1G, 2G 1G,1R + γ 1G, 2G 1G,1G = 0 Thus, α 1G 1R 2G 1G + β 1G 1G 2G 1R + γ 1G 1G 2G 1G = 0

(2)

Since 1G 1R = 0, 1G 1G = 1 , (2) yields:

β 2G 1R + γ 2G 1G = 0 (3) Similarly: α 2G 1R 1G 1G + β 2G 1G 1G 1R + γ 2G 1G 1G 1G = 0

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Thus:

α 2G 1R + γ 2G 1G = 0

(4)

b)

2G = q1/2 1G + 1 − q 1R

(5)

2R = − 1 − q 1G + q1/2 1R Thus:

2G2G ψ = α 2G 1R 2G 1G + β 2G 1G 2G 1R + γ 2G 1G 2G 1G = α 1 − qq1/2 + β 1 − qq1/2 + γ q From (3), (4) we have β (1 − q)1/2 = −γ q1/2 , α (1 − q)1/2 = −γ q1/2 . Thus:

2G2G ψ = −γ q − γ q + γ q = −γ q so 2G2G ψ

2

= γ q2 ≡ p 2

(6)

However since ψ ψ = 1, we have

α +β +γ 2

Thus,

γ and

2

2

2

=

2q ⎞ 2⎛ = γ ⎜1 + =1 ⎝ 1 − q ⎟⎠

1− q 1+ q

1− q (1 − q ) p=q = q2 1+ q 1 − q2 2

(

2

)

(7)

We maximize p by varying q, and find that this occurs when q 2 + q − 1 = 0 , thus when:

(

)

1 5 −1 ≡ z 2 Inserting this in (7) we obtain p = z 5 . The remaining probabilities are then calculated easily. For this purpose we use: 1− q γ = = q 3/2 = z 3/2 1+ q q=

α = β = −γ

q = −z 3/2 z −1/2 = −z 1− q

Hence

ψ = −z 1G1R − z 1R1G + z1/2 1G1G

(8)

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Making use of (8) and (5) and computing the absolute values squared of the relevant matrix elements with ψ we find the following probabilities: 1G1R : z 2

2G1G : 0

2

2G1R : z 3

1G1G : z 3

2G2G : z 5

1G1G : 0

2G2G : z 4

1G2R : z 1R1R : 0

2R2G : z 4 2R1G : z

1R2g : z 3

2R1R : z 4

1R2R : z 4

2R2R : z

1R1G : z

17

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4.1.The commutation relation ⎡⎣ q̂, p̂ n ⎤⎦ = nip̂ n −1 holds for n = 1 . We now show that if it holds for an arbitrary positive integer n, it holds for n+1, by writing: ⎡⎣ q̂, p̂ n +1 ⎤⎦ = q̂p̂ n p̂ − p̂ n q̂p̂ + p̂ n q̂p̂ − p̂ n p̂q̂

(

=

) (

nip̂ n −1 p̂

+

)

ip̂ n

= (n + 1)ip̂ n

The proof of the relation ⎡⎣ p̂, q̂ m ⎤⎦ = −miq̂ m −1 proceeds in a similar way.

4.2. From Sec. 4.2 of the text we have the inequality:

( ΔA )2 ( ΔB )2 = u | A0 A0 | u u | B0 B0 | u ≥ u | A0 B0 | u

2

(1) In the present case, we let A = q̂, B = f (q̂, p̂) . The right hand side of (1) is greater than: 2 1 2 u [ q̂, f ] u = 4 4

Thus,

ΔqΔf ≥

∂f ∂p̂

2

 ∂f 2 ∂p̂

4.3. The Cauchy-Schwarz inequality: u u w w ≥ u w

2

becomes an equality if and

only if u = C w where C is a constant. In the case of the uncertainty relation:

u | q̂0 q̂0 | u u | p̂0 p̂0 | u ≥ u | q̂0 p̂0 | u

2

where q̂0 = q̂ − q̂ , p̂0 = p̂ − p̂ the inequality reduces to an equality when: q̂0 u = Cp̂0 u .

(1) For simplicity we assume in what follows that q̂ = p̂ = 0 , and we go to the coordinate representation. In this case (1) becomes: ∂ψ qψ = −iC (2) ∂q We want ψ to represent a localized wave packet, so we require C to be negative imaginary. Then (2) becomes: ∂ψ = −α qψ (3) ∂q where α is a positive real number. Equation (3) is readily integrated to yield:

ψ = constie−α q /2 which is a Gaussian. ev 1 e 1 4.4. We have B1 ≈ and B2 ≈ . The requirement B2 >> B1 implies: 2 4π c r 2mec 4π r 3  (1) >> me vr 2 The second requirement is: r >> Δr (2) 2

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

Combining (1) and (2), we obtain:  (3) >> me vΔr 2 However, me v is the electron momentum, which cannot be less than the uncertainty in momentum Δp . Hence (3) implies  ΔpΔr << 2 which violates the uncertainty principle.

4.5. After the result of the neutral atom Stern-Gerlach experiment became known, several persons suggested that the method could be used to determine the electron spin magnetic moment. However, soon after Heisenberg discovered the uncertainty principle in 1927, N. Bohr presented the following argument to show that it would be impossible.

Pole tip y Lines of force

z

x Beam (out of page)

Pole tip

Fig.4.1S(=Fig.4.1) Let x=y=0 correspond to the center of an undeflected electron beam moving in the z direction. If we have a well-defined electron beam, we must have: pz = me vz >> Δpx   However, the uncertainty principle implies that Δpx ≥ . Therefore, Δx >> .

2Δx 2me vz Now, ∂B Bx ( Δx )  Bx (0) + x Δx ∂x ∂By ∂Bx =− Also, Bx (0) = 0 and since ∇iB = 0, . (We can assume that Bz = 0.) Hence, ∂x ∂y ∂By ∂By  Bx ( Δx ) = Δx >> ∂y ∂y 2me vz

19

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∴

evz e ∂By Bx ( Δx ) >> c 2me c ∂y

(1)

e due to 2me c the field gradient. The left hand side of (1) is the uncertainty in the Lorentz force in the y direction, which arises from the uncertainty Δx . Inequality (1) implies that the experiment is impossible.

The right hand side of (1) is the deflection force on the magnetic moment

4.6. a) We write: ψ A = A ψ + ΔA ψ 1A and B ψ = B ψ + ΔB ψ 1B which are valid if ψ ψ = ψ 1A ψ 1A = ψ 1B ψ 1B = 1,

ψ ψ 1A = ψ ψ 1B = 0 From (1), (2) we obtain: ψ AB ψ = A B + ΔAΔB ψ 1A ψ 1B and similarly, ψ BA ψ = A B + ΔAΔB ψ 1B ψ 1A Thus: [ A, B ] = 2iΔAΔBiIm ψ 1A ψ 1B and {A, B} = 2 A B + 2ΔAΔBiRe ψ 1A ψ 1B

(1) (2)

(3) (4) (5) (6)

Suppose that [ A, B ] = i . Then since ψ 1A ψ 1B ≤ 1 , (5) implies that:  2 Also (6) implies that: 1 ΔAΔB ≥ {A, B} − A B 2 ΔAΔB ≥

(7)

(8)

20

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4.7.a) 2 d ψ ( t ) ψ ( 0 ) = ψ ( t ) ψ ( 0 ) ψ ( 0 ) ψ ( t ) + ψ ( t ) ψ ( 0 ) ψ ( 0 ) ψ ( t ) dt i = ⎡⎣ ψ ( t ) H ψ ( 0 ) ψ ( 0 ) ψ ( t ) − ψ ( t ) ψ ( 0 ) ψ ( 0 ) H ψ ( t ) ⎤⎦  i = − ΔE ⎡⎣ ψ ( t ) ψ ( 0 ) ψ ( 0 ) ψ 1 ( t ) − ψ 1 ( t ) ψ ( 0 ) ψ ( 0 ) ψ ( t ) ⎤⎦  2ΔE =− Re i ψ ( t ) ψ ( 0 ) ψ ( 0 ) ψ 1 ( t ) 

(

)

Therefore, 2 d 2ΔE ψ (t ) ψ ( 0 ) ≥ − ψ (t ) ψ ( 0 ) ψ ( 0 ) ψ 1 (t ) dt 

(1)

Now let us write the following expansion of ψ ( 0 ) :

ψ (0) = ψ ( t ) ψ ( t ) ψ ( 0 ) + ψ 1 ( t ) ψ 1 ( t ) ψ ( 0 ) + α ψ 2 ( t ) where we assume that: ψ ( 0 ) ψ ( 0 ) = ψ (t ) ψ (t ) = ψ 1 (t ) ψ 1 (t ) = ψ 2 (t ) ψ 2 (t ) = 1

(2)

and

ψ (t ) ψ 1 (t ) = ψ (t ) ψ 2 (t ) = ψ 1 (t ) ψ 2 (t ) = 0 and also that α is a complex number. Then from (2), ψ ( 0 ) ψ ( 0 ) = 1 = ψ ( 0 ) ψ (t )

2

+ ψ ( 0 ) ψ 1 (t )

2

+α

2

Hence,

ψ ( 0 ) ψ 1 (t )

2

≤ 1 − ψ ( 0 ) ψ (t )

2

Thus, employing (1), we have: 2 2 d 2ΔE ψ (t ) ψ ( 0 ) ≥ − ψ (t ) ψ ( 0 ) 1 − ψ (t ) ψ ( 0 ) dt  Let f (t) = ψ ( t ) ψ ( 0 ) . Then equation (4) reduces to:

df

≥−

(3) (4)

ΔE dt 

1− f 2 and the minimum time T is found by the following integration:  1 df π T= = ∫ 2 ΔE 0 1 − f 2ΔE b) For a spin magnetic moment precessing in the xy plane about a uniform magnetic field in the z direction, the Hamiltonian is H = aσ z where a is a real constant, and the initial

21

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⎛ 1⎞ ⎛ 0⎞ 1 ( χ + + χ − ) where χ + = ⎜ 0⎟ , χ − = ⎜ 1⎟ . It is very easy ⎝ ⎠ ⎝ ⎠ 2 to show that E = 0, ΔE = a . The time development operator is:

state can be chosen as ψ (0) =

⎛ aσ t ⎞ U(t, 0) = exp ⎜ −i z ⎟ ⎝  ⎠ π π ⎛ π ⎞ At time T = , we obtain: U = exp ⎜ −i σ z ⎟ = −iσ z ; hence: = ⎝ 2 ⎠ 2ΔE 2a ⎛ χ+ ⎞ ψ (T ) = −iσ zψ ( 0 ) = −i ⎜ ⎝ − χ − ⎟⎠ which is orthogonal to ψ ( 0 ) .

4.8.a)

∫ W (x, p,t)dp =

∞ 1 ∞ dp dyψ * (x + y,t)ψ (x − y,t)e2ipy = ∫ ∫ −∞ −∞ π ∞ 1 ∞ 2ipy ∫−∞ ψ * (x + y,t)ψ (x − y,t)dy π ∫−∞ e dp = ∞

2 ∫ ψ * (x + y,t)ψ (x − y,t)δ (2y)dy = −∞

∞

y

y

∫ ψ * (x + 2 ,t)ψ (x − 2 ,t)δ (y)dy = ψ (x,t)

2

−∞

b)

∫ W (x, p,t)dx = ∫∫ ψ * ( x + y,t )ψ (x − y,t)e

2ipy

dydx

1 e2ipy dy ∫ dx ∫ φ * ( p2 )e−i( x + y) p2 dp2 ∫ φ ( p1 )ei( x − y) p1 dp1 2π 2 ∫ 1 = e2ipy dy ∫ φ * ( p2 ) dp2 ∫ φ ( p1 )dp1 ∫ e−iy ( p1 + p2 )eix ( p1 + p2 )dx 2 ∫ 2π 1 = ∫ e2ipy dy ∫ φ * ( p2 ) dp2 ∫ φ ( p1 )dp1e−iy( p1 + p2 )δ ( p1 − p2 ) π 2 1 = ∫ e2ipy dy ∫ φ ( p1 ) e−2iyp1 dp1 π 2 1 = ∫ φ ( p1 ) dp1 ∫ e2i( p − p1 )y dy π 2 1 2 = φ ( p1 ) δ ( p − p1 ) dp1 = φ ( p ) ∫ 2π =

22

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2π ∫∫ W1* ( x, p,t )W2 ( x, p,t ) dxdp =

2 dx ∫ dy1 ∫ dy2ψ 1* ( x − y1 ,t )ψ 2 ( x − y2 ,t )ψ 2* ( x + y2 ,t )ψ 1 ( x + y1 ,t ) ∫ dpe2ip( y2 − y1 ) ∫ π

= 4 ∫ dx ∫ dy1 ∫ dy2δ ⎡⎣ 2 ( y2 − y1 ) ⎤⎦ ψ 1* ( x − y1 ,t )ψ 2 ( x − y2 ,t )ψ 2* ( x + y2 ,t )ψ 1 ( x + y1 ,t ) y ⎞ ⎛ y ⎞ ⎛ y ⎞ ⎛ y ⎞ ⎛ = ∫ dx ∫ dyψ 1* ⎜ x − ,t ⎟ ψ 2 ⎜ x − ,t ⎟ ψ 2* ⎜ x + ,t ⎟ ψ 1 ⎜ x + ,t ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ = ∫ duψ 1* ( u,t )ψ 2 ( u,t ) ∫ dvψ 2* ( v,t )ψ 1 ( v,t ) = ψ 1 ψ 2

2

y y In the last expression, u = x − , v = x + . The absolute value of the Jacobian of this 2 2 transformation is unity.

4.9. We start with the expression: ∞ ⎡ (x − a)2 + (y − a)2 ⎤ 1 − a 2 /(2 σ 2 ) (1) M (x, y) = e exp ⎢− ⎥ da 2πσ s ∫−∞ 4s 2 ⎣ ⎦ The exponent in the integrand is: ⎡ ⎤ (x 2 + y 2 ) ⎛ 1 1 ⎞⎢ 2 x+y ⎥ (2) E=− −⎜ 2 + a⎥ ⎟ ⎢a − ⎝ 2s s2 ⎥ 4s 2 2σ 2 ⎠ ⎢ 1+ 2 ⎢⎣ σ ⎥⎦ Completing the square for the factor in square brackets on the right hand side of (2) we obtain: 2

⎡ ⎤ 2 ⎢ ⎥ 2 2 (x + y ) ⎛ 1 1 ⎞⎢ x+y ⎥ ⎛ 1 1 ⎞ ( x + y) E=− −⎜ 2 + +⎜ + ⎟ a− ⎟ 2 ⎝ 2s 4s 2 2σ 2 ⎠ ⎢ ⎛ s 2 ⎞ ⎥ ⎝ 2s 2 2σ 2 ⎠ ⎛ s2 ⎞ 2 1 + 4 ⎜1 + 2 ⎟ ⎢ ⎥ ⎜⎝ σ 2 ⎟⎠ ⎦ σ ⎠ ⎝ ⎣ Thus (1) can be written: ⎧ ⎡ 2 ⎤⎫ ⎪ ⎢ (1 + 2s2 )(x 2 + y 2 ) − 2xy ⎥ ⎪ 1 ⎪ σ ⎥ ⎪⎬i M (x, y) = exp ⎨− ⎢ 2 ⎥⎪ 2πσ s ⎛ s ⎞ ⎪ ⎢ 8s 2 ⎜ 1 + 2 ⎟ ⎢ ⎥⎪ ⎪⎩ ⎣ σ ⎠ ⎝ ⎦⎭ 2 ⎧ ⎛ ⎞ ⎫ ⎪ ∞ 1 ⎞⎜ x + y ⎟ ⎪⎪ ⎪ ⎛ 1 ⎬da ∫−∞ exp ⎨− ⎜⎝ 2s 2 + 2σ 2 ⎟⎠ ⎜ a − s 2 ⎟⎟ ⎪ ⎪ ⎜ 2(1 + 2 ⎝ ⎪⎩ σ ⎠ ⎪⎭ 23

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The integral in this expression is a Gaussian integral, which is immediately evaluated to yield: ⎧ ⎡ 2 ⎤⎫ ⎪ ⎢ (1 + 2s2 )(x 2 + y 2 ) − 2xy ⎥ ⎪ 1 1 ⎪ ⎢ σ ⎥ ⎪⎬ (3) M (x, y) = 2 exp ⎨ − 2 2 ⎥⎪ s ⎞ 2π s 1 + σ 2⎛ ⎪ ⎢ 8s ⎜ 1 + 2 ⎟ ⎥⎪ ⎪⎩ ⎢⎣ s2 ⎝ σ ⎠ ⎦⎭ Making the following substitutions in (3):

σ2 4z 1+ z 1+ z = , x = 21/2 s X, y = 21/2 s Y 2 2 s (1 − z) 1− z 1− z and performing some simple algebra, we arrive at: ⎧ ⎡ (1 + z 2 )(X 2 + Y 2 ) − 4zXY ⎤ ⎫ 1 1− z M (x, y) = exp ⎨− ⎢ ⎥⎬ 2(1 − z 2 ) ⎪⎩ ⎣ 2π s 2 1 + z ⎦ ⎪⎭ It is easy to show from (4) that: ∞

tr ρ = ∫ M (x, x)dx = 1 −∞

(4)

(5)

24

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5.1. The equation of motion of e in the field of g is: e eg r × r m r = r × B = c 4π c r 3 Thus r × r  (1) r=a 3 r eg where a = − . The “spin” angular momentum of the charge-monopole pair is: 4π mc r (2) S = ma r The orbital angular momentum of e about the monopole is: L = mr × r (3) Obviously, LiS = 0 . Now, r⎤ ⎡ J = L + S = m ⎢ r × r + a ⎥ (4) r⎦ ⎣ Then: r r ⎤ ⎡ J = m ⎢ r ×  r + a − a 2 r ⎥ r r ⎦ ⎣ However, from (1), a a r ×  r = 3 r × ( r × r ) = 3 ⎡⎣ ri rr − r 2 r ⎤⎦ r r a = 3 ⎡⎣ rrr − r 2 r ⎤⎦ r a a  − r = 2 rr r r   r = r i r = 0 . Thus, Thus J = 0 . Now from (1), ri 2 d dv =0 ( ri r ) = dt dt so v is a constant. Also, d ( ri r ) = rir + ri r = v2 = constant dt hence 1 d 2 (5) r = ri r = v2t + c 2 dt where c is a constant. Now let e arrive at the distance of closest approach r0 at t = 0. Then c = 0, hence from (5), (6) r 2 = v2t 2 + r02 Next, consider the following diagram:

25

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 L P

J

S

θ

φ

Fig. 5.1S Since LiS = 0, JiJ = m 2 r × rir × r + a 2 . However,

(

r × rir × r = r 2 v2 − ( ri r )

)

2

= v4 t 2 + v2 r02 − v4 t 2 = v2 r02 Thus,

(

J 2 = m 2 v2 r02 + a 2

)

(7)

and cosθ =

a a 2 + v2 r02

=const.

Hence the orbit of e is confined to the cone with half-angle θ.

Finally, from (1), xy − yx  z=a r3 Also, y tan φ = x Thus  yx − xy sec 2 φφ = 2 x and therefore r 3 cos 2 φ r z   φ= z= 2 ax a sin 2 θ However, z = r cosθ , and since cosθ is a constant, (9) can be written: 1 r r cosθ φ = a sin 2 θ Also, from (5) and (6), we easily obtain:

(8)

(9)

(10)

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v2 r02 r3 so (10) becomes: 1 v2 r02 cosθ φ = a v2t 2 + r02 sin 2 θ Hence, r02 cosθ t dt φ= 2 2 ∫ −∞ r0 a sin θ + t2 2 v r v cosθ ⎡ π vt ⎤ = 0 ⎢ + a tan ⎥ 2 a sin θ ⎣ 2 r0 ⎦ With the help of (8), this is easily reduced to: π /2 + a tan ( vt /r0 ) φ= sin θ  r=

(11)

5.2 The phase shift between the two parts of the electron wave function is: eΔVt δ=  The intensity at the detector is proportional to: I = 2(1 + cos δ ) 5.3The Lagrangian is: mv2 µ L= + σ iE × v 2 c The generalized (canonical) momentum is: ∂L µ pi = = mvi + σ j Ek ε ijk ∂vi c The Hamiltonian is: mv2 H = ∑ vi pi − L = 2 i Writing this in terms of the canonical momenta we have: 2 2 µ µ ⎛ ⎞ ⎛ ⎞ ⎜⎝ px + Eyσ z ⎟⎠ ⎜⎝ py − Exσ z ⎟⎠ c c (1) H= + 2m 2m Since we assume the particle spin is in the ±z direction, we can replace σ z in (1) by ±1 wherever it appears. Hence: 2 2 µ ⎞ µ ⎞ ⎛ ⎛ ⎜⎝ px ± Ey ⎟⎠ ⎜⎝ py  Ex ⎟⎠ c c H= + 2m 2m 27

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Comparing this to the Aharonov-Bohm Hamiltonian for a particle of charge e, we have the correspondence:

µ e Ey ↔ − Ax c c µ e  Ex ↔ − Ay c c Now, the phase shift in the Aharonov-Bohm experiment is: e ∫ Ax dx + Ay dy c  Thus the phase shift in the Aharonov-Casher experiment is: µ (2) δ =  ∫ ⎡⎣Ey dx − Ex dy ⎤⎦ c λ λ In plane polar coordinates, Ey = sin θ , Ex = cosθ , dx = −r sin θ dθ , and 2π r 2π r dy = r cosθ dθ . Thus (2) becomes: λµ 2 π λµ (3) δ =± dθ = ± ∫ 0 2π c c ±

(

)

5.4 We start with: ν e = cosθ ν1 − sin θ ν 2

ν µ = sin θ ν1 + cosθ ν 2 Let Ψ(0) = ν e = cosθ ν1 − sin θ ν 2 . Then,

Ψ(t) = cosθ ν1 e−iE1t / − sin θ ν 2 e−iE2 t / = cosθ ⎡⎣ cosθ ν e + sin θ ν µ ⎤⎦ e−iE1t / − sin θ ⎡⎣ − sin θ ν e + cosθ ν µ ⎤⎦ e−iE2 t /

Hence:

1 = ⎡⎣ cos 2 θ e−iE1t / + sin 2 θ e−iE2 t / ⎤⎦ ν e + sin 2θ ⎡⎣ e−iE1t / − e−iE2 t / ⎤⎦ ν µ 2

ν µ Ψ(t) =

1 sin 2θ e−iE1t / ⎡⎣1 − ei ( E1 − E2 )t / ⎤⎦ 2

so

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

Pν e→νµ = ν µ Ψ(t)

2

2 1 2 sin 2θ 1 − ei ( E1 − E2 )t / 4 2 1 2 ⎡⎛ (E1 − E2 )t ⎞ (E − E2 )t ⎤ = sin 2θ ⎢⎜ 1 − cos + sin 2 1 ⎥ ⎟ ⎠ 4   ⎢⎣⎝ ⎥⎦

=

1 2 (E − E2 )t ⎞ ⎛ sin 2θ i2 ⎜ 1 − cos 1 ⎟⎠ ⎝ 4  (E − E2 )t = sin 2 2θ sin 2 1 2 =

(

)

(1)

⎡⎛ m2c4 ⎞ ⎛ m2c4 ⎞ ⎤ m − m c 2 . Thus (1) p 2 c 2 + m1,2 c 4  pc ⎢⎜ 1 + 12 2 ⎟ − ⎜ 1 + 22 2 ⎟ ⎥ = 2p c ⎠ ⎝ 2p c ⎠⎦ 2p ⎣⎝

Now, E1,2 = becomes:

Pν e →ν µ = sin 2 2θ sin 2

2 1

2

3

(m − m )c t 2 1

2 2

3

4p

(m − m )c L = sin 2θ sin 2

2 2

2 1

2 2

3

(2)

4E

The factor

( m − m ) c L is dimensionless and can be expressed in any consistent set of 2 1

2 2

3

4E units (e.g., cgs units). However for practical experimental work it is convenient to use Δm 2 in eV2/c4, L in kilometers, and E in GeV. Thus we have to use the conversion factors: 1 gram = 5.609•1032 eV/c2 1 erg=625 GeV 1 cm = 10-5 km and c 3 /  = 2.56i10 58 in cgs units. Hence:

( 2.56i10 i625 ) = 1.27 ( 5.609 • 10 ) i10 i4 58

32 2

−5

NGF . The eigenvalues λ of the new Hamiltonian matrix: ω 2 sin 2θ ⎛ cos 2θ + q ⎞ Hm = ω ⎜ −(cos 2θ + q)⎟⎠ ⎝ sin 2θ are found by solving the secular equation: ω ( cos 2θ + q ) − λ sin 2θ =0 sin 2θ −ω ( cos 2θ + q ) − λ

b) Let q =

This yields:

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

λ± = ±ω (cos 2θ + q)2 + sin 2 2θ Let the new eigenvectors be: ν1m = cosθ m ν e + sin θ m ν µ ν 2 m = − sin θ m ν e + cosθ m ν µ Then the angle θ m is found from the matrix equation: sin 2θ ⎛ cos 2θ + q ⎞ ⎛ cosθ m ⎞ ⎛ cosθ m ⎞ ω⎜ = λ+ ⎜ ⎟ ⎜ ⎟ −(cos 2θ + q)⎠ ⎝ sin θ m ⎠ ⎝ sin 2θ ⎝ sin θ m ⎟⎠ ⎛ cosθ m ⎞ = ω (cos 2θ + q)2 + sin 2 2θ ⎜ ⎝ sin θ m ⎟⎠ This yields the two equations: ( cos 2θ + q ) cosθ m + sin 2θ sin θ m = λ+ cosθ m

sin 2θ cosθ m − ( cos 2θ + q ) sin θ m = λ+ sin θ m Dividing the second of these equations by the first we have: sin 2θ cosθ m − ( cos 2θ + q ) sin θ m tan θ m = ( cos 2θ + q ) cosθ m + sin 2θ sin θ m

sin 2θ − ( cos 2θ + q ) tan θ m ( cos 2θ + q ) + sin 2θ tan θ m which yields the desired result: sin 2θ 2 tan θ m = = tan 2θ m cos 2θ + q 1 − tan 2 θ m =

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6.1. To prove that Δx 2 is a quadratic function of the time for any square integrable free particle wave packet it is most convenient to work in the momentum representation. Let the state vector at time t=0 be Ψ(0) . Then:

Ψ ( 0 ) = ∫ p p Ψ ( 0 ) dp = ∫ p φ ( p, 0)dp

Hence

⎡ i p̂ 2 ⎤ ⎡ i p̂ 2 ⎤ Ψ(t) = exp ⎢ − t ⎥ Ψ(0) = ∫ exp ⎢ − t ⎥ p φ ( p, 0)dp ⎣  2m ⎦ ⎣  2m ⎦ ⎡ i p2 ⎤ = ∫ exp ⎢ − t ⎥ p φ ( p, 0)dp ⎣  2m ⎦ Therefore

⎡ i p2 ⎤ φ ( p ',t) = p ' Ψ(t) = ∫ exp ⎢ − t ⎥ p ' p φ ( p, 0)dp ⎣  2m ⎦ ⎡ i p2 ⎤ = ∫ δ ( p '− p) exp ⎢ − t ⎥ φ ( p, 0)dp ⎣  2m ⎦ ⎡ i p '2 ⎤ = exp ⎢ − t ⎥ φ ( p ', 0) ⎣  2m ⎦ or, dropping the primes, and choosing units where m = 1,  = 1 for convenience in writing from now on: φ ( p,t) = exp −ip 2t / 2 φ ( p, 0) (1)

(

)

2

Now, Δx 2 = x 2 − x 2 = ∫ φ * ( p,t)x 2φ ( p,t)dp − ⎡⎣ ∫ φ * ( p,t)xφ ( p,t)dp ⎤⎦ and in momentum representation, x = i

∂ ∂2 , x 2 = − 2 . Thus: ∂p ∂p

(

2

)

2 2 ∂2 ⎡ ∂ ⎤ ( Δx(t)) = − ∫ φ * ( p,t) 2 (φ ( p,t)) dp + ⎢ ∫ φ * ( p, 0)eip t /2 e−ip t /2φ ( p, 0) dp ⎥ ∂p ∂p ⎣ ⎦ (2) Using integration by parts we transform the first term on the right hand side of (2) as follows: ∂2 ∂ ⎡ ∂ ⎤ ∂ ∂ − ∫ φ * ( p,t) 2 (φ ( p,t)) dp = − ∫ ⎢φ * ( p,t) φ ( p,t) ⎥dp + ∫ φ * ( p,t) ⋅ φ ( p,t)dp ∂p ∂p ⎣ ∂p ∂p ∂p ⎦ However the first integral on the right hand side of this expression vanishes, so we have:

2

(

)

2 2 ∂ ∂ ⎡ ∂ ⎤ ( Δx(t)) = + ∫ φ * ( p,t) ⋅ φ ( p,t)dp + ⎢ ∫ φ * ( p, 0)eip t /2 e−ip t /2φ ( p, 0) dp ⎥ ∂p ∂p ∂p ⎣ ⎦ Now:

2

2

(3)

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∂ ⎡ ∂ ⎤ 2 φ * ( p,t) = ⎢iptφ * (0) + φ * (0) ⎥ eip t /2 ∂p ∂p ⎣ ⎦ ∂ ⎡ ∂ ⎤ 2 φ ( p,t) = ⎢ −iptφ (0) + φ (0) ⎥ e−ip t /2 ∂p ∂p ⎣ ⎦ Thus the first integral on the right hand side of (3) is: I1 = t 2 p 2 t = 0 + terms of order t while the second integral on the right hand side of (3) is: 2 I 2 = −t 2 p t = 0 + terms of order t Therefore, finally, in units where m =  = 1 , ( Δx(t))2 = t 2 Δpt2= 0 + terms of order t Since the wave packet is square integrable, Δx must be finite; hence by the uncertainty principle, Δp 2 must be greater than zero.

6.2 We divide the x axis into 3 regions: I: x<a II : a ≤ x ≤ b III : x > b Consider a plane wave incident from the left. Then, in region I, (1) ψ = eikx + Be−ikx while in region III, (2) ψ = Ceikx where B,C are numerical coefficients. Also, if we have a plane wave incident from the right with the same value of k, the wave function in region I is: (3) χ = C 'e−ikx while in region III it is: (4) χ = e−ikx + B'eikx where B',C ' are numerical coefficients.

Now, for all x, ψ "+ ⎡⎣ k 2 − U(x) ⎤⎦ ψ = 0

χ "+ ⎡⎣ k 2 − U(x) ⎤⎦ χ = 0

Therefore, ψχ "− χψ " = 0; hence ψχ '− χψ ' = K, a constant. Hence in region I:

( −ikC 'e )( e + Be ) − C 'e (ike − ikBe ) = −ikx

while in region III:

ikx

−ikx

−ikx

ikx

−ikx

− 2ikC ' = K

(5)

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

(

)

(

)

Ceikx −ike−ikx + ikB'eikx − ikCeikx e−ikx + B'eikx =

(6) − 2ikC = K 2 2 Hence C = C ' ; therefore C = C ' , so the transmission coefficients are the same in both directions.

6.3 The Schroedinger equations for this problem are: 2m a ψ "+ 2 Eψ = 0 x ≤  2 2m a ψ "+ 2 (E − V0 )ψ = 0 x >  2

(1) (2)

Fig.6.1S For any given energy E, we can make the well sufficiently deep so that V0 >>E. The average force exerted by the right hand wall of the well on the particle is: ⎛ ∂V ⎞ f =− ⎜ ⎝ ∂x ⎟⎠ a /2 Thus the average force exerted by the particle on the right hand wall is: ∞ ∂V F = − f = ∫ ψ 2 (x) dx −∞ ∂x a /2 ∞ a = V0 ∫ ψ 2 (x)δ (x − )dx (3) −∞ 2 a = V0ψ 2 ( ) 2 The even-parity solutions to (1), (2) are: 2mE a ψ = A cos(kx) k= x≤ 2  2 2m a ψ = Be−κ x κ= V − E) x> 2 ( 0  2 Equating the logarithmic derivatives of these two functions at x = a/2 , we obtain:

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

cot

ka E E =  << 1 2 V0 − E V0

This implies that: sin

ka ka E ≈ ±1 and cos ≈ ± 2 2 V0

(4)

hence that

π 5π 9π , , ,... [ + signs in (4)] 2 2 2 (5) 3π 7π ka ≈ , ,...

[- signs in (4)] 2 2 ⎛ a⎞ To calculate ψ 2 ⎜ ⎟ in (3), we need to find the normalization factor A2, determined as ⎝ 2⎠ follows. We assume that ka ≈

∞

a /2

−∞

− a /2

1 = ∫ ψ 2 (x)dx  A 2 ∫

cos 2 kxdx

A 2 ka /2 cos 2 ydy ∫ − ka /2 k A 2 (2n +1)π /2 = cos 2 ydy ∫ − 2n +1 π /2 k ( ) 2n + 1 A 2 = π 2 k =

where n = 0,1, 2, 3,... Thus, 2ka 2 A2 = = ( 2n + 1) π a a Hence, using (4) we obtain the desired result:

2E (6) a independent of V0. When we take the limit as V0 goes to infinity, the answer remains the same. Of course, the same answer is obtained for the odd-parity solutions.

In classical mechanics, we have a particle of mass m and constant velocity v shuttling back and forth between the left- and right-hand walls. Each time the particle strikes the right hand wall and rebounds, it delivers an impulse of momentum: Δp = 2mv v to the wall. The number of impulses per second is . Thus according to classical 2a mechanics the average force exerted by the particle on the right hand wall is: v 2mv2 2E Δp = = (7) 2a 2a a F = V0ψ 2 ( a/2 ) = V0 A 2 cos 2 ( ka/2 ) =

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

which is the same result as obtained in (6).

6.4 For the case V (x) = k x , − ∞ < x < ∞ , the energy eigenvalues: E0 < E2 < E4 < ... correspond to even parity solutions, while the eigenvalues: E1 < E3 < E5 < ... correspond to odd-parity solutions. All the odd parity solutions vanish at x=0. Hence these, and only these, are also solutions for the case: 3/2

V (x) = ∞, x < 0 V (x) = kx 3/2 , x > 0

p2 + mgx , the classical equation of motion is  x = −g , and 2m the wave function at x and at time T is given by: 6.5 The Hamiltonian is H =

∞

ψ (x,T ) = ∫ G(x,T ; x0 , 0)ψ ( x0 , 0 )dx0 −∞

where

G(x,T ; x0 , 0) = and

⎧⎪ im ⎡ ( x − x )2 m g 2T 3 ⎤ ⎫⎪ 0 exp ⎨ ⎢ − g(x + x0 )T − ⎥⎬ 2π iT 2 T 12 ⎥⎦ ⎪ ⎩⎪ ⎢⎣ ⎭

ψ (x0 , 0) =

(

⎛ x02 ⎞ exp 1/ 4 ⎜⎝ − 4a 2 ⎟⎠ 2π a 2 1

)

Thus:

ψ (x,T ) =

1

m 2π iT

( 2π a )

2 1/ 4

⎧⎪ im ⎡ ( x − x )2 ⎛ x02 ⎞ g 2T 3 ⎤ ⎫⎪ 0 exp − g(x + x )T − exp ⎢ ⎥ ⎨ ⎬ 0 ∫−∞ ⎜⎝ − 4a 2 ⎟⎠ dx0 12 ⎥⎦ ⎪ ⎪⎩ 2 ⎢⎣ T ⎭ The exponent in the integrand in (1) is:

(1)

∞

⎛ im 2 imgT im ⎞ 2 ⎛ imx imgT ⎞ img 2T 3 ⎞ ⎛ 1 −⎜ 2 − x − + x + x − x − = ⎟ 0 ⎜⎝ ⎟ 0 ⎜ ⎝ 4a 2T ⎠ T 2 ⎠ 2 24 ⎟⎠ ⎝ 2T

(2)

− Ax02 − Bx0 + C where the coefficients A,B,C are defined by (2). We complete the square by writing:

35

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 2

B B⎞ B2 ⎛ −A(x + x0 ) = −A ⎜ x0 + + ⎟ ⎝ A 2A ⎠ 4A Thus the integral in (1) is: 2 0

⎛ B2 ⎞ ∞ B 2⎤ ⎡ I = exp ⎜ + C ⎟ ∫ exp ⎢ −A(x0 + ) dx0 2A ⎥⎦ ⎝ 4A ⎠ −∞ ⎣

(3)

⎛ B2 ⎞ π = exp ⎜ + C⎟ A ⎝ 4A ⎠

Substituting into the right hand side of (3) the values of A,B,C defined in (2) we obtain: ⎡ ⎛ imx imgT ⎞ 2 ⎤ + ⎢ ⎜⎝ ⎟ ⎥ 1 m π T 2 ⎠ ⎥ ⎢ ψ (x,T ) = exp i 1/ 4 1 im 1 im ⎞ ⎥ ⎢ 2π iT ⎛ 2π a 2 − (4) ⎢ 4 ⎜⎝ 4a 2 − 2T ⎟⎠ ⎥ 4a 2 2T ⎣ ⎦

(

)

⎛ im 2 imgT img 2T 3 ⎞ exp ⎜ x − x− 2 24 ⎟⎠ ⎝ 2T

This expression can be written as: 2 ⎡ ⎛ ⎤ gT 2 ⎞ ⎢ −⎜ x − ⎥ 2 ⎟⎠ ⎛ ⎝ 1 1 2ima 2 ⎞ ⎥ ⎢ ψ (x,T ) = exp 1+ i ⎢ ⎛ 2  2T 2 ⎞ ⎜⎝ T ⎟⎠ ⎥ ( 2π )1/ 4 a + iT ⎢ ⎜ 4a + 2 2 ⎟ ⎥ 2ma ma ⎠ ⎢⎣ ⎝ ⎥⎦

⎛ im 2 imgT img T ⎞ exp ⎜ x − x− 2 24 ⎟⎠ ⎝ 2T 2

(5)

3

Thus:

ψ (x,T ) = 2

2 ⎡ ⎛ gT 2 ⎞ ⎤ ⎢ −⎜ x + ⎥ 2 ⎟⎠ ⎥ ⎝ 1 ⎢ exp ⎢⎛ ⎡ 2 ⎛ T ⎞ 2 ⎤  2T 2 ⎞ ⎥ 2 2a + ⎢ ⎥ 2π ⎢ a + ⎜ 2 2⎟ ⎟⎠ ⎥ ⎜⎝ 2m a ⎝ ⎠ 2ma ⎢ ⎥⎦ ⎣ ⎦ ⎣

(6)

If g were equal to zero (6) would reduce to equation (6.12) of Chapter 6, which describes the spreading of a free-particle Gaussian wave packet. Let’s now make the Galilean transformation:

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

x' = x +

gT 2 2

(7)

T'=T The Schroedinger equation in the laboratory frame is:  2 ∂ 2ψ ∂ψ − + mgxψ = i (8) 2 2m ∂x ∂T In the accelerated (primed) frame there is no gravitational force and the Schroedinger equation is:  2 ∂2 χ ∂χ − = i (9) 2 2m ∂x ' ∂T ' Using the analysis found in Sec 4.8 of Chapter 4, we write ψ (x,T ) = eif ( x ',T ') χ ( x ',T ') .

Also, ∂ ∂ ∂x ' ∂ ∂ ∂ ∂x ' ∂ ∂T ' ∂ ∂ = = , = + = + gT ' ∂x ∂x ' ∂x ∂x ' ∂T ∂x ' ∂T ∂T ' ∂T ∂T ' ∂x ' Thus, (8) can be written as:

(

)

2 if ⎛ 2 ∂ e χ gT '2 ⎞ if ∂ ⎞ ⎛ ∂ − + mg x '− e χ = i ⎜ + gT ' ⎟ eif χ 2 ⎜ ⎟ ⎝ ∂T ' 2m ∂x ' 2 ⎠ ∂x ' ⎠ ⎝ which reduces to:

(

)

(

)

⎛ 2 gT '2 ⎞ χ =  ( − f χ + i χ − gT ' f ' χ + igT ' χ ') (10) ( χ "+ 2if ' χ '− f " χ ) + mg ⎜ x '− 2m 2 ⎟⎠ ⎝ In (10) the primes or dots on f , χ refer to differentiation with respect to x ',T ' respectively. Subtracting (9) from (10) we obtain: −

⎛ 2 gT '2 ⎞ χ =  ( − f χ − gT ' f ' χ + igT ' χ ') ( 2if ' χ '− f " χ ) + mg ⎜ x '− ⎟ 2m 2 ⎝ ⎠ which yields: −

⎛ 2 ⎞ ⎛ 2 ⎞ mg 2T '2 −i ⎜ f '+ gT '⎟ χ '+ ⎜ f "+ mgx '− + f + gT ' f '⎟ χ = 0 2 ⎝m ⎠ ⎝ 2m ⎠

(11)

mgT ' mgx 'T ' . Then, f = − + w(T ')   where w is a function of T’ to be determined. So that the coefficients of χ ' and of χ in

A solution to this equation exists if we choose f ' = −

mg 2T '3 (11) both vanish, we must choose w = . Thus, 2 37

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

mgx 'T ' mg 2T '3 +  2 Taking (12), (7), and (5) into account, we then obtain: ⎡ ⎤ ⎢ 2 2 ⎥ ⎛ 1 1 −x ' 2ima ⎞ ⎥ χ (x ',T ') = exp ⎢ 1+ i 1/ 4 2 2 ⎜ T ' ⎟⎠ ⎥ ( 2π ) a + iT ' ⎢ ⎛ 4a 2 +  T ' ⎞ ⎝ ⎢⎜ ⎥ 2ma m 2 a 2 ⎟⎠ ⎣⎝ ⎦ f =−

(12)

(13)

⎛ imx '2 img 2T '3 ⎞ exp ⎜ − 6 ⎟⎠ ⎝ 2T '

χ (x ',T ') = 2

Thus,

⎡ ⎤ ⎢ ⎥ 2 1 x' ⎢ ⎥ exp − ⎢ ⎛ 2  2T ' 2 ⎞ ⎥ ⎡ 2 ⎛ T ' ⎞ 2 ⎤ 2π ⎢ a + ⎜ ⎢ ⎜ 2a + ⎥ ⎟⎠ ⎥ 2m 2 a 2 ⎟⎠ ⎦ ⎝ 2ma ⎣ ⎝ ⎣ ⎦

(14)

which describes a free particle wave packet that does not accelerate but spreads in the normal way.

6.6 The potential V (x) = −V0 [δ (x − a) + δ (x + a)] is inversion-symmetric, and boundstate solutions are non-degenerate (although the delta function is infinite). Thus the bound states have definite parity, even or odd. The Schroedinger equation can be written: (1) ψ "− k 2ψ + U 0 [δ (x − a) + δ (x + a)]ψ = 0 where k 2 = − have:

2mE 2m E 2mV0 = , and U 0 = . For x > a , ψ "− k 2ψ = 0, and we must 2 2 2   

(2) ψ = Ae− kx (x>a) Thus for x < −a we have: (3) ψ = ±Aekx , (x < −a, ± for even, odd parity) kx − kx For x < a , ψ = B(e ± e ) . To determine B we must use the boundary conditions at x = ±a . At x = a, ψ is continuous, ψ ' is discontinuous, and ψ " is infinite. Since ψ is continuous, we have: B ( eka ± e− ka ) = Ae− ka and therefore A (4) = e2 ka ± 1 B To find a second constraint, we integrate (1) from x = a − ε to a + ε where ε is very small: 38

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 a+ε

a+ε

a+ε

a−ε

a−ε

a−ε

2 ∫ ψ "dx − k ∫ ψ dx + U0 ∫ δ (x − a)dx = 0

This yields: ψ '(a + ε ) − ψ '(a − ε ) − 2k 2εψ (a) + U 0ψ (a) = 0 In the limit as ε → 0 , we obtain: ψ '(a+) = ψ '(a−) − U 0ψ (a) Hence: (5) −kAe− ka = kB(eka  e− ka ) − U 0 Ae− ka which, for even parity, yields: 2ka e2 ka + 1 Even parity (6) = 2 ka = 1 + e−2 ka U0a e This equation can be solved graphically for given U 0 and a , as we see in Fig. 6.2S.

(

)

2ka/U0a

1+e-2ka

Fig.6.2S The figure shows that for any U 0 there is one and only one even-parity solution. For odd parity, (5) results in the transcendental equation: 2ka Odd parity (7) = 1 − e−2 ka U0a Here we have the graph: 1-e-2ka 2ka/U0a ka Fig.6.3S A solution exists if and only if the slope of 1 − e−2 y is greater than that of 2y / U 0 a at y=0. Thus we require that U 0 a > 1 for an odd-parity bound state. The following schematic figures reveal the general nature of the even and odd parity bound state wave functions as a function of the separation 2a between the delta functions.

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

Even parity! large a

2a

Odd parity! large a

2a Even parity! small a

Odd parity! small a

Fig. 6.4S For sufficiently large a, the even and odd parity solutions are determined by the asymptotic form of equations (6, 7):

2k (8) →1 U0 Hence both energies become the same (we have degeneracy). Physically this happens because the probability of tunneling between the regions x ≈ -a and x ≈ a goes to zero.

6.7a) We assume that ψ ψ = 1. Then,

∂ψ ∂ψ ψ + ψ =0 ∂λ ∂λ Now E(λ ) = ψ H (λ ) ψ , so:

∂E ∂ψ ∂ψ ∂H = Hψ + ψ H + ψ ψ ∂λ ∂λ ∂λ ∂λ ⎡ ∂ψ ∂ψ ⎤ ∂H = E⎢ ψ + ψ ψ ⎥+ ψ ∂λ ⎦ ∂λ ⎣ ∂λ ∂H = ψ ψ ∂λ

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

b) We have shown that any observable A with no explicit time dependence satisfies the equation: d 1 (1) ψ Aψ = ψ [ A, H ] ψ dt i Of course, if ψ is an eigenstate of Hamiltonian H, then both sides of (1) must vanish.

p̂ 2 + V (x), and assume that ψ is an eigenstate of H . Then we have: 2m ⎛ p̂ 2 ∂V ⎞ ∂V [ x̂p̂, H ] = i ⎜ − x ⎟ = i ⎛⎜⎝ 2T − x ⎞⎟⎠ ∂x ⎠ ∂x ⎝m

Let A = x̂p̂ , H =

and thus

ψ 2T ψ = ψ x

∂V ψ ∂x

(2)

p̂ 2 is the kinetic energy operator. Now, from the Feynman-Hellmann 2m theorem, we have: ∂E ∂H 1 1 ∂V = ψ ψ =− ψ T ψ =− ψ x ψ ∂m ∂m m 2m ∂x ∂V The energy differences are independent of particle mass if x = c is a constant; for in ∂x ∂E const ∂V this case , the same for each eigenstate. The equation x = = c implies that: ∂m m ∂x V (x) = b + cn ( x ) (3) where b is another constant. where T =

6.8

( Δx ) = x − x = mω (Q − Q ) 2

and

2

2

2

(

2

( Δp )2 = p 2 − p 2 = mω P 2 − P 2

)

Also, for any number eigenstate n , Q = P = 0. Thus,

ΔxΔp =  Q 2 ⋅ P 2

(1)

Now,

41

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

( a + a )( a + a ) n = 1 n aa + aa + a a + a a n Q = n †

†

2

†

2

= P = n 2

(

†

† †

2

) (

−i a − a † ⋅ −i a − a † 2

) n = − 1 n aa − aa − a a + a a n †

†

1 1 n 2a †a + 1 n = n + 2 2

1 ΔxΔp = (n + ) 2

6.9 Since ψ

† †

2

= Thus,

1 1 n 2a †a + 1 n = n + 2 2

(2)

is normalized to unity, we can write: A0 = reiα , B0 = 1 − r 2 eiβ where

(

)

0 ≤ r ≤ 1, α , β are real numbers. Since Q = 2 −1/2 a + a † we have:

ψ (0) Q ψ (0) = =

r 1 − r 2 ⎡ i( β − α ) e 0 a 1 + ei(α − β ) 1 a † 0 ⎤⎦ ⎣ 2 r 1 − r 2 i( β − α ) i(α − β ) ⎡e ⎤⎦ +e 2 ⎣

(1)

= 2r 1 − r 2 cos (α − β )

Now, r 1 − r 2 is maximum when r = 2 −1/2 , in which case r 1 − r 2 = 1/2 . Also, cos(α − β ) is maximum (=1) when α = β . Thus: 1 ψ (0) Q ψ (0) max = , 2 and this is achieved for A0 = B0 = 2 −1/2 (apart from an arbitrary common phase factor).

For t > 0, we have: e−iω t /2 e−iω Nt ψ (t) = U(t, 0) ψ (0) = ⎡⎣ 0 + 1 ⎤⎦ 2 e−iω t /2 ⎡⎣ 0 + e−iω t 1 ⎤⎦ = 2 Thus: 1 ψ (t) Q ψ (t) = 3/2 ⎡⎣ e−iω t 0 a 1 + eiω t 1 a † 0 ⎤⎦ 2 (2) 1 = cos ω t 2

42

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Also,

1 ψ (t) (a + a † )(a + a † ) ψ (t) 2 1 = ψ (t) aa + a †a † + aa † + a †a ψ (t) 2 For ψ (t) as given, the terms in aa and a †a † make no contribution. Also we can write:

ψ (t) Q 2 ψ (t) =

aa † + a †a = 2a †a + 1 = 2N + 1

Hence,

1 ⎡ 0 2N + 1 0 + 1 2N + 1 1 ⎤⎦ 4⎣ 1 = [1 + 3] = 1 4

ψ (t) Q 2 ψ (t) =

Therefore, Q2 t − Q t = 1 − 2

1 cos 2 ω t 2

(3)

6.10 Following the analysis of the coherent state given in Sec. 6.13.5 , we write: 2 † S(s) n = exp(−iP̂s) n = e− s / 4 esa / 2 e− sa / 2 n ≡ Ψ(t = 0)

(

)(

)

∴ Ψ(t) = U(t, 0) Ψ(0) = e− s / 4 Uesa / 2U † Ue− sa / 2U † U n 2

†

Now, † ⎛ s † −iω t ⎞ Uesa / 2U † = exp ⎜ ae ⎟ ⎝ 2 ⎠

⎛ s ⎞ Ue− sa / 2U † = exp ⎜ − aeiω t ⎟ ⎝ ⎠ 2 1 ⎤ ⎡ U n = exp ⎢ −iω (n + )t ⎥ 2 ⎦ ⎣ Therefore: 2 ⎡ s † −iω t ⎤ ⎡ s ⎤ Ψ(t) = e−iω (n +1/2)t e− s / 4 exp ⎢ a e ⎥ exp ⎢ − aeiω t ⎥ n ⎣ 2 ⎦ ⎣ 2 ⎦ ⎡ s ⎤ = e−iω (n +1/2)t exp ⎢ a †e−iω t − aeiω t ⎥ n ⎣ 2 ⎦ However, s a †e−iω t − aeiω t = −is Q̂ sin ω t + P̂ cost 2 Thus, (1) becomes:

(

(

)

(

)

(1)

)

43

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

⎛ s2 ⎞ Ψ(t) = e exp ⎜ i sin 2ω t ⎟ e−isQ̂ sin t e−isP̂ cos ω t n ⎝ 4 ⎠ This is also an oscillating coherent state.

6.11 For convenience in writing, we choose  = m = ω = 1. Then: 1 H = p̂x2 + p̂y2 + x̂ 2 + ŷ 2 2 with [ x̂, p̂x ] = ⎡⎣ ŷ, p̂y ⎤⎦ = i, −iω (n +1/2)t

(

Also,

(2)

)

(1)

⎡⎣ x̂, p̂y ⎤⎦ = [ ŷ, p̂x ] = 0

Lz = xpy − ypx .

Let: a=

(

)

(

1 1 1 1 ŷ + ip̂y , b † = ŷ − ip̂y ( x̂ + ip̂x ) , a† = ( x̂ − ip̂x ); b = 2 2 2 2

)

Then:

⎡⎣ a, a † ⎤⎦ = ⎡⎣b,b † ⎤⎦ = 1; ⎡⎣ a,b † ⎤⎦ = [ a,b ] = 0 Also it is easy to verify that: (2) Lz = i(ab † − ba † ) The Hamiltonian can be written as: 1 1 (3) H = a †a + + b †b + = a †a + b †b + 1 2 2 Since ⎡⎣ a †a,b †b ⎤⎦ = 0 , the energy eigenvalues are just the sums of the eigenvalues of

a †a and b †b , plus unity. Let us denote the eigenvalues of a †a and of b †b by n1, n2 respectively. Then E = n1 + n2 + 1 and we have the following table: n1 n2 E Degeneracy 0 0 1 1 1 0

0 1

2 2

2

2 0 1

0 2 1

3 3 3

3

3 2 1 0

0 1 2 3

4 4 4 4

4

…

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The Hamiltonian can be written in two other forms. Define: 1 1 † c= a − ib † ( a + ib ) , c† = 2 2 Then 2c†c = a † − ib † ( a + ib )

(

(

)

(

)

= a †a + b †b − i ab † − a †b

)

= a †a + b †b − Lz Thus the Hamiltonian can be written as: H = 1 + 2c†c + Lz Alternatively we can define: 1 1 † d= a + ib † ( a − ib ) , d † = 2 2 Then it is easy to verify that: 2d †d = a †a + b †b + Lz Thus the Hamiltonian can be written as: H = 1 + 2d †d − Lz

(

(4)

(5)

)

(6)

It is easy to show that ⎡⎣ c, c† ⎤⎦ = ⎡⎣ d, d † ⎤⎦ = 1 and that ⎡⎣ c†c, Lz ⎤⎦ = ⎡⎣ d †d, Lz ⎤⎦ = 0 . Since the eigenvalues of c†c and of d †d are each 0,1,2,… we can make the following table, taking into account equations (5) and (6): E

Eigenvalue of: cc d †d Lz

1

0

0

0

2 2

1 0

-1 1

0 1

†

3 3 3

2 -2 0 1 0 1 0 2 2 ….

Thus the possible eigenvalues of Lz are as follows: E Eigenvalues of Lz 1 0 2 -1,1 3 -2,0,2 4 -3,-1,1,3

45

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When E=2, it is obvious that the eigenstates of Lz are: 1 { n1 = 1, n2 = 0 ± i n1 = 0, n2 = 1 } ≡ ± 2 The corresponding wave functions are: ⎛ x 2 + y2 ⎞ ψ ± = N ( x ± iy ) exp ⎜ − 2 ⎟⎠ ⎝ where N is a normalization factor.

6.12 Choose the vector potential to be axially symmetric: B (1) A = ( −yx̂ + xŷ ) 2 where x̂, ŷ are the usual unit vectors in the x,y directions respectively. The Hamiltonian is: 2 2 pz2 1 ⎛ eB ⎞ 1 ⎛ eB ⎞ H= p + y + p − x + (2) ⎜ x ⎟ ⎜ y ⎟ 2m ⎝ 2c ⎠ 2m ⎝ 2c ⎠ 2m pz2 Since commutes with the Hamiltonian, free particle motion in the z direction is 2m p2 eB and H 0 = H − z . Then: independent of motion in the xy plane. Let Ω = 2mc 2m

(

)

1 2 1 2 mΩ2 2 H0 = px + py + x + y 2 − Ω xpy − ypx 2m 2m 2 (3) 1 2 1 2 mΩ2 2 = px + py + x + y 2 − ΩL0z 2m 2m 2 where L0z = xpy − ypx . As in the case of the simple harmonic oscillator it is convenient to define: 1/2 ⎛ mΩ ⎞ X=⎜ x ⎝  ⎟⎠

⎛ mΩ ⎞ Y =⎜ ⎝  ⎟⎠

(

)

(

)

1/2

y

⎛ 1 ⎞ Px = ⎜ ⎝ mΩ ⎟⎠

1/2

⎛ 1 ⎞ Py = ⎜ ⎝ mΩ ⎟⎠

1/2

px py

Then: H0 =

(

)

2 Ω ⎡ 2 Px + Y ) + Py − X ⎤ ( ⎦ 2 ⎣

(4)

46

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Note that this Hamiltonian is invariant if Px → Px − β , Y → Y + β Py → Py + α , X → X + α , where α , β are constants.

(

)

and/or if

Define a = 2 −1/2 ( X + iPx ) , b = 2 −1/2 Y + iPy , c = 2 −1/2 ( a + ib ) . Then,

( ) ( 1⎞ ⎛ = Ω ( 2c c + 1) = ω ⎜ c c + ⎟ ⎝ 2⎠

H 0 = Ω a †a + b †b + 1 − iΩ ab † − a †b †

where ω =

†

)

(5)

eB . It is easy to check that: mc 1 c = ⎡⎣ X + iPx + i(Y + iPy ) ⎤⎦ 2 1⎡ ∂ ∂ ⎤ = ⎢X + + i(Y + ) 2⎣ ∂X ∂Y ⎥⎦

Also it can be verified that ⎡⎣ c, c† ⎤⎦ =1 and that c is invariant under the transformation Px → Px − β , Y → Y + β , and/or the transformation Py → Py + α , X → X + α . The ground state is found from the condition c 0 = 0. In coordinate representation this yields:

∂ ∂ ⎤ ⎡ (6) ⎢⎣ X + ∂X + i(Y + ∂Y ) ⎥⎦ ψ 0 ( X,Y ) = 0 We try a solution of the form: ψ 0 = F(x)G(y) . Substitution in (6) yields: ∂F ∂G XF + YG + ∂X = −i ∂Y (7) F G Since the left hand side of (7) is a function of X, and the right hand side is a function of Y, both are equal to a constant k: 1 ∂F X+ =k F ∂X 1 ∂F =k−X F ∂X nF = kX − X 2 /2 + const Thus, 2 F = constie− X /2 ekX and similarly, 2 G = constie−Y /2 e−ikY Therefore: 2 2 (8) ψ = constie− X /2 e−Y /2 ekX e−ikY Let k = α + iβ where α , β are real. Then the exponent on the right side of (8) is: 47

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

− X 2 /2 + α X − Y 2 /2 + βY + iβ X − iαY = 1 α2 1 β2 2 2 − (X − α ) + − (Y − β ) + + iβ X − iαY 2 2 2 2 ⎡1 ⎤ Since the quantity exp ⎢ (α 2 + β 2 ) ⎥ can be absorbed in the constant appearing on the ⎣2 ⎦ right side of (8), we have: 2 2 (9) ψ = constie− ( X − α ) /2 e− (Y − β ) /2 eiβ X e−iαY Note that α , β are just the shifts under which the Hamiltonian H0 is invariant: they correspond to the arbitrariness of the center of classical circular motion. Thus there is a large amount of degeneracy for the ground state (and for excited states as well).

Concerning this last remark, we could have chosen a different form of the vector potential by making a gauge transformation. For example, we could have chosen: A = −Byx̂ Here our ground state wave function would consist of plane waves in the x and z directions and a one dimensional harmonic oscillator solution in the y direction. Such a solution could be constructed from a suitable linear combination of solutions of the form (9) for ground and excited states.

If the charged particle is confined to a finite region in the xy plane, we can calculate the degree of degeneracy in a simple way as follows. First assume that B=0 and that we have periodic boundary conditions at the perimeter of a square box of edge L in the XY plane.

Then the motion in the XY plane is free particle motion described by the wave function: eik1 x eik2 y 2π n1 2π n2 where k1 = and n1,2 are positive or negative integers. The energy is: , k2 = L L

 2 k 2  2 4π 2 2 = n1 + n22 2 2m 2m L In how many ways can we choose n1,2 to give an energy less than or equal to E? We can imagine n1 and n2 as the coordinates of points on a square lattice in a plane. Then to a very good approximation for n1 and n2 sufficiently large, mL2 E n12 + n22 ≤ R 2 = 2π 2  2 Here, R is the radius of a circle in the lattice plane. Thus the number of distinct lattice vertices is: mL2 E π R2 = 2π  2 and the number of distinct states per unit of energy is: mL2 ρ= 2π  2 E=

(

)

48

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When B ≠ 0 , the separation between energy levels is ω =

e B . In this interval, there mc

e mL2 eBL2 B = states, which is the degeneracy. Note that BL2 and 2π c/e mc 2π  2 2π c are magnetic fluxes; (the latter is the “quantum unit of magnetic flux”). Thus Z is the number of units of flux that can be accommodated in area L2.

Finally the wave function (9), which corresponds to the ground state energy, is like a harmonic oscillator eigenstate in 2 dimensions. Thus we cannot obtain a “classical” rotating wave packet from it. To do so, we must form a suitable coherent state; that is, a superposition of eigenstates of c†c . This can be done in essentially the same way as for a one dimensional simple harmonic oscillator. are Z =

6.13. P n = ( −1) n = e n

iπ n

∞

n =∑

(iπ n )k n = ∞ (iπ N )k n = eiπ N n

∑

k! k! k=0 Also, since the eigenstates n of N form a complete set, any arbitrary one dimensional k=0

state Ψ can be expressed as a superposition of n states: Thus,

Ψ =∑ n nΨ P Ψ = ∑ P n n Ψ = e iπ N Ψ

6.14a) We write: 2

2 r 1 ⎛ r⎞ ⎡r ⎤ S = exp ⎢ a 2 − a † 2 ⎥ = I + a 2 − a † 2 + ⎜ ⎟ a 2 − a † 2 + ...

2 2! ⎝ 2 ⎠ ⎣2 ⎦

(

)

(

)

2

(

)

2 r 1 ⎛ r⎞ ⎡ r ⎤ ∴ S = I − a 2 − a † 2 + ⎜ ⎟ a 2 − a † 2 − ... = exp ⎢ − a 2 − a † 2 ⎥ = S −1 2 2! ⎝ 2 ⎠ ⎣ 2 ⎦ so S is unitary.

Now, a 2 − a † 2 = (a − a † )(a + a † ) − 1 = (a + a † )(a − a † ) + 1 †

(

)

(

= 2iP̂Q̂ − 1

)

(

)

= 2iQ̂P̂ + 1

Therefore:

S = S1 = er /2 eirQ̂P̂ = s1/2 eirQ̂P̂

(1) = S2 = e− r /2 eirP̂Q̂ = s −1/2 eirP̂Q̂ b) In coordinate representation it is obviously most convenient to use S1 which becomes: ⎛ ⎛ ∂⎞ ∂ ⎞ (2) S1 = s1/2 exp ⎜ rQ = s1/2 exp ⎜ r ⎟ ⎟ ⎝ ∂Q ⎠ ⎝ ∂y ⎠ where y = n Q . c) The ground state coordinate wave function of the simple harmonic oscillator is: 49

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ψ 0 (Q) = π −1/ 4 e−Q /2 ≡ φ (y) 2

and: ⎛ ∂⎞ S1ψ (Q) = S1φ (y) = s1/2 exp ⎜ r ⎟ φ (y) ⎝ ∂y ⎠ =s

1/2

⎛ ⎞ ∂ 1 2 ∂2 1/2 ⎜⎝ 1 + r ∂y + 2! r ∂y 2 + ...⎟⎠ φ (y) = s φ (y + r)

(3)

= s1/2ψ 0 ( sQ ) = π −1/ 4 s1/2 e− s Q /2 2

2

The ground state momentum wave function is χ 0 (P) = π −1/ 4 e− P /2 ≡ η(z) where z = n P .

Here it is obviously most convenient to use S2: ∂ ⎞ ∂⎞ ⎛ ⎛ S2 = s −1/2 eirPQ → s −1/2 exp ⎜ −rP ⎟ = s −1/2 exp ⎜ −r ⎟ (4) ⎝ ⎝ ∂z ⎠ ∂P ⎠ Thus, S2η(z) = s −1/2η(z − r) = s −1/2 χ 0 s −1P 2

(

)

(5)

⎛ P2 ⎞ = π −1/ 4 s −1/2 exp ⎜ − 2 ⎟ ⎝ 2s ⎠

From (3, 5) we see that for the squeezed state r ≡ S(r) 0 , 1 s ΔQ = , ΔP = (6) s 2 2 d)We start with the squeezed wave function (3) and calculate how it evolves in time by using the Green function for the harmonic oscillator, which is derived in Chapter 6 (see 1/2 1/2 ⎛ mω ⎞ ⎛ mω ⎞ x ', Q = equation 6.97). Let Q = ⎜ ⎜⎝ ⎟ x0 . Then we obtain: 0 ⎝  ⎟⎠  ⎠

ψ (Q,t) =

∞ ⎛ s 2Q02 ⎞ 1 2QQ0 ⎤ ⎫ ⎧i ⎡ 2 2 cos ω t exp (Q + Q ) − exp ⎬ 0 ∫ ⎨⎩ 2 ⎢⎣ ⎜⎝ − 2 ⎟⎠ dQ0 2π i sin ω t −∞ sin ω t sin ω t ⎥⎦ ⎭

s1/2 π 1/ 4

(7) The integral is evaluated in the usual way by completing the square in the exponent and using the standard formula for a Gaussian integral. We obtain: ψ (Q,t) =

⎡ ⎤ ⎡ iQ 2 sin 2ω t(s 4 − 1) ⎤ s 2Q 2 exp ⎢ − ⎥ exp ⎢ ⎥ π 1/ 4 1 2 cos 2 ω t + s 4 sin 2 ω t ⎥⎦ 4 cos 2 ω t + s 4 sin 2 ω t ⎥⎦ ⎢ ⎢ ⎣ ⎣ cos ω t + is sin ω t s (8) Note that when s = 1 (8) reduces to: 2 1 (9) ψ (Q,t) = 1/ 4 e−iω t /2 e−Q /2 π which is just what we would expect by applying the time development operator to the harmonic oscillator ground state wave function. From (8) with s ≠ 1 we obtain: 1

1

(

)

(

)

50

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

⎡ ⎤ ⎢ ⎥ 1 1 Q2 ψ (Q,t) = 1/2 ⎥ (10) 1/2 exp ⎢ − 1 π ⎡ 2 2 2 2 2 1 ⎤ 2 ⎢ ⎥ cos ω t + s sin ω t ⎢⎣ s sin ω t + s 2 cos ω t ⎥⎦ ⎣ s2 ⎦ 2

Thus:

1 s2 2 2 cos ω t + sin ω t (11) 2s 2 2 Since the Hamiltonian is symmetric in Q̂ 2 and P̂ 2 we can perform a similar calculation to find the time evolution of the momentum wave function (5), by constructing the momentum space analog of the Green function (7). This results in: s2 1 (12) ( ΔP )2 = cos2 ω t + 2 sin 2 ω t 2 2s Therefore, 1⎡ 1 1 ⎤ (13) ( ΔQ )2 ( ΔP )2 = ⎢ cos 4 ω t + sin 4 ω t + ⎛⎜⎝ 4 + s 4 ⎞⎟⎠ sin 2 2ω t ⎥ 4⎣ 4 s ⎦

( ΔQ )2 = Q̂ 2 =

This implies that ΔQΔP = and any s ≠ 1 .

1 π 3π 1 π 3π for ω t = 0, , π , ,... , but ΔQΔP > for ω t = , ,...

2 2 2 2 4 4

(

)

e) Next consider T (r) = exp ⎡⎣ r ab − a †b † ⎤⎦ where: 1 1 ( x + ipx ) , b = (y + ipy ) 2 2 † † Thus, ab − a b = i(xpy + ypx ) . Let: a=

1 1 (x − y), x2 = (x + y) 2 2 ∂ 1 ⎛ ∂ ∂ ⎞ ∂ 1 ⎛ ∂ ∂ ⎞ Then, . Thus we can write: = + , = − + ⎜ ⎟ ⎜ ∂x 2 ⎝ ∂x1 ∂x2 ⎠ ∂y 2 ⎝ ∂x1 ∂x2 ⎟⎠ x1 =

ab − a †b † =

(14)

⎛ ∂ ⎛ ∂ 1 ∂ ⎞ 1 ∂ ⎞ ( x1 + x2 ) ⎜ − + ⎟ + ( −x1 + x2 ) ⎜ + ⎟ 2 ⎝ ∂x1 ∂x2 ⎠ 2 ⎝ ∂x1 ∂x2 ⎠

1 ∂ 1 ∂ = − x1 + x2 2 ∂x1 2 ∂x2

Consequently, ⎡ r ∂ r ∂ ⎤ T (r) = exp ⎢ − x1 + x2 ⎥ ⎣ 2 ∂x1 2 ∂x2 ⎦ ∂ ∂ Since x1 we can write (15) as: commutes with x2 ∂x1 ∂x2

(15)

51

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⎡r ⎡ r ∂ ⎤ ∂ ⎤ T (r) = exp ⎢ x2 ⎥ exp ⎢ − x1 ⎥ ⎣ 2 ∂x2 ⎦ ⎣ 2 ∂x1 ⎦

Now, ⎛ x 2 + y2 ⎞ ⎛ x12 + y22 ⎞ −1/2 ψ 0 ( x, y ) = π −1/2 exp ⎜ − = π exp ⎜⎝ − 2 ⎟⎠ 2 ⎟⎠ ⎝

Therefore ⎛ x2 ⎞ ⎛ x2 s2 ⎞ T (r)ψ 0 ( x, y ) = π −1/2 exp ⎜ − 1 2 ⎟ exp ⎜ − 2 ⎟ 2 ⎠ ⎝ 2s ⎠ ⎝

Hence:

s s 2 2 s 1 Δx2 = , Δpx 2 = 2 s 2 Δx1 =

1

, Δpx1 =

6.15 We start with: u0 = N 0 ( sec hβ x ) . Then: p −1 β sinh β x u0 ' = − pN 0 ( sec hβ x ) = − pβ u0 itanh β x cosh 2 β x Hence pβ W= tanh β x 2 pβ 2 W'= sec h 2 β x 2 1 p2β 2 pβ 2 V− = W 2 − W'= tanh 2 β x − sec h 2 β x 2 2 2 p2β 2 pβ 2 = 1 − sec h 2 β x − sec h 2 β x 2 2 1 p2β 2 pβ 2 V+ = W 2 + W'= 1 − sec h 2 β x + sec h 2 β x 2 2 2 p

(

)

(

)

p 2 β 2 ( p − 1)pβ 2 − sec h 2 β x 2 2 Clearly we can construct a sequence of potentials: V− ,V+ = V1 ,V2 ,V3 ,...,Vn ,... where =

52

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

p 2 β 2 ( p − n ) ( p − n + 1) 2 − β sec h 2 β x 2 2 ⎡ ( p − n)2 β 2 ( p − n ) ( p − n + 1) 2 ⎤ p 2 − ( p − n)2 2 2 =⎢ − β sec h β x ⎥ + β 2 2 2 ⎣ ⎦ Thus we can write: n Vn = V− ( p − n, β ) + n( p − )β 2 2 The lowest eigenstate of Vn is the function: Vn =

u0(n) (x) = const ( sec hβ x ) For this function to represent a bound state we must have p > n (so that the function vanishes at x = ±∞ ). The energy of this state is: n E0(n) = n( p − )β 2 2 1⎞ ⎛ p −1 For n = 1, E0(1) = ⎜ p − ⎟ β 2 and u0(1) = const ( sec hβ x ) . Thus, ⎝ 2⎠ p−n

⎛ ∂ u '⎞ p −1 u1 = constiA†u0(1) = const ⎜ + 0 ⎟ [ sec hβ x ] ⎝ ∂x u0 ⎠ = consti( sec hβ x )

p −1

tanh β x

Finally consider the potential: V = −V0 sec h 2 β x p 2 β 2 p( p + 1) 2 − β sec h 2 β x . Let us assume that: We compare it to V− ( p; x) = 2 2 2 p( p + 1)β V0 = 2 which is equivalent to: 1 1 8V p=− + 1 + 20 2 2 β

p2β 2 Then, V = V− − . Hence the eigenvalues of V are: 2 p2β 2 n p2β 2 En = E0(n) − = n( p − )β 2 − 2 2 2 2 β β2 =− p 2 − 2np + n 2 = − ( p − n)2 2 2

(

)

β2 ⎛ 8V ⎞ = − ⎜ −1 − 2n + 1 + 20 ⎟ 8 ⎝ β ⎠

2

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7.1. a) We write J x =

1 1 ( J + + J − ) , J y = ( J + − J − ) . Then, since: 2 2i

J + jm =

j( j + 1) − m(m + 1) j, m + 1

J − jm =

j( j + 1) − m(m − 1) j, m − 1

it is obvious that jm J x jm = jm J y jm = 0 . b) For any operator A we have: ⎡⎣ A,(J x J y − J y J x ) ⎤⎦ = i ⎡⎣ A, J z ⎤⎦ Now,

(

) (

)

⎡⎣ A,(J x J y − J y J x ) ⎤⎦ = AJ x J y − J x J y A − AJ y J x − J y J x A (1) If A commutes with J x and with J y , both terms in parentheses on the right hand side of

(1) vanish. Hence A commutes with J z .

c) Let ψ = ∑ a jm jm and let us assume that this state is normalized to unity, which j,m

implies that:

∑a

2 jm

=1

(2)

j, m

We require ΔJ x = ΔJ y = ΔJ z = 0. If ΔJ z = 0, we must have: 2

ψ J z2 ψ = ψ J z ψ hence:

∑a

2 jm

jm

m 2 = ∑ a jm m ∑ a j ' m ' m ' 2

2

jm

(3)

j'm'

Using (2) we can write (3) in two different ways:

∑a

jm

∑a

jm

2

jm j'm'

a j ' m ' m 2 = ∑ a jm m ∑ a j ' m ' m '

(4)

a j ' m ' m '2 = ∑ a jm m ∑ a j ' m ' m '

(5)

2

2

jm

2

j'm'

and 2

jm j'm'

2

2

jm

2

j'm'

Adding (4) to (5) and transposing we obtain:

∑a

2 jm

aj 'm'

2

( m − m ' )2 = 0

(6)

jm j'm'

This is a sum of non-negative terms; hence we require m = m ' if a jm ≠ 0, a j ' m ' ≠ 0.

Thus only a single value of m can contribute to ψ :

54

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ψ = ∑ a jm jm

(7)

j

2

2

J x2 = J x , J y2 = J y . However, we know that for a single

We also require that value of m: J x = J y = 0.

Thus

ψ J x2 + J y2 ψ = ψ J 2 − J z2 ψ = 0 which gives:

∑a

2 jm

j

⎡⎣ j( j + 1) − m 2 ⎤⎦ = 0

(8)

2

Since for a jm ≠ 0 we require m 2 ≤ j 2 , (8) can only be satisfied for j = 0 . Thus we can have ΔJ x = ΔJ y = ΔJ z = 0 only in the case j = m = 0. d) For an arbitrary state ψ which could be a superposition of different jm states, it is convenient to define the operators: J x 0 = J x − J x , J y0 = J y − J y .

Clearly, ⎡⎣ J x 0 , J y0 ⎤⎦ = iJ z (where here we exhibit  explicitly), and also:

(9)

( ΔJ x )2 = J x20 , ( ΔJ y ) = J y02 2

(10)

From the Cauchy-Schwarz inequality we have:

J

2 x0

J

≥ J x 0 J y0

2 y0

2

 = α + i Jz 2

2

where α is a real number. Thus,

 Jz 2 e) From the obvious inequality: ΔJ x ΔJ y ≥

(11)

( ΔJ − ΔJ ) ≥ 0 2

x

we obtain:

y

( ΔJ x )2 + ( ΔJ y ) ≥ 2ΔJ x ΔJ y ≥  J z 2

(12)

55

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

( ) = J

It follows that if ( ΔJ x ) + ΔJ y 2

2

z

, we must have ΔJ x ΔJ y =

 Jz .

2

f) For any state jm , we have jm J x jm = jm J y jm = 0 . Thus,

( ΔJ x )2 + ( ΔJ y ) = jm J x2 + J y2 jm 2

= jm J 2 − J z2 jm

(13)

=  2 ⎡⎣ j( j + 1) − m 2 ⎤⎦

If and only if m = ± j , (13) becomes:

( ΔJ x )2 + ( ΔJ y ) =  2 ⎡⎣ j( j + 1) − j 2 ⎤⎦ 2

= 2 j = 2 m =  J z

7.2a) We start with: γ Hψ = − σ iBψ = iψ 2 or: ⎛ a ⎞ ⎧ iγ B1 iγ B0 ⎫ ⎛ a ⎞ ⎜⎝ b ⎟⎠ = ⎨ 2 ⎡⎣σ x cos ω t − σ y sin ω t ⎤⎦ + 2 σ z ⎬ ⎜⎝ b ⎟⎠ ⎩ ⎭

iγ B1 iω t ⎞ ⎛ iω 0 e ⎜ ⎟ ⎛ a⎞ 2 2 =⎜ ⎟⎜ ⎟ iω ⎝ b⎠ ⎜ iγ B1 e−iω t − 0 ⎟ ⎝ 2 2 ⎠ Thus we have the two coupled first order equations: ω γB a = i 0 a + i 1 eiω t b 2 2 ω γB b = −i 0 b + i 1 e−iω t a 2 2 It is convenient to make the following substitutions: ⎛ iω t ⎞ ⎛ iω t ⎞ A = a exp ⎜ − 0 ⎟ , B = b exp ⎜ + 0 ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Then, equations (1) result in: iγ B1 i (ω − ω 0 )t A = Be 2 iγ B1 −i (ω − ω 0 )t B = Ae 2 We now differentiate both sides of (3) with respect to t :  = iγ B1 ⎡ A − i (ω − ω 0 ) A ⎤ e−i(ω − ω 0 )t B ⎦ 2 ⎣

(1)

(2) (3)

(4)

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In the right hand side of (4), we substitute the right hand side of (2) for A , and we use (3) to express A in terms of B . Thus after simple algebra we obtain the decoupled second order differential equation: γ 2 B12   (5) B + iΔB + B=0 4 iλt where Δ = ω − ω 0 . We try a solution to (5) of the form e . This results in the quadratic equation: γ 2 B12 (6) λ 2 + Δλ − =0 4 which has the solutions: Δ 1 λ=− ± Δ 2 + γ 2 B12 2 2 Thus (7) B = α e−iΔt /2 ei Δ + γ B1 t /2 + β e−iΔt /2 e−i Δ + γ B1 t /2 where α , β are coefficients to be determined by initial conditions. First, we note that at t = 0 , B(0) = α + β = 0 ; hence Ωt B = 2iα e−iΔt /2 sin (8) 2 2

2

2

2

2

2

where Ω = Δ 2 + γ 2 B12 . Next we differentiate both sides of (8) and use (3) to obtain:

Ωt Ωt ⎤ γ B ⎡ α ⎢ Δ sin + iΩ cos ⎥ = i 1 Ae−iΔt /2 2 2 ⎦ 2 ⎣ γB At t = 0, A = 1 , hence α = 1 . Therefore, 2Ω γ B1 −iΔt /2 Ωt B=i e sin Ω 2 and iγ B1 −iω t /2 Ωt b = Be−iω 0 t /2 = e sin Ω 2 The transition probability is: 2 γ B1 ) ( Ωt 2 b = sin 2 2 Ω 2 which is indeed Rabi’s formula. b) The rotation matrix d for J=1 is:

(9)

(10)

(11)

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sin β ⎛ 1 + cos β − ⎜ 2 2 ⎜ sin β d=⎜ cos β ⎜ 2 ⎜ ⎜ 1 − cos β sin β ⎜⎝ 2 2 Applying this to the m = 0 3-spinor: ⎛ 0⎞ ψ 10 = ⎜ 1 ⎟ ⎜ ⎟ ⎝ 0⎠ we obtain the transition probabilities:

1 − cos β ⎞ ⎟ 2 ⎟ sin β ⎟ − 2 ⎟ ⎟ 1 + cos β ⎟ ⎟⎠ 2

1 2 sin β 2 β β = 2 sin 2 cos 2 2 2 β β⎤ ⎡ = 2 ⎢sin 2 − sin 4 ⎥ 2 2⎦ ⎣

P0→1 = P0→−1 =

where, as usual,

β (γ B1 ) Ωt sin = sin 2 2 2 Ω 2 2

2

7.3 In the coordinate representation, L = r × p = −ir × ∇ r In the momentum representation, L = r × p = i∇ p × p = −ip × ∇ p Thus the eigenfunctions of L2 , Lz in momentum representation are also the spherical

()

harmonics: Ym k̂ where k̂ is a unit vector in the direction of momentum.

7.4 The relevant formula for part a) is eq. (7.153) of Chapter 7:

gJ =

1 ⎧ ⎡ j( j + 1) + ( + 1) − s(s + 1) ⎤ ⎡ j( j + 1) + s(s + 1) − ( + 1) ⎤ ⎫ + gs ⎢ ⎨ g ⎢ ⎥ ⎥⎦ ⎬ j( j + 1) ⎩ ⎣ 2 2 ⎦ ⎣ ⎭

(1)

For the nucleus 15 N we have an odd proton in a p1/2 state, so j = 1 / 2,  = 1, s = 1 / 2.

Also, g = 1, gs = 2i2.79 = 5.58. Inserting these numbers in (1), we obtain g j = −.527 ,

(

= −.263 nm. The observed value is ) −.527 2

hence µshell model 15 N =

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µobs ( 15 N ) = −.28313 nm . The discrepancy is not too surprising, given the naïve simplicity of the nuclear shell model. b) For the nucleus 17 O we have one neutron in a d5 /2 state outside of closed shells. Thus, j = 5 / 2,  = 2, s = 1 / 2. Also, g = 0 since the neutron has no charge and thus no orbital magnetic moment, while gs = −1.91i2 = −3.82. Inserting these numbers in (1) we

( )

( )

obtain µshell model 17 O = −1.91 nm. The observed value is µobs 17 O = −1.89370 nm. c) Here the deuteron magnetic moment, in units of the nuclear Bohr magneton, is given by the formula: ⎡L ⎤ ⎢⎣ 2 + g p S p + gn Sn ⎥⎦iJ (2) µ= J +1 The term L/2 in the square brackets is due to the orbital motion of the proton, with orbital angular momentum one half of the total orbital angular momentum L. Since the neutron has no charge it gives no analogous contribution. Since: 1 1 g p S p + gn Sn = g p + gn S + g p − gn S p − Sn 2 2 and since S p − Sn = 0 for triplet states, (2) can be written as:

(

(

)

(

)(

)

)

1 ⎡⎣ L + g p + gn S ⎤⎦iJ µ= 2 J +1

(

)

g p + gn ⎛ J(J + 1) + S(S + 1) − L(L + 1) ⎞ 1 ⎛ J(J + 1) + L(L + 1) − S(S + 1) ⎞ = ⎜ + ⎟⎠ ⎜⎝ ⎟⎠ 4⎝ J +1 4 J +1

(3)

Using g p + gn = 5.58 − 3.82 = 1.76 we obtain the following results:

µ ( 3 S1 ) = .888, µ ( 3 P1 ) = .698, µ ( 3 D1 ) = .318

7.5 The given facts concerning this reaction are as follows: a) The incoming deuteron is polarized in the m =+1 state. b) The triton target is unpolarized: there are equal numbers of tritons in the m =±1/2 states. c) The reaction proceeds in the J=3/2 channel. d) The orbital angular momentum of relative motion of the deuteron and triton is  = 0. e) Parity is conserved in the reaction. f) The neutron has spin ½, and the alpha particle has zero spin.

From these facts we can immediately draw the following conclusion, without any calculation: Since the orbital angular momentum of the initial state is zero, the parity, (−1) , is positive. Therefore the parity of the final state is also positive. Hence the orbital angular momentum of relative motion of the neutron and the alpha particle must be even: 59

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 = 0, 2, 4,... a priori. However, since angular momentum is conserved, J = 3 / 2 for the final states. The only way to construct a final state with J = 3/2 from a neutron spin of ½ and an even orbital angular momentum is with  = 2 ; therefore the orbital angular momentum of relative motion in the final state is  = 2.

Now consider the possible initial states, which must have J=3/2. If the triton has spin up (down) along z, it is described by the spinor α T , ( βT ) . We have: 3 3 , = 1,1 D α T (1) 2 2 and: 3 1 2 1 , = 10 D α T + 11 D βT 2 2 3 3

1 1 1 2 , = 10 D α T − 11 D βT 2 2 3 3

Thus,

1 31 2 11 − (2) 3 22 3 22 Of course the reaction cannot proceed in the J =1/2 channel; thus only the first term on the right hand side of (2) is relevant for this problem. So far, we see from (1) that if the 33 triton spin is up, the final state must be , and from (2) we see that if the triton spin 22 31 is down, the final state must be with a relative probability 1/3.

22 Now consider the possible final states which have orbital angular momentum  = 2 . It is easy to find the various Clebsch-Gordan coefficients in the following formulae, by using raising and lowering operators: 33 4 2 1 1 = Y2 β n − Y2 α n 22 5 5 (3) 31 3 1 2 0 = Y2 β n − Y2 α n 22 5 5 Hence the angular intensity distribution for the neutron with spin up is proportional to: 1 12 12 02 neutron spin up (4) Y2 + Y2 5 35 while the angular intensity distribution for neutron spin down is proportional to: 4 22 13 12 neutron spin down (5) Y2 + Y2 5 35 Now: 11 D βT =

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15 sin 4 θ 32π 2 15 2 Y21 = sin θ cos 2 θ 8π 2

Y22 =

(6) 2

5 ⎛3 1⎞ 2 Y = ⎜⎝ cos θ − ⎟⎠ 4π 2 2 Employing (6) in (4), (5) we obtain the results: S+ = A(1 + 3cos 2 θ ) 0 2 2

S− = A(9 sin 2 θ ) where A is a constant.

7.6 a) We start with: jm = ∑ j1m1 j2 m2

j1m1 j2 m2 jm

(1)

and the inverse relation: j1m1 j2 m2 = ∑ jm j1m1 j2 m2 jm

(2)

m1 , m2

j

(Of course, j1m1 j2 m2 jm = 0 unless m = m1 + m2 ). In (2) apply the rotation operator R(α , β , γ ) to both sides:

∑D

j1 m1 ' m1

m1 ' m2 '

Dmj22 ' m2 j1m1 ' j2 m2 ' = ∑ Dmj ' m jm ' j1m1 j2 m2 jm

(3)

j, m '

On the right hand side of (3), expand jm ' as in (1). Then, (3) becomes:

∑D

j1 m1 ' m1

Dmj22 ' m2 j1m1 ' j2 m2 ' =

m1 ' m2 '

∑ D

j m'm

j1m1 " j2 m2 " j1m1 " j2 m2 " jm ' j1m1 j2 m2 jm

(4)

j, m ' m1 ", m2 "

On the right hand side of (4) the first Clebsch-Gordan coefficient is non-zero only if m1 "+ m2 " = m ' , so the sum is really over j, m1 ", and m2 " only. In (4) take the scalar product of both sides with j1m1 ''' j2 m2 ''' . Then we obtain:

Dmj11 ''' m1 Dmj22 ''' m2 = ∑ Dmj ' m j1m1 j2 m2 jm

j1m1 ''' j2 m2 ''' jm '

(5)

j

Replacing the symbols m1,2 ''' by m1,2 ' in (5) we finally obtain:

Dmj11 ' m1 Dmj22 ' m2 = ∑ Dmj ' m j1m1 j2 m2 jm

j1m1 ' j2 m2 ' jm '

(6)

j

We can also apply the rotation operator to both sides of (1) as follows: ∑ Dmj ' m jm ' = ∑ Dmj11 ' m1 Dmj22 ' m2 j1m1 ' j2 m2 ' j1m1 j2 m2 jm m'

(7)

m1 , m2 m1 ', m2 '

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Now we use (2) to rewrite (7) as: ∑ Dmj ' m jm ' = ∑ Dmj11 ' m1 Dmj22 ' m2 j ', m1 '+ m2 ' j1m1 j2 m2 jm m'

j1m1 ' j2 m2 ' j ', m1 '+ m2 '

m1 , m2 m1 ', m2 ' j'

Taking the scalar product of this expression with symbols by j, m ' we finally obtain:

∑ D

Dmj ' m =

j1 m1 ' m1

Dmj2'− m1 ', m − m1 j1m1 j2 m2 jm

(8) j"m ''' and replacing the latter (9)

j1m1 ' j2 m '− m1 ' j ', m '

m1 ', m2 '

1 ; and employ (9). We calculate 4 of the 9 matrix elements. The 2 other 5 are calculated in a similar way.

2 2 1111 β 1 1/2 1/2 1/2 D1,1 = D1/2,1/2 D1/2,1/2 11 = D1/2,1/2 = exp [ −i(α + γ )] cos 2 2222 2

b) Let j = 1, j1 = j2 =

(

)

⎛ 1 + cos β ⎞ = exp [ −i(α + γ )] ⎜ ⎟⎠ ⎝ 2 1 1/2 1/2 D1,0 = 2D1/2,1/2 D1/2,−1/2

1111 11 2222

1111 10 2222

1/2 1/2 = 2D1/2,1/2 D1/2,−1/2

β ⎤⎡ β⎤ ⎡ = 2 ⎢ e−i(α + β )/2 cos ⎥ ⎢ −e−i(α − γ )/2 sin ⎥ 2 ⎦⎣ 2⎦ ⎣ −iα e sin β =− 2 2 2 1 2 11 1 1 1/2 1/2 D1,−1 = D1/2,−1/2 − − 1 − 1 = D1/2,−1/2 2 22 2

(

)

(

)

β⎞ 1 − cos β ⎛ = ⎜ −e−i (α − γ ) /2 sin ⎟ = e−i (α − γ ) ⎝ ⎠ 2 2 2

1 0,0

D

= 2D

1/2 1/2,1/2

= cos 2

1/2 −1/2,−1/2

D

111 1 − 0 222 2

2 1/2 1/2 + 2D1/2,−1/2 D−1/2,1/2

111 1 − 0 222 2

1 111 − 0 2 222

β β − sin 2 = cos β 2 2

c) Consider the quantity: 

I = ∑ Ym* (θ 2 , φ2 )Ym (θ1 , φ1 )

(10)

m=−

Let us rotate the coordinate system through Euler angles α , β , γ . Then:

62

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and

Ym (θ1 , φ1 ) = R −1Ym [ R (θ1 , φ1 )] = R −1Ym (θ1 ', φ1 ')

Ym* (θ 2 , φ2 ) = ⎡⎣ R −1Ym (θ 2 ', φ2 ') ⎤⎦ * Since the rotation matrix is unitary,

(

R −1Ym (θ1 ', φ1 ') = ∑ Dm ' m m'

) Y (θ ',φ ') −1

m' 

1

1

* m' =∑ Dmm 'Y (θ1 ', φ1 ' )

(11)

m'

and

 m "* ⎡⎣ R −1Ym (θ 2 ', φ2 ') ⎤⎦ * = ∑ Dmm (θ 2 ',φ2 ') "Y

(12)

m"

Thus,

I=

∑ D

* mm '

 m "* Dmm (θ 2 ',φ2 ')Ym ' (θ1 ',φ1 ') "Y

(13)

m, m ', m "

The unitarity of D implies that:

∑D

* mm '

 Dmm " = δ m ', m "

m

and thus that:

I = ∑ Ym* (θ 2 ', φ2 ')Ym (θ1 ', φ1 ') m

In other words, I is invariant under rotations of the coordinate system. Let us choose α , β , γ so that θ1 ' = 0 and φ2 ' = 0. Then, θ 2 ' = θ is the angle between the rays defined by θ1 , φ1 and θ 2 , φ2 . Also,

⎛ 2 + 1 ⎞ Ym ( 0, φ1 ') = δ m 0 ⎜ ⎝ 4π ⎟⎠ Therefore: 1/2  ⎛ 2 + 1 ⎞ 0 I =⎜ Y θ , 0 = Ym* (θ 2 , φ2 )Ym (θ1 , φ1 ) ( ) ∑ ⎟  ⎝ 4π ⎠ m=− which is the spherical harmonic addition theorem.

1/2

(14)

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8.1The radial Schroedinger equation for H (Z=1) in atomic units is: d 2 R 2 dR ⎡ 2  (  + 1) ⎤ (1) + + ⎢ 2E + − R=0 2 dr r dr ⎣ r r 2 ⎥⎦ F (η ) 1 Let r = λη 2 and R = . Then, after straightforward manipulations, (1) becomes: 2 η 2 ⎡ d 2 F 1 dF λ 2η 2 ( 2 + 1) ⎤ + + 2 ⎢ 2λ − − ⎥F = 0 dη 2 η dη 2n 2 2η 2 ⎥⎦ ⎢⎣

where we have used the Balmer formula: E = −

(2)

1 . Now consider a 2-dimensional 2n 2

ω2 2 η . Here ω is a constant and η is the radial 2 coordinate in plane polar coordinates. Assuming the oscillator has definite angular momentum M, the wave function is: ψ = f (η ) eiM φ and f satisfies the radial equation: ⎡ d 2 f 1 df ω 2η 2 M 2 ⎤ (3) + + 2 ⎢ E0 − − 2⎥f =0 dη 2 η dη 2 2η ⎦ ⎣ harmonic oscillator with potential V =

where E0 = ω ( n0 + 1) and n0 = 0,1, 2,... . Also, for given n0 , M = n0 , n0 − 2,..., −n0 .

(Recall problem 6.11.) Equations (2) and (3) are formally identical if we make the following choices: λ ω= n0 = 2n − 1 M = ± ( 2 + 1) n Thus we obtain the following correspondences: H atom 2d harmonic oscillator n =1 = 0 n0 =1 M = ±1 n = 2 =1 n0 =3 M = ±3 n=2 =0 n0 =3 M = ±1 n=3 =2 n0 =5 M = ±5 n = 3 =1 n0 =5 M = ±3 n=3 =0 n0 =5 M = ±1 For n0 = 1, M = ±1 the normalized 2d oscillator radial wave function is: ⎛ ωη 2 ⎞ . The corresponding H atom wave function is: f (η ) = π −1/2η exp ⎜ − ⎝ 2 ⎟⎠ ⎛ ωη 2 ⎞ 1 −1/2 ψ 10 ( r ) = f (η ) = π exp ⎜ − = π −1/2 e− r ⎟ η ⎝ 2 ⎠

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8.2 a) The one-dimensional radial equation for this problem is: 2µ (1) χ "+ 2 ⎡⎣ E + V0 e− r /a ⎤⎦ χ = 0  r Let z = exp(− ) . Then: 2a 1 z dz = − e− r /2a dr = − dr 2a 2a Thus, ∂ z ∂ ∂2 1 ∂⎛ ∂⎞ =− , = 2 z ⎜z ⎟ 2 ∂r 2a ∂z ∂r 4a ∂z ⎝ ∂z ⎠ hence (1) can be written: ∂⎛ ∂⎞ 8 µa 2 z ⎜ z ⎟ χ + 2 ⎡⎣ E + V0 z 2 ⎤⎦ χ = 0 (2) ∂z ⎝ ∂z ⎠  8 µa 2V0 2 8µa 2 , ν = 2 E , and x = κ z. Then, assuming that E < 0, (2) becomes: Define κ 2 = 2  ∂⎛ ∂⎞ x ⎜ x ⎟ χ + x2 − ν 2 χ = 0 (3) ∂x ⎝ ∂x ⎠ which is Bessel’s differential equation. The boundary conditions are: χ (r = 0) = χ (r = ∞) = 0 , which are equivalent to: χ (x = 0) = χ (x = κ ) = 0 The solution to (3) satisfying these conditions is: χ = AJν (x) = AJν (κ z ) (4) The Bessel function Jν (x) has the following properties:

(

)

• J 0 (0) = 1 but Jν ( 0 ) = 0 for all ν > 0.

• J 0 (2.405) = 0 (The first non-zero root of J 0 is 2.405.)

• The first root of Jν is an increasing function of ν .

Therefore the first bound state appears when 1/2

⎛ 8 µa 2V0 ⎞ κ =⎜ = 2.405 ⎝  2 ⎟⎠ Hence we require: ( 2.405 )2  2 = .723  2 for E ≤ 0 .

V0 a 2 ≥ (5) 8µ µ In the case of interaction between a neutron and a proton, both particles have very nearly the same mass m p , so µ = m p /2 and (5) becomes: V0 a 2 ≥ 1.45

2 mp

(6)

b) Choosing a = 1.4i10 −13 cm in (6), we have in cgs units: 65

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

V0 ≥

(

1.45i 1.054i10 −27

)

(1.4i10 ) 1.67i10 −13 2

2

−24

= 4.92i10 −5 erg

= 30.8 MeV c) We start from the approximation that the magnetic moment of the deuteron is constructed from the vector sum of the spin magnetic moments of the proton and the neutron, plus the orbital magnetic moment of the proton. Thus the deuteron magnetic moment operator is: 1 (7) µ D = µ N g p S p + µ N gn Sn + µ N gLp L 2 Here the first term on the right hand side of (18) refers to the proton spin magnetic moment, the second to the neutron spin magnetic moment, and the third to the proton orbital magnetic moment. In the latter term, gLp = 1 is the orbital g-value, and the factor of ½ must be included because the proton’s orbital motion is about the center of mass, which is located midway between the two nucleons. Now, according to the analysis of the Lande g-factor, the actual magnetic moment of the deuteron (that is, the expectation value of µ D in the state I = 1, mI = +1 where I is the nuclear spin) is given by the formula: µ iΙ µ 1 µD = D = N g p S p + gn Sn iI + LiI I +1 I +1 2 (8) µN 1 1 1 = g p + gn S p + Sn iI + g p − gn S p − Sn iI + LiI I +1 2 2 2 This expression simplifies because the expectation value of S p − Sn vanishes for triplet states, and S p + Sn = S . Hence (8) becomes:

(

(

)

)(

)

(

)(

)

µN 1 1 g p + gn SiI + LiI I +1 2 2 µ ⎧1 1 ⎫ = N ⎨ g p + gn [ I(I + 1) + S(S + 1) − L(L + 1)] + [ I(I + 1) + L(L + 1) − S(S + 1)]⎬ 2 ⎩4 4 ⎭ (9) 3 For the S1 state, I = 1, S = 1, L = 0 . Hence: µ (10) µ D ( 3 S1 ) = N g p + gn = µ p + µn 2 For the 3 D1 state, I = 1, S = 1, L =2. Thus (9) gives: µ ⎧ 1 3⎫ 1 3 µ D ( 3 D1 ) = N ⎨− g p + gn + ⎬ = − µ p + µn + µ N (11) 2 ⎩ 2 2⎭ 2 4 Since the probabilities that the deuteron is in the 3 S1 , 3 D1 states are 1 − w, w respectively, (10) and (11) together give:

(

µD =

(

)

)

(

)

(

)

(

)

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3 ⎤ ⎡3 µ D = µ P + µn − w ⎢ ( µ P + µn ) − µ N ⎥ 4 ⎦ ⎣2 The experimental values of the magnetic moments are: µ p = 2.79274 µN

(12)

µn = −1.91314 µ N µ D = .857438 µN Inserting these in (12) and solving for w , we obtain: .857438 − (2.79274 − 1.91314) w= = .03892 (13) 3 3 − (2.79274 − 1.91314) 4 2 Actually, various relativistic effects, meson exchange current effects, etc, make calculation of the deuteron magnetic moment a good deal more complicated than our simple model would suggest. However the model is valid as a first approximation.

8.3 In units where  = m = 1 the radial Schroedinger equation for the function χ = rR is: 2q − ( + 1) ⎤ ⎡ χ "+ ⎢ −2 E + (1) ⎥⎦ χ = 0 r2 ⎣ For given finite E, (1) becomes the following at sufficiently small r: γ (2) χ "+ 2 χ = 0 r where γ = 2q − ( + 1) . We try a solution of the form: χ = r p . Then: p( p − 1) + γ = 0 which has the solutions: 1 1 1 1 (3) p1 = + 1 − 4γ , p2 = − 1 − 4γ .

2 2 2 2 Thus we write: (4) χ = Ar p1 + Br p2 γ To find a relationship between the coefficients A, B we replace in (2) by the r2 following: γ 0 ≤ r ≤ r0 r02 (5) γ r ≥ r0 r2 (Ultimately we shall take the limit as the constant r0 goes to zero.) It is easy to see that with the replacement (5) the solution is:

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χ (r < r0 ) = C sin

γ 1/2 r r0

χ ( r ≥ r0 ) = Ar + Br where C is another constant. Since χ and χ ' are continuous at r0, we have: p1

(6)

p2

Ar0 p1 + Br0 p2 = C sin γ 1/2

γ 1/2 cos γ 1/2 r0 Dividing the second equation by the first we obtain: Ap1r0p1 + Bp2 r0 p2 1/2 1/2 γ cot γ = Ar0p1 + Br0 p2 B p1 + p2 r0p2 − p1 A = B p2 − p1 1 + r0 A Now substituting p1,2 from (3) in (7), we obtain: Ap1r0p1 −1 + Bp2 r0p2 −1 = C

(7)

B −D ⎞ ⎛ 1 − r0 1 D⎜ ⎟ 1/2 1/2 A (8) γ cot γ = + ⎜ 2 2 1 + B r−D ⎟ ⎜⎝ ⎟ A 0 ⎠ where D = 1 − 4γ . This equation can be re-arranged to yield: B (9) = F(γ )r0 1− 4 γ A D + 1 − 2γ 1/2 cot γ 1/2 F( γ ) = where . Of course, for given γ , F(γ ) is just a constant.

D − 1 + 2γ 1/2 cot γ 1/2 Now, in (9) we take the limit as r0 goes to zero. Here we distinguish between the three cases: γ < 1/4, γ = 1/4, and γ >1/4. a) γ < 1/4 Here, we see from (9) that lim B/A = 0 . Hence for sufficiently small r , r0 →0

1/2 +1/2 1− 4 γ ) χ = Ar (

(10) Indeed, if E = 0 this solution is valid for the entire range of r from 0 to ∞. However this E = 0 solution has no nodes. Therefore there are no solutions for E < 0 ; that is, there are no bound states. b) γ = 1/4 1 Here, p1 = p2 = , so χ = Ar1/2 for sufficiently small r, and this solution is valid for 2 arbitrarily large r if E = 0. Thus there are no bound states.

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c) γ > 1/4 .

Here, 1 − 4γ = i 4γ − 1 , so (9) becomes: (11) B/A = F(γ )ei 4 γ −1 n r0 This ratio approaches no definite limit as r0 → 0 ; so here we cannot conclude that B = 0.

1 i 1 i Indeed, we now have p1 = + 4γ − 1, p2 = − 4γ − 1 = p1 * ; thus if χ is to be a 2 2 2 2 real function, it must be of the form: ⎡1 ⎤ χ = Ar1/2 cos ⎢ 4γ − 1 n r ⎥ (12) ⎣2 ⎦ This function oscillates infinitely rapidly as r approaches zero, for any given negative energy E. Thus there is an infinite number of bound states, and there is no lowest bound state: the lowest bound state would have E=-∞.

8.4a) The potential is: k k k V = x 2 + y2 + z 2 = ρ 2 + z 2 = r 2 2 2 2

(

)

(

)

where k = mω 2 and ρ = x 2 + y 2 is the radial coordinate in cylindrical polar coordinates, while r = x 2 + y 2 + z 2 is the radial coordinate in spherical polar coordinates. Thus it is obvious that the Schroedinger equation can be separated in Cartesian, cylindrical polar, and spherical polar coordinates. b) In Cartesian coordinates it is obvious that the energy eigenvalues are: En = ω nx + ny + nz + 3/2 (1)

(

)

where the number operator eigenvalues are nx , ny , nz . Given the value of n, we can choose

nx = j where j = 0,1,2,…,n. Then, ny + nz = n − j . In this case we can choose ny = k where k = 0,1,2,…, n – j, and nz is then fixed. For any given j there are n − j + 1 possible values of k. Therefore the degeneracy is: n

n

j=0

j=0

gn = ∑ ( n − j + 1) = (n + 1)2 − ∑ j n(n + 1) 2 (n + 1)(n + 2) = 2 In the following table we list several values of n, the values of orbital angular momentum  corresponding to each n for wave functions in spherical polar coordinates, and the corresponding degeneracies gn. = (n + 1)2 −

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n



gn

0 1 2 3 4 5 6

0 1 0,2 1,3 0,2,4 1,3,5 0,2,4,6

1 3 6=1+5 10=3+7 15=1+5+9 21=3+7+11 28=1+5+9+13

For n = 3, the wave functions with  = 1, 3 are constructed from linear combinations of the 10 possible wave functions in Cartesian coordinates. Choosing units with  = m = ω = 1 , the Cartesian wave functions are:

ψ1 =

1 4x 3 − 6x ⎡ 1 ⎤ iexp ⎢ − x 2 + y 2 + z 2 ⎥ 3/ 4 π 6 ⎣ 2 ⎦

ψ2 =

1 4y 3 − 6y ⎡ 1 ⎤ iexp ⎢ − x 2 + y 2 + z 2 ⎥ 3/ 4 π 6 ⎣ 2 ⎦

(

)

(

)

1 4z 3 − 6z ⎡ 1 ⎤ iexp ⎢ − x 2 + y 2 + z 2 ⎥ 3/ 4 π 2 6 ⎣ ⎦ 1 ⎡ 1 ⎤ ψ 4 = 3/ 4 (2x 2 − 1)yiexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦

(

ψ3 =

)

(

)

⎡ 1 ⎤ (2x 2 − 1)ziexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦ 1 ⎡ 1 ⎤ ψ 6 = 3/ 4 (2y 2 − 1)ziexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦

ψ5 =

1

3/ 4

(

)

(

)

⎡ 1 ⎤ (2y 2 − 1)xiexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦ 1 ⎡ 1 ⎤ ψ 8 = 3/ 4 (2z 2 − 1)xiexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦

ψ7 =

1

3/ 4

(

)

(

)

⎡ 1 ⎤ (2z 2 − 1)yiexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦ 3/ 4 2 ⎡ 1 ⎤ ψ 10 = 3/ 4 xyziexp ⎢ − x 2 + y 2 + z 2 ⎥ π ⎣ 2 ⎦

ψ9 =

(

1

3/ 4

(

)

)

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To find the radial functions Rn = 3,=1 , R3, 3 identified with  = 1, 3 respectively we write the 9 one dimensional radial Schroedinger equation satisfied by χ = rR for En = 3 = : 2 2 ⎛ r ( + 1) ⎞ χ "+ 2 ⎜ E3 − − χ=0 2 2r 2 ⎟⎠ ⎝ Thus: 2 (2) χ 3,1 "+ (9 − r 2 − 2 ) χ 3,1 = 0 r and: 12 (3) χ 3, 3 "+ (9 − r 2 − 2 ) χ 3, 3 = 0 r It is easy to verify that: 2 5 χ 3,1 = const(r 4 − r 2 )e− r /2 2

(4)

and

χ 3, 3 = constir 4 e− r /2 Thus the 7 wave functions identified with n = 3,  = 3 are: 2

ψ 3, 3, m ( r,θ , φ ) = constir 3e− r /2Y3m (θ , φ ) while the 3 wave functions identified with n = 3,  = 1 are: 2 5 ψ 3,1, m (r,θ , φ ) = consti(r 3 − r)e− r /2Y1m 2 Of course, in each case the constant is fixed by normalization.

2

8.5 We start with: χ "+ f (r) χ = 0 where:  2  (  + 1) ⎤ 2µ ⎡ f (r) = 2 ⎢ E − V (r) − ⎥  ⎣ 2 µr 2 ⎦

(5) (6) (7)

(1) (2)

We assume that χ → a r +1 as r → 0 . Multiply both sides of (1) by r q χ ' and integrate from ε to ∞. Then we have: ∞

∞

ε

ε

− ∫ r q χ ' χ "dr = ∫ r q χχ ' fdr

Integration of the left hand side of (3) by parts yields: ∞ 1 1 ∞ − r q χ '2 ε + q ∫ r q −1 χ '2 dr 2 2 ε Integration of the right hand side of (3) by parts gives:

(3) (4)

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1 q 2 ∞ q ∞ q −1 2 1 ∞ r f χ ε − ∫ r f χ dr − ∫ r q f ' χ 2 dr 2 2 ε 2 ε

(5)

For bound states, r q f χ 2 = r q χ '2 = 0 at r = ∞. Thus: 1 q 2 1 ∞ 1 q ∞ 1 ∞ r χ ' ε + q ∫ r q −1 χ '2 dr = − r q f χ 2 ε − ∫ r q −1 f χ 2 dr − ∫ r q f ' χ 2 dr 2 2 ε 2 2 ε 2 ε

(6)

However, ∞

∞

∞

∞

ε

ε

q ∫ r q −1 χ '2 dr = qr q −1 χ χ ' ε − q ∫ r q −1 χχ "dr − q(q − 1) ∫ r q − 2 χχ 'dr ε

(7)

If q > −2 the first term on the right hand side of (7) vanishes in the limit as ε → 0 since:

χ = a r +1

χ ' = a (  + 1) r 

χχ ' = a2 (  + 1) r 2+1 The second term on the right hand side of (7) is: ∞

∞

ε

ε

−q ∫ r q −1 χχ "dr = q ∫ r q −1 χ 2 fdr = q r q −1 f

Integrating the last term on the right hand side of (7) by parts we obtain: ∞ q(q − 1) q − 2 2 ∞ q(q − 1)(q − 2) q − 3 −q(q − 1) ∫ r q − 2 χχ 'dr = − r χ ε + r ε 2 2 Thus, (6) yields: q(q − 1) q − 2 2 q(q − 1)(q − 2) q − 3 r q χ '2 ε + q r q −1 f + r χ ε+ r = 2 2 = −r q f χ 2 ε − q r q −1 f − r q f '

(8) (9)

(10)

Once again we use the fact that χ → a r +1 as r → 0 . Then, q(q − 1) q − 2 2 r q χ '2 ε + r χ ε + rq f χ 2 ε = 2 q 2 ⎡ ⎤ 2 q + 2 = δ q,−2 a2 (2 + 1)2 ⎢⎣(  + 1) + 2 ( q − 1) +  (  + 1) ⎥⎦ a r ε Thus finally, (10) yields:

δ q,−2 a2 (2 + 1)2 = − 2qr q −1 f − r q f ' −

q(q − 1)(q − 2) q − 3 r 2

(11)

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Choosing units where µ =  = e = 1 from now on, we employ the very useful formula (11) as follows: 1) For q =  = 0 we have: ∂V a02 = − f=' 0 = 2 (12) ∂r However, R(0) = χ (0)/r = a0 . Thus: ∂V R(0)2 = 2 ∂r Hence, 1 1 ∂V 2 ψ = 0 ( 0 ) = R(0)2 = (13) 4π 2π ∂r The Coulomb potential is V = −Z /r . Thus: Z 1 2 Coulomb potential ψ = 0 ( 0 ) = 2 r 2 = 0

(14)

2) q = 1 Here, (11) yields: 0 = 2 f + rf '

1 ∂V 1 ⎤ ⎡ = 2 ⎢ 2E − 2 V − ( + 1) 2 − r + ( + 1) 2 ⎥ r ∂r r ⎦ ⎣ ∂V ⎤ ⎡ = 2 ⎢ 2E − 2 V − r ∂r ⎥⎦ ⎣ ∂V ⎤ ⎡ = 2 ⎢2 T − r ∂r ⎥⎦ ⎣ Hence:

∂V ∂r which is just the virial theorem. c) q = 2 Here (11) gives: 4rf + r 2 f ' = 0 2 T = r

This yields:

1 ∂V − r2 =0 r ∂r For the Coulomb potential in atomic units (15) becomes: 4E r − 4 rV −  (  + 1)

(15)

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4E r + 4Z −  (  + 1)

1 −Z =0 r

which yields:

−3Z ( + 1) 1 + (16) 4E 4E r 1 1 2 = − V = − E . Thus, (16) can be written: However, r Z Z 3Z ( + 1) 1 ⎡⎣ 3n 2 −  (  + 1) ⎤⎦ (17) r =− − = 4E 2Z 2Z where we have used the Balmer formula. To summarize, the basic formula (11) can be used to obtain the following results for the Coulomb potential: r =

q =  = 0 : The wave function at the origin expressed in terms of r −2 = 0 Z r −2 >0 ( + 1) q = 1: The Virial Theorem 1 ⎡⎣ 3n 2 − ( + 1) ⎤⎦ q=2: r = 2Z n2 ⎡ 5n 2 − 3 (  + 1) + 1⎤⎦ q = 3 : r2 = 2Z 2 ⎣ ...

2 2Z ⎡ 3n − ( + 1) ⎤⎦ −3 q = −1 : r −4 = 2 ⎣ r n 4( + 1) − 3

q = 0,  ≠ 0 : r −3 >0 =

(18) (19) (20) (21) (22)

(23)

We see that all quantities r p where p is a positive integer can be obtained from (11) with q = p + 1 . If p = −3, −4,... , and  > 0 we can obtain

r p recursively from r −2 ,

by using q = 0, −1, −2,... However (11) by itself does not yield r −2 for any value of  .

8.6 a) We start with: L = r × p = −i ∫ (ψ * r × ∇ψ )dτ Now

 i r × ψ * ∇ψ − r × ψ∇ψ *] − [ r × ψ * ∇ψ + r × ψ∇ψ *] [ 2mi 2 i = mr × j − r × ∇ (ψ * ψ ) 2 i = mr × j − r × ∇ρ 2

−i (ψ * r × ∇ψ ) = m

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However, r × ∇ρ = −∇ × ( r ρ ) . Thus: i L = m ∫ r × jdτ + ∫ ∇ × ( r ρ )dτ 2 i = m ∫ r × jdτ + ∫ ρ r × n̂dS 2 The surface integral vanishes if ρ → 0 more rapidly than r −1 as r → ∞ . This is always the case for bound states. Hence for bound states we have: L = m ∫ r × jdτ (1) b) We want to demonstrate the identity: 1 1 1 (2) ψ †σψ = r × ⎡⎣∇ × ψ †σψ ⎤⎦ − ∇ ⎡⎣ riψ †σψ ⎤⎦ + ∂ j ⎡⎣ x jψ †σψ ⎤⎦ 2 2 2 where here and in what follows we use the repeated index summation convention. On the right hand side of (2): ⎡ r × ∇ × ψ †σψ ⎤ = xi ∂ ψ †σ mψ ε mnε ink ⎣ ⎦

(

)

k

( ) = x ( ∂ ψ σ ψ ) (δ δ − δ δ ) = x (∂ ψ σ ψ ) − x (∂ ψ σ ψ ) †

i



i

k

m

k mi

i

i

†

i mk

†

i

−∇ k ⎡⎣ riψ †σψ ⎤⎦ = −∂ k ⎡⎣ x jψ †σ jψ ⎤⎦

(

= −ψ †σ kψ − xi ∂ k ψ †σ iψ and:

(

k

)

)

∂i xiψ †σ kψ = 3ψ †σ kψ + xi ∂iψ †σ kψ Collecting these terms, we see that the right hand side of (2) is indeed equal to the left hand side. Now, from (2):  S = ∫ ψ †σψ dτ 2 (3)    † † † = ∫ r × ⎡⎣∇ × ψ σψ ⎤⎦ dτ − ∫ ∇ ⎡⎣ riψ σψ ⎤⎦ dτ + ∫ ∂ j ⎡⎣ x jψ σψ ⎤⎦ dτ 4 4 4 However, † † (4) ∫ ∇ ⎡⎣ riψ σψ ⎤⎦ dτ = ∫ riψ σψ n̂ds

(

)

which vanishes if ψ is a bound state. Similarly, the third term on the right hand side of (3) can also be converted to a vanishing surface integral. Thus:  S = ∫ r × ⎡⎣∇ × ψ †σψ ⎤⎦ dτ 4 (5) = m ∫ r × js dτ

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 † ψ σψ . Note that ∇i js = 0 so js makes no contribution 4m to the motion of particles. In this it is fundamentally different from the ordinary probability current density. ⎛ 1⎞ c) We write ψ = F(r) χ + , where F is the spatial part of the wave function and χ + = ⎜ ⎟ is ⎝ 0⎠ 2 2 a two-component spinor. Then ψ †σψ = F χ † σ iˆ + σ ĵ + σ k̂ χ = F k̂ . Thus,

where js = ∇ × Vs and Vs =

+

js =

(

x

( )

y

z

)

+

1 1 1∂F 2 ∇ × ψ †σψ = ∇ F × k̂ = r̂ × k̂ 4 4 4 ∂r

(

)

2

1∂F = sin θφ̂ 4 ∂r 1⎞ e− r ⎛ 3r r 2 ⎞ ⎛ ∴ js ⎜ 2 2 s1/2 , ms = + ⎟ = − 1 − + ⎟ sin θφ̂ ⎝ 2⎠ 16π ⎜⎝ 4 8⎠ 2

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9.1Choose units with  = m = 1. Including the correct normalization factor, the trial function is: 1/ 4 2 ⎛ 2λ ⎞ ψ = ⎜ ⎟ e− λ x ⎝ π ⎠ The expectation value of the kinetic energy is: 2 1 ∞ T = ∫ [ψ (x)] dx 2 −∞ ∞ 2 3/2 = 1/2 λ 5 /2 ∫ x 2 exp −2 λ x 2 dx (1) −∞ π λ = 2 The expectation value of the potential energy is: 2 λ ∞ −2 λ x 2 V =− e V (x) dx (2) π ∫−∞ Thus the energy calculated for this trial function is: λ 2 ∞ −2 λ x 2 E(λ ) = − λ 1/2 e V dx 2 π ∫−∞ E(λ ) < 0 if:

(

)

π 1/2 1/2 λ (3) ∫−∞ 2 3/2 This inequality is obviously satisfied if λ = 0 . As λ increases from zero, the left hand side of (3) decreases continuously, and the right hand side increases continuously.

Therefore, for given V (x) there is a positive real value of λ ( = λ0 ) such that the inequality of (3) is replaced by an equality; and for all values λ < λ0 the inequality (3) holds. Thus there exists at least one bound state in the potential V (x) , no matter how weak it is. ∞

e−2 λ x V dx > 2

9.2 We assume that ψ is a function of r only, and write ψ = Y00 iu/r . We want to prove that if u has a radial node (that is, if u(r0 ) = 0 for 0 < r0 < ∞ ) then ψ cannot be the f (r) 0 ground state. To this end we construct a trial function φ (r) = Y0 and show that: r φHφ ψ Hψ (1) < φφ ψψ For convenience assume that ψ ψ = 1, and also m =  = 1. Choose an interval r0 − ε − δ < r < r0 + ε + δ , where ε , δ are small positive quantities. (See Fig. 9.1S).

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Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 u,w,f

w,f f u,w

r0-ε-δ

r0-ε

r0

r0+ε

r0+ε+δ

u

Fig.9.1S First make a Taylor expansion of u(r) about r0: 1 1 3 (2) u(r) ≈ u(r0 ) + u '(r0 )(r − r0 ) + u"(r0 )(r − r0 )2 + u ''' ( r0 ) ( r − r0 ) 2!

3!

Now u ''+ 2(E − V )u = 0 and u(r0 ) = 0 ; hence the first and third terms on the right hand side of (2) are zero; thus we have: (3) u(r) = −a(r − r0 ) + c(r − r0 )3 where a > 0 and c are constants. In fact, for sufficiently small ε, δ we can ignore the term in c on the right hand side of (3), so we can just assume that: u(r) = −a(r − r0 ) It is also convenient to construct the following two functions shown in Fig. 9.1S: w=u r < r0 and:

w = −u

r > r0

f =w f = aε

r < r0 − ε r0 − ε < r < r0 + ε

f =w

r > r0 + ε

Now,

φφ =∫

r0 − ε

f 2 dr + ∫

∞

r0 + ε

0

= 1+ ∫

r0 + ε

= 1+ ∫

r0 + ε

r0 − ε

r0 − ε

f 2 dr + ∫

r0 + ε

r0 − ε

f 2 dr

( f − u ) dr 2

2

(a ε − a (r − r ) ) dr 2 2

(4)

2

2

0

= 1 + O(ε 3 ) Therefore, ignoring terms of order ε 3 , we have:

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φHφ − ψ Hψ = φHφ − ψ Hψ φφ Now

Q≡ φ H φ − ψ H ψ

(

)

∞ ∞ 1 ∞ = − ⎡⎢ ∫ ff "dr − ∫ uu"dr ⎤⎥ + ∫ f 2 − u 2 Vdr 0 ⎦ 0 2⎣ 0 ∞ ∞ 1 ∞ = − ⎡⎢ ∫ ff "dr − ∫ ww"dr ⎤⎥ + ∫ f 2 − w 2 Vdr 0 ⎦ 0 2⎣ 0 r0 + ε 1 r0 + ε + δ = − ⎡∫ ff "− ww") dr ⎤ + ∫ f 2 − w 2 Vdr ( ⎦⎥ r0 − ε 2 ⎣⎢ r0 − ε − δ

(

)

(

(

(5)

)

)

r0 + ε 1 r0 + ε + δ ff "dr + ∫ f 2 − w 2 Vdr ∫ r0 − ε 2 r0 − ε − δ The second integral on the right hand side of the last line of (5) is V(r0) multiplied by a quantity of order ε 3 , and can therefore be neglected. However, the first integral on the right hand side is significant. Here, since f ' is discontinuous at r = r0 ± ε , we have:

=−

f " = a ⎡⎣δ ( r − r0 + ε ) + δ ( r − r0 − ε ) ⎤⎦ Thus, taking the limit as δ → 0, and keeping ε very small but non-zero, we obtain: (6) Q = −a 2ε which proves that ψ cannot be the ground state.

A nodeless ground state cannot be degenerate, for if it were there would exist at least two linearly independent functions ψ 1,2 without nodes. Any arbitrary linear combination of these would also be an eigenstate of the Hamiltonian with the same energy. We could choose this linear combination in such a way that it would vanish at any given r.

9.3a) According to the variational principle, H = T + V is minimized by a ψ ( r ) that satisfies Schroedinger’s equation. Suppose that we have found such a function ψ ( r ) , and we replace it by means of the norm-preserving transformation: ψ ( r ) → a 3/2ψ ( ar ) .

Then H → H ( a ) . The extremum property requires that:

d H (a) = 0 . We now da a =1

prove that this yields the virial theorem. First, write: 2 2 2 H (a) = ∇ rψ ( ar ) a 3d 3 r + ∫ ψ ( ar ) V ( r ) a 3d 3 r (1) ∫ 2m Let r ' = ar . Then ∇ rψ ( ar ) = a∇ r' ( r ') and a 3d 3 r = d 3 r ' . Thus (1) can be written:

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H (a) = a 2

2 2 ⎛ r '⎞ 2 ∇ r 'ψ ( r ') d 3 r '+ ∫ ψ ( r ') V ⎜ ⎟ d 3 r ' ∫ ⎝ a⎠ 2m

⎛ r⎞ = a2 T + V ⎜ ⎟ ⎝ a⎠ Now,

(2)

d ⎛ r '⎞ r 'i∇ r '/a ⎛ r ' ⎞ d ⎛ r '⎞ V⎜ ⎟ = − V ⎜ ⎟ . Thus, lim V ⎜ ⎟ = −r 'i∇ r 'V ( r ') . Therefore the 2 a→1 ⎝ a⎠ da ⎝ a ⎠ a da ⎝ a ⎠

requirement

d H (a) = 0 yields 2 T − ri∇V = 0 which is the virial theorem. da a =1

⎡i ⎤ b) Next we show that the unitary operator: U(a) = a 3/2 exp ⎢ ( ln a ) ri p ⎥ satisfies the ⎣ ⎦ equation: (3) U(a)ψ ( r ) = a 3/2ψ ( ar ) Proof: consider a function β (a) and an operator function D ( r, p = −i∇ ) , and write: ⎡i ⎤ ψ ( ar ) = exp ⎢ β ( a ) D ⎥ ψ ( r ) (4) ⎣ ⎦ Differentiate both sides of (4) with respect to a: dψ ( ar ) i d β (5) = Dψ ( ar ) da  da dψ ( ar ) 1 However, = ri∇ rψ ( ar ) . Therefore (5) can be written: da a i dβ ⎞ ⎛1 D ⎟ ψ ( ar ) = 0 (6) ⎜⎝ ri∇ r − a  da ⎠ which is satisfied by D = −iri∇ r and d β /da = 1/a , hence β = ln(a) since β (1) = 0. c) Define A = ln(a) , so that: ⎡i ⎤ U = e3A /2 exp ⎢ Ari p ⎥ (7) ⎣ ⎦ Then, H (a) = ψ U † ( a ) HU(a) ψ (8) = e3A ψ exp ( −iApir/ ) H exp ( iAri p/ ) ψ The extremum condition in terms of A is: d H (A) =0 dA A= 0 Applying this to (8) and using pir = ri p − 3i we obtain:

[ ri p, H ] = 0

(9)

which is the starting point for the usual derivation of the virial theorem.

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9.4 Here we make use of the following integrals: ∞ π I 0 = ∫ exp −α 2 x 2 dx = −∞ α ∞ π I 2 = ∫ x 2 exp −α 2 x 2 dx = −∞ 2α 3 ∞ 3 π I 4 = ∫ x 4 exp −α 2 x 2 dx = −∞ 4α 5 ∞ 15 π I 6 = ∫ x 6 exp −α 2 x 2 dx = −∞ 8α 7 Start with the trial ground state function: α 1/2 ψ = 1/ 4 exp −α 2 x 2 (1) π Here: 1 ∞ ∂ 2ψ ψ T ψ = − ∫ ψ 2 dx 2 −∞ ∂x 1 α ⎡α 4 I 2 − α 2 I 0 ⎤⎦ (2) =− 1/2 ⎣ 2π α2 = 4 and: α 3 (3) ψ V ψ = 4 1/2 I 4 = 4 π α Thus: α2 3 E= T + V = + 4 (4) 4 α and so α 12 ∂E/∂α = − 5 2 α E thus reaches a minimum with respect to variation of α when α = 31/6 21/2 . At this value, E in (4) becomes: 34 / 3 E= = 1.08168718 (5) 4 which is to be compared with E0=1.060362.

Next consider the trial function: 21/2 α 3/2 −α 2 x 2 /2 ψ= xe π 1/ 4 Now the expectation value of kinetic energy is:

(

)

(

(

)

(

)

(

)

)

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T =− while

1 2α 3 4 3α 2 2 ⎡ ⎤ α I − 3 α I = 4 2⎦ 2 π 1/2 ⎣ 4

(6)

2α 3 15 I6 = 4 α π

(7)

V = 4i Thus

E= hence

3α 2 15 + 4 4 α

(8)

∂E 3α 60 = − ∂α 2 α5 which vanishes when α = 51/6 21/2 . Inserting this value into the right hand side of (8) we obtain: 9i51/ 3 E= = 3.847446 (9) 4 which should be compared with E1 = 3.799673 .

9.5We start with: H n = En n Now E≡ vH v =∑ vm m H n nv n, m

= ∑ v m δ nm En n v n, m

= ∑ En v n

2

n

and:

D = v HH v = ∑ En2 v n

2

n

Also, since v v = 1, ∑ v n

2

= 1 . Now we have:

n

Δ = D − E 2 = D − 2E 2 + E 2 = ∑ En2 v n

2

n

= ∑ ( En − E )

− 2E ∑ En v n n

2

vn

2

+ E2 ∑ v n

2

(1)

n

2

n

Suppose that E lies closer to Ek than it does to any other level; or at least that E lies as close to Ek as it does to any other level. Then, for any n: ( E k − E )2 ≤ ( E n − E )2 Hence, 82

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

Δ ≥ ( Ek − E ) ∑ v n 2

2

= ( Ek − E )

2

(2)

n

If Ek ≥ E,

Δ ≥ Ek − E , hence: E + Δ ≥ Ek ≥ E

However, if Ek < E, then

(3)

Δ ≥ E − Ek ; hence:

Δ ≥ E − Ek , or equivalently, Ek − E ≥ − Δ

Hence, E − Δ ≤ Ek ≤ E Therefore, in general: E + Δ ≥ Ek ≥ E − Δ

(4) (5)

9.6 a)Start with: x1 1 (n + )π  = ∫ 2m(E − kx a )dx − x 1 2

(

)

= 2 2mkx

⎛ x⎞ Let ⎜ ⎟ ⎝x ⎠

1/2

a 1/2 1

⎡ ⎛ x ⎞a⎤ ∫0 ⎢⎢1 − ⎜⎝ x1 ⎟⎠ ⎥⎥ dx ⎣ ⎦ x1

(1)

a /2

= sin θ , so x = x1 sin 2 /a θ . Then:

1

2x1 2a −1 sin θ cosθ dθ a

dx =

Hence (1) yields: 1⎞ ⎛ ⎜⎝ n + ⎟⎠ π a 2

(

2 2mkx1a + 2

)

1/2

= 2∫

π /2

0

cos θ sin 2

2 −1 a

θ dθ

3 ⎛ 1⎞ Γ( )Γ ⎜ ⎟ 2 ⎝ a⎠ = ⎛ 3 1⎞ Γ⎜ + ⎟ ⎝ 2 a⎠ (Here we have used Euler’s famous formula: π /2 Γ(x)Γ(y) = 2 ∫ cos 2 x −1θ sin 2 y−1 θ dθ ) 0 Γ(x + y) Equation (2) yields:

(2)

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a +1 2 1

x

⎛ 3 1⎞ Γ⎜ + ⎟ ⎝ 2 a⎠ 1⎞ a ⎛ = ⎜ n + ⎟ π 1/2 1/2 ⎝ 2⎠ ⎛ 1⎞ ( 2mk ) Γ⎜ ⎟ ⎝ a⎠

Thus,

⎛ E⎞ ⎜⎝ ⎟⎠ k

1 1 + 2 a

⎛ 3 1⎞ Γ⎜ + ⎟ ⎝ 2 a⎠ 1⎞ a ⎛ = ⎜ n + ⎟ π 1/2 1/2 ⎝ 2⎠ ⎛ 1⎞ ( 2mk ) Γ⎜ ⎟ ⎝ a⎠

or: 2a

⎡ 1/2 ⎛ 3 1 ⎞ ⎤ 2 + a a 2a Γ⎜ + ⎟ ⎥ 2a 2 ⎢ 2π 2 2 + a ⎝ 2 a⎠ ⎛ n 1 ⎞ 2+a ⎛  ⎞ 2+a 2+a ⎢ ⎥ E=⎜ + ⎟ ⎜ a k ⎟ ⎝ 2 4 ⎠ ⎝ 2m ⎠ ⎛ 1⎞ ⎢ ⎥ Γ⎜ ⎟ ⎢⎣ ⎥⎦ ⎝ a⎠

b) When a = 1 (3) yields the following for 1⎞ ⎤ ⎡ 3π ⎛ E = ⎢ ⎜n + ⎟⎥ 2⎠ ⎦ ⎣ 4 ⎝

(3)

2 k 2 = 1: 2m

2/3

(4)

n=

Exact result

1 2.33811 3 4.08795 5 5.52056 7 6.78671 9 7.94413 11 9.02265 13 10.04017 The agreement between the 2nd and 3rd columns is very good.

From (4) 2.32025 4.08181 5.51716 6.78445 7.94249 9.02137 10.03914

c) When a = 2, (3) becomes: 1/2

1 ⎞ ⎛ 2k ⎞ 1⎞ ⎛ ⎛ E = ⎜ n + ⎟  ⎜ ⎟ = ⎜ n + ⎟ ω ⎝ ⎠ ⎝ ⎠ ⎝ 2 m 2⎠ which agrees with the exact result. d) When k = a =4, µ =  = 1 (3) becomes:

E = 16π

2/3

⎡ Γ(3/4) ⎤ ⎢ Γ(1/4) ⎥ ⎣ ⎦

4/3

⎛ n 1⎞ ⎜⎝ + ⎟⎠ 2 4

(5)

4/3

(6)

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We compare this to the “exact” energies in the following table: n=

E”exact”

E from (6)

0 1.060362 1.272555 1 3.799673 5.506030 2 7.455703 10.88020 The agreement between the second and third columns is not good. The reason is that V(x) varies quite rapidly near the classical turning points.

9.7 Let V (x) = V (−x) and assume V (0) = 0. The quantization condition is: x2 1⎞ ⎛ ∫− x2 pdx = ⎜⎝ n + 2 ⎟⎠π  or x2 1⎞ π ⎛ (1) ∫0 2m(En − V (x)dx = ⎜⎝ n + 2 ⎟⎠ 2 Differentiate both sides of (1) with respect to n, regarded as a continuous variable:

π  ∂ x2 = 2m(En − V (x)dx 2 ∂n ∫0 ⎤ ∂x 1 ∂En x2 dx 1/2 ⎡ = ( 2m ) ⎢ En − V (x2 ) 2 + ⎥ ∂n 2 ∂n ∫0 En − V (x) ⎥⎦ ⎢⎣ 1/2 2m ) ∂En ( =

(2)

dx 2 En − V (x) since En = V (x2 ) . Equation (2) can be written: ∂n ∫

( 2m )1/2 ∂En =

x2

0

1 dx π  ∂n ∫0 En − V (x) From the WKB approximation we have: A π⎤ ⎡ 1 x2 ψ (x) = p dx + ⎥ 1/4 sin ⎢ ∫ 4⎦ ⎣ x [(2m)(En − V (x))] x2

(3)

Hence:

ψ 2 (x) =

A2

[(2m)(E − V (x))]

1/2

n

π⎤ ⎡ 1 x2 sin 2 ⎢ ∫ pdx + ⎥ 4⎦ ⎣ x

The normalization condition is:

85

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 x2 1 = ∫ ψ 2 dx 0 2 x2 A2 dx π⎤ ⎡ 1 x2 = sin 2 ⎢ ∫ pdx + ⎥ 1/2 ∫0 (2m) 4⎦ En − V (x) ⎣ x

π⎤ ⎡ 1 x2 Now make the following approximation: replace sin 2 ⎢ ∫ pdx + ⎥ by its average 4⎦ ⎣ x value: ½. Then we have: 2m A2 = x (4) dx 2 ∫0 En − V (x) Comparison of (3) with (4) yields: 2m ∂En (5) A2 = π  ∂n Now, x2 A2 π⎤ 2 2 ⎡1 ψ (0) = pdx + ⎥ 1/2 sin ⎢ ∫ 0 4⎦ ⎣ [ 2mEn ] 1 ⎛ 2m ⎞ = π  ⎜⎝ En ⎟⎠

1/2

1 ⎛ 2m ⎞ = π  ⎜⎝ En ⎟⎠

1/2

∂En 2 ⎡⎛ 1⎞ π π ⎤ sin ⎢⎜ n + ⎟ + ⎥ ∂n 2⎠ 2 4 ⎦ ⎣⎝

(6)

∂En nπ cos 2 ∂n 2

where we have used (1) to go from the first to the second line. Also we have: ∂ψ (x) A p π⎤ ⎡ 1 x2 =− cos ⎢ ∫ pdx + ⎥ + 1/ 4 ∂x 4⎦ [ 2m(En − V (x))]  ⎣  x

∂V (x) A 1 π⎤ ⎡ 1 x2 ∂x pdx + ⎥ 5 / 4 sin ⎢ ∫ 1/ 4 4 (2m) ( En − V ) 4⎦ ⎣ x Since

∂V = 0 this gives: ∂x x = 0

ψ '(0) = 2

A2

( 2m )

1/2

2mEn 1⎞ π π ⎤ 2 ⎡⎛ cos n + ⎜ ⎟ + 2 ⎢ 2 ⎠ 2 4 ⎥⎦ En1/2  ⎣⎝

(2m)3/2 1/2 ∂En 2 nπ = En sin π 3 ∂n 2

(7)

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We now compare results (7) and (6) with the simple harmonic oscillator wave functions for n=3 and n=4, respectively. For simplicity, we choose units where  = m = ω = 1. The exact solutions are: 1 4x 3 − 6x − x 2 /2 ψ 3 (x) = 1/ 4 e (8) π 12 and: 1 4x 4 − 12x 2 + 3 − x 2 /2 ψ 4 (x) = 1/ 4 e (9) π 24 From (9) we obtain: 9 = .211571 24π 1/2

(10)

ψ 42 (0)WKB =

2 = .2122066 3π

(11)

(ψ ) (ψ )

16 = 1.0030 9π 1/2

(12)

ψ 42 (0)exact = while from (6) we have:

Thus,

2 4 WKB 2 4 exact

=

Also, from (7) and (8) we find:

[ψ '(0)] [ψ '(0)]

2

3

WKB 2

3

exact

=

2i71/2 = .99514 3π 1/2

(13)

The WKB approximations (6), (7) improve rapidly with increasing n.

9.8 From eq. (9.28) we have: ⎡ 2 3/2 m1/2 x2 ⎤ T = exp ⎢ − V (x) − Edx ⎥ ∫ x1  ⎣ ⎦ In the present case, V (x) = µ + φ − eFx, E = µ, x1 = 0, x2 =

(1)

φ . Thus after routine eF

evaluation of a definite integral, (1) becomes: 3/2 ⎡ 2 5 /2 m1/2 φ 3/2 ⎤ ⎡ ⎤ 7 φ T = exp ⎢ − = exp −6.8i10 ⎥ ⎢ 3e F ⎦ F ⎥⎦ ⎣ ⎣ where φ is expressed in eV, and F is in V/cm.

(2)

9.9 a) We start with the formula: ⎡ 2 3/2 m1/2 r2 2Ze2 ⎤ T = exp ⎢ − − E dr ⎥  ∫R 4 π r ⎢⎣ ⎥⎦

(1)

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where E =

2Ze2 . We wish to evaluate the integral: 4π r2

I=∫

r2 R

1/2

∫

r2 R

1−

r dr r2 r1/2

(2)

r = sin 2 θ the right hand side of (2) becomes: r2

With the substitution

I=

⎛ 2Ze2 ⎞ 2Ze2 − E dr = ⎜ 4π r ⎝ 4π ⎟⎠

4Ze2 π /2 2 R cos θ dθ E 1/2 ∫Arc sin r2

2Ze2 π /2 = 1/2 [ sin θ cosθ + θ ] R Arc sin E r

(3)

2

b) We employ the last expression of (3) in (1) for each of the cases listed in the problem statement. Multiplying each calculated value of T by v0 /R , where v0 = 10 9 cm/s , we obtain the following results: Initial nucleus Observed trans. prob/s Calculated trans. prob/s −18 -1 232 Γ = 1.73 • 10 s 1.8•10-19 s-1 90 Th 230 Γ = 3.01 • 10 −13 s-1 5.6•10-14 s-1 90 Th 228 Γ = 1.178i10 −8 s-1 3.0•10-9 s-1 90 Th 226 Γ = 4.26i10 −4 s-1 7.0•10-6 s-1 90 Th Considering the great sensitivity of Γ to the numerical value of the exponent in (1), the agreement between our crude calculations and the observed values is not too bad.

9.10. The zeroth order Hamiltonian for the isotropic 3-dimensional harmonic oscillator is: 3 ⎡ pi2 mω 2 2 ⎤ H0 = ∑ ⎢ + xi ⎥ 2 i =1 ⎣ 2m ⎦ (1) 3 3 ω ⎡ ⎤ = ⎢ ∑ ω ai†ai ⎥ + 2 ⎣ i =1 ⎦ where 1 ai = (Qi + iPi ) 2 1/2 1/2 ⎛ mω ⎞ ⎛ 1 ⎞ xi , Pi = ⎜ pi . The perturbing Hamiltonian is: H ' = −eEz . and Qi = ⎜ ⎝  ⎟⎠ ⎝ mω  ⎟⎠ Hence: H = H0 + H '

(

)

p33 mω 2 2 ⎡ 2 ⎤ † = ⎢ ∑ ω ai ai ⎥ + ω + + z − eEz 2m 2 ⎣ i =1 ⎦

(

)

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Now,

mω 2 2 mω 2 ⎛ 2 2eE ⎞ z − eEz = z⎟ ⎜z − 2 2 ⎝ mω 2 ⎠

(2)

mω 2 ⎛ eE ⎞ e2 E 2 = ⎜z − ⎟ − 2 ⎝ mω 2 ⎠ 2mω 2 This equation reveals two features: 2

• The average position of the charge is now at z0 =

eE rather than z =0. Thus the mω 2

e2 E induced dipole moment is: d = ez0 = . mω 2 • The total energy of the oscillator is shifted by: 1 e2 ΔE = − E2 2 2 mω e2 This means that the polarizability is α = . mω 2 Now we do the same problem with second-order perturbation theory. We have: ΔE

(2)

=∑

m≠n

m H'n

2

(3)

En − E m

Also, ⎛  ⎞ H ' = −eEz = −eE ⎜ ⎝ mω ⎟⎠

1/2

⎛  ⎞ = −eE ⎜ ⎝ mω ⎟⎠

1/2

Q3 a3 + a3† 2

Now, we write: n = n1 , n2 , n3 . Then:

a3 + a3† 1 ⎡ n3 n1 , n2 , n3 − 1 + n3 + 1 n1 , n2 , n3 + 1 ⎤ n = ⎦ 2 2⎣ Therefore, for given n the sum on the right hand side of (3) can be written: (2) ΔEn1,n2,n 3 =

⎤ e2 E 2 ⎛  ⎞ ⎡ n3 n3 + 1 + ⎥ ⎜⎝ ⎟⎠ ⎢ 2 mω ⎣ ω [ n3 − (n3 − 1)] ω [ n3 − (n3 + 1)] ⎦

e2 E 2 =− 2mω 2 This shows that the energy shift is the same for all levels, independent of n3, and the result is in agreement with the exact solution.

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9.11

The Hamiltonian is: H = AS1 iS2 + BS2 iS3 + CS3 iS1 = AS1z S2 z + BS2 z S3z + CS3z S1z

A B C ( S1+ S2 − + S1− S2 + ) + ( S2 + S3− + S2 − S3+ ) + ( S3+ S1− + S3− S1+ ) 2 2 2 We write the matrix (H) of H in the m1m2 m3 representation, where the rows, and columns, are arranged in the following order: ( m1, m2 , m3 ) = 1/2,1/2,1/2; -1/2,1/2,1/2; 1/2,-1/2,1/2; 1/2,1/2,-1/2; +

⎛a ⎜0 ⎜ ⎜0 ⎜0 (H ) = ⎜ ⎜0 ⎜ ⎜0 ⎜0 ⎜ ⎝0

Here,

1/2,-1/2,-1/2; -1/2,1/2,-1/2; -1/2,-1/2,1/2; -1/2,-1/2,-1/2 0 0 0 0 0 0 0⎞ b e f 0 0 0 0⎟ ⎟ e c g 0 0 0 0⎟ f g d 0 0 0 0⎟ ⎟ 0 0 0 b e f 0⎟ ⎟ 0 0 0 e c g 0⎟ 0 0 0 f g d 0⎟ ⎟ 0 0 0 0 0 0 a⎠

(1)

A+ B+C 4 −A + B − C b= 4 −A − B + C c= 4 A− B−C d= 4 a=

and e=

A B C , f = , g= 2 2 2

A+ B+C is two-fold degenerate so far. To find the remaining 4 eigenvalues we must diagonalize the matrix:

The eigenvalue a =

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A ⎛ −A + B − C ⎜ 4 2 ⎜ A −A − B+C M =⎜ ⎜ 2 4 ⎜ C B ⎜ ⎝ 2 2 The secular determinant is: −A + B − C −λ 4 A 2 C 2

A 2 −A − B + C −λ 4 B 2

C ⎞ ⎟ 2 ⎟ B ⎟ ⎟ 2 A− B−C⎟ ⎟ ⎠ 4

C 2 B =0 2 A− B−C −λ 4

We add the second row to the first row, and the third row to the first row. Then:

A+ B+C A+ B+C A+ B+C −λ −λ −λ 4 4 4 A −A − B + C B −λ =0 2 4 2 C B A− B−C −λ 2 2 4 Now we subtract the first column from the third column, and subtract the first column from the second column: A+ B+C −λ 0 0 4 A −3A − B + C B− A −λ =0 2 4 2 C B−C A − B − 3C −λ 2 2 4 A+ B+C Thus we find the eigenvalue a = again, and since the matrix M appears 4 twice, we conclude that a is 4-fold degenerate. As for the remaining 4 eigenvalues, consider the remaining secular determinant:

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3A + B − C ⎞ ⎛ 3C + B − A ⎞ 1 ⎛ ⎜⎝ λ + ⎟⎠ ⎜⎝ λ + ⎟⎠ − ( B − C ) ( B − A ) = 0 4 4 4 After some simple algebra, we obtain:

λ=−

(A + B + C) ± 1

A 2 + B 2 + C 2 − AB − AC − BC 4 2 and each of these eigenvalues is two-fold degenerate.

9.12 Start with n H m = δ nm Em . Differentiating both sides with respect to λ we obtain:

∂n ∂m ∂E ∂H H m + nH = δ nm m − n m ∂λ ∂λ ∂λ ∂λ which gives: Em

∂n ∂m ∂E ∂H m + En n = δ nm m − n m ∂λ ∂λ ∂λ ∂λ

However, n m = δ nm , so

(1)

∂ ∂n ∂m nm = m + n = 0. Therefore (1) becomes: ∂λ ∂λ ∂λ

( En − E m ) n

∂m ∂E ∂H = δ nm m − n m ∂λ ∂λ ∂λ

(2)

If n = m , (2) becomes:

∂H ∂E n = n ∂λ ∂λ Now differentiate (3) with respect to λ and use completeness to obtain: ∂ 2 En ∂2 H ∂n ∂H ∂H ∂n − n n = n + n 2 2 ∂λ ∂λ ∂λ ∂λ ∂λ ∂λ n

∂n ∂H ∂H =∑ m m n +∑ n m ∂λ ∂λ m ≠ n ∂λ m≠n Now for m ≠ n , we obtain with the help of (2): ∂H n m ∂n ∂m ∂λ m =− n = ∂λ ∂λ En − E m and: ∂H m n ∂n ∂λ m = ∂λ En − E m Using (5) and (6) in (4) we get:

∂n m ∂λ

(3)

(4)

(5)

(6)

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∂H m n ∂λ En − E m

2

∂ 2 En ∂2 H − n n = 2∑ ∂λ 2 ∂λ 2 m≠n Now we write: H = H0 + λH '

(7)

(8)

En = En0 + λ En1 + λ 2 En2 + ...

(9)

n = n0 + λ n1 + λ n2 + ...

(10) We use these expressions in lowest non-vanishing order to recover the standard formulae of Rayleigh-Schroedinger perturbation theory from (3), (6), and (7). Starting with (3) we employ (8), (9), (10) to obtain: En1 = n0 H ' n0 (11) Equation (6) yields: m H'n (12) m n1 = En − E m which leads to the first order correction to the wave function. Finally, the left hand side of (7) gives 2En2 ; hence: 2

En2 = ∑

m≠n

9.13 a)

∂H m n ∂λ En − E m

2

(13)

H 0Qn0 = H 0 ( I − n0 n0 ) = H 0 − En0 n0 n0 = H 0 − n0 n0 Eno = Qn0 H n0

b)Start with: (En − H 0 ) n = H ' n ∴Qn0 (En − H 0 ) n = Qn0 H ' n ∴(En − H 0 )Qn0 n = Qn0 H ' n ∴Qn0 n = (En − H 0 )−1 Qn0 H ' n = (En − H 0 )−1 ∑ m0 m0 H ' n m0 ≠ n0

= ∑ m0 m0 ≠ n0

m0 H ' n En − E m 0

= Rn H ' n

c) From Qn0 n = Rn H ' n we have n0 + Qn0 n = n0 + Rn H ' n .

(1)

If n0 n = 1 , the left hand side of (1) is:

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n0 + ( I − n0 n0 ) n = n

Thus n = n0 + ∑

m0 m0 H ' n

(2)

En − E m 0

m0 ≠ n0

or iterating,

n = n0 + ∑

m0 ≠ n0

m0 m0 H ' n0 En − E m 0

+ ∑

m0 ≠ n0 k0 ≠ n0

m0 m0 H ' k0 k0 H ' n0

( En − E m 0 ) ( En − E k 0 )

+ ...

(3)

To calculate corrections to the energy, go back to: (En − H 0 ) n = H ' n Now take the scalar product of both sides with n0 and employing n0 n = 1 , use (3):

En = En0 + n0 H ' n0 + ∑

m0 ≠ n0

m0 H ' n0 En − E m 0

2

+ ...

(4)

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10.1 To calculate the hyperfine splitting in the 12s1/2 ground state of the deuterium atom, we start with equation (10.29) of Chapter 10 for the first order energy shift: ΔEhfs (ns) = a IiS (1) where a is given in atomic units by equation (10.30). For n = 1 we have: 2 m (2) a(2 H ,12 s1/2 ) = gD gsα 2 e 3 mp Since the nuclear magnetic moment of the deuteron is 0.857438 nuclear Bohr magnetons, (the most current value) and I = 1, we have gD = 0.857438. The other constants are: gs = 2.002319

α = 7.297353i10 −3 me = 5.446170i10 −4 mp where we have included radiative corrections in gs . Also the conversion factor from atomic units of energy to Hz is: 1 au = 6.57968•1015 Hz. Thus (2) yields: a = 218.4096 MHz (3) Now since I =1, S =1/2, we have F = 3/2 or 1/2 . Also, F(F + 1) − I(I + 1) − S(S + 1) IiS = 2 1 =+ F = 3/2 2 = −1 F = 1/2 Thus we predict the hyperfine splitting to be: 3 (4) δ calc ( 2 H ,12 s1/2 ) = a = 327.614 MHz 2 The experimental value is: δ exp t ( 2 H ,12 s1/2 ) = 327.384 MHz (5) The discrepancy between (4) and (5) is about 200 parts per million, and is resolved when we take into account reduced mass, relativistic, nuclear structure, and further quantum electrodynamic corrections. For the remainder of this problem we ignore all these small effects, and assume that a is given by (3), and that gs and gD are as given above.

The hyperfine-Zeeman Hamiltonian is: a (6) H = ( I + S− + I − S+ ) + aI z Sz + k1Sz − k2 I z 2 where k1 = gs µ B B = 2.80244B MHz m (7) k2 = gD e µ B B = .0006536B MHz mp 95

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and B is the external magnetic field in Gauss. It is convenient to write out the matrix of the Hamiltonian (6) in the mIms representation. Labeling the rows, and columns in the following order: mI ms = (1)α , (-1)β , (0)α , (1)β , (0)β , (-1)α where α , β refer to ms = ±1/2 respectively, the Hamiltonian matrix is: ⎛ a k1 + − k2 ⎜2 2 ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ (H ) = ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎜ ⎜ 0 ⎜ ⎝

0

0

0

0

a k1 − + k2 2 2

0

0

0

k1 2 a 2

a 2 a k − − 1 − k2 2 2

0

0

0

0

0

0

0 0

0 0 k − 1 2 a 2

⎞ ⎟ ⎟ ⎟ 0 ⎟ ⎟ ⎟ 0 ⎟ ⎟ 0 ⎟ ⎟ ⎟ a ⎟ 2 ⎟ ⎟ a k1 − + + k2 ⎟ ⎠ 2 2 0

Diagonalizing the two 2x2 submatrices, we obtain the following eigenvalues, which are readily associated with quantum numbers F, mF: 3 3 a k λ (F = , mF = ) = + 1 − k2 2 2 2 2 3 3 a k λ ( , − ) = − 1 + k2 2 2 2 2 3 1 a k 3a 4 4 λ( , ) = − − 2 + 1 + x + x2 2 2 4 2 4 9 9 3 1 a k 3a 4 4 λ( , − ) = − + 2 + 1 − x + x2 2 2 4 2 4 9 9 1 1 a k 3a 4 4 λ( , − ) = − + 2 − 1 − x + x2 2 2 4 2 4 9 9 1 1 a k 3a 4 4 λ( , ) = − − 2 − 1 + x + x2 2 2 4 2 4 9 9 where x = (k1 + k2 )/a . These eigenvalues are plotted versus B in Fig. 10.1S.

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600 mF

Frequency, MHz

400

3/2

200

1/2 -1/2

F=3/2 -3/2

0

-1/2

-200 F=1/2

1/2

-400

-600 0

100

200

300

400

B, gauss

Fig.10.1S At B=200 Gauss, the eigenvalues in MHz are: F, mF 3 3 , 389.32 2 2 3 1 , 314.14 2 2 3 1 ,− 218.95 2 2 3 3 ,− -170.91 2 2 1 1 ,− -328.02 2 2 1 1 , -423.47 2 2

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10.2 The Hamiltonian is: H = aIiS + uSz − wI z (1) where a is the hyperfine constant, and is related to the hyperfine splitting as shown below, while u = gs µ B B0 and w = gI µ N B0 . As usual we write (1) as: a (2) H = ( I + S− + I − S+ ) + aI z Sz + uSz − wI z 2 Now write the Hamiltonian matrix (H) in the mI mS basis, with the order of the rows, and columns, specified by the following sequence: ( mI , mS ) = ( 3/2,1/2 ); ( 3/2,-1/2 ); (1/2,1/2 ); (1/2,-1/2 );

(-1/2,1/2 ); (-1/2,-1/2 ); (-3/2,1/2 ); (-3/2,-1/2 )

It is to verify that with this ordering, (H) is in block-diagonal form: ⎛ M1 ⎜ M2 ⎜ (H ) = ⎜ ⎜ ⎜ ⎜⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ M 5 ⎟⎠

M3 M4

where M1 =

Also:

3a u 3w + − 4 2 2

mI = 3/2 mS = -1/2 ⎛ -3a - 2u - 6w ⎜ 4 M2 = ⎜ a ⎜ 3 ⎝ 2

mI = 3/2, mS = 1/2

(3)

1/2 1/2 a ⎞ 3 ⎟ 2 ⎟ a + 2u − 2w ⎟ ⎠ 4 (4)

mI = 1/2

1/2

mS = 1/2

-1/2

⎛ −a − 2u − 2w ⎜ 4 M3= ⎜ ⎜ a ⎝

⎞ ⎟ ⎟ −a + 2u + 2w ⎟ ⎠ 4 a

(5)

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mI = -1/2 mS = -1/2

-3/2 1/2

⎛ a - 2u + 2w ⎜ 4 M4 = ⎜ a ⎜ 3 ⎝ 2

a ⎞ 3 ⎟ 2 ⎟ −3a − 2u + 6w ⎟ ⎠ 4 (6)

3a u 3w − + mI = −3/2, mS = −1/2 4 2 2 We now write the secular equations for the 2x2 matrices M 2, M 3, M 4 : M2: 3a u 3w ⎞ ⎛ a u w ⎞ 3a 2 ⎛ λ + + + λ − − + ⎟− =0 ⎜⎝ ⎟⎜ 4 2 2 ⎠⎝ 4 2 2⎠ 4 a ∴ λ = − − w ± a 1 + x + x2 4 u+w where x = .

2a M4: a λ = − + w ± a 1 − x + x2 4 M3: a λ = − ± a 1 + x2 4 M5 =

(7)

(8)

(9)

(10)

Next we determine the ordering of these eigenvalues. When B0 = 0, we have u = w =x =0 and: From (3): λ = 3a/4 “ (7): λ = 3a/4 “ (8): λ = 3a/4, − 5a/4 “ (9): λ = 3a/4, − 5a/4 “ (10): λ = 3a/4, − 5a/4 The 5 eigenvalues λ = 3a/4 correspond to F = 2 while the 3 eigenvalues λ = −5a/4 correspond to F = 1 . The splitting between F = 2 and F =1 at B0 = 0 is: Δν = 1772 MHz =2a Thus, a = 886 MHz (11) Next consider a weak magnetic field (x << 1) and also recall that u >> w. The eigenvalues for F = 2 are:

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F = 2, mF = 1 F = 2, mF = 0 F = 2, mF = −1 F = 2, mF = −2 while for F = 1 we have: F = 1, mF = 1

F = 1, mF = 0 F = 1, mF = −1 Thus we obtain: F, mF

3a u + 4 2 3a u λ≈ + 4 4 3a λ≈ 4 3a u λ≈ − 4 4 3a u λ≈ − 4 2

λ≈

F = 2, mF = 2

from (3) from (8) from (10)

(12)

from (9) from (7)

5a u − from (8) 4 4 5a λ≈− from (10) 4 5a u λ≈− + from (9) 4 4

λ≈−

(13)

3a u 3w + − 4 2 2 a 2,1 λ = − − w + a 1 + x + x2 4 a 2, 0 λ = − + a 1 + x2 4 a 2, −1 λ = − + w + a 1 − x + x 2 4 3a u 3w 2, −2 λ= − + 4 2 2 a 1,1 λ = − − w − a 1 + x + x2 4 a 1, 0 λ = − − a 1 + x2 4 a 1, −1 λ = − + w − a 1 − x + x2 4 Next we find numerical values: 2, 2

λ=

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µ B = 1.396 MHz/Gauss m µN = µ B e = 7.60 •10 −4 MHz/Gauss mp I = 3/2 gI µ N I = 2.2175 µ N 2 ∴ gI = 2.2175 = 1.4783 3 ∴w = gI µ N B0 = 1.124 i10 −3 MHz i B0 u = gS µ B B0 = 2 i1.396 B0 MHz = 2.792 B0 MHz u+w x= = .00158B0 2a Thus when B0 = 500 Gauss we have: E22 = 1361.7 MHz E21 = 1154.5 MHz E20 = 907.6 MHz E2 −1 = 588.2 MHz E2 − 2 = −32.7 MHz E11 = −1598.7 MHz E10 = −1350.6 MHz E1−1 = −1030.1 MHz 10.3 The Fermi contact interaction Hamiltonian for positronium s-states is: 8π 2 2 (+ ) (− ) 3 H Fermi = µ B gs S i S δ ( r ) 3 α2 In atomic units, µ B2 = , and gs = 2.002319 .Thus: 4 8π H Fermi = (1.00232)α 2 S (+ ) iS (− ) δ 3 ( r ) 3 Also, 3 + + − − F(F + 1) − F(F + 1) − S S + 1 − S S + 1 2 S (+ ) iS (− ) = = 2 2 Thus,

(

)

(

(1)

)

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1 if F = 1 4 3 = − if F = 0 4

S (+ ) iS (− ) =

Furthermore, δ 3 ( r ) = ψ (0) = 2

2 m 1 a ' = , where: and µ = e . Thus in atomic 0 2 3 µe 2 π a0 '

units, 1 (2) 8π Therefore the calculated energy splitting between F=1 and F=0 in atomic units is: α2 δ EFermi = 1.00232 3 To convert to Hz, we write: α2 δ EFermi = 1.00232i i6.57968i1015 Hz (3) 3 = 1.1706i1011 Hz The annihilation portion of the Hamiltonian in atomic units is: (4) H ann. = 2πα 2δ S1δ 3 ( r ) Using (2), we have: α2 H ann. = δ S1 4 or in Hz, α2 δ Eann. = 6.57968i1015 Hz (5) 4 11 = .87594i10 Hz Combining (3) and (5) we obtain: (6) δ Etheo = 2.047i1011 Hz The experimental value is: δ Eexp t = 2.034i1011 Hz (7)

δ 3 (r) =

After small QED corrections to (6) there is good agreement between theory and experiment. Next we consider the hyperfine-Zeeman Hamiltonian: H = aS (+ ) i S (− ) − µ (+ ) i B − µ (− ) i B = aS (+ ) i S (− ) + gs µ B B0 ( Sz(− ) − Sz(+ ) )

(8)

a (+ ) (− ) ( S+ S− + S−(+ )S+(− ) ) + aSz(+ )Sz(− ) + k ( Sz(− ) − Sz(+ ) ) 2 where k = gs µ B B0 . We write the matrix for H in the ms(+ ) ms(− ) basis. Ordering the rows, and columns in the following sequence: ms(+ ) , ms(− ) = (1/2,1/2 ) , ( -1/2,-1/2 ) , (1/2,-1/2 ) , ( -1/2,1/2 ) =

(

)

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we obtain:

⎛a ⎜4 ⎜ ⎜0 ⎜ (H ) = ⎜ ⎜0 ⎜ ⎜ ⎜⎝ 0

0 a 4 0 0

⎞ ⎟ ⎟ 0 0 ⎟ ⎟ a a ⎟ ⎟ − −k 4 2 ⎟ ⎟ a a − + k⎟ ⎠ 2 4 0

0

(9)

The first two eigenvalues are degenerate and independent of B0: a (10) E ( F = 1, mF = 1) = E (1, −1) = 4 Next we diagonalize the 2x2 matrix on the lower right hand side of (9) to obtain the eigenvalues: a a (11) λ± = − ± 1 + x2 4 2 2k 2gs µ B B0 where x = . Inserting gs = 2.002319, µ B = 1.3996 MHz/Gauss, and = a a a = 2.03i10 5 MHz , we find x = .276 at B0 = 10 4 Gauss.

Next we calculate the eigenvectors associated with λ± . Let: ⎛ c⎞ ψ + = ⎜ ⎟ = cα (+ )β (− ) + dα (− )β (+ ) ⎝ d⎠

⎛ e⎞ ψ − = ⎜ ⎟ = eα (+ )β (− ) + f α (− )β (+ ) ⎝ f⎠ After simple algebra, we obtain: 1 c= 2 1 + x + 1 + x2

(

d=

)

x + 1 + x2

(

2

)

2

)

1+ x + 1+ x e=

(

1

1+ x − 1+ x

2

2

(12)

(13)

(14)

and

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x − 1 + x2

f =

(

1+ x − 1+ x

2

)

(15)

2

Also taking into account that: 1 α (+ )β (− ) = ( F = 1, mF = 0 + 0, 0 ) 2 1 α (− )β (+ ) = ( 1, 0 − 0, 0 ) 2 we obtain:

ψ+ =

(

1

2 ⎡⎢1 + x + 1 + x 2 ⎣ 1

)

2

⎡(1 + x + 1 + x 2 ) 10 + (1 − x − 1 + x 2 ) 00 ⎤ ⎦ ⎤⎣ ⎥⎦

⎡(1 + x − 1 + x ) 10 + (1 − x + 1 + x ) 00 ⎤ ⎦ 2 ⎣ ⎡ ⎤ 2 ⎢1 + x − 1 + x 2 ⎥ ⎣ ⎦ When x = .276 these formulae yield: ψ + = .991 10 − .134 00

ψ− =

(

2

)

ψ − = .134 10 + .991 00 The decay rate of positronium is proportional to: nγ T ψ

(16)

2

(17) (18)

2

where T is some transition operator. Thus: 2 1 = A 3 γ T 10 τ 3 S1

( )

2 1 = A 2γ T 00 τ S0

( ) 1

( )

( )

where A is a common constant. Since τ 3 S1 = 1.4i10 −7 s, τ 1 S0 = 1.25i10 −10 s we have:

3γ T 10

2

2γ T 00

2

= 8.93i10 −4

We see from (17) that introduction of even the relatively small admixture of 00 into

ψ + (with coefficient -.134) greatly increases its decay rate at B0 = 10 4 Gauss compared to the decay rate at zero magnetic field. However from (18) we see that the decay rate of ψ − is hardly affected. The numbers are as follows:

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1 (.991) + (.134 ) = 3 τ (ψ + ) τ S1 τ 1 S0 2

2

( )

( )

= 7.01i10 6 + 1.436i10 8 = 1.506i10 8

Therefore, τ (ψ + , x = .276) = 6.64i10 −9 s.

Also, 1 (.134)2 (.991)2 = + 1 τ (ψ − ) τ 3 S1 τ S0

( )

(19)

( )

= 1.283i10 5 + 7.857i10 9 = 7.857i10 9

Therefore, τ (ψ − , x = .276 ) = 1.273i10 −10 s.

(20)

10.4 The second order Stark shift for the 6 2 P1/2 , F = 1, mF state of the thallium atom in external electric field E = E ẑ is given by the formula:

δE

=eE ∑

7S, F ' mF z 6 2 P1/2 ,1, mF

2

(1) E(6P1/2 ,1, mF ) − E(7S, F ', mF ) If mF = ±1 , we must have F ' = 1 . Also in the absence of a magnetic field, the shifts for mF = 1 and mF = −1 are identical, because: (2)

2

2

F'

(

)

7S,1,1 z 6P,1,1 = 7S,1,1 Ry−1 (π ) Ry (π )zRy−1 (π ) Ry (π ) 6P,1,1 = 7S,1, −1 −z 6P,1, −1 = − 7S,1, −1 z 6P,1, −1

If mF = 0, we must have F ' = 0 because 7S,1, 0 z 6P,1, 0 is proportional to the vector coupling coefficient: 1, 0,1, 0 1, 0 = 0 .

It only remains to determine the ratio of Now,

7S,1,1 z 6P1/2 ,1,1 to 7S,0,0 z 6P1/2 ,1, 0 .

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7S,1,1 = α eα N Y00 R7S ⎛ α β − α N βe ⎞ 0 7S, 0, 0 = ⎜ e N ⎟⎠ Y0 R7S ⎝ 2 ⎡ 1 2 1 ⎤ 6P1/2 ,1,1 = ⎢ Y10α e − Y1 β e ⎥ α N R6 P 3 ⎣ 3 ⎦ ⎡ 1 α β − α N βe ⎤ 2 1 2 −1 6P1/2 ,1, 0 = ⎢ Y10 e N − Y1 β e β N + Y1 α eα N ⎥ R6 P 6 6 2 ⎣ 3 ⎦ where in the last two equations, we have used equations (7.110, 7.111) of Chapter 7.

Thus, 1 0 7S,1,1 z 6P1/2 ,1,1 = Y0 R7S z Y10 R6 P 3 and

7S, 0, 0 z 6P,1, 0 =

1 0 Y0 R7S z Y10 R6 P 3

which are identical.

Therefore, the second order Stark shifts are in the ratio:

δ E (2) ( 6P,1, 0 ) E ( 6P, F = 1) − E ( 7S, F = 1) −ΔE = = (2) δ E ( 6P,1, ±1) E ( 6P, F = 1) − E ( 7S, F = 0 ) −ΔE + a ' ΔE a' =  1+ ΔE − a ' ΔE a' In other words, δ ' = δ << δ . ΔE Now apply a weak magnetic field: B = By ŷ + Bz ẑ , where By << Bz . Then, if we ignore the very small contribution of the nuclear magnetic moment, the Zeeman energy of the 6P1/2, F=1 state is given to a very good approximation by: H Zeeman = gF µ B ⎡⎣ Fy By + Fz Bz ⎤⎦ ⎛ 0 −i 0 ⎞ ⎛1 0 0 ⎞ 1 ⎜ where Fy = i 0 −i ⎟ , Fz = ⎜ 0 0 0 ⎟ ⎟ ⎜ ⎟ 2⎜ ⎝0 i 0 ⎠ ⎝ 0 0 −1⎠ and the rows, and columns, of these matrices are ordered by the sequence mF = 1, 0, −1 ; while:

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F(F + 1) + J(J + 1) − I(I + 1) 2(F + 1) g 1 = J = 2 3 Now including both Zeeman and Stark effects, the F=1 Hamiltonian matrix takes the form: −ik2 0 ⎞ ⎛ k1 − δ (2) (H ) = ⎜ ik2 −δ − δ ' −ik2 ⎟ ⎜ ⎟ ik2 −k1 − δ ⎠ ⎝ 0 1 1 µ B By << k1 . The secular determinant is: where k1 = µ B Bz , and k2 = 3 3 2 k1 − δ − λ −ik2 0 k1 − Λ −ik2 0 (3) ik2 −δ − δ '− λ −ik2 = ik2 −δ '− Λ −ik2 = 0 gF = g J

0

ik2

−k1 − δ − λ

0

ik2

−k1 − Λ

where Λ = λ + δ . When k2 = 0 , we obviously have the three eigenvalues: Λ + = k1 , Λ 0 = −δ ', Λ − = −k1 Thus to find approximate solutions to (3) we try Λ + = k1 + a, Λ − = −k1 + b , where a,b are small quantities. Substituting Λ + = k1 + a into the second determinant of (3), writing out the resulting cubic equation, and dropping all terms of order a2 and higher, we find after simple algebra that: k22 a= k1 + δ ' and similarly, that: −k22 b= k1 − δ ' Thus, ⎛ 1 1 ⎞ λ+ − λ− = Λ + − Λ − = 2k1 + k22 ⎜ + ⎝ k1 + δ ' k1 − δ ' ⎟⎠

⎛ k k2 ⎞ = 2 ⎜ k1 + 2 1 2 2 ⎟ k1 − δ ' ⎠ ⎝ 10.5 Let H = H 0 + V (x) where H 0 u1,2 = E1,2 u1,2 . Also write: H ψ =Eψ

where

ψ = c1 (x) u1 + c2 (x) u2

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Then, u1 H ψ = c1E1 + c1V11 + c2V12 = Ec1

(1)

u2 H ψ = c2 E2 + c1V21 + c2V22 = Ec2

(2)

and Equations (1, 2) can be written as: ( E1 + V11 − E ) c1 + V12 c2 = 0 V21c1 + (E2 + V22 − E)c2 = 0 These two homogeneous linear equations have a solution only if the determinant of the coefficients vanishes: E1 − E + V11 V12 =0 V12 * E2 − E + V22 This yields: 2 (3) E 2 − (E1 + E2 + V11 + V22 )E + ⎡⎣( E1 + V11 ) ( E2 + V22 ) − V12 ⎤⎦ = 0 This quadratic equation has the solutions:

1 1 ( E1 + E2 + V11 + V22 ) ± ( E1 + E2 + V11 + V22 )2 − 4 ( E1 + V11 )( E2 + V22 ) + 4 V12 2 2 2 1 1 = ( E1 + E2 + V11 + V22 ) ± ( E1 − E2 + V11 − V22 )2 + 4 V12 2 2 2 The two energies E+ , E− can become equal only when the discriminant vanishes, which requires that each term inside the square root vanish simultaneously, since each is nonnegative. In the case of the Zeeman effect caused by a magnetic field in the z direction, the matrix element V12 is in general non-zero if the states 1,2 have the same value of mF .

Thus we obtain the “no-crossing theorem” which yields the following rule for the Zeeman effect: Levels with the same value of mF do not cross.

Consider the example of an atom with J = 1, I = 3/2, where the nuclear magnetic moment is assumed to be positive. When B0 = 0 , the energy levels are determined by the hyperfine Hamiltonian: H = aIiJ Thus: F(F + 1) − I(I + 1) − J(J + 1) H = a IiJ = a 2 Since the nuclear magnetic moment is positive, a > 0. Thus we have: E± =

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F = 5/2

H =

a⎛5 7 3 5 ⎞ 3a − − 1i2 ⎟ = ⎜⎝ ⎠ 2 2 22 22

F = 3/2

H =

a⎛ 3 5 3 5 ⎞ − − 1i2 ⎟ = −a ⎜⎝ ⎠ 2 22 22

a⎛1 2 3 5 5a ⎞ − − 1i2 ⎟ = − ⎜⎝ ⎠ 2 22 22 2 In weak magnetic field we treat the Zeeman effect as a perturbation and use as a zeroorder basis the states F, mF . Then, the first-order energies are given by: F = 1/2

H =

ΔE = F, mF − µez B0 − µ Nz B0 F, mF

(4)

However, µez = −gJ µ B J z where gJ is the Lande g-factor, and µez >> µ Nz , so to a good approximation we can ignore the nuclear magnetic moment contribution. From the Wigner-Eckart theorem: g Bµ ΔEF,mF  gJ B0 µ B F, mF J z F, mF = J 0 B F, mF J i F Fz F, mF F(F + 1) (5) gJ B0 µ B mF = F, mF J i F F, mF F(F + 1) Now, F(F + 1) − I(I + 1) − J(J + 1) JiF = Ji( J + I ) = J(J + 1) + 2 F(F + 1) + J(J + 1) − I(I + 1) = 2 Therefore m ⎡ F(F + 1) + J(J + 1) − I(I + 1) ⎤ (6) ΔEF,mF = B0 µ B F ⎢ gJ ⎥ F ⎣ 2(F + 1) ⎦ The factor in brackets on the right hand side of (6) is called gF . It has the same sign as gJ for F = 5/2, 3/2, but the opposite sign for F = ½. Therefore, in weak magnetic field, the levels are ordered as is shown in Fig. 1 below.

For strong magnetic field it is more convenient to use the basis mI mJ . Here the energies are given by: E = mI mJ aIiJ − µez B0 − µ Nz B0 mI mJ

= amI mJ + gJ µ0 mJ B0 − gI µ N mI B0 Fig. 10.2S shows the connection between the levels at zero field, weak field, strong field, and extremely strong field. Here we use the fact that levels of the same mF cannot cross.

The diagram is not to scale and is only intended to show the ordering of the various levels, not the detailed dependence of each energy on magnetic field.

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mI mJ 3/2 1 1/2 1 -1/2 1 Zero B

Weak B

-3/2 1

mF 5/2 3/2 1/2 F=5/2

Strong B

Very strong B

-1/2 -3/2 -5/2 3/2 1/2

F=3/2

-1/2 -3/2

-3/2 0 -1/2 0 1/2 0 3/2 0

-1/2 F=1/2

1/2

-3/2 -1 -1/2 -1 1/2 -1 3/2 -1

Fig. 10.2S

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11.1 No solution is given here since the answer and abundant hints are stated in the problem.

11.2 We start with the exchange integral:

K = ∫∫ R1s (1)Y00 (1)iRn (2)Y0 (2)

1 R1s (2)Y00 (2)iRn (1)Y0 (1)dτ 1dτ 2 r1 − r2

(1) 1 = ∫ R1s (1)Y (1)Rn (1)Y (1)dτ 1 ∫ R1s (2)Y (2)Rn (2)Y (2) dτ 2 r1 − r2 From the addition theorem for spherical harmonics we have the formula: ∞  1 1 r< m* (2) = 4π ∑ ∑ Y (1)Ym (2) +1  r1 − r2 = 0 m = −  2 + 1 r> Inserting (2) in (1), and making use of the orthonormality of spherical harmonics, we find after some simple algebra that (1) reduces to: 0 0

K=

0 

1

0 0

0 

⎡ ∫ ∞ R1s (1)Rn (1)r11−  dr1 ∫ r R1s (2)Rn (2)r2 + 2 dr2 + ∫ ∞ R1s (1)Rn (1)r1 + 2 dr1 ∫ ∞ R1s (2)Rn (2)r21−  dr2 ⎤ 0 0 r ⎦ 2 + 1 ⎣ 0 1

1

=

1

⎡ ∫ ∞ R1s (1)Rn (1)r11−  dr1 ∫ r R1s (2)Rn (2)r2 + 2 dr2 + ∫ ∞ R1s (2)Rn (2)r2 + 2 dr2 ∫ ∞ R1s (1)Rn (1)r11−  dr1 ⎤ 0 0 r ⎦ 2 + 1 ⎣ 0 1

2

=

2

⎡ ∫ R1s (1)Rn (1)r11−  dr1 ∫ R1s (2)Rn (2)r2 + 2 dr2 ⎤ 0 ⎦ 2 + 1 ⎣ 0 ∞

r1

(3)

Now: R1s (Z = 2, r) = 2 5 /2 e−2r 1 R2 p (Z = 1, r) = re− r /2 2 6 8 r R3 p (Z = 1, r) = r(1 − )e− r / 3 6 27 6

Thus: K(1sZ = 2 , 2 pZ =1 ) =

x 8 ∞ −5 x /2 xe dx ∫ y 4 e−5 y /2 dy = .00382 ∫ 0 9 0

(4)

and x 211 ∞ x y x(1 − )e−7 x / 3dx ∫ y 4 (1 − )e−7 y / 3dy = .001236 (5) 8 ∫0 0 3 6 6 (Evaluation of the integrals in (4) and (5) is routine and we skip the details). Thus we predict that: ΔE(1s2 p) = E 1s2 p 1P − E 1s2 p 3 P = 2K = .00764 (6) which should be compared with: ΔE(1s2 p)exp t = .00933 (7)

K(1sZ = 2, 3pZ =1 ) =

(

)

(

)

and:

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(

)

(

)

ΔE(1s3p) = E 1s3p 1P − E 1s3p 3 P = 2K = .00247 which should be compared with: ΔE(1s3p)exp t = .00294

(8)

(9)

Obviously, the agreement between theory and experiment improves as n increases. The reason is that the average distance of the 3p electron from the nucleus is larger than that of the 2p electron. Hence the 1s electron is more effective at screening the nucleus from the 3p electron, and also the 3p electron is less effective at screening the nucleus from the 1s electron. Therefore the assumption Z=2 for 1s, Z=1 for np is better for n=3 than for n=2.

11.3 If an observable were described by an irreducible spherical tensor of half-integral rank, it would have matrix elements connecting states of integral angular momentum to half-integral angular momentum. This would imply the existence of matrix elements between symmetric and anti-symmetric states; hence the operator in question could not be symmetric with respect to exchange, and this is impossible.

1 11.4 a)For given mI = the 3 H wave function is obtained from the corresponding 3 He 2 wave function by applying the operator T− = ∑ ti − : i

3

H , mI =

1 1 = ∑ ti − 3 He, mI = 2 2 i

Thus from eq. (11.83) we get: ψ 3 H , mI = 1/2

(

=−

)

1 ⎡ n1n2 p3 (α1β 2 − α 2 β1 )α 3 + n1 p2 n3 (α 3β1 − α1β 3 )α 2 + p1n2 n3 (α 2 β 3 − α 3β 2 )α1 ⎤⎦ 6⎣ (1)

b) Obviously from a),

3

He, mI = 1/2 ∑ ti +

3

H , mI = 1/2 = +1 . We also calculate:

i

3

He, mI = 1/2

3 ∑ σ iz ti + 3 H , mI = 1/2 = − 3

(2)

i

c) To a good approximation we can ignore the 16O core, in which case the initial 19Ne wave function is given by (11.83) and the final 19F wave function is given by eq. (1) in part a) of this problem. The requested matrix elements are therefore: 19

F, mI = 1/2 ∑ ti − 19 Ne, mI = 1/2 = +1

(3)

i

19

F, mI = 1/2

3 ∑ σ iz ti − 19 Ne, mI = 1/2 = − 3

(4)

i

The predicted magnetic moments in this scheme are:

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µTheo (19 Ne) = µn = −1.91nm µTheo ( 19 F ) = µ p = 2.79nm

(

)

( )

The observed moments are: µ 19 Ne = −1.89 nm, µ 19 F = 2.63 nm , which is quite good agreement, considering the simplicity of the theoretical model.

11.5 Since pions have zero spin, J = 0 implies that the total orbital angular momentum  of the 3 pion state is Ltotal = 0 . Now Ltotal = L +  , where  is the orbital angular momentum of two pions about their center of mass, and L is the orbital angular momentum of the third pion about the 3-pion center of mass. Thus: L+ = 0 and therefore L =  . Since pions are bosons, the total 3-pion wave function must be symmetric in spatial coordinates and charge coordinates, with respect to exchange of any pair of pions. Now the spatial wave function is symmetric when  = L = 0, 2, 4,... Thus, for these values of  = L , the isospin function must also be symmetric.

Now, the isospin of a pion is T=1. Therefore, a priori, the isospin of the 3 pion system can be: T=3 1 multiplet (7 states) T=2 2 multiplets (2x5 states =10 states) T=1 3 multiplets (3x3 states= 9 states) T=0 1 state (1 state) The T=3 multiplet is exchange symmetric, as can be seen by starting with the state: T = 3,Tz = 3 = ( + + + ) From this we obtain: 6 3, 2 = 2 ⎡⎣ + + 0 + +0 + + 0 + + ⎤⎦ or: 1 3, 2 = (1) ⎡⎣ + + 0 + +0 + + 0 + + ⎤⎦ 3 and: 1 3,1 = (2) ⎡⎣ + + − + + − + + − + + + 2 +00 + 2 0 + 0 + 2 00 + ⎤⎦ 15 We can construct a symmetric T=1, Tz =1 state 1,1 by requiring that it be orthogonal to the 3,1 state in (2), and that T+ 1,1 = 0 . After some straightforward algebra, we find:

1 (3) ⎡ 2 + + − + 2 + − + + 2 − + + − 00 + − 0 + 0 − +00 ⎤⎦ 15 ⎣ Applying T− to this state, we obtain: 1 1, 0 = (4) ⎡⎣ +0 − + 0 + − + + − 0 + 0 − + + − + 0 + −0 + − 3 000 ⎤⎦ 15 By similar arguments we show that states with T = 2 or 0 cannot be exchangesymmetric. Thus if  = L is even, T must be odd, and vice versa.

1,1 =

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11.6 The basic idea is to choose a beta radioactive source that radiates reasonably lowenergy beta particles and no X-rays or gamma rays, and surround it with an absorber of moderately high Z. If the beta rays and ordinary atomic electrons are distinguishable in any way, the beta particle will be able to occupy various states of the absorber atom as it loses energy, and will ultimately be captured in the innermost K shell of the atom, even though that K shell is already occupied by 2 atomic electrons. However if beta rays are indistinguishable from electrons, then the Pauli principle must prevent the beta particle from occupying any occupied atomic orbits. Thus if beta rays and electrons are distinguishable one should be able to observe characteristic X rays from the absorber, especially the Kα X ray. The most sensitive experiment of this type was done by Davies and Grace at Oxford, U.K. in 1951. They employed a powerful tritium source and an absorber of Zirconium (Z=40) for which the Kα line occurs at 15.8 keV. They observed no such X-ray, and were thus able to conclude that less than 1 in 100,000 beta rays absorbed in the zirconium resulted in an X-ray with energy ≈ 15.8 keV.

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12.1 a)We start with the functional: Ze2 n e2 n(r)n(r ') h [ n ] = c ∫ n 5/3dτ − d τ + dτ dτ '− λ ∫ n dτ 4π ∫ r 2 ∫∫ 4π r − r '

(

)

(1)

35 / 3 π 4 / 3  2 and λ is an undetermined multiplier. We make small variations 10 me of n and determine the condition for which h reaches an extremum: δ h = 0 : 5 Ze2 δ n e2 n(r)δ n(r ') e2 δ n(r)n(r ') δ h = 0 = c ∫ n 2/3δ ndτ − d τ + d τ d τ '+ dτ dτ ' 3 4π ∫ r 2 ∫∫ 4π r − r ' 2 ∫∫ 4π r − r ' (2) where c =

(

− λ ∫ δ n dτ

)

In this expression the 3rd and 4th terms on the right hand side are the same. Since the variations δ n are arbitrary, we conclude from (2) that:

5 2/3 Ze2 e2 n(r)dτ cn − + −λ =0 3 4π r 4π ∫ r − r ' Now n ( r ) is spherically symmetric. Thus in:

(3)



1 1 ∞ ⎛r ⎞ = ∑ ⎜ < ⎟ P ( cosθ ) r − r ' r> = 0 ⎝ r> ⎠ we can replace the right hand side by r>−1 . Hence the integral on the left hand side of (3) becomes: ∞ n(r)dτ 1 r 2 = n(r ')r ' dr ' + ∫ 4π r − r ' r ∫0 ∫r n(r ')r 'dr ' and (3) can be written as: ∞ 5 2/3 Ze2 e2 r cn − + ∫ n(r ')r '2 dr ' + e2 ∫ n(r ')r 'dr ' − λ = 0 (4) r 3 4π r r 0 Now differentiate (4) with respect to r to obtain: 5 dn 2/3 Ze2 e2 r c + − n(r ')r '2 dr ' = 0 (5) 3 dr 4π r 2 r 2 ∫0 Multiplying (5) by r2 we have: r 5 2 dn 2/3 Ze2 cr + − e2 ∫ n(r ')r '2 dr ' = 0 (6) 0 3 dr 4π Now differentiate (6) and divide by r 2 to obtain: 5 1 d ⎛ 2 dn 2/3 ⎞ (7) c r = e2 n 3 r 2 dr ⎜⎝ dr ⎟⎠ In (7) it is convenient to make the substitution: 3/2 2me e) ( n= Φ 3/2

3π 2  3

(8)

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which yields:

3/2 2mee) ( ∇ Φ=e Φ 3/2

(9) 3π 2  3 This is the same equation as (12.22), and it leads directly to the Thomas-Fermi equation. b) Under the norm preserving transformation n(r) → β 3n(β r) , 2

E1 = const ∫ n 5 / 3r 2 dr → const ∫ β 5 n 5 / 3 ( β r ) r 2 dr

= constiβ 2 ∫ n 5 / 3 ( y )y 2 dy

where y = β r . Thus E1 → β 2 E1 . Also,

E2 = const ∫ n(r)rdr → constiβ ∫ n(y)ydy

hence E2 → β E2 . Finally,

n(r)n(r ')d 3 rd 3 r ' n(β r)n(β r ')d 3β 3 rd 3β 3 r ' , E3 = const ∫∫ → const β ∫∫ r − r' βr − βr ' hence E3 → β E3 . Therefore: E → E ( β ) = β 2 E1 + β E2 + β E3 Since E(β ) must reach a minimum when β → 1 we must have:

dE(β ) = 2E1 + E2 + E3 = 0 d β β =1

(10)

(11)

which is the virial theorem for the Thomas-Fermi model. Equation (11) agrees with (12.40) of the text.

12.2 Start with equations (10.53-10.55) for the magnetic induction field generated by diamagnetic currents. In atomic units α2 x '2 + y'2 B' ( 0 ) = − B0 ẑ ∫ ρ ( r ') d 3r ' (1) 3 2 r' Here, B0 ẑ is the initial applied magnetic field, and ρ is the electron probability density.

To employ the Thomas-Fermi model, make the replacement ρ ( r ') → ne ( r ') : π ∞ α2 B0 ẑ i 2π ∫ sin 3 θ dθ ∫ ne ( r )rdr 0 0 2 2 3/2 α Z ∞φ =− B0 ẑ i ∫ 1/2 dx 3 a 0 x

B' ( 0 ) = −

Thus we have:

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α2 Z 4/3 ∞ B0 ẑ i φ "dx 3 .885 ∫0 α2 Z 4/3 =+ B0 ẑ φ '(0) 3 .885 α2 1.58875Z 4/3 1.795Z 4/3 2 =− B0 ẑ =− α B0 ẑ 3 .885 3

B'(0) = −

(2)

12.3a) 1 − ψ "− Zδ (x)ψ = Eψ 2 For x ≠ 0, ψ "+ 2Eψ = 0 Thus, ψ = N exp ( −α x )

(1)

where α = −2E and N = α 1/2 . To find α , we integrate (1) across the singularity: ε ε 1 ε − ∫ ψ "dx − Z ∫ δ (x)ψ dx = E ∫ ψ dx −ε −ε 2 −ε where ε is a positive real infinitesimal. This yields: 1 1 − ψ '(0+) + ψ '(0−) = Zψ (0) 2 2 3/2 which implies: α = Zα 1/2 or: α = Z = −2E hence: Z2 E=− (2) 2 Now, ∞

V = −Z ∫ δ (x)ψ 2 (x)dx −∞

= −Z ∫ Z exp ( −2Z x )δ (x)dx = −Z 2 ∞

−∞

Thus

1 1 (3) T = E − V = − Z 2 + Z 2 = Z 2 =-E 2 2 Alternatively, we can calculate T directly using integration by parts. Consider: ε ∞ 1 ∞ 1 −ε T = − ∫ ψψ "dx = − ⎡⎢ ∫ ψψ "dx + ∫ ψψ "dx + ∫ ψψ "dx ⎤⎥ −ε ε ⎦ 2 −∞ 2 ⎣ −∞ Integrate each of these by parts:

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Let

∞ 1 −ε ε ∞ T = − ⎡⎢ψ ψ ' −∞ + ψ ψ ' − ε + ψ ψ ' ε − ∫ ψ '2 dx ⎤⎥ −∞ ⎦ 2⎣ ∞ ∞ 1 = ∫ ψ '2 dx = ∫ ψ '2 dx 0 2 −∞

ψ = λ 1/2 e− λ x in the range 0 ≤ x < ∞ . Then ψ '2 = λ 3e−2 λ x and

λ2 T = λ ∫ e dx = 0 2 for any λ > 0 , and in particular when λ = Z . b) The Schroedinger equation is: 1 ∂ 2ψ 1 ∂ 2ψ (4) − − − Zδ ( x1 )ψ − Zδ ( x2 )ψ + δ (x1 − x2 )ψ = Eψ 2 ∂x12 2 ∂x22 In the absence of the last term on the left hand side of (4), the zero-order solution to (4) is: ψ 0 = Z exp ⎡⎣ −Z ( x1 + x2 ) ⎤⎦ (5) 3

∞

−2 λ x

and the energy is E0 = −Z 2 The first-order perturbation energy shift due to H ' = δ (x1 − x2 ) is:

ΔE (1) = Z 2 ∫

∞

∞

−∞

∫ e

−2 Z x1

−∞

(6)

e−2 Z x2 δ (x1 − x2 )dx1dx2

∞

= Z 2 ∫ e−4 Z x1 dx1 −∞

∞

= 2Z 2 ∫ e−4 Zx1 dx1 0

= Thus,

Z 2

Z 1 1 (7) = −(Z − )2 + 2 4 16 We recall that for 3d helium, use of first order perturbation theory yields: 5 (8) E = E0 + ΔE (1) = −Z 2 + Z 8 c) Now let ψ = ηe−η x1 e−η x2 where η is a real parameter. Then repeating the calculation that led to (7) we obtain: η (9) E(η ) = η 2 − 2Zη + 2 We now vary η to find the minimum in (9): E = E0 + ΔE (1) = −Z 2 +

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∂E 1 = 2η − 2Z + = 0 ∂η 2 1 which yields η = Z − , and thus: 4 1 (10) E (η ) = −(Z − )2 4 d) Now consider the Hartree equation: ∞ 1 ∂ 2φ − − Zδ (x)φ (x) + ⎡⎢ ∫ φ 2 ( y )δ (x − y)dy ⎤⎥ φ (x) = εφ (x) 2 ⎣ −∞ ⎦ 2 ∂x which is: 1 ∂ 2φ − − Zδ (x)φ (x) + φ 3 (x) = εφ (x) (11) 2 2 ∂x Multiply both sides of (11) by φ ' : 1 (12) − φ 'φ "− Zδ (x)φφ '+ φ 3φ ' = εφφ ' 2 If x ≠ 0 we have: 1 (13) − φ 'φ "+ φ 3φ ' = εφφ ' 2 Integrating both sides of (13) we obtain: 1 1 1 (14) − φ '2 + φ 4 = εφ 2 + k 4 4 2 where k is a constant of integration. However, if φ is a bound state we require that: φ → 0, φ ' → 0 as x → ∞.

Thus k = 0 and (14) yields: dφ (15) I≡∫ = ±x + C φ φ 2 − 2ε where C is another constant of integration. Since ε < 0 we make the substitution: (16) λ 2 = −2ε and we thus obtain: dφ (17) I=∫ φ φ2 + λ2 Let φ = λ tan θ . Then:

λ sec 2 θ dθ 1 dθ 1 θ = ∫ = n tan λ tan θλ sec θ λ sin θ λ 2 Thus (15) yields: θ n tan = ± λ x + const 2 or I=∫

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tan

θ = Ke± λ x 2

Therefore,

θ ±λx 2 = 2 λ Ke φ = λ tan θ = θ 1 − K 2 e±2 λ x 1 − tan 2 2 2 λ tan

(18)

Obviously the ± sign applies for x < 0, (x > 0) respectively. The constants λ and K are determined by integrating (11) across the singularity, and by normalization, respectively.

After straightforward calculation, we find: 2 1 1⎛ 1⎞ λ = (Z − ); hence ε = − ⎜ Z − ⎟ 2 2⎝ 2⎠ and: 1/2 1⎞ ⎛ 1 ⎞ 1 ⎤ ⎛ ⎡ 2⎜ Z − ⎟ ⎜ ⎟⎠ exp ⎢ −(Z − ) x ⎥ ⎝ ⎠ ⎝ 2 4Z − 1 2 ⎦ ⎣ (19) φ= 1 1 ⎤ ⎡ 1− exp ⎢ −2(Z − ) x ⎥ 4Z − 1 2 ⎦ ⎣ Finally, 2 1⎞ ⎛ H = 2ε − δ (x1 − x2 ) = − ⎜ Z − ⎟ − δ (x1 − x2 ) (20) ⎝ 2⎠ After straightforward calculation employing (19), we obtain the result: 1 1 (21) δ (x1 − x2 ) = (Z − ) 2 3 Thus, 2 1⎞ 1⎛ 1⎞ ⎛ H = −⎜ Z − ⎟ − ⎜ Z − ⎟ ⎝ 2⎠ 2⎝ 3⎠ Z 1⎤ ⎡ (22) = − ⎢Z 2 − + ⎥ 2 12 ⎦ ⎣ 1 1 = −(Z − )2 − Hartree 4 48 This should be compared with the perturbation theory result (7), and the perturbationvariation result (10).

12.4a) A d 2 configuration has: 10! g(d 2 ) = = 45 distinct states 2!8! and for each of these there are 2 possible s states. Thus the total number of distinct states for the configuration ns(n 'd)2 is 90.

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Now consider the possible terms of (n 'd)2 . Ignoring the requirement of antisymmetrization for the moment, we see that the possible values of L are 4,2,3,1,0; while the possible values of S are 1,0. Thus we have the terms: 3 G, 3 F, 3 D, 3 P, 3 S 1

G,1 F,1 D,1 P,1 S However, from the symmetry properties of the vector coupling coefficients, triplet (singlet) spin states of 2 electrons are exchange- symmetric (antisymmetric) respectively.

Also, G,D,S orbital angular momentum states are exchange symmetric, whereas F, P states are exchange antisymmetric. Thus the Pauli principle allows 1 G, 3 F,1 D, 3 P,1 S terms but prohibits 3 G,1 F, 3 D,1 P, 3 S terms. We now consider possible terms of the configuration ns(n 'd)2 . These are obtained easily by adding a spin ½ vectorially to each

of the allowed terms: 1 G, 3 F,1 D, 3 P,1 S . Thus we obtain: 2 G 18 2

F, 4 F

14, 28

2

D

10

2

4

P, P

6. 12

2

S 2 Here the multiplicity of each term is listed on the right. As we expect, the total number of states is 90. Hund’s rules yield the following list (in order of increasing energy): 4 F, 4 P,2 G,2 F,2 D,2 P,2 S This prediction is not in good agreement with observations. b) The possible terms for (nd)7 are the same as those for (nd)3 . In each case the number of independent states is: 10! = 120 3!7!

We construct the usual table, where mi ≤ 2 :

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m1 m2 m3 ms1 ms 2 ms 3 M L M S L S 2 2 2 Impossible ________________________________________________________________________ 2 H (22 states) 2 2 1 + + 5 ½ 5 ½ 2 2 0 + + 2 H , 2G : (18 states) 4 ½ 4,5 ½ 2 1 1 + + -

( )

2 2

+ + 3 1/2 (2 H ,2 G) 4 F : (28 states) + + + 3 3/2 2 F : (14 states) + + 3 ½ + + 3 ½ + + 3 ½ Here there are 5 independent Slater determinants. Thus, in addition to 2 H ,2 G we must have 4 F and 2 F .

2 2 -2 + + 2 ½ (2 H , 2G, 4 F, 2 F) 2 1 -1 + + + 2 3/2 + + 2 ½ Two 2D : (20 states) + + 2 ½ + + 2 ½ 2 0 0 + + 2 ½ 1 1 0 + + 2 ½ Here there are 7 independent Slater determinants. Thus in addition to 2 H , 2G, 4 F, 2 F we must have two independent 2 D terms. Including these, we have so far accounted for 102 states. We need 18 more, all of which must have M L ≤ 1 . This is achieved with 4 P (12 states), and 2 P (6 states). Altogether we have: 2 H 22 2 G 18 4 F 28 2 F 14 2 D 10 2 D 10 4 P 12 2 P 6 Total

2 1

-1 0

120

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12.5 The possible multiplets in a p3 configuration are 2 D, 4 S, and 2 P. The 2 P multiplet consists of the terms 2 P3/2 and 2 P1/2 . Consider first the state 2 P3/2 , M J = 3/2. Since S = 1/2, L = 1 , we must have M L = 1, M S = 1/2. There are two possible Slater determinants that can give these values. They are: ( m1, ms1, m2 , ms 2 , m3 , ms 3 ) = (1+,1−, −1+)

( m1, ms1, m2 , ms 2 , m3 , ms 3 ) = (1+, 0+, 0−)

Thus,

P3/2 , M J = 3/2 = A(1+,1−, −1+) + B(1+, 0+, 0−) (1) where A,B are coefficients to be determined. We apply the raising operator J+ to both sides of (1): J + 3/2,3/2 = 0 = A(1+,1+, −1+) + 2A(1+,1−, 0+) + (2 B(1+, 0+, 0+) + 2B(1+,1+, 0−) + 2B(1+, 0+,1−) Since a determinant vanishes if 2 rows or 2 columns are equal, (1+,1+, −1+) = (1+, 0+, 0+) = (1+,1+, 0−) = 0 Thus (2) yields: 2A(1+,1−, 0+) + 2B(1+, 0+,1−) = 0 However, (1+,1−, 0+) = −(1+, 0+,1−) since a determinant changes sign if any two rows, or columns, are interchanged. Therefore A = B , and since we require the state in (1) to be normalized: 1 1 2 P3/2 , M J = 3/2 = (1+,1−, −1+) + (1+, 0+, 0−) (3) 2 2 By similar reasoning, 1 ⎡ 1 ⎤ 2 P3/2 , M J = −3/2 = − ⎢ (−1−, −1+,1−) + (−1−, 0−, 0+) ⎥ (4) 2 ⎣ 2 ⎦ 2

where the overall minus sign on the right hand side can be deduced from the behavior of individual spin ½ and spin 1 states under rotation of (3) by 180 deg about the y axis.

To find the composition of 2 P3/2 , M J = 1/2, we apply the lowering operator to both sides of (3). Then, discarding all determinants that are equal to zero, and re-arranging, we obtain: 3 1 1 ⎡(1+,1−, −1−) − (0+,1−, 0−) + 2(0+,1−, −1+) − 2(1+, −1−, 0+) ⎤ , = (5) ⎦ 2 2 6⎣ Similarly, by applying the raising operator to both sides of (4) we obtain: 3 1 1 ⎡(−1−, −1+,1+) − (0−, −1+, 0+) + 2(0−,1+, −1−) − 2(1−, −1+, 0−) ⎤ ,− =− ⎦ 2 2 6⎣ (6)

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Now consider J = 1/2,M J = 1/2 . This state must be constructed from those Slater determinants with M L + M S = 1/2 . The possibilities are : (1+,1−, −1+),

(1−, 0+, 0−), (1+, −1−, 0+), (1−, −1+, 0+), (1+, −1+, 0−).

Thus we try: 1 1 , = a(1+,1−, −1−) + b(1−, 0+, 0−) + c(1+, −1−, 0+) + (7) 2 2 + d(1−, −1+, 0+) + e(1+, −1+, 0−) Applying the raising operator to both sides of this equation and discarding all determinants that vanish we obtain: 0 = a(1+,1−, −1+) + 2a(1+,1−, 0−) + b(1+, 0+, 0−) + 2b(1−,1+, 0−) + c(1+, −1+, 0+) + 2c(1+, 0−, 0+) + d(1+, −1+, 0+) + 2d(1−, −1+,1+) + e(1+, −1+, 0+) + 2e(1+, 0+, 0−) + 2e(1+, −1+,1−) or, rearranging: a − 2e + 2d (1+,1−, −1+) + 2a − 2b (1+,1−, 0−)

(

(

)

)

(

)

+ b − 2c + 2e (1+, 0+, 0−) + ( c + d + e) (1+, −1+, 0 + ) = 0 Since these 4 determinants are linearly independent, their coefficients must vanish; hence: a = − 2d = 2c = b 1 1 and to achieve normalization of the state , we must require a = (3)−1/2 . Hence: 2 2 1 1 1 ⎡ 2(1+,1−, −1−) − 2(0+,1−, 0−) − (0+,1−, −1+) + (1+, −1−, 0+) ⎤ , = (8) ⎦ 2 2 6⎣ 3 1 1 1 , and , Comparing (8) to (5) we see that the states are orthogonal. We can 2 2 2 2 1 1 ,− easily find the Slater determinant composition of by performing a rotation of 2 2 both sides of (8) about the y axis by 180 deg. This yields: 1 1 1 ⎡ 2(−1−, −1+,1+) − 2(0−, −1+, 0+) − (0−, −1+,1−) + (−1−,1+, 0−) ⎤ ,− = ⎦ 2 2 6⎣ (9) 12.6 a) J =

9 5 3 , , 2 2 2

b) J = 8, 6, 5, 4, 4, 2, 2, 0

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13.1 The Hamiltonian is: p2 p2 p2 k 2 2 H = 1 + 3 + 2 + ⎡( x2 − x1 − a ) + ( x3 − x2 − a ) ⎤ ⎣ ⎦ 2M 2M 2m 2 It is convenient to make the following changes in the variables: y1 = M 1/2 (x1 + a)

(1)

y3 = M 1/2 (x3 − a) y2 = m1/2 x2 Then after simple algebra, the Hamiltonian in (1) becomes: 2 2 1 k ⎡⎛ y2 y1 ⎞ ⎛ y3 y2 ⎞ ⎤ H = y12 + y22 + y33 + ⎢⎜ 1/2 − 1/2 + − ⎟ ⎜ ⎟ ⎥ 2 2 ⎢⎣⎝ m M ⎠ ⎝ M 1/2 m1/2 ⎠ ⎥⎦

(

=

)

1 2 k ⎡ y 2 2y 2 y 2 yy yy ⎤ y1 + y22 + y33 + ⎢ 1 + 2 + 3 − 2 1 2 − 2 2 3 ⎥ 2 2 ⎣M m M mM mM ⎦

(

)

(2) The potential energy term in (2), that is, the term proportional to k, may be written as: V = yiW i y

where y = ( y1 , y2 , y3 ) and W is the second-rank symmetric tensor:

1 1 ⎛ ⎞ − 0 ⎜ M ⎟ mM ⎜ ⎟ k⎜ 1 2 1 ⎟ W = − − (3) 2⎜ m mM mM ⎟ ⎜ ⎟ 1 1 ⎟ ⎜ 0 − ⎜⎝ ⎟ M ⎠ mM We can diagonalize this symmetric matrix by a suitable orthogonal transformation, and this leaves the kinetic energy term diagonal. To find the eigenvalues of W we write the secular determinant: k k −λ − 0 2M 2 mM k k k − −λ − =0 (4) m 2 mM 2 mM k k 0 − −λ 2M 2 mM After simple algebra we obtain the eigenvalues and corresponding normal modes:

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1 λ ≡ ω 12 = 0 (Center-of-mass motion) 2 1 k λ ≡ ω 22 = (x1 = − x 3 , x2 = 0) 2 2M 1 k k λ ≡ ω 32 = + ( x1 = + x3 ) 2 2M m Ignoring the center-of-mass motion, we quantize the other two normal modes and thus obtain the energy eigenvalues: 1⎞ 1⎞ ⎛ ⎛ E = ω 2 ⎜ n2 + ⎟ + ω 3 ⎜ n3 + ⎟ (5) ⎝ ⎝ 2⎠ 2⎠ where n2 , n3 independently can take on the values 0,1,2,3,… . Also, ⎛ 2k ⎞ ω3 = ⎜ ⎟ ⎝ m⎠

1/2

1/2

m ⎤ ⎡ ⎢⎣1 + 2M ⎥⎦

1/2

m ⎤ ⎡ ⎢⎣1 + 4M ⎥⎦

1/2

⎛ k ⎞ +⎜ ⎟ ⎝M⎠

⎛ 2k ⎞ ≈⎜ ⎟ ⎝ m⎠ ⎛ 2k ⎞ =⎜ ⎟ ⎝ m⎠

1/2

⎛ m ⎞ ⎜⎝ ⎟ 8M ⎠

1/2

Thus, 1/2 ⎡⎛ 1⎞ ⎛ m ⎞ ⎛ 1⎞ ⎤ (6) n + ⎢⎜ n2 + ⎟ + ⎜ ⎟ ⎜ 3 ⎟⎥ 2 ⎠ ⎝ 8M ⎠ ⎝ 2 ⎠ ⎥⎦ ⎢⎣⎝ This completes the first part of the analysis. We now follow the Born-Oppenheimer approach, by fixing the protons at ±R/2 . Then the Hamiltonian (1) reduces to: 2 2 p22 k ⎡⎛ R ⎞ ⎛R ⎞ ⎤ H= + ⎢ ⎜ x2 + − a ⎟ + ⎜ − a − x2 ⎟ ⎥ ⎠ ⎝2 ⎠ ⎥⎦ 2m 2 ⎢⎣⎝ 2

⎛ 2k ⎞ E  ⎜ ⎟ ⎝ m⎠

=

1/2

1⎞ ⎛ ⎛ k⎞ ⎜⎝ n3 + ⎟⎠ +  ⎜⎝ ⎟⎠ 2 M

1/2

2 p22 k ⎡ 2 ⎛R ⎞ ⎤ + ⎢ 2x2 + 2 ⎜ − a ⎟ ⎥ ⎝2 ⎠ ⎥⎦ 2m 2 ⎢⎣

⎛ p 2 mω 2 x22 ⎞ ⎛R ⎞ =⎜ 2 + + k ⎜ − a⎟ ⎟ ⎝ ⎠ 2 ⎠ 2 ⎝ 2m

(7) 2

⎛ p22 mω 2 x22 ⎞ 2k where ω ≡ . Now quantizing the harmonic oscillator described by ⎜ , + m 2 ⎟⎠ ⎝ 2m we obtain the following eigenvalues of the Hamiltonian in (7):

2k ⎛ 1⎞ ⎛R ⎞ En (R) =  ⎜⎝ n3 + ⎟⎠ + k ⎜⎝ − a ⎟⎠ m 2 2

2

(8)

Next we take into account the motion of the protons by writing the equation:

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p12 p2 2k 1 k 2 + 2 + (n3 + ) + ( R − 2a ) = H ' (9) 2M 2M m 2 4 Ignoring the center of mass motion, we find that the first two terms on the left hand side of (9) reduce to: p2 2µ where µ = M /2 and p is the relative momentum of the two protons. When we quantize that portion of H ' pertaining to nuclear motion, namely: p2 k 2 + ( R − 2a ) 2µ 4 1/2

1⎞ ⎛ k ⎞ ⎛ we obtain energies of the form  ⎜ ⎟ ⎜ n2 + ⎟ . Thus in the Born-Oppenheimer ⎝M⎠ ⎝ 2⎠ approximation the energies are given by the expression: 1/2 1⎞ 2k ⎛ 1⎞ ⎛ k⎞ ⎛ E ' =  ⎜ ⎟ ⎜ n2 + ⎟ +  n3 + ⎟ (10) ⎜ ⎝M⎠ ⎝ 2⎠ m ⎝ 2⎠ Comparing E ' of (10) to E of (6), we see that E ' lacks the term: 1/2 1/2 1⎞ ⎛ k⎞ ⎛ m ⎞ ⎛ T = ⎜ ⎟ ⎜ n3 + ⎟ (11) ⎟ ⎜ ⎝ M ⎠ ⎝ 8M ⎠ ⎝ 2⎠ m Compared to the main term in n3 + 1/2 , T is of relative size .

4M 13.2 We utilize the following integrals several times in this solution: n ∞ ax ) ⎤ n!e− ax ⎡ ( n − ay (1) ∫x y e dy = a n +1 ⎢⎢1 + ax + ... + n! ⎥⎥ ⎣ ⎦ 1 ∞ R 3 2π f λ , µ , φ d τ = dφ ∫ d µ ∫ f ( λ, µ, φ ) λ 2 − µ 2 d λ (2) ( ) ∫ ∫ −1 1 8 0 where in (2) λ, µ, φ are the usual prolate spheroidal coordinates defined in Sec. 13.3. If function f in (2) is independent of φ , (2) reduces to:

(

)

∞ π R3 1 f λ , µ d τ = d µ f ( λ, µ ) ( λ 2 − µ 2 ) d λ ( ) ∫ ∫ ∫ −1 1 4 a) Normalization of the wave function.

We write: ψ = N ( e−α ra + e−α rb ) r +r r −r 2 where λ = a b , µ = a b , and N is chosen so ∫ ψ dτ = 1 . Then: R R −2 −2 α ra N = ∫ ⎡⎣ e + e−2α rb + 2e−α ( ra + rb ) ⎤⎦dτ

(3)

(4)

(5)

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On the rhs of (5) the first two terms are equal and are easily evaluated in spherical coordinates. Both of these terms together give: ∞ 16π 2π (6) 8π ∫ exp ( −2α r )r 2 dr = = 3 3 0 ( 2α ) α From (3), the third term on rhs of (5) is: ∞ π R3 1 d µ ∫ exp [ −α Rλ ] λ 2 − µ 2 d λ = ∫ 1 2 −1 ⎡ 2e−α R ⎛ π R3 1 (α R )2 ⎞ − µ 2 e−α R ⎤ = d µ 1 + α R + ⎢ ⎥ 2 ∫−1 ⎢⎣ (α R )3 ⎜⎝ 2 ⎟⎠ α R ⎥⎦ 2 −α R ⎛ α R ) ⎞ π R 3 e− α R ( 3 e (7) 2π R 1+ αR + − 2 ⎟⎠ 3 αR (α R )3 ⎜⎝ Combining the contributions from (6), (7) we obtain: 2π (8) N −2 = 3 (1 + S ) α where 2 ⎛ α R) ⎞ ( −α R (9) S = e ⎜1 + α R + 3 ⎟⎠ ⎝ b) Calculation of T .

1 T = − ∫ ψ∇ 2ψ dτ 2 ⎡⎛ d 2 e−α ra 2 de−α ra ⎞ ⎛ d 2 e−α rb 2 de−α rb ⎞ ⎤ N2 − α ra − α rb ⎡ ⎤ =− e + e ⎦ ⎢⎜⎝ dr 2 + r dr ⎟⎠ + ⎜⎝ dr 2 + r dr ⎟⎠ ⎥ dτ 2 ∫⎣ a a a b b b ⎣ ⎦

(

)

⎡⎛ 2 2α ⎞ −α ra ⎛ 2 2α ⎞ −α rb ⎤ N2 − α ra − α rb ⎡ ⎤ =− e + e + ⎜α − e ⎥ dτ ⎦ ⎢⎜⎝ α − r ⎟⎠ e 2 ∫⎣ rb ⎟⎠ ⎝ a ⎣ ⎦ − α ra N 2α 2 e−α rb ⎞ − α ra − α rb 2 2 − α ra − α rb ⎛ e ⎡ ⎤ =− e + e ⎦ dτ + N α ∫ e +e ⎜⎝ r + r ⎟⎠ dτ 2 ∫⎣ a b

(

=−

)

∞ ⎛ 1 1⎞ α2 + 2N 2α i4π ∫ e−2α r rdr + N 2α ∫ ⎜ + ⎟ e−α (ra + rb )dτ 0 2 ⎝ ra rb ⎠

3 − α Rλ 1 ∞ 4λ e α 2 2π N 2 2 πR =− + +N α dµ ∫ λ 2 − µ 2 ) dλ 2 2 ( ∫ −1 1 2 α 4 R λ −µ

(10) The last term on the rhs of the last line of (10) is: − α Rλ ∞ 4λ e π R3 1 2π 2 −α R N 2α d µ λ 2 − µ 2 dλ = N e (1 + α R ) 2 2 ∫ ∫ 1 R λ − µ 4 −1 α

(

)

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Thus:

α 2 2π N 2 ⎡⎣1 + e−α R (1 + α R ) ⎤⎦ + 2 α Now making use of (8) and (9) in (11) we obtain: T = α 2 F1 (α R ) where: ⎛ α 2 R2 ⎞ 1 + e− α R ⎜ 1 + α R − 3 ⎟⎠ ⎝ 1 F1 (α R ) = 2 ⎛ α 2 R2 ⎞ 1 + e− α R ⎜ 1 + α R + 3 ⎟⎠ ⎝ T =−

c) Calculation of − −

(11) (12)

(13)

1 1 . − ra rb

⎛ 1 1⎞ 1 1 + = −N 2 ∫ ⎜ + ⎟ e−2α ra + e−2α rb + 2e−α ( ra + rb ) dτ ra rb ⎝ ra rb ⎠

(

)

(14)

= I1 + I 2 + I 3 + I 4 + I 5

Here,

1 I1 = −N 2 ∫ e−2α ra dτ ra 1 I 2 = −N 2 ∫ e−2α rb dτ = I1 rb 1 I 3 = −N 2 ∫ e−2α rb dτ ra 1 I 4 = −N 2 ∫ e−2α ra dτ = I 3 rb

(15)

⎛ 1 1⎞ I 5 = −2N 2 ∫ ⎜ + ⎟ e−α ( ra + rb )dτ ⎝ ra rb ⎠ I1 = I 2 is easily evaluated in spherical polar coordinates. I 3 = I 4 may be evaluated in spherical polar coordinates or in prolate spheroidal coordinates. I 5 must be evaluated in prolate spheroidal coordinates. Here we use the formulae: (λ + µ ) R , r = (λ − µ ) R ra = b 2 2 The detailed calculation of (14) is routine and is similar to what is shown above for parts a), b). The final result is: 1 1 (16) − + = α F2 (α R ) ra rb where 129

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1 ⎞⎤ ⎡ −α R −2 α R ⎛ ⎜⎝ 1 + ⎟ ⎢ 2e (1 + α R ) + 1 − e α R ⎠ ⎥⎦ F2 (α R) = − ⎣ (1 + S)

(

)

(17)

13.3 The total energy is: 1 1 (1) E= T − + = α 2 F1 (α R ) + α F2 (α R ) ra rb Let w = α R . Then: ∂E ∂F ∂w ∂F ∂w = 2α F1 + α 2 1 + F2 + α 2 ∂α ∂w ∂α ∂w ∂α ∂F ∂F = 2α F1 + α 2 R 1 + F2 + α R 2 ∂w ∂w ∂F ⎞ ⎛ ∂F ⎞ ⎛ = α ⎜ 2F1 + w 1 ⎟ + ⎜ F2 + w 2 ⎟ ⎝ ∂w ⎠ ⎝ ∂w ⎠ Thus E reaches an extreme value (which is easily shown to be a minimum) when: ∂F F2 + w 2 ∂w (2) α=− ∂F1 2F1 + w ∂w w This gives α for any w. Then, R = gives R for any w. Hence we obtain α as a α function of R. 13.4

N 2 : 1Σ , 3 Σ ,5 Σ , 7 Σ O2 : 3 1Σ , 3 3 Σ , 3 5 Σ ; 2 1Π , 2 3Π , 2 5Π ; 1Δ, 3 Δ,5 Δ Cl2 ,TlF : 3 1Σ , 3 3 Σ ; 2 1Π , 2 3Π ; 1Δ, 3 Δ HCl :

1

Σ , 3 Σ ; 1Π , 3 Π

CN : 2 Σ , 4 Σ ,6 Σ ; 2Π , 4 Π ,6 Π The number appearing in front of a molecular symbol indicates the number of independent states of that type, when that number is greater than unity.

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14.1We start with the equation: 5 A 7/3 2 ∇ 2ψ + ψ = Aψ 3B 3 2 where A = ∫ ( ∇ψ ) dτ , B = ∫ ψ 10 / 3dτ . Let φ = rψ . Then (1) becomes:

d 2φ 5 A φ 7 / 3 2 + = Aφ dr 2 3 B r 4 / 3 3 Now let r = ax, φ = b χ . Then (2) becomes:

χ "+

5 A 2/3 4/3 χ 7/3 2 2 a b = Aa χ 3B x4/3 3

(1)

(2)

(3)

We choose

⎛ 3 ⎞ a=⎜ ⎝ 2A ⎟⎠

1/2

(4)

3B , hence: 5A

and a 2 / 3b 4 / 3 =

31/2 21/ 4 B 3/ 4 b= 5 3/ 4 A1/2

(5)

⎡⎛ χ ⎞ 4 / 3 ⎤ χ "+ χ ⎢⎜ ⎟ − 1⎥ = 0 ⎢⎣⎝ x ⎠ ⎥⎦

(6)

so that (3) reduces to:

Now: d 2φ A = −4π ∫ ψ∇ ψ r dr = −4π ∫ φ 2 dr dr 2

2

⎛ dφ ⎞ = 4π ∫ ⎜ ⎟ dr ⎝ dr ⎠ 2

(7)

b2 χ '2 dx ∫ a Using (4) and (5) on the right hand side of the last line of (7), and rearranging, we obtain: 3/2 8π i31/2 ⎛ A⎞ χ '2 dx = 3.893548 ∫ χ '2 dx (8) ⎜⎝ ⎟⎠ = 3/2 ∫ B 5 or: 2/3 A (9) = 2.474934 ∫ χ '2 dx B = 4π

(

)

Also,

4π b10 / 3 χ 10 / 3 ⎛φ⎞ 2 B = ∫ ψ dτ = 4π ∫ ⎜ ⎟ r dr = dx ⎝ r⎠ a1/ 3 ∫ x 4 / 3 Employing (4) and (5) on the far rhs of (10), and rearranging, we obtain: 10 / 3

10 / 3

(10)

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8π i33/2 χ 10 / 3 ⎛ A⎞ = dx (11) ⎜⎝ ⎟⎠ B 5 3/2 ∫ x 4 / 3 Now divide (8) by (11) to obtain: 3 χ 10 / 3 2 χ ' dx = dx (12) ∫ 5 ∫ x4/3 Numerical integration of (6) yields evaluation of the rhs of (9) or (11), and thus gives: A (13) = 9.578 B 3/2

14.2 If the energy of a bulk sample of N bosons is: e2 E=− N 7/5 4π a0

e2 the energy of two such samples before they are united is: Einitial = −2 N 7/5 , whereas 4π a0 the energy after they are united is: E final = −2

ΔE = ( 2

7/5

7/5

e2 N 7/5 . The energy released is thus: 4π a0

e2 − 2) N 7/5 4π a0

(1)

e2 Now = 27.2 eV and N = 6.02i10 23 . Inserting these in (1) we obtain: 4π a0 (2) ΔE = 3.4i10 34 eV = 5.44i1015 joules 235 U nuclei. It is For comparison we compute the energy release from fission of 2N known that each fissioning nucleus releases 170 MeV. Thus: ΔE fission = 2i1.7i10 8 i6.02i10 23 eV = 2.05i10 32 eV = 3.27i1013 joules (3) 14.3a) The data of interest are as follows: Element Mass density, g/cm3 Mass number Li .534 7 Na ,971 23 K .862 39 Rb 1.532 85 Cs 1.873 133 Cu 8.96 64 Au 19.32 197

Conduction electron density, 1022 cm-3 4.57 2.53 1.32 1.08 .843 8.38 5.87

b) We use the following formulae for pressure (P) and bulk modulus (B):

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∂E ∂rs ⎛ −∂E ⎞ P=⎜ =− ⎟ ⎝ ∂V ⎠ N ∂rs ∂V ⎛ ∂P ⎞ B = −V ⎜ ⎝ ∂V ⎟⎠ N After straightforward calculation we obtain the results: ⎡ 2.21 .458 ⎤ e2 P= − 4 ⎥ 2 4 ⎢ rs ⎦ ( 4π ) a0 ⎣ rs5 which vanishes at the equilibrium value rs = 4.83 , and:

(1)

ne e2 ⎡ 11.05 1.832 ⎤ (2) − 9 4π a0 ⎢⎣ rs2 rs ⎥⎦ which takes the value: e2 (3) B = .01049 ne = 6.52 i10 9 dyne/cm 2 4π a0 at rs = 4.83. This is on the low side compared to the following experimental data for liquid metals: Element Bexpt, 109 dyne/cm2 Li 115 Na 53 K 26 Rb 20 Cs 14 B=

14.4a)The nuclear radius is: Rnuc = kA1/ 3 , k = 1.4i10 −13 cm. The nuclear volume is: 4π 3 Vnuc = k A , and the nuclear mass is M nuc = Am p , where m p is the proton mass. Thus 3 the nuclear density is: 3m p M (1) ρ = nuc = = 1.45i1014 g/cm 3 3 Vnuc 4π k The star radius R is given by: 3 M star M star 3 R3 = = k 4π ρ mp Thus, 1/ 3

⎛ 2i10 33 ⎞ R=⎜ k = 1.49i10 6 cm. −24 ⎟ 1.67i10 ⎝ ⎠ b) The model neutron star is in equilibrium when: GM 2 P= R4

(2)

(3)

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Since M ≈ ρ R 3 , (3) can be written:

P = Gρ 2 R2 The pressure is assumed to be non-relativistic degeneracy pressure, so 2 5 / 3 2 P≈ nn = 8 / 3 ρ 5 / 3 mn mn Thus (4) is written: 2 5 / 3 ρ = Gρ 2 R2 8/3 mn or:

(4) (5)

 1 (6) 4/3 G mn ρ1/6 Using the value of ρ from (1) in (6), we obtain: R ≈ 9i10 5 cm (7) Comparison of (7) with (2) suggests that a neutron star of solar mass might be stabilized by non-relativistic degeneracy pressure.

R≈

1/2

14.5 Start with Newtonian equation of hydrostatic equilibrium: ∂P M (r)ρ = −G ∂r r2 We have 1 1 c2 p2 P = vp = 3 3 E However,

E = mc

2

⎛ p⎞ 1+ ⎜ ⎝ mc ⎟⎠

= where x =

4

(2)

2

Thus (2) can be written as: pF c2 p 4 dp P= 2 3∫ 2 3π  0 ⎛ p⎞ mc 2 1 + ⎜ ⎝ mc ⎟⎠ 4 5

(1)

(3)

4 5

m c x z dz m c = f (x) ≡ Af (x) = 6.01i10 22 f (x) 2 3 ∫0 2 3 2 3π  24π  1+ z

pF and: mc

f (x) = x(2x 2 − 3) 1 + x 2 + 3sinh −1 x

(4)

Now

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∂P ∂P ∂x 8x 4 ∂x GM (r)ρ = =A =− 2 ∂r ∂x ∂r r2 1 + x ∂r

(5)

and

ne = Also,

pF3 m 3c 3 3 = x 3π 2  3 3π 2  3

ρ = µe m p ne =

µ e m p m 3c 3

(6)

x 3 ≡ Bx 3

(7)

3π  Using (7) for ρ in the rhs of (5) we obtain: 8A ∂ M (r) 1 + x2 = − 2 BG ∂r r 2

3

(8)

Let y = 1 + x 2 . Then (8) becomes: 8A 2 ∂y r = −M (r) BG ∂r and so, employing (7), we have: 8A ∂ ⎛ 2 ∂y ⎞ 2 3 ⎜⎝ r ⎟⎠ = −4π r Bx BG ∂r ∂r Therefore: 3/2 1 ∂ ⎛ 2 ∂y ⎞ π GB 2 2 y −1 ⎜⎝ r ⎟⎠ = − 2 r ∂r ∂r 2 A Let y = y0φ with φ (0) = 1 . Then:

(

)

(9) 3/2

1 ∂ ⎛ 2 ∂φ ⎞ π GB 2 2 ⎛ 2 1 ⎞ r = − y0 ⎜ φ − 2 ⎟ ⎜ ⎟ r 2 ∂r ⎝ ∂r ⎠ 2 A y0 ⎠ ⎝ To simplify (10) make the substitution: r = aη , where: ⎛ 2A ⎞ a=⎜ ⎝ π G ⎟⎠

1/2

1 By0

(10)

(11)

(Note that for µe = 2, a = 3.855i10 8 y0−1 ). Then (10) becomes: 3/2

⎛ 1 ∂ ⎛ 2 ∂φ ⎞ 1⎞ (12) η = − ⎜φ2 − 2 ⎟ 2 ⎜ ⎟ η ∂η ⎝ ∂η ⎠ y0 ⎠ ⎝ which is Chandrasekhar’s equation. We have already noted that φ (0) = 1 . At 1 η = 0, x = y02 − 1 . The radius of the star is defined by η1 , where φ (η1 ) = and thus y0 where the rhs of (12) vanishes.

The mass of the star is calculated as follows:

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η1

0

0

M = 4π ∫ ρr 2 dr = 4π a 3 ∫ Bx 3η 2 dη ⎛ 2A ⎞ = 4π ⎜ ⎝ π G ⎟⎠

3/2

3/2

1 η1 ⎛ 2 1 ⎞ φ − 2 ⎟ η 2 dη 2 ∫0 ⎜ B y0 ⎠ ⎝

⎛ 2A ⎞ = −4π ⎜ ⎝ π G ⎟⎠

3/2

1 η1 ∂ ⎛ 2 ∂φ ⎞ η dη B 2 ∫0 ∂η ⎜⎝ ∂η ⎟⎠

⎛ 2A ⎞ = −4π ⎜ ⎝ π G ⎟⎠

3/2

1 ⎡ 2 ∂φ ⎤ η B 2 ⎢⎣ ∂η ⎥⎦η1

(13)

Although this expression does not contain y0 explicitly, φ does depend on y0 , as is evident from Chandrasekhar’s equation. Thus M is also a function of y0 . In particular, when y0 → ∞ , (12) reduces to: 1 ∂ ⎛ 2 ∂φ ⎞ (14) η = −φ 3 η 2 ∂η ⎜⎝ ∂η ⎟⎠ which is the Lane-Emden equation for a polytrope of index n=3. In this case one reaches the following result by numerical integration of (14): ⎡ ∂φ ⎤ − ⎢η 2 ⎥ = 2.01824 ⎣ ∂η ⎦η1

Employing this value and A=6.01•1022, B=9.82•105 µe in (13) we obtain:

M C = 5.75 µe−2 M  .

Assuming µe ≅ 2 , which holds for almost all degenerate white dwarfs, the limiting mass is: (15) M C = 1.4M  .

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15.1a) We start with the formulae: i ω Ei (r,t) = 3/2 ∑ (ε̂ kα )i ⎡⎣ akα ei(kir−ω t ) − ak†α e−i(kir−ω t ) ⎤⎦ L k,α 2

(

)

i ω ' B j (r ',t) = 3/2 ∑ k̂ '× ε̂ k 'α ' ⎡⎣ ak 'α 'ei( k'ir'−ω 't ) − ak†'α 'e−i( k'ir'−ω 't ) ⎤⎦ j L k',α ' 2 Using the commutation relations: ⎡⎣ akα , ak† ' α ' ⎤⎦ = δ k, k 'δα ,α '

[ a , a ] = ⎡⎣ a , a

⎤⎦ = 0 we see that equations (1) yield the result:  ⎡⎣Ei ( r,t ) , B j ( r ',t ) ⎤⎦ = 3 ∑ ω ( ε̂ kα )i k̂ × ε̂ kα eiki(r−r ') − e−iki(r−r ') j 2L k,α kα

k 'α '

† kα

(1)

† k 'α '

(

)(

)

Let R = r − r ' and let a particular k̂ vector be: k̂ = sin θ cos φ x̂ + sin θ sin φ ŷ + cosθ ẑ Also choose the two linear polarization vectors orthogonal to k̂ as: ε̂1 = − cosθ cos φ x̂ − cosθ sin φ ŷ + sin θ ẑ and ε̂ 2 = sin φ x̂ − cos φ ŷ

(2)

(3) (4) (5)

Then k̂ × ε̂1 = ε̂ 2 and k̂ × ε̂ 2 = −ε̂1 . We now let i = x, j = y in (2), and replace the sum over k by an integral. Then, (2) yields:  L3d 3k ⎡⎣Ex ( r,t ) , By ( r ',t ) ⎤⎦ = 3 ∫ ω ⎡⎣ε̂1x ε̂ 2 y − ε̂1yε̂ 2 x ⎤⎦ ( eikiR − e−ikiR ) (6) 2L ( 2π )3 Also, note that: 3 3  ikiR −ikiR L d k ⎡⎣Ex ( r,t ) ,Ey ( r ',t ) ⎤⎦ = 3 ∫ ω ⎡⎣ε̂1x ε̂1y + ε̂ 2 x ε̂ 2 y ⎤⎦ ( e − e (7) ) ( 2π )3 2L Now, from (4) and (5), 1 ε̂1x ε̂1y + ε̂ 2 x ε̂ 2 y = − sin 2 θ sin 2φ 2 −kx ky = − k̂x k̂y = k2 hence (7) can be written as: ∂2 1 ikiR −ikiR d 3k ⎡⎣Ex ( r,t ) ,Ey ( r ',t ) ⎤⎦ = c ( e − e ) ( 2π )3 ∂x ∂y ∫ k In a polar coordinate system in which the vector R lies along the polar axis, the integral in this expression can be written as:

137

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 K π 1 ik i R d 3k 1 −ik i R e − e lim kdk ∫ eikR cos θ − e−ikR cos θ sin θ dθ 3 = 2 K →∞ ∫0 ∫k 0 ( 2π ) 4π

(

)

(

=

(

)

)

K 1 1 lim kdk ∫ eikRu − e−ikRu du = 0 2 K →∞ ∫0 −1 4π

which vanishes because:

∫ (e 1

−1

ikRu

)

− e−ikRu du = 0

Hence ⎡⎣Ex ,Ey ⎤⎦ = 0 and similarly ⎡⎣Ex ,Ez ⎤⎦ = 0 . On the other hand,

ε̂1x ε̂ 2 y − ε̂1yε̂ 2 x = cosθ =

kz ckz = k ω

Thus, (6) can be written: c d 3k ikiR −ikiR ⎡⎣Ex ( r,t ) , By ( r ',t ) ⎤⎦ = kz ( e − e ) ( 2π )3 2 ∫

=

c ∂ d 3k ikiR −ikiR e + e ( ) 2i ∂z ∫ ( 2π )3

(8)

∂ 3 δ ( R) ∂z The commutators ⎡⎣Ey ( r,t ) , Bz ( r ',t ) ⎤⎦ and ⎡⎣Ez ( r,t ) , Bx ( r ',t ) ⎤⎦ are easily found from (8) = −ic

by cyclic permutation. Also, it is easy to show from (8) that: ∂ ⎡⎣ By ( r,t ) ,Ex ( r ',t ) ⎤⎦ = −ic δ 3 ( R ) ∂z b) We start with the expression: 1 H = ∫ ( E 2 + B2 )d 3r ' 2 Consider the commutator: 1 ⎡⎣Ex ( r ) , H ⎤⎦ = ∫ ⎡⎣Ex ( r ) ,E 2 ( r ') ⎤⎦ + ⎡⎣Ex ( r ) , B2 ( r ') ⎤⎦ d 3r ' 2 Now, E 2 = Ex2 + Ey2 + Ez2

{

also,

1

[E , H ] = 2 ∫ ⎡⎣E , B ⎤⎦d r ' x

2

(11)

similarly

for

3

x

(

(10)

}

⎡⎣Ex ,Ey2 ⎤⎦ = ⎡⎣Ex ,Ey ⎤⎦ Ey + Ey ⎡⎣Ex ,Ey ⎤⎦ = 0 , and ⎡⎣Ex ,Ez2 ⎤⎦ and ⎡⎣Ex ,Ex2 ⎤⎦ . Thus ⎡⎣Ex ,E 2 ⎤⎦ = 0 , so (11) reduces to:

and

(9)

)

1 = ∫ ⎡⎣Ex , Bx2 + By2 + Bz2 ⎤⎦d 3r ' 2 2 Now, ⎡⎣Ex , Bx ⎤⎦ = [ Ex , Bx ] Bx + Bx [ Ex , Bx ] = 0 . However, from (8),

(12)

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∂ ∂ 3 ⎡⎣Ex , By2 ⎤⎦ = ⎡⎣Ex , By ⎤⎦ By + By ⎡⎣Ex , By ⎤⎦ = −2icBy δ 3 ( R ) = 2icBy δ ( R) ∂z ∂z '

Therefore, 1 ∂ ⎡⎣Ex , By2 ⎤⎦d 3r ' = ic ∫ By ( r ') δ 3 ( R ) d 3r ' ∫ 2 ∂z ' ∂B ( r ) = −ic y ∂z Similarly, ∂B ( r ) 1 ⎡⎣Ex , Bz2 ⎤⎦d 3r ' = +ic z ∫ 2 ∂y Therefore, ⎛ ∂B ∂B ⎞ [Ex , H ] = ic ⎜⎝ ∂yz − ∂zy ⎟⎠ = ic (∇ × B)x However, from Heisenberg’s equation, we have: dEx 1 = [ Ex , H ] dt i (12) = c ( ∇ × B )x and similarly for the y and z components. Of course, this is the third Maxwell equation.

By a similar analysis starting with [ Bx , H ] and using (9) and cyclic permutation, we obtain the fourth Maxwell equation (Faraday’s law).

15.2 Let X =

4π 3 n and let Y be the exact answer. From a simple computer program we 3

obtain: n X 1 4.19 2 33.51 3 113.10 4 268.08 5 523.60 6 904.78 7 1436.76 8 2144.66 9 3053.63 10 4188.79 11 5575.28 12 7238.23 13 9202.77 14 11494.04

Y 6 32 122 256 514 924 1418 2108 3070 4168 5574 7152 9170 11512

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15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

14137.17 17157.29 20579.53 24429.03 28730.91 33510.32 38792.39 44602.24 50965.01 57905.84 65449.85 73622.18 82447.96 91952.32 102160.40 113097.34 124788.30 137258.30 150532.60 164636.21 179594.40 195432.20 212174.80 229847.31 248474.90 268082.60 288695.62 310339.10 333038.20 356817.90 381703.50 407720.10 434892.80 463246.71 492807.00 523598.80

14146 17076 20478 24404 28670 33400 38910 44472 50882 57776 65266 73524 82518 91964 101942 113080 124486 137064 150554 164516 179578 195268 212094 229548 248438 267760 288358 310176 332778 356636 381914 407596 434550 462780 492566 523304

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15.3 The Euler-Maclaurin sum formula is: N −1 N 1 1 ∑ f (n) = ∫0 f (x)dx − 2 [ f (N ) − f (0)] + 12 ⎡⎣ f (1) ( N ) − f (1) ( 0 )⎤⎦ 0 1 1 ⎡⎣ f ( 3) ( N ) − f (3) ( 0 ) ⎤⎦ + ⎡ f ( 5 ) ( N ) − f (5) ( 0 ) ⎤⎦ − ... − 720 30240 ⎣ where f

( j)

(1)

≡ d f /dx . Suppose that: j

j

lim f (N ) = lim f ( j ) ( N ) = 0

N →∞

N →∞

for j =1,3,5, … . Then, ∞

1 1 1 1 ∑ f (n) = ∫ f (x)dx + 2 f (0) − 12 f (0) + 720 f ( 0 ) − 30240 f (0) + ...

0

∞

(1)

(3)

(5)

0

(2)

or ∞

1 1 1 1 ∑ f (n) = ∫ f (x)dx − 2 f (0) − 12 f (0) + 720 f ( 0 ) − 30240 f (0) + ...

1

∞

(1)

(3)

(5)

0

(3)

Now consider the zero-point energy in the region between the plates. It is: 1 U= ∑ ω 2 all modes Here, 1/2

⎡⎛ π n1 ⎞ 2 ⎛ π n2 ⎞ 2 ⎛ π n3 ⎞ 2 ⎤ 2 2 (4) ω = kc = c ⎡⎣ k1 + k2 + k ⎤⎦ = ⎢⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎥ ⎢⎣⎝ d ⎠ ⎝ L ⎠ ⎝ L ⎠ ⎥⎦ where n1,2, 3 can be zero or positive integers for standing waves. Note that the vector potential is given by: Ax = A1 cos k1 x sin k2 ysin k3 z 2 1/2 3

Ay = A2 sin k1 x cos k2 ysin k3 z

(5)

Az = A3 sin k1 x sin k2 y cos k3 z where A1,2,3 are constants, and k1 A1 + k2 A2 + k3 A3 = 0 is implied by the Coulomb gauge condition ∇i A = 0 . If k1 = 0 , Ay = Az = 0 and the linear polarization is necessarily along x. If k1 ≠ 0 there are two possible polarizations. Hence, 1/2

2 2 1 ⎧⎪ ∞ ∞ ⎡⎛ π n2 ⎞ ⎛ π n3 ⎞ ⎤ U = c ⎨ ∑ ∑ ⎢⎜ ⎟ +⎜ ⎟ ⎥ 2 ⎩⎪ n2 = 0 n3 = 0 ⎢⎣⎝ L ⎠ ⎝ L ⎠ ⎥⎦

+

(6) 1/2 ⎡⎛ π n1 ⎞ 2 ⎛ π n2 ⎞ 2 ⎛ π n3 ⎞ 2 ⎤ ⎫⎪ 2 ∑ ∑ ∑ ⎢⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎥ ⎬ n1 =1 n2 = 0 n3 = 0 ⎢ ⎣⎝ d ⎠ ⎝ L ⎠ ⎝ L ⎠ ⎥⎦ ⎪⎭ Since L is very large we may replace the sums over n2, n3 by integrals with adequate accuracy. Hence the triple sum on the right hand side of (6) becomes: ∞

∞

∞

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∞ ⎡ 2 d2 2 ⎤ π c ∞ ∞ (7) S3 = dy ∫ dz ⎢ n1 + 2 ( y + z 2 ) ⎥ ∑ ∫ 0 0 d n1 =1 L ⎣ ⎦ d 2 Let r = y + z 2 . Then (2.7) becomes: L 1/2 π 2 cL2 ∞ ∞ 2 2 (8) S3 = ∑ rdr ⎡⎣ n1 + r ⎤⎦ 2d 3 n1 =1 ∫0 Actually, to take into account the fact that the plates become transparent at extremely high frequencies we should multiply the integrand in (8) by a cutoff function g ( n12 + r 2 ) such that g(0)=1 but g → 0 for extremely large values of its argument. Then:

π 2 cL2 ∞ S3 = ∑ f (n1 ) 2d 3 n1 =1

(9)

where ∞

(

f ( n1 ) = ∫ rdr ⎡⎣ n12 + r 2 ⎤⎦ g n12 + r 2 1/2

0

)

(10)

We use the Euler-Maclaurin sum formula (3) to write: ∞ ∞ 1 1 1 ∑ f (n1 ) + 2 f (0) − ∫0 f (x)dx = − 12 f '(0) + 720 f (3) (0) − ... n1 =1 Now,

(11)

∞

f (x) = ∫ r x 2 + r 2 gdr 0

Thus, ignoring the variation of g with x for small x, we have: ∞ ∂f x ∂f = ∫ rg dr; =0 2 2 0 ∂x ∂x x = 0 x +r

(12)

⎡ ⎤ ∞ ∞ ∂2 f 1 x2 gr 3dr ⎢ ⎥ = rg − dr = 3/2 ∫0 x 2 + r 2 3/2 ∂x 2 ∫0 ⎢ x 2 + r 2 ⎥ x2 + r2 ⎣ ⎦ ∞ ∂3 f gr 3 x = −3 ∫0 x 2 + r 2 5 /2 dr ∂x 3

(

(

)

)

(

)

(13)

∂3 f . In (13) it is convenient to make the substitution: x→0 ∂x 3

We must be careful about lim

u = x2 + r2 Then it is easy to show that: ∂3 f 3 ∞ gx 3 ∞ gx 3 = − du + du ∂x 3 2 ∫x 2 u 3/2 2 ∫x 2 u 5 /2 1 1 → −3x + x 3 3 = −2 x x 142

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Therefore,

∂3 f = −2 x→0 ∂x 3

(14)

lim Hence,

∞

1

∞

2

∑ f (n ) + 2 f (0) − ∫ f (x)dx  − 720

n1 =1

1

0

and consequently, the final result is: π 2 c L2 Δ≅− (15) 720 d 3 b) Let the separation between the plates be x. Then the force of attraction per cm2 of one plate for another is: ∂Δ π 2 c ∂ 1 = ∂x 720 ∂x x 3 π 2 c 1 a =− =− 4 4 240 x x 2 π c where a = . The mass per unit area of each plate is m = .1ρ = .89 g/cm 2 . The 240 equation of motion is: a (16) µ  x=− 4 x 1 where µ = m . Thus: 2 a 1 b  (17) x = −2 =− 4 4 mx x where b = 2.92•10-18. Multiplying both sides of (17) by x and integrating once with respect to time we obtain: 2 ⎛ 1 1⎞ x 2 = b ⎜ 3 − 3 ⎟ 3 ⎝x x0 ⎠ where x0 = .1 cm is the initial separation. From this equation, we obtain for the time T to make contact: 3 x0 x 3/2 dx 2 3 5 /2 π /2 2 / 3 T= = x0 ∫ sin θ dθ 0 2b ∫0 x 3 3 2b 1− 3 x0 F=−

⎛ 5⎞ ⎛ 1⎞ Γ⎜ ⎟ Γ⎜ ⎟ ⎝ 6⎠ ⎝ 2⎠ x = = 1.19 • 10 6 s = 13.8 days ⎛ 4⎞ 6b Γ⎜ ⎟ ⎝ 3⎠ 5 /2 0

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15.4 We start with the following general principle from statistical mechanics: if electronpositron pairs are in equilibrium with black-body radiation, then we must have: µ + + µ − = µγ (1) where µ+ , µ− , µγ are the chemical potentials of the positrons, the electrons, and the radiation respectively. Since µγ = 0 , µ+ = − µ− . Also, from Fermi-Dirac statistics, the number densities of electrons and positrons are given by the formulae: ∞ 1 p 2 dp n− = 2 3 ∫ π  0 ⎛ E − µ− ⎞ exp ⎜ +1 ⎝ kT ⎟⎠ n+ =

∞ 1 2 3 ∫0 π 

p 2 dp ⎛ E − µ+ ⎞ exp ⎜ +1 ⎝ kT ⎟⎠

(2)

(3)

where in each integral, E = p 2 c 2 + me2 c 4 . Now, by charge conservation, we must have n+ = n− , so the integrals on the right hand sides of (2) and (3) must be equal. Therefore we must have µ+ = µ− . However, we have already shown that µ+ = − µ− . Therefore, µ+ = µ− = 0 and (2, 3) reduce to: n± =

∞ 1 2 3 ∫0 π 

p 2 dp ⎛ E⎞ exp ⎜ ⎟ + 1 ⎝ kT ⎠

The integral in (4) is easily evaluated in the two limiting cases

(4)

kT kT << 1, >> 1 .

2 me c me c 2

kT << 1 me c 2 Here, the only momenta p that contribute significantly to the integral are so small that: p2 E = p 2 c 2 + me2 c 4 ≈ me c 2 + 2me ⎛ m c2 ⎞ Since exp ⎜ e ⎟ >> 1 , we can ignore the 1 in the denominator of the integrand, and (4) ⎝ kT ⎠ becomes: ⎛ ⎛ m c2 ⎞ ∞ 1 p2 ⎞ n± = 2 3 exp ⎜ − e ⎟ ∫ p 2 exp ⎜ − dp π  ⎝ kT ⎠ 0 ⎝ 2me kT ⎟⎠ (5) 3/2 ⎛ me c 2 ⎞ ⎛ me kT ⎞ = 2⎜ exp ⎜ − ⎝ 2π  2 ⎟⎠ ⎝ kT ⎟⎠

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kT >> 1 me c 2 Here a large range of momenta contribute significantly to the integral in (4), and for most of this range, p >> me c . Hence, we can write: ∞ 1 p 2 dp n± ≈ 2 3 ∫ π  0 ⎛ pc ⎞ exp ⎜ ⎟ + 1 ⎝ kT ⎠ k 3T 3 ∞ x 2 dx k 3T 3 ∞ ∞ = 2 3 3∫ x = 2 3 3 ∑ ∫ (−1)n +1 x 2 e− nx dx 0 π c e + 1 π  c n =1 0

(6)

1.8k 3T 3 ≈ 2 3 3 π c Next we calculate the entropy of the combined system: photons plus electron-positron pairs, in the high temperature limit: kT >> me c 2 . Let’s start with the thermal energy of the electrons and positrons. According to Fermi-Dirac statistics: Ep 2 dp ⎛ E⎞ exp ⎜ ⎟ + 1 ⎝ kT ⎠ In the high temperature limit, we can write E = pc , so (7) becomes: ∞ V cp 3dp U± = 2 3 ∫ π  0 ⎛ pc ⎞ exp ⎜ ⎟ + 1 ⎝ kT ⎠ ∞ V U± = 2 3 ∫ π  0

(7)

Vk 4T 4 ∞ x 3dx Vk 4T 4 ∞ ∞ (8) = ∑ (−1)n +1x 3e− nx dx π 2  3c 3 ∫0 e x + 1 π 2  3c 3 n =1 ∫0 7 = aVT 4 8 2 π k4 where a = is the Stefan-Boltzmann constant; (recall that the thermal energy of 15  3c 3 the photons is Uγ = aVT 4 ). Hence the total thermal energy is: =

7 7⎞ 11 ⎛ U = ⎜ 1 + + ⎟ aVT 4 = aVT 4 (9) ⎝ 8 8⎠ 4 The pressure of each component (photons, electrons, positrons) of the system is one third of the thermal energy density, since each component is ultra-relativistic. Thus the total pressure is: 1 11 4 (10) P= aT 34 The thermodynamic identity is:

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TdS = dU + PdV 11 ⎡ 1 ⎤ ad VT 4 + aT 4 dV ⎥ ⎢ 4⎣ 3 ⎦ The factor in brackets on the right hand side of this equation is the same as TdS for the radiation alone. Hence, 11 11 4 11 (11) Stotal = Sγ = aVT 3 = aVT 3 4 4 3 3 =

(

)

15.5 a) The polarization 4-vector satisfies the condition: ε µ ε µ * = −1 . From the Lorenz gauge condition ∂ µ Aµ = 0 we also have k µ ⋅ ε µ = 0 . Now the 4-momentum of a massive photon is: E pµ = , p c 1 1⎛ E ⎞ and the 4-vector k µ is: k µ = p µ = ⎜ , p ⎟ . Thus, in the rest frame of the massive  ⎝ c ⎠

⎛ mc ⎞ , 0, 0, 0 ⎟ or in units where  = c = 1, k µ → ( m, 0, 0, 0 ) . Since k µ ε µ = 0 photon: k µ → ⎜ ⎝  ⎠

we conclude that in the rest frame, ε µ → ( 0, −ε̂ ) where ε̂ is a unit three-vector. If the massive photon is linearly polarized along the ±z axis in the rest frame, then: ⎛ 0⎞ ⎜ 0⎟ (1) εµ = ⎜ ⎟ ⎜ 0⎟ ⎜⎝ 1⎟⎠ If the massive photon is circularly polarized along the ±z axis, then: ⎛ 0⎞ 1 ⎜ −1⎟ ⎜ ⎟ (2) εµ = 2 ⎜ i ⎟ ⎜⎝ 0 ⎟⎠ 5b) We make a Lorentz transformation to a new frame F’ where the massive photon has velocity v along the +z axis. Such a Lorentz transformation is described by the following 4x4 matrix: 0 0 − βγ ⎞ ⎛ γ ⎜ 0 1 0 0 ⎟ ⎟ (3) a=⎜ 0 1 0 ⎟ ⎜ 0 ⎜⎝ − βγ 0 0 γ ⎟⎠

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(

)

−1/2

where, as usual, β = v/c and γ = 1 − β 2 . We obtain the polarization 4-vector ε µ ' in the frame F’ by applying matrix a to ε µ in (2) for circular polarization, or to ε µ in (1) for linear polarization. It is obvious that in the case of circular polarization, ε µ ' = ε µ .

However, for linear polarization: ⎛ ± βγ ⎞ ⎜ 0 ⎟ ⎟ εµ ' = ⎜ ⎜ 0 ⎟ ⎜⎝ γ ⎟⎠ In the limit where the massive photon is ultra-relativistic, β → 1 and: ⎛ ±γ ⎞ ⎜ 0⎟ εµ ' → ⎜ ⎟ ⎜ 0⎟ ⎜⎝ γ ⎟⎠

(4)

In that same limit, p = mc β γ → mcγ , E = mc 2γ , so in units where  = c = 1 , ⎛ mγ ⎞ ⎜ 0 ⎟ ⎟ kµ ' → ⎜ ⎜ 0 ⎟ ⎜⎝ −mγ ⎟⎠

(5)

Comparing (5) to (4) we see that in the ultra-relativistic limit, ε µ ' = ±m −1kµ ' for linear polarization along the ±z axis in the rest frame.

15.6a) In Sec.15.8 we show that if the photon has non-zero mass, the first Maxwell equation must be replaced by: (1) ∇ i E + κ 2Φ = ρ In the static case, this is the same as: (2) ∇2Φ − κ 2Φ = − ρ which is the modified Poisson equation. We integrate (1) over a finite volume V bounded by a surface S. Then from Gauss’ theorem:

∫ E i n̂ dS = ∫ ρ dV − κ ∫ Φ dV (3) = Q − κ ∫ Φ dV where Q = ∫ ρdV is the total charge. Equation (3) is Gauss’ law modified for non-zero 2

2

photon mass. Inside any solid conductor we must have E = 0 for static equilibrium, hence

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Φ must be constant throughout the conductor, regardless of the latter’s shape. Thus from (3), (4) ρ = κ 2Φ which means that ρ is also constant throughout the conductor, regardless of its shape. b) Consider a conducting sphere of radius a with total charge Q. Outside the sphere, (2) yields: ∇2Φ − κ 2Φ = 0 which has the spherically symmetric solution: exp ( −κ r ) (5) Φ=C C = const, r ≥ a r Inside the sphere, Φ = const. Therefore: e −κ a Φ=C r<a (6) a Thus the volume charge density is: e −κ a ρ = κ 2C (7) a and the total volume charge is: 4π 3 4πκ 2 a 2 −κ a QV = ρa = Ce (8) 3 3 Now we employ (5.3) at r = ∞ where the surface integral vanishes, since E → 0 faster than r-2. Then the total charge is:

Q = κ 2 ∫ Φ dV ∞ 4πκ 2 a 2 −κ a Ce + 4πκ 2C ∫ re−κ r dr a 3 ⎡ κ 2a2 ⎤ = 4π C ⎢1+ κ a + exp ( −κ a ) 3 ⎥⎦ ⎣

=

(9)

The surface charge is: QS = Q − QV

= 4π C [1+ κ a ] e−κ a (10) Thus the ratio of surface charge to total charge is: Q 1+κa R= S = κ 2a2 Q 1+κa + 3 mγ c → 0 , R → 1 and QV → 0 .

Of course, in the limit when κ = 

(11)

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16.1 The zeroth order Hamiltonian is: H 0 = −g p µ N Sz B0 The energy eigenvalues for H 0 are as follows: 1 ω m=+ : E+ = − 0 2 2 1 ω m=− : E− = + 0 2 2 where: g p µ N B0 ω0 = = γ B0  The perturbing Hamiltonian is: H ' = −g p µ N SiB1 [ cos ω tx̂ − sin ω tŷ ] g p µ N B1

⎡ cos ω tσ x − sin ω tσ y ⎤⎦ 2 ⎣ g p µ N B1 ⎛ 0 eiω t ⎞ =− 2 ⎜⎝ e−iω t 0 ⎟⎠ =−

(1)

(2)

(3)

We write: Ψ(t) = a+ (t) + + a− ( t ) − Then, according to first order time dependent perturbation theory: i ⎡i ⎤ a− = − a+ (0) − H ' + exp ⎢ ( E− − E+ ) t ⎥  ⎣ ⎦ (4) i = − − H ' + exp [ iω 0t ]  where we have used (2) and assumed that a+ ( 0 ) = 1, a− ( 0 ) = 0 . The matrix element in (4) is: g p µ N B1 g p µ N B1 −iω t ⎛ 0 eiω t ⎞ ⎛ 1 ⎞ − H' + =− =− e ( 0 1) ⎜ −iω t ⎜ ⎟ ⎟ 2 2 0 ⎠ ⎝ 0⎠ ⎝e Thus, (4) becomes: γB a− = i 1 ei (ω 0 − ω )t 2 We integrate this equation subject to the initial condition a− (0) = 0 to obtain:

γ B1 ei (ω 0 − ω )t − 1 a− (t) = 2 (ω 0 − ω ) The transition probability (not the transition probability per unit time) is:

(5)

(6)

149

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014 2 2 γ B1 ) ( i (ω − ω ) t a− (t) = −1 2 e 4 (ω 0 − ω ) 2 γ B1 ) ( 2 (ω 0 − ω ) t = 2 sin 2 (ω 0 − ω ) 2

0

We now compare (7) with the exact result given by Rabi’s formula: 2 γ B1 ) ( 2 2 2⎛ 2 t ⎞ a− (t) exact = sin γ B + ( ω − ω ) ( ) ⎜ ⎟ 1 0 2 2⎠ (γ B ) + (ω − ω )2 ⎝ 1

(7)

(8)

0

Eq (8) is valid for all t ≥ 0 , and for all values of ω 0 − ω . However, (7) is a good approximation only when a− (t) << 1 . For example when ω 0 = ω , (8) reduces to: 2 ⎛γ Bt⎞ a− (t) exact = sin 2 ⎜ 1 ⎟ (9) ⎝ 2 ⎠ and (7) reduces to: 2 2 ⎛ γ B1t ⎞ a− (t) = ⎜ (10) ⎝ 2 ⎟⎠ Clearly, these two expressions are approximately equal only when γ B1t /2 <<1.

16.2 We start with the interaction Hamiltonian: σ iσ H ' = a(t)S1 iS2 = a(t) 1 2 4 a) The final state ψ f is:

ψ (t = +∞) = U ( +∞, −∞ ) ψ i = U ( +∞, −∞ ) + − where the time development operator is: U = exp [ −iK σ 1 iσ 2 ] and 1 ∞ K= a(t)dt 4 ∫−∞ Now, 1 10 = ( +− + −+ ) 2 1 00 = ( +− − −+ ) 2 so 1 +− = ( 10 + 00 ) 2 Also,

(1)

(2) (3)

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σ 1 iσ 2 10 = 10 σ 1 iσ 2 00 = −3 00 Therefore,

1 ⎡ e− iK σ 1 i σ 2 10 + e− iK σ 1 i σ 2 00 ⎤⎦ 2⎣ 1 ⎡⎣ e− iK 10 + e3iK 00 ⎤⎦ = 2 1 = ⎡⎣ e3iK + e− iK + − − e3iK − e− iK − + ⎤⎦ 2

U +− =

(

Hence,

)

(

)

ψ f = eiK ⎡⎣ cos 2K + − − i sin 2K − + ⎤⎦

(4)

Thus,

− + ψ f = −ieiK sin 2K and the transition probability is:

P= −+ ψf

2

= sin 2 2K

(5)

For small K, 2 ∞ 1 (6) a(t)dt 4 2 ∫−∞ b) The same problem can be treated by first order time dependent perturbation theory (TDPT). Let: ψ (t) = c1 (t) + − + c2 (t) − +

P = sin 2 2K → 4K 2 =

where c1 (−∞) = 1, c2 (−∞) = 0. According to first-order TDPT,

c2 = −

i a − + σ 1 iσ 2 + −  4

i a 1 ⎡ − + 10 − 3 − + 00 ⎤⎦ 4 2⎣ i a ⎡ 1 3 ⎤ =− + ⎢ 2 4 ⎣ 2 2 ⎥⎦ ia =− 2 =−

Thus,

c2 ( +∞ ) = −

i ∞ a(t)dt 2 ∫−∞

so

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P = c2 (∞) =

2 ∞ 1 a(t)dt 4 2 ∫−∞

(7)

which agrees with (6). c) Finally, let: H = H0 + H ' where

H 0 = −γ 1S1z B − γ 2 S2 z B Then:

γ 2 − γ1 B 2 γ −γ2 − + H0 − + = 1 B 2 γ − γ1 ω 12 = −ω 21 = 2 B 

(8)

+ − H0 + − =

Now we have:

(9)

ψ = c1 + − e− iω1t + c2 − + e− iω 2 t and:

i ⎡ ⎛ a⎞ ⎤ ⎛ a⎞ c1 = − ⎢ c1 ⎜ − ⎟ + c2 ⎜ ⎟ eiω12 t ⎥ ⎝ 2⎠  ⎣ ⎝ 4⎠ ⎦ i ⎡ ⎛ a⎞ ⎤ ⎛ a⎞ c2 = − ⎢ c2 ⎜ − ⎟ + c1 ⎜ ⎟ e− iω12 t ⎥ ⎝ 2⎠  ⎣ ⎝ 4⎠ ⎦

(10)

On the right hand side of equations (10) choose c1 = 1, c2 = 0 in TDPT. Then: i c2 = − aeiω 21t 2 Thus, 2 2 i ∞ c2 ( ∞ ) = − a0 e− t /τ eiω 21t dt ∫ −∞ 2 Completing the square in the exponent of the integrand, we obtain: ∞ 2 2 ⎡ (t − iω 21τ 2 / 2)2 ⎤ i − ω 21 τ /4 c2 (∞) = − a0 e ⎥dt ∫−∞ exp ⎢⎣ − 2 τ2 ⎦

=−

2 2 i a0τ π e− ω 21τ / 4 2

Thus, 2 2 ⎛ ω 21 π a02τ 2 τ ⎞ c2 (∞) = exp ⎜ − 2 4 2 ⎟⎠ ⎝ 2

(11)

Taking the last of equations (9) into account in (11) we see that as B increases, c2 (∞) decreases.

2

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16.3 The interaction Hamiltonian is: gs µ B SiB  which couples the electron spin magnetic moment to the magnetic radiation field B. (We neglect a contribution from the coupling of the proton magnetic moment to B since it is 3 orders of magnitude smaller). The matrix element for spontaneous emission of a magnetic dipole photon is: µ 1 (1) M = −i B 3/2 2ω ψ f k̂ × ε̂ i S ψ i  L where we assume gs = 2 . For given photon polarization, the transition probability per unit time is: 3 2 2π 2 Lω dW = M dΩ (2)  8π 3c 3 We choose the initial and final atomic states as: ψ i = 12 s1/2 , F = 1, mF = 0 (3) H'=

and

ψ i = 12 s1/2 , F = 0, mF = 0

(4)

(The rates for the transitions (1,1 → 0, 0 ) , (1, 0 → 0, 0 ) , and (1,-1 → 0,0 ) are identical).

Then, S 1 (5) M = −i µ B 3/2 2ω k̂ × ε̂ 00 z 10 z L  Now, Sz 1 Sz 1 10 = α eβ p + α p β e = α eβ p − α p β e  2  2 2 1 = 00 2 Therefore, 2 ω 2 (6) M = 3 µ B2 k̂ × ε̂ z 2L We choose: k̂ = sin θ cos φ x̂ + sin θ sin φ ŷ + cosθ ẑ

(

(

(

)

)

(

)

)

ε̂1 = − cosθ cos φ x̂ − cosθ sin φ ŷ + sin θ ẑ ε̂ 2 = sin φ x̂ − cos φ ŷ Thus,

( k̂ × ε̂ )i ẑ = 0 ( k̂ × ε̂ )i ẑ = − sinθ 1

2

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Hence, M ε̂1 = 0 and: 2

ω 2 2 µ B sin θ 2L3 Summing over both polarizations, we have from (2): 2π ω 2 2 L3ω 2 dW = µ B sin θ 3 3 dΩ  2L3 8π c 2 3 1 µ Bω = sin 2 θ dΩ 8π 2 c 3 The total transition probability per unit time is found by integrating over the solid angle.

Since M ε̂ 2 = 2

8π

∫ sin θ dΩ = 3 2

we obtain:

µ B2ω 3 W= 3π c 3

Now we insert numerical values: ω = 2π i (1.420 i10 9 ) rad/s

(7)

(Hyperfine splitting)

µ B / 4π = 9.27 i10 −21 erg/G =1.054 i10 -27 erg s c=3i1010 cm/s to obtain: or:

W = τ −1 = 2.865 i10 −15 s-1

τ = 3.49 i1014 s = 1.11i10 7 y

(8)

16.4 The differential transition probability per unit time for the 3d → 2s electric quadrupole transition in atomic hydrogen is given by the formula: 3 2 2 Lω 2π (1) dW = ∑ H ' fi 8π 3c 3 dΩγ  pol where 2 2 1 e2ω 3 H ' fi = 3 π  k̂ ε̂ 1s xi x j 3d (2) 2 i j L 8π c Ignoring electron spin, we can calculate the spontaneous emission rate for any initial m value of the 3d state, since the rates are the same for all m values. Consider the quantity:

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(

)

⎡1 ⎤⎡1 ⎤ k̂irε̂ ir = ⎢ k̂+ x− + k̂− x+ + k̂z z ⎥ ⎢ ( ε̂ + x− + ε̂ − x+ ) + ε̂ z z ⎥ ⎣2 ⎦⎣2 ⎦ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ = ⎜ k̂+ ε̂ + x− x− ⎟ + ⎜ k̂− ε̂ − x+ x+ ⎟ + ⎜ ⎡⎣ k̂+ ε̂ z + k̂z ε̂ + ⎤⎦ x− z ⎟ + ⎜ ⎡⎣ k̂− ε̂ z + k̂z ε̂ − ⎤⎦ x+ z ⎟ ⎝4 ⎠ ⎝4 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ + ⎜ ⎡⎣ k̂+ ε̂ − + k̂− ε̂ + ⎤⎦ x+ x− + k̂z ε̂ z z 2 ⎟ ⎝4 ⎠ On the far right side of (3) the terms in large parentheses transitions respectively: 3d, m = 2 → 2s, m = 0;

(3) refer to the following

3d, m = −2 → 2s, m = 0; 3d, m = 1 → 2s, m = 0; 3d, m = −1 → 2s, m = 0; 3d, m = 0 → 2s, m = 0 Note that:

x− x− = r 2 sin 2 θ e−2iφ =

32π 2 −2 r Y2 (θ , φ ) 15

x+ x+ = r 2 sin 2 θ e2iφ =

32π 2 2 r Y2 (θ , φ ) 15

x− z = r 2 sin θ cosθ e−iφ =

8π 2 −1 r Y2 (θ , φ ) 15

x+ z = r 2 sin θ cosθ eiφ = −

8π 2 1 r Y2 (θ , φ ) 15

We choose: k̂ = sin θγ cos φγ x̂ + sin θγ sin φγ ŷ + cosθγ ẑ

ε̂1 = − cosθγ cos φγ x̂ − cosθγ sin φγ ŷ + sin θγ ẑ ε̂ 2 = sin φγ x̂ − cos φγ ŷ Thus, ±iϕ k̂z = cosθγ , k̂± = sin θγ e γ ,

ε̂1z = sin θγ , ε̂ 2 z = 0, ε̂1± = − cosθγ e

±iφγ

, ε̂ 2± = ie

±iφγ

Hence,

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1 1 32π ⎡ − sin θγ cosθγ e2iφγ ⎤ r 2Y2−2 (θ , φ ) k̂+ ε̂1+ x− x− = ⎦ 4 4 15 ⎣ 1 1 32π ⎡ −i sin θγ e2iφγ ⎤ r 2Y2−2 (θ , φ ) k̂+ ε̂ 2 + x− x− = ⎦ 4 4 15 ⎣ 1 1 8π ⎡(sin 2 θγ − cos 2 θγ )eiφγ ⎤ r 2Y2−1 (θ , φ ) k̂+ ε̂1z + k̂z ε̂1+ x− z = ⎦ 2 2 15 ⎣ 1 1 8π ⎡ −i cosθγ eiφγ ⎤ r 2Y2−1 (θ , φ ) k̂+ ε̂ 2 z + k̂z ε̂ 2 + x− z = ⎦ 2 2 15 ⎣

(

)

(4) (5)

(

)

(6)

(

)

(7)

Also, since k̂iε̂ = 0 in Coulomb gauge, we can write:

1 1 ⎡ k̂+ ε̂1− + k̂− ε̂1+ ⎤ x+ x− + k̂z ε̂1z z 2 = k̂z ε̂1z ⎛⎜ z 2 − x+ x− ⎞⎟ ⎣ ⎦ ⎝ ⎠ 4 2 ⎛ x 2 + y2 ⎞ = k̂z ε̂1z ⎜ z 2 − 2 ⎟⎠ ⎝ 1 ⎞ ⎛3 = k̂z ε̂1z ⎜ z 2 − r 2 ⎟ ⎝2 2 ⎠

(8)

4π ⎡⎣ cosθγ sin θγ ⎤⎦ r 2Y20 (θ , φ ) 5 and there is no contribution analogous to (8) from ε̂ 2 .

We now employ the foregoing expressions to calculate the differential transition probability per unit time for the decay 3d, m = 1 → 2s, m = 0 . From (1, 2, 6, 7) we have: = k̂z ε̂1z r 2 P2 (cosθ ) =

2π L3ω 2 1 e2ω 3 π  i  8π 3c 3 L3 8π c 2 1 ⎡⎣(sin 2 θγ − cos 2 θγ )2 + cos 2 θγ ⎤⎦ dΩγ i (9) 4 2 8π ⎡ ∞ 4 0 −1 1 ⎤ r R (r)R (r)dr Y Y Y dΩ 2s 3d ∫ 0 2 2 ⎥⎦ 15 ⎢⎣ ∫0 1 e2 0 Now employing Y0 = , ∫ Y2−1Y21dΩ = −1, and α = , we see that (9) reduces to: 4π c 4π dW =

2 1 ω5 ⎡ dW = α 4 ⎢ sin 2 θγ − cos 2 θγ + cos 2 θγ ⎤⎥ dΩγ • ⎦ 16 • 15π c ⎣

(

)

∞

⎡ R R r dr ⎤ ⎢⎣ ∫0 2 s 3d ⎥⎦

2

(10)

4

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The angular distribution of radiation is evident from the first bracketed factor on the right hand side of (10). The total transition probability per unit time W is obtained by integrating over the photon solid angle. We have:

∫ ⎡⎣sin θ + cos θ − 2 sin θ cos θ + cos θ ⎤⎦dΩ = 2π ∫ (1 - 3x + 4x )dx 4

4

γ

2

γ

1

2

2

γ

2

γ

γ

γ

4

-1

8 ⎞ 16π ⎛ = 2π ⎜ 2 − 2 + ⎟ = ⎝ 5⎠ 5

Thus: 2 αω 5 ⎡ ∞ 4 ⎤ W= R2 s R3d r dr ⎥ ⎦ 75c 4 ⎢⎣ ∫0

(11)

To proceed further it is convenient to employ atomic units, where c = α −1 , 1⎛ 1 1⎞ 5 ω = ⎜ 2 − 2⎟ = from the Balmer formula, and: 2⎝2 3 ⎠ 72

1 r (1 − )e− r /2 2 2 4 R3d = r 2 e− r / 3 81 30 Thus (11) becomes: R2 s =

5 α 5 ⎛ 5 ⎞ 16 ⎡ ∞ ⎛ 6 r 7 ⎞ −5r /6 ⎤ W= r − ⎟e dr ⎥ ⎢ ⎜ ⎟ 75 ⎝ 72 ⎠ 9 4 60 ⎣ ∫0 ⎜⎝ 2⎠ ⎦

2

(12)

The integral is:

⎛ 6 r 7 ⎞ −5r /6 1 ⎛ 6⎞ ⎛ 6⎞ ∫0 ⎜⎝ r − 2 ⎟⎠ e dr = ⎜⎝ 5 ⎟⎠ 6!− 2 ⎜⎝ 5 ⎟⎠ 7! ∞

7

8

7

⎛ 6⎞ ⎡ 1 6 • 7⎤ = ⎜ ⎟ 6! ⎢1 − ⎝ 5⎠ ⎣ 2 5 ⎥⎦ 7

32 ⎛ 6 ⎞ = − ⎜ ⎟ 6!

10 ⎝ 5 ⎠ Thus:

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α 5 ⎛ 5 ⎞ 16 ⎛ 32 ⎞ ⎛ 6 ⎞ W= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 75 ⎝ 72 ⎠ 9 4 60 ⎝ 10 ⎠ ⎝ 5 ⎠ 5

2

14

( 720 )2

(13)

= 5.96523 • 10 −5 α 5 To go back to ordinary (cgs) units we have to multiply the right hand side by α c/a0 : αc W = 5.96523 • 10 −5 α 5 i a0 (14) 6 1 ⎞ 3i1010 −5 ⎛ -1 = 5.96523 • 10 ⎜ = 51.08 s ⎝ 137.036 ⎟⎠ 5.29i10 −9 We compare this result with the transition probability per unit time for the electric dipole transition 3d → 2 p , in which a photon of the same frequency is emitted as in the transition 3d → 2s : W (3d → 2s) = 51 s-1 W (3d → 2 p) = 6.4 • 10 7 s-1 The ratio of these rates is: 51 R= = 7.97 • 10 −7 = .015α 2 7 6.4 • 10 We expect the rate for an atomic E2 transition to be roughly of order α 2 compared to that for an atomic E1 transition of the same frequency. The factor .015 is not too surprising, however, since the matrix element depends on details of the atomic wave functions.

16.5 In a uniform magnetic field B = B0 ẑ an electron with speed v in the xy plane executes circular motion with angular frequency: eB 1 ω0 = 0 (1) mc γ and radius pc mvγ c (2) ρ= = eB0 eB0 In the instantaneous rest frame of the electron there exists a magnetic field: (3) B' = γ B = γ B0 ẑ (There is also an electric field E ' = βγ B0 ŷ .). Since the electron has a spin magnetic moment, there are two separate energy states in B' : 1 γ Spin up: E+ = gs µ B B' = gs µ B B0 2 2 γ Spin down: E− = − gs µ B B0 2

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The frequency difference between these two states is: γ g µ B γ g eB ω= s B 0= s 0  2mec The matrix element for M1 radiation is: −i ω M = 3/2 gs µ B ψ f k̂ × ε̂ i S ψ i L 2 Here, ψi = α , ψ f = β Hence,

(

(4)

(5)

)

 (6) k̂ × ε̂ x+iy 2 The transition probability per unit time in the rest frame, into solid angle dΩ and summed over photon polarizations, is: 3 2 2π 2 Lω (7) dW ' = ∑ M dΩ 8π 3c 2 pols  From (5-7) we find: 2 2π 2 2 1 ω 1 L3ω 2 W'= gs µ B 3 k̂ × ε̂ dΩ i ∑ x+iy  L 2 4 ∫ pols 8π 3c 3 (8) 2 gs2 µ B2ω 3 = ∑ k̂ × ε̂ x+iy dΩ 32π 2 c 3 pols ∫ Now, 2 π 16π k̂ × ε̂ dΩ = 2π ∫ (1 + cos 2 θ )sin θ dθ = ∑∫ 0 x +iy 3 pols Thus making use of (4) we find that (8) can be written: 2gs2 µ B2ω 3 1 gs5 µ B2γ 3e3 B03 W'= = (9) 3c 3 12 m 3c 6

β k̂ × ε̂ i S α =

(

(

(

Since µ B2 =

)

)

Now for v ≈ c , ρ =

B03 =

)

mvγ c mγ c 2 . Hence: ≈ eB0 eB0

m 3γ 3c 6 e3 ρ 3

e2  2 , (9) becomes: 16π m 2 c 2

gs5 e2 γ 6 192π m 2 c 2 ρ 3 Finally, the transition probability per unit time in the laboratory frame is: W'=

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W' gs5 e2 γ 5 Wlab = = = t −1 2 2 3 γ 192π m c ρ

(10)

mγ c 2 Substituting ρ = and gs = 2 into this expression, we finally obtain: eB0 3 m 5c8 γ = 2 e5 B03t where in the last expression all quantities are in cgs units. Inserting numerical values we find γ = 2.52 • 10 3 , hence E = mc 2γ = 1.29 GeV if t = 1 hour, B0 = 10 4 Gauss.

2

16.6 a) Since the transition probability per unit time for the 2s-1s M1 transition is:

Γ M 1 = constiω 3 H ' M 1

2

and ω = constiZ 2 for 2s → 1s while H ' M 1 = constiZ 2α 2 , we have:

Γ M 1 ∝ Z 6 Z 4 = Z 10 b) Consider the state 2s1/2 , m j = 1/2 = R2 sY00α

(1) (2)

This can decay to:

and to

⎡ 1 2 1 ⎤ 2 p1/2 , m j = 1/2 = R2 p ⎢ Y10α − Y1 β ⎥ 3 3 ⎣ ⎦

⎡ 1 2 −1 ⎤ 2 p1/2 , m j = −1/2 = R2 p ⎢ − Y10 β + Y1 α ⎥ 3 ⎣ 3 ⎦ The matrix elements for these two transitions are: 1 ∞ ε̂ ir 1 1 = ε̂ (0) R2 s R2 p r 3dr ∫ Y10Y00 cosθdΩ ∫ 0 3 22 1 ∞ = ε̂ (0) ∫ R2 s R2 p r 3dr 3 0 and: 2 ∞ ε̂ ir 1 − 1 = −ε̂ (+1) R2 s R2 p r 3dr ∫ Y1−1*Y00 cosθdΩ ∫ 0 3 2 2

(3)

(4)

2 ∞ = ε̂ R2 s R2 p r 3dr ∫ 0 3 In each transition there are two polarization states, a priori. These are: (+1)

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ε̂1(0) = sin θγ , ε̂ 2(0) = 0 ε̂1(+1) = −

1 i cosθγ , ε̂ 2(+1) = − 2 2

Thus: 2

∑ ε̂ ir 1 1 = sin 2 θγ 22

pol

∑ ε̂ ir pol

2 1 1 − 2 2

(

)

2 1 ∞ 3 R R r dr 2s 2 p 9 ∫0

(

= 1 + cos 2 θγ

) 19 ( ∫ R R r dr ) ∞

2

3

2s

0

2p

To find the total decay rate we add these two contributions, integrate over the photon

α 3ω 3 solid angle, and multiply by an overall factor . Thus we find the decay rate in 2π atomic units:

(

)

2 4 3 3 ∞ 3 Γ = α ω ∫ R2 s R2 p r dr 0 9

Now: ⎛ Z⎞ Ri = R2 s = ⎜ ⎟ ⎝ 2⎠

3/2

⎛ Z⎞ R f = R2 p = ⎜ ⎟ ⎝ 2⎠

( 2 − Zr ) e− Zr /2

3/2

Zr − Zr /2 e 3

Thus,

∫

∞

0

Z4 ∞ 4 r ( 2 − Zr )e− Zr dr ∫ 0 8 3 Z 4 ⎛ 2i4! 5! ⎞ 9 = ⎜⎝ 5 − 5 ⎟⎠ = − Z 8 3 Z Z 3

R2 s R2 p r 3dr =

Hence, in atomic units: 4 9 2 12α 3ω 3 Γ = α 3ω 3 2 = 9 3Z Z2 The Lamb shift is S  2π i1.058i10 9 Z 4 rad/s = 1.61i10 −7 Z 4 in atomic units .

In ordinary units,

Γ=

12α 4 c (1.61i10 −7 )3 Z10 s-1 a0

= 8i10 −10 Z 10s-1

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c) The 2-photon differential transition probability per unit time is given by the golden rule formula:

dΓ = consti M ω 12 dω 1dΩ1ω 22 dω 2 dΩ2δ ( Δ − E1 − E2 ) 2

(5)

where Δ = E2 s − E1s = ω 0 and E1 = ω 1 , E2 = ω 2 are the photon energies. Integrating over E2 and using the delta function we find the differential transition probability per unit time for photon #1 to have frequency between ω 1 and ω 1 + dω 1 :

dΓ = consti M ω 12 dω 1 (ω 0 − ω 1 ) dΩ1dΩ2 2

2

(6)

The matrix element is: H ' fn H 'ni H ' fn H 'ni ⎫ ⎧ M = − ⎨∑ +∑ ⎬ ⎩ n En − E2 s + ω 1 n En − E2 s + ω 2 ⎭

(7)

In the first sum photon #1 appears in the intermediate state; in the second it is photon #2.

(Since these photons are indistinguishable, it turns out that we must multiply the transition probability by ½). In the first sum, e 1  f e−ikir ε̂ i p n 3/2 mL 2ω 2 and similarly for each of the factors H ' fn , H 'ni . That is to say, each such factor makes a H ' fn =

contribution proportional to: 1 Z ε̂ i p ∝ = Z 0 1/2 ω Z Also, each energy denominator in M is proportional to Z 2 . Therefore, in (7) we have: 2 dΓ ∝ Z −4ω 12 (ω 0 − ω 1 ) dω 1 (8) ∝ Z −4 Z 4 Z 4 Z 2 = Z 6 d) In the presence of a static electric field the 22s1/2 state is mixed with p states by the Stark effect. Since 22p1/2 lies closest to 22s1/2 it dominates. Consider E = E0 ẑ and the m j = 1/2 components of 22s1/2 and 22p1/2:

i iω t a2 s = − a2 p H '2 s,2 p e sp  i γ −iω t a2 p = − a2 s H '2 p,2 s e sp − a2 p  2 To save writing, let a2 s = a1 , a2 p = a2 , ω = ω s, p , and  −1 H '2 s,2 p = V. Then: 162

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a1 = −ia2Veiω t a2 = −ia1V * e−iω t −

(9)

γ a2 2

Thus we have: a1 = −iV ( a2 + iω a2 ) eiω t

γ ⎛ ⎞ = −iV ⎜ −ia1V * e−iω t − a2 + iω a2 ⎟ eiω t ⎝ ⎠ 2 iγ ⎞ 2 ⎛ = − V a1 + i ⎜ ω + ⎟ a1 ⎝ 2⎠ which yields: ⎛γ ⎞ 2 a1 + ⎜ − iω ⎟ a1 + V a1 = 0 ⎝2 ⎠

(10)

Assuming a1 (t = 0) = 1 we try a solution to (10) of the form a1 (t) = e− λt . Substitution into (10) gives: ⎛γ ⎞ 2 λ 2 − ⎜ − iω ⎟ λ + V = 0 ⎝2 ⎠ which has the solutions:

1⎛γ ⎞ 1 ⎛γ ⎞ 2 λ± = ⎜ − iω ⎟ ± − iω ⎟ − 4 V ⎜ ⎠ 2 ⎝2 ⎠ 2⎝ 2 When V = 0, we want no 2s decay: λ = 0 . Thus we choose λ− in (10): 2

⎡ ⎤ 2 ⎢ ⎥ 4V 1⎛γ ⎞⎢ ⎥ λ = ⎜ − iω ⎟ 1 − 1 − 2 ⎠⎢ 2⎝ 2 γ ⎛ ⎞ ⎥ ⎢ ⎜⎝ − iω ⎟⎠ ⎥ 2 ⎣ ⎦ Assuming that V is small we expand the square root in (12) to obtain: γ 2 + iω V 2 2 λ =V γ γ2 − iω ω2 + 2 4 Thus, ⎡ ⎤ 2 ⎢ γ V t ⎥ 2 a1 = exp ⎢ − 2 ⎥ ⎢ ω2 + γ ⎥ ⎢⎣ 4 ⎥⎦

(11)

(12)

(13)

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Hence the rate of 2s decay is:

τ

−1

=

γ V

2

γ H '2 s,2 p

2

= γ2  2γ 2 ω + S2 + 4 4 where S = ω is the Lamb shift. Now,

(14)

2

2s, m j = 1/2 = Y00 R2 sα e ⎡ 1 2 1 ⎤ 2 p1/2 , m j = 1/2 = R2 p ⎢ − Y10α e + Y1 β e ⎥ 3 ⎣ 3 ⎦ Thus, (see solution to part b) above): 1 0 eE a H '2 p,2 s = − Y1 R2 p eE0 z Y00 R2 s = 3 0 0 3 Z

(15)

Therefore (14) becomes:

τ −1 =

3γ e2E02 a02 ⎛  2γ 2 ⎞ Z 2 ⎜ S2 + 4 ⎟⎠ ⎝

(16)

or, assuming Z=1 (hydrogen):

⎛ 2  2γ 2 ⎞ −1 ⎜⎝ S + 4 ⎟⎠ τ E0 = 3γ e2 a02 Inserting the following numbers (cgs units) into (17): γ = 6.25i10 8

(17)

S = 1.058i10 9 i6.625i10 −27 e = 4.8i10 −10 a0 = 5.29i10 −9

τ −1 = 8 we obtain: (18) E0 = 1.81i10 −4 esu (statvolts)/cm = 0.054 V/cm This shows that even a very small external electric field shortens the mean lifetime of the 2s state quite drastically.

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16.7 All of the sum rules in this problem, and many others in addition, can be treated by a single general method, the identities for which are derived in problem 2.15. They are summarized as follows: (1) If S p ≡ ∑ ω knp n A† k k A n where n , k are eigenstates of a Hamiltonian H, k

then:

S p =  − p n A† Ap n

(1)

where Ap ≡ ⎡⎣ H , Ap −1 ⎤⎦ , A0 ≡ A , and A is an arbitrary operator.

(2)

n A† Ap n = n Am† Ap − m n

(2)

for m ≤ p , (and where Am ≡ ⎡⎣ Am −1 , H ⎤⎦ .) †

†

We now discuss the sum rules mentioned in the statement of the problem. In each case we use A = A† = x. Also, we choose p = 0,1,2,3.

p=0:

S0 = ∑ n x k k x n = n x 2 n

(3)

k

p = 1:

S1 = ∑ ω kn n x k k x n = k

1 n x[ H , x] n 

(4)

⎡ p2 ⎤ i Now, [ H , x ] = − [ x, H ] = − ⎢ x, x ⎥ = − px = x1 . Therefore, m ⎣ 2m ⎦

∑ω k

2

kn

xkn = −

i i n xpx n = − n xpx − px x n m 2m −i  = i = 2m 2m

Hence,

2m 2 ω kn xkn = 1 ∑  k

(5)

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1 n x1† x1 n 2  1 = 2 n px2 n m

2 S2 = ∑ ω kn xnk = 2

p=2:

k

=−

(6)

2 ∂2 n n m2 ∂x 2

If n is spherically symmetric,

− 2 ∂2 − 2 1 n 2 n = n ∇2 n = n p2 n 2 2 2 m ∂x 3m 3m 2 nT n 3m 2 = n H −V n 3m =

Therefore:

∑ω x 2 kn

2 nk

=

k

(

2 En − n V n 3m

).

(7)

p = 3:

S3 = ∑ ω kn3 xnk = 2

k

where:

1 n A1† A2 + A2† A1 n 3 2

(8)

A0 = x A1 = [ H , x ] = −i A1† = i

 px = x1 m

 px m

  2 ∂V p , H = [ x ] m ∂x m  2 ∂V † A2 = A2 = m ∂x A2 = [ H , x1 ] = i

Thus,

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S3 = ∑ ω kn3 xnk = 2

k

1 i 3 ∂V ∂V n px − xn 3 2 2 m ∂x ∂x

=

i ⎡ ∂V ⎤ n ⎢ px , n 2 2m ∂x ⎥⎦ ⎣

=

 ∂ 2V n n 2m 2 ∂x 2

(9)

If n is spherically symmetric:  2 S3 = ∑ ω kn3 xnk = n ∇ 2V n 2 6m k 2 2 Now, ∇ V = + ρ e where ρ is the probability density. Thus:

e2 S3 = n ρ (r) n 6m 2

(10)

(11)

16.8 Consider an atom with polarization initially along x. The polarization vector precesses in the xy plane with angular velocity ω proportional to the magnitude of the π magnetic field B. After time t, the polarization vector is at angle φ =  ω t with respect 2 to the y axis. The probability that a right circularly polarized photon is emitted along the y direction between t and t + dt is proportional to:

(1+ cosφ )2 γ e−γ t dt = (1 ± sin ω t ) γ e−γ t dt 2

Thus the total probability for a count in the detector between t = 0 and t = ∞ is proportional to: ∞

I = ∫ (1 ± sin ω t ) e−γ t dt = 0

2

3 2ω γ ± 2 − 2 2 2γ γ + ω 2 (γ + 4ω 2 )

16.9 Since photons are bosons, the state of 2 identical bosons must be symmetric with respect to their exchange. Any given photon propagating along a given axis (say the z axis) can have only 2 possible polarization states, which we can choose to be positive helicity: + , and negative helicity: − . From these states the only symmetric twophoton states that can be formed are the following:

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j = 2, m = 2 = +1 + 2

(

)

1 1 2,0 + 2 0,0 = ( +1 − 2 + −1 + 2 ) 3 2 2,−2 = −1 − 2

where the subscripts 1,2 refer to photon #1,#2 respectively. We see that we can form 2photon plane-wave states that have total angular momentum j = 2 or a linear combination of j = 2 and j = 0 . No 2-photon state with j = 1 is possible. Therefore: The transitions F = 0 → F = 0 and F = 1 → F = 1 can occur, but the transitions F = 0 → F = 1 and F = 1 → F = 0 cannot occur.

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17.1 The original Hamiltonian is: ω 2 1⎤ ⎡ ⎡⎣ P + Q 2 ⎤⎦ = ω ⎢ a †a + ⎥ H0 = 2 2⎦ ⎣ 1 1 where a = (Q + iP ) , a† = (Q + iP ) . Let: 2 2

(1)

2 ω ⎡ P − 2x1 + (Q − 2x2 )2 ⎤ ⎥⎦ 2 ⎢⎣ (2) 1⎤ ⎡ † = ω ⎢ a − α * ( a − α ) + ⎥ 2⎦ ⎣ where α = ( x1 + ix2 ) and x1 , x2 are two slowly varying real parameters that together describe a closed curve in the x1 x2 plane. We now show that:

H=

(

)

(

)

H = D(α )H 0 D † (α ) where

(

D(α ) = exp α a † − α * a

(

(3)

)

D † (α ) = exp α * a − α a †

(4)

)

To demonstrate (3) we use the identity: 1 e A Be− A = B + [ A, B ] + ⎡⎣ A, [ A, B ]⎤⎦ + ...

2! † where A = α a − α * a and B = a †a . Now, ⎡⎣ a † , a †a ⎤⎦ = −a †

(5)

⎡⎣ a, a †a ⎤⎦ = a

Hence,

[ A, B ] = −α a† − α * a

from which it follows that: ⎡⎣ A, [ A, B ]⎤⎦ = − ⎡⎣ α a † − α * a , α a † + α * a ⎤⎦

(

)(

)

=2α Thus (5) can be written: 2 D(α )a †aD † (α ) = a †a − α a † − α * a + α which proves (3). Now, the eigenstates of H 0 are obviously the number eigenstates n , 2

1⎞ ⎛ with corresponding eigenvalues En = ω ⎜ n + ⎟ . Thus we have: ⎝ 2⎠ H 0 n = En n

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and therefore, D(α )H 0 D † (α )D(α ) n = HD n = En D n Thus the eigenvectors of H corresponding to the energies En are: n, α ≡ D(α ) n

(6) Now we have shown in Chapter 17 that the geometric phase associated with the state n is:

γ n = − Im ∫ nα ∇ nα id R However,

(7)

† = − Im  ∫ n D ∇D n id R

∂D ∂D (8) x̂1 + D † x̂2 ∂x1 ∂x2 To evaluate the right hand side of (8) we make use of the following identity, which was the subject of problem 2.14: D †∇D = D †

e−iA

∞ d iA ( −i ) A n ⎧ dA ⎫ e = i∑ ⎨ ⎬ dx ⎩ dx ⎭ n = 0 ( n + 1)! n

( )

where A0 { B} = A, A1 { B} = [ A, B ], A2 { B} = ⎡⎣ A, [ A, B ]⎤⎦ , ...

(

)

We choose A = −i α a † − α * a . Then, D†

⎤ ∂D ∂ i [ 0,1] A ∂ 1⎡ ∂ = e−iA e = α a† − α * a − ⎢ α a† − α * a , α a † − α * a ⎥ + ... ∂x1 ∂x1 ∂x1 2! ⎣ ∂x1 ⎦

(

)

(

(

)

(

)

) 2!1 ⎡⎣(α a − α * a ), ( a − a )⎤⎦ + ... = −ix + ( a − a ) = a† − a −

†

†

†

2

(9) and similarly, ∂D D† = ix1 + i(a † + a) ∂x2 Inserting (9) and (10) in (8) we obtain: D †∇D = ⎡⎣ −ix2 + (a † − a) ⎤⎦ x̂1 + ⎡⎣ix1 + i(a † + a) ⎤⎦ x̂2 Thus (7) becomes: γ n = ∫ ( x2 dx1 − x1dx2 )

(10)

(11)

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which is the same for all values of n. The geometric interpretation of (11) is easily obtained from Stokes’ theorem. Given a vector field V, we have:

∫ V idl = ∫ ∇ × V in̂dS C

where C is a closed path, and the surface integral on the right hand side is taken over a surface bounded by C. In (11) let

V = x2 x̂1 − x1 x̂2 Then the component of ∇ × V in the direction normal to the x1x2 plane is:

∂V2 ∂V1 − = −2 ∂x1 ∂x2 Hence, in (11) γ n is equal to -2 times the area of the closed loop in the x1 x2 plane.

17.2 In cgs units, the magnitude of magnetic field By generated at the entrance or exit of the electric field by each single turn coil is: 2π I By = Gauss = .001 Gauss 10a The magnetic field at the exit (entrance) generated by the coil at the entrance (exit) is negligible. The electric field between the plates is: 20, 000 V V Ez = = 10 5 .2 cm cm 5 10 esu esu = = 333 300 cm cm The motion of the beam through this field causes a motional magnetic field in the atom rest frame of: v 333i1i10 4 Bx = Ez = = 1.11i10 −4 Gauss c 3i1010 This field can be assumed to turn on or off adiabatically at the entrance or exit of the electric field region, because the electric field goes from zero to full magnitude in a distance of the order of the separation between the plates: ≈.2 cm. Thus, in the rest frame of an atom, the motional magnetic field turns on or off in a time of order t ≈ .2/ 1i10 4 s =2i10 -5 s . During this time, the precession of the magnetic moment in the magnetic field Bz is about 15 radians, which is several orders of magnitude larger than the precession caused by Bx in these transition regions at either end of the electric field plates.

(

)

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Let the value of y just before the entrance to the E field be y=A; Just after the entrance to the E field: y=B; Just before the exit from the E field: y=C Just after the exit from the E field: y=D Then at P, B = Bz ẑ , at A, B = Bz ẑ + By ŷ , Ev x̂ , c Ev at C, B = Bz ẑ − By ŷ + x̂ , c at D, B = Bz ẑ − By ŷ

at B, B = Bz ẑ + By ŷ +

at Q, B = Bz ẑ This figure illustrates how the magnetic field vector traces out a closed curve.

B A

P,Q

C D

Fig.17.1S The solid angle subtended by the closed curve traced out by B at its point of origin is: ByEv 10 −3 i1.11i10 −4 Ω= = = 2.78i10 −6 sterad 2 cBz .04 1 Thus, the magnitude of the phase in question is: φ = m Ω = Ω = 1.39i10 −6 rad .

2

πv µ B0 ω2 , ω0 = , Ω = ω 02 + L  4 In the laboratory frame, the Schroedinger equation is: 17.3Let z = vt, L = vT , ω =

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ψ = iω 0 ⎡⎣ cos ω tσ z + sin ω tσ x ⎤⎦ψ

(1)

in the region 0 < z < L where:

⎛ a+ ⎞ ψ =⎜ ⎟ . ⎜⎝ a− ⎟⎠ We perform a rotation about the –y axis by amount −ω t during the interval 0 < t < T . In the rotating frame the applied magnetic field is constant with magnitude B0 in the z direction. In addition, since a rotation is equivalent to a magnetic field (see Chapter 7, ω Sec. 7.6) we have a magnetic field of magnitude B0 in the –y direction. The 2ω 0 Schroedinger equation in the rotating frame is: ∂ψ ' ⎛ ω ⎞ = i ⎜ ω 0σ z − σ y ⎟ ψ ' ⎝ ∂t 2 ⎠ where ψ ' is the wave function in the rotating frame. Eq. (2) can be written as:

ω a'− 2 ω  − + iω 0 a'− = + a'+ a' 2  + − iω 0 a'+ = − a'

(2)

(2)

The solution to these equations, given the initial conditions a'+ ( 0 ) = 1, a'− ( 0 ) = 0 , is easily found to be: a'+ (z = L) = cos a'− (z = L) =

ΩL ω 0 ΩL + i sin = a− ( z = L ) v Ω v

ω ΩL sin = a+ ( z = L ) 2Ω v

ω2 . Thus a− ≅ 1 and a+ << 1. In the 8ω 0 ω sudden approximation limit, ω >> ω 0 and Ω ≅ . Here, 2 In the adiabatic limit, ω << ω 0 and Ω ≅ ω 0 +

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ωL π = cos 2 = 0 2v 2 2 ωL π a+ ( L ) ≅ sin 2 = sin 2 = 1 2v 2 a− ( L ) ≅ cos 2 2

17.4a) When a tritium nucleus decays in a tritium atom, the atomic number changes suddenly from Z=1 to Z=2. The beta electron is emitted with an average energy of about 10 keV. The kinetic energy of the atomic electron in the ground state is of order 10 eV. Thus the beta electron has a velocity about 30 times greater than that of the atomic electron, so it exits from the atom in a time which is about 1/30 of a period of atomic motion. Thus, the sudden approximation is adequate to describe what happens to the atomic electron. According to that approximation, we can write in atomic units: 1 ψ initial = 1/2 e− r = ∑ anψ n ' π n where the ψ n ' are the eigenstates of the final Hamiltonian, with Z=2, and:

an = ∫ ψ n 'ψ i dτ 2

2

Now,

2 3/2 −2r e π 1/2 1 ψ 2 s ' = 1/2 (1 − r ) e− r π

ψ 1s ' =

Thus,

∞ 2 3/2 2 9 /2 4π ∫ r 2 e−3r dr = 0 π 27 ∞ 1 a2 s = 4 ∫ r 2 (1 − r)e−2r dr = − 0 2 a2 p = 0

a1s =

Hence we find:

a1s

2

29 = 6 = .702 3

2

a2 s = .25 2

a2 p = 0

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b) For zero neutrino mass, the number of beta rays emitted in the energy range Ee to Ee + dEe from a population of decaying tritium nuclei is:

dN e = Ai Ee2 − me2 Ee ( Δ − Ee ) dEe ≡ Ai f idEe (1) where we choose units where c=1, and A is a constant.

If the neutrino mass is non-zero, we must take into account the difference between the magnitude of neutrino linear momentum and its energy, and thus replace f in (1) with the formula: 2

f = Ee2 − me2 Ee ( Δ − Ee )

( Δ − Ee )2 − mν2

(2)

Note two facts about (2): the right hand side vanishes at Ee = Δ − mν , which is the new endpoint. Also:

df = −∞ dEe at Ee = Δ − mν .

The following figure shows the shape of the function f near the end point, for the two cases (1) and (2).

-12

8x10

6 f

mν=0 4 2

mν=2 eV/c 2

0 -3.0

-2.5

-2.0

-1.5 Ee-Δ, eV

-1.0

-0.5

0.0

Fig.17.2S c) Returning to (1) we choose units from now on where c = me =1, and we also choose

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Ee = Δ − ε where ε is a very small energy. Also we write Δ = me c 2 + Δ 0 = 1 + Δ 0 where 18.6 keV Δ0 = = .0364 . Then, 511 keV f = ε 2 (Δ − ε )

2

( Δ − ε )2 − 1

= ε 2 (1 + Δ 0 − ε )

2

(1 + Δ 0 − ε )2 − 1

(3)

 .2924 ε 2

The number of beta electrons radiated in the entire spectrum is:

Ne = A ∫

1+ Δ 0

= A∫

1.0364

1

1

fdEe E E 2 − 1 (1.0364 − E ) dE 2

(4)

= 2.0097i10 −6 A where the last step is obtained by numerical evaluation of the integral on the right hand side of the second line. The number of beta electrons radiated in the energy interval between Δ − ε and ε is:

ne = A ∫

1+ Δ 0

1+ Δ 0 − ε

ε

fdE = Ai.2924 ∫ ε 2 dε = Ai.0975ε 3 0

(5)

If we choose ε = 5 eV = 9.78i10 −6 in the present units, we obtain:

ne = Ai9.12i10 −17

(6)

Comparing this with (4) we obtain the result:

ne = 4.54i10 −11 Ne This ratio is so small that the experiment is extremely difficult.

(7)

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18.1a) The total cross section is:

4π ∞ 2 + 1) sin 2 δ  2 ∑( k =0 If this sum is cut off for  > L the maximum cross section is 2

σ T = ∫ f (θ ) dΩ =

4π L 4π 4π 2 2 + 1) = 2 [ (L + 1) + L(L + 1)] = 2 ( L + 1) 2 ∑( k =0 k k 2π C b) Consider the potential U(r) such that U(r) → 2 s for large r , where C is a  r constant, which can be positive (repulsive potential) or negative (attractive potential). If s < 2 , then for sufficiently large  the total potential including the centrifugal part is dominated by the latter for small r , in both attractive and repulsive cases. According to the WKB approximation, the radial wave function χ = rR is given by: const π⎤ ⎡1 r χ= sin ⎢ ∫ p dr + ⎥ r 4⎦ p ⎣ 0 where

σ T ,max =

 (  + 1) r2 and r0 is the inner classical turning point. Thus, p = 2m(E − Veff ) =  k 2 − U −

⎡ r const ( + 1) π⎤ sin ⎢ ∫ k 2 − U − dr + ⎥ r r2 4⎦ p ⎣ 0 The corresponding radial function for U = 0 is:

χ (r ) =

(1)

⎡ r const ( + 1) π⎤ sin ⎢ ∫ k 2 − dr + (2) ⎥ r0 ' r2 4⎦ ( + 1) 2 ⎣ k − r2 Here r0 ' is the inner classical turning point for U = 0 . For very large  the difference r0 − r0 ' is negligible because of the dominance of the centrifugal potential in both cases.

The phase shift δ  is then given by: s ( r ) =

δ ≈ ∫

r→∞

r0

⎡ 2 2 2 ⎤ 2 ⎢ k − U − 2 − k − 2 ⎥dr r r ⎥⎦ ⎢⎣

(3)

In (3),

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⎧ 2 ⎫ 2 ⎪ 2 2 2 ⎪ k −U − 2    ⎪ r − 1⎪ k2 − U − 2 − k2 − 2 = k2 − 2 ⎨ ⎬ r r r ⎪ 2 2 ⎪ k − 2 ⎪⎭ r ⎩⎪ ⎧ ⎫ 2 ⎪ ⎪⎪  U ⎪ = k 2 − 2 ⎨ 1− 2 − 1⎬  r ⎪ ⎪ k2 − 2 ⎪⎩ ⎪⎭ r U ≈− 2 2 2 k − 2 r

Thus

δ ≈ −

Let r =

µC ∞ dr µC ∞ dr = −  2 ∫r 2 ∫  0 s 2 2 k k s 2 r k − 2 r 1− 2 2 r k r

(4)

 . Then (4) is written as: ku1/2

µC ⎛ k ⎞ δ ≈ − ⎜ ⎟ 2k 2 ⎝  ⎠ =−

µC ⎛ k ⎞ ⎜ ⎟ k 2 ⎝  ⎠

s−1

s−1

u (s−3)/2 ∫0 (1− u)1/2 du 1

⎡ ⎛ s − 1⎞ ⎤ ⎢ π Γ ⎜⎝ 2 ⎟⎠ ⎥ ⎢ ⎥ ⎢ 2Γ ⎛ s ⎞ ⎥ ⎜⎝ ⎟⎠ ⎥ ⎢⎣ 2 ⎦

(5)

The factor in brackets on the rhs of (5) is of order unity. Also, since V (r) = C / r , s

⎛ ⎞ ⎛ k⎞ V (r0 ) = V ⎜ ⎟ = C ⎜ ⎟ ⎝ k⎠ ⎝ ⎠

s

Thus,

δ ≈ −

µ  ⎛ ⎞ V⎜ ⎟ 2 k k ⎝ k ⎠

(6)

Now in

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4π (2 + 1)sin 2 δ  2 ∑ k δ  is small for large  , so sin 2 δ  ≈ δ 2 Thus,

σT =

4πµ 2C 2 2(s−3) σT ≈ k ( 2 + 1)2(1−s ) ∑ 4  8πµ 2C 2 2(s−3) 3−2 s ≈ k ∫  d 4 The integral converges if s>2, but diverges if s<2. Thus we obtain an infinite contribution to σ for 2>s>1even though each phase shift is finite. The divergence in σ comes from large  (in other words, large impact parameter=small scattering angle). However, in any real scattering process the potential is screened by surrounding matter and generally falls to zero more rapidly than any power of r at great distance. Thus the infinities we have found are not real, but just an artifact of the way we have idealized the problem.

2µ ⎡ E + V0 e− r /a ⎤⎦ χ = 0 2 ⎣ r Let z = exp(− ) . Then: 2a 1 z dz = − e− r /2a dr = − dr 2a 2a Thus,

18.2

χ "+

(1)

∂ z ∂ ∂2 1 ∂⎛ ∂⎞ =− , = 2 z ⎜z ⎟ 2 ∂r 2a ∂z ∂r 4a ∂z ⎝ ∂z ⎠ hence (1) can be written: ∂⎛ ∂⎞ 8 µa 2 ⎡ E + V0 z 2 ⎤⎦ χ = 0 z χ + (2) ⎜ ⎟ ∂z ⎝ ∂z ⎠ 2 ⎣ 2 k 2 8 µa 2V0 2 8 µa 2 E > 0 . Define κ 2 = , λ = − , and x = κ z.

In the case of scattering, E = 2 2 2µ Then λ = ±2iak and (2) becomes: z

∂⎛ ∂⎞ 2 2 ⎜⎝ x ⎟⎠ χ + x − λ χ = 0 ∂x ∂x The general solution to (3) is:

)

(3)

χ = AJ 2iak (x) + BJ −2iak ( x )

(4)

x

(

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When r = 0, z = 1, hence x = κ ; and also χ (r = 0) = 0. Hence (4) gives: AJ 2iak (κ ) + BJ −2iak (κ ) = 0 and therefore: ⎡ ⎤ J (κ ) χ = A ⎢ J 2iak ( x ) − 2iak J −2iak ( x ) ⎥ J −2iak (κ ) ⎣ ⎦ or apart from an overall multiplicative constant:

χ = J 2iak (κ ) J −2iak ( x ) − J −2iak (κ ) J 2iak ( x )

(5) λ

1 ⎛ x⎞ . Therefore for r → ∞, x → 0. However, for small x, J λ (x) → ⎜ ⎟ ⎝ 2 ⎠ Γ (1+ λ ) sufficiently large r , (5) becomes:

When

−2iak

2iak

⎛ x⎞ ⎛ x⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 2 2 χ → J 2iak (κ ) − J −2iak (κ ) Γ (1 − 2iak ) Γ (1 + 2iak ) ⎡ J (κ ) ⎛ κ ⎞ = ⎢ 2iak ⎜ ⎟ ⎢⎣ Γ (1 − 2iak ) ⎝ 2 ⎠

−2iak

⎤ ikr ⎡ J −2iak (κ ) ⎛ κ ⎞ ⎥e − ⎢ ⎜ ⎟ ⎥⎦ ⎢⎣ Γ (1 + 2iak ) ⎝ 2 ⎠

(6) 2iak

⎤ −ikr ⎥e ⎥⎦

We compare this result with the general formula for the asymptotic form of the wave function for scattering from a potential:

eikr ψ =e + f (7) r For s- wave (  = 0 ) scattering, this may be written in terms of the scattering amplitude f in the form: ikz

(8) χ = reikz + feikr where we limit ourselves to the  = 0 components in the expansions of eikz and f in partial waves. Thus:

reikz → rj0 (kr) =

sin(kr) eikr − e−ikr = k 2ik

(9)

)

(10)

and:

(

1 2iδ 0 e −1 2ik Substituting (9, 10) in (8) we obtain: f =

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eiδ 0 i ( kr + δ 0 ) ⎡e − e−i ( kr + δ 0 ) ⎤⎦ (11) ⎣ 2ik Apart from an overall constant, this is the same as (6) provided that we have: −4iak Γ (1 + 2iak ) J 2iak (κ ) ⎛ κ ⎞ (12) e2iδ 0 = ⎜ ⎟ Γ (1 − 2iak ) J −2iak (κ ) ⎝ 2 ⎠ We can establish a connection between the foregoing analysis and the  = 0 bound states in the same potential as follows. Let us continue the function on the right hand side of (12) into the complex k plane. When k = i k the energy is negative, and bound states

χ=

correspond to those values of k for which the coefficient of e−ikr in (6) or (11) is zero: J 2a k (κ ) = 0

This is precisely the condition for bound states in the same potential. Why is it that poles of Γ (1 + 2ika ) do not also yield bound states? The reason is that such poles occur for:

1 + 2iak = 0, −1, −2, −3,... hence for: k=

ni where n = 1,2,3,…

2a

In this case,

χ = J 2iak (κ ) J −2iak ( x ) − J −2iak (κ ) J 2iak ( x ) becomes: χ = J − n (κ ) J n ( x ) − J n (κ ) J − n ( x )

(13)

However, J − n (x) = (−1) J n (x) . Therefore the right hand side of (13) vanishes identically. n

18.3 If the 2 identical particles are fermions (bosons), the total wave function of relative motion must be anti-symmetric (symmetric) with respect to exchange. If the spin function of the fermions is symmetric : S, (antisymmetric : A) the spatial function must be A (S).

In the CM frame the spatial function is: eikr ikz −ikz ψ S, A = e ± e + [ f (θ ) ± f (π − θ )] (1) r mv µv Here, k = (not ) . The cross section for either particle to be scattered into solid   angle dΩ at angle θ is:

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dσ 2 = f (θ ) ± f (π − θ ) dΩ = f (θ ) + f (π − θ ) ± f * (θ ) f (π − θ ) ± f (θ ) f * (π − θ ) For Coulomb scattering, θ −inn sin 2 Γ(1 + in) in 2 fC (θ ) = e Γ(1 − in) 2θ 2k sin 2 Thus, n2 2 fC (θ ) = θ 4k 2 sin 4 2 n2 n2 2 fC (π − θ ) = = π −θ θ 4k 2 sin 4 4k 2 cos 4 2 2 and n2 θ θ⎞ ⎛ f * (θ ) f (π − θ ) = exp ⎜ inn sin 2 − inn cos 2 ⎟ θ θ ⎝ 2 2⎠ 4k 2 sin 2 cos 2 2 2 2 n θ θ⎞ ⎛ f (θ ) f * (π − θ ) = exp ⎜ −inn sin 2 + inn cos 2 ⎟ θ θ ⎝ 2 2⎠ 4k 2 sin 2 cos 2 2 2 Also, n2 Z4 e4 = 4k 2 16E 2 ( 4π )2 2

2

Therefore, dσ Z4 e4 1 ⎡ θ θ θ⎞⎤ ⎛ = 1+ tan 4 ± 2 tan 2 cos ⎜ nn tan 2 ⎟ ⎥ 2 2 ⎢ ⎝ dΩ S,A 16E ( 4π ) sin 4 θ ⎣ 2 2 2⎠ ⎦ 2

(2)

In the classical limit,  → 0 so n → ∞. Thus the interference term in (2) oscillates extremely rapidly as θ varies. Thus when averaged over a small but non-zero range of angles it gives no contribution. For scattering of two alpha particles, only the + sign would appear in front of the interference term.

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18.4 The rate of reactions per cm3 is: 1 (1) λ = nA nB σ v or λ = n02 σ v 2 where nA, B are the number densities of distinct reactants A,B ; or n0 is the number density when the two reactants are the same. Also, ∞

σ v = ∫ σ (v)vφ ( v) dv 0

(2)

where v is the relative velocity of the reactants, σ (v) is the cross section for the reaction, and φ (v) is the normalized Maxwell relative velocity distribution, given from kinetic theory by: 3/2 ⎛ µ v2 ⎞ ⎛ µ ⎞ 2 φ (v) = 4π ⎜ v exp (3) ⎜⎝ − 2kT ⎟⎠ ⎝ 2π kT ⎟⎠ mA mB Here k is Boltzmann’s constant, T is the absolute temperature, and µ = is the mA + mB reduced mass. Therefore: 3/2 ∞ ⎛ − µ v2 ⎞ ⎛ µ ⎞ 3 (4) σ v = 4π ⎜ v σ ( v) exp ⎜ dv ⎝ 2π kT ⎟⎠ ∫0 ⎝ 2kT ⎟⎠ 1 We make the substitution E = µv2 to write (4) as: 2 3/2 ∞ 8π ⎛ 1 ⎞ ⎛ −E ⎞ σ v = 1/2 ⎜ Eσ ( E ) exp ⎜ dE (5) ⎟ ∫ 0 ⎝ kT ⎟⎠ µ ⎝ 2π kT ⎠ Now ⎡ 2π n −2 π n ⎤ σ (E) = S(E) ⎢ e ⎥ (6) ⎣ v ⎦ where the bracketed quantity is the Gamow factor due to the Coulomb barrier, with: Z Z e2 n= 1 2 (7) 4π v while S(E) is a factor associated with the nuclear physics of the reaction. Generally, S varies very slowly over the energy range of interest, and for our purposes we can take it as a constant S0 ' . Now defining:

S0 = S0 '

Z1Z 2 µe2 4

we have:

σ (E) =

S0 −bE −1/2 e E

(8)

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where

Z1Z 2 e2 µ1/2 (9) 2 3/2  Thus, assuming the two nuclear reactants are the same, and defining θ = (kT )−1 , we have: b=

4π n 2 ⎛ 1 ⎞ λ = 1/20 ⎜ ⎟ µ ⎝ 2π kT ⎠

3/2

∞

(

)

S0 ∫ exp −θ E − bE −1/2 dE 0

(10)

To evaluate the integral in (10), consider the function: f (E) = θ E + bE −1/2 For large E, f ≈ θ E ; while for small E, f ≈ bE −1/2 . Thus somewhere in between, f must have a minimum, and therefore exp(− f ) has a very sharp maximum. Our strategy is to find the location of that minimum in f, expand f in a Taylor series about the location of the minimum to second order, and approximate the integral in (10) by the resulting Gaussian integral.

We have: f (E) = θ E + bE −1/2

∂f b = θ − E −3/2 ∂E 2 Therefore the location of the minimum in f is 2/3 ⎛ b⎞ E = E0 = ⎜ ⎟ ⎝ 2θ ⎠ and f (E0 ) = 3i2 −2 / 3θ 1/ 3b 2 / 3 ≡ τ The Taylor expansion to second order is: 1 ∂2 f f (E) = f (E0 ) + ( E − E 0 )2 2 2! ∂E E

(11) (12) (13)

0

since the first derivative vanishes at E = E0 . The second derivative is:

∂2 f = 3i2 −1/ 3θ 5 / 3b −2 / 3 2 ∂E E

(14)

0

Hence (13) becomes: f (E) = τ + β (E − E0 )2 where β = 3i2 −4 / 3θ 5 / 3b −2 / 3 Thus the integral in (10) is:

(15) (16)

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∫ e e

− τ − β (E−E0 )2

∞

dE ≅ e−τ ∫ e− β x dx 2

(17)

−∞

0

1/2

−1/2 − τ

⎛π⎞ ⎜⎝ β ⎟⎠

1/2

=π β

e

Therefore we obtain:

4π n 2 ⎛ θ ⎞ λ = 1/20 S0 ⎜ ⎟ ⎝ 2π ⎠ µ

3/2

e− τ (18)

24  = 5/2 n S τ 2 e− τ 2 0 0 3 Z1Z 2 µe where, as already noted:

τ = 3i 2 −2/3θ 1/3b 2/3 1/3

3 ⎛ Z 2 Z 2 e4 µ ⎞ = 5/3 ⎜ 1 2 2 2 ⎝  kT ⎟⎠ Since the quantity exp(− f (E)) has a sharp maximum at E = E0 practically all of the reactions occur in a narrow band of energies about:

⎛b ⎞ E0 = ⎜ kT ⎟ ⎝2 ⎠

2/3

⎛ kT Z Z e2 µ1/2 ⎞ = ⎜ 5/2 1 2 ⎟⎠  ⎝2

2/3

For the proton-proton reaction in the sun, Z1 = Z 2 = 1 and µ =

⎛ e2 m1/2 ⎞ p k E0 = ⎜ T⎟ ⎝ 8 ⎠

mp 2

. Thus:

2/3

With e = 4.8 4π i10 −10 , m p = 1.67 i10 −24 g, k = 1.38 i10 −16 erg/deg ,  = 1.054i10 −27 erg s, and T = 1.2i10 7 deg K, we obtain E0 = 8.14i10 −9 erg . Compare this to the most probable kinetic energy of relative motion, which is kT = 1.656i10 −9 erg . The ratio is: E0 = 4.9 kT Thus the protons that contribute most to the reaction are far out on the high-velocity tail of the Maxwell distribution.

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18.5 We have shown that the Golden rule and the Born approximation are equivalent.

Thus we start with the Golden Rule formula: 2π 2π L3 p f dp f dΩ 2 2 dW = M ρ= M   ( 2π  )3 dE f 2

(1)

where p f , E f = p 2f /(2me ) refer to the ejected electron, which is assumed to be nonrelativistic. Since dE f = p f dp f /me , (1) can be written:

dW = M

2

L3 me p f dΩ

(2)

4π 2  4 The matrix element is: M=

c 2 φ f eik0 z px φi 2ω

e n1/2 mc L3/2

(3)

where k0 is the incoming photon wave vector with the linear polarization along x. Also,

φi =

1

(π a )

3 1/2 0

e− r /a0 (Hydrogen ground state)

1 ik f i r e (Ejected electron) L3/2 Inserting these expressions in (3) and re-arranging, we obtain:

φf =

−ie ⎛ 1 ⎞ M= mc ⎜⎝ π a03 ⎟⎠

1/2

n1/2 L3

c 2 I 2ω

(4)

where

⎛ ∂ − r /a0 ⎞ 3 ⎜⎝ e ⎟⎠ d r ∂x Integrating I by parts, we obtain: ⎛ ∂ −ik i r ⎞ I = − ∫ e− r /a0 ⎜ eik0 z e f ⎟ d 3 r ⎝ ∂x ⎠ I = ∫ eik0 z e

−ik f i r

= ik fx ∫ eik0 z e

−ik f i r − r /a0

e

d 3r

(5)

= ik fx ∫ eiΚ i r e− r /a0 d 3 r where K = k0 ẑ − k f . We evaluate the integral in the last line of (5) in the usual way, as follows:

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π

0

0

I = ik fx ∫ r 2 e− r /a0 dri2π ∫ eiKr cos θ sin θ dθ r − r /a0 iKr ⎡⎣ e − e−iKr ⎤⎦ dr e 0 iK

= 2π ik fx ∫

∞

2 2 2π k fx ⎡⎛ a0 ⎞ ⎛ a0 ⎞ ⎤ = ⎢ − ⎥ K ⎢⎣⎜⎝ 1 − iKa0 ⎟⎠ ⎜⎝ 1 + iKa0 ⎟⎠ ⎥⎦ 8π ik fx a03 = 2 1 + a02 K 2

(

)

Inserting this in (4) we find:

M = 2

32π ne2  3a03k 2fx

L6 m 2ω (1+ a02 K 2 )

4

Hence, from (2) we have: 8ne2 a03k 2fx k f dW = 3 dΩ L π mω (1+ a02 K 2 )4

(6)

The differential cross section is: dσ 1 dW L3 dW = = dΩ j dΩ nc dΩ

(7)

8e2 a03k 2fx k f = π mcω (1+ a02 K 2 )4 Now,

(

)(

K 2 = k0 ẑ − k f i k0 ẑ − k f

)

= k02 + k 2f − 2k0 k f cosθ

(8)

where θ is the scattering angle, while from conservation of energy we have:

mc 2α 2 (9) 2m 2 where the second term on the right hand side is the hydrogen ground state binding energy. Now we assume that mc 2α 2 << ω << mc 2 . Then from (9) we have: ω = k0 c =

 2 k 2f

k0 c 

+

 2 k 2f 2m

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Hence,

k f

Also, from (9), k 2f a02 >>

⎡

kf =

vf

k f << k f 2mc 2c since the final electron is non-relativistic. Therefore (8) becomes: vf ⎛ ⎞ K 2  k 2f ⎜ 1 − cosθ ⎟ ⎝ ⎠ c k0 

m 2 c 2α 2 a02 = 1 ; hence 2 ⎤

(1 + a K )  ⎢1 + a k ⎛⎜⎝ 1 − c cosθ ⎞⎟⎠ ⎥  a k ⎛⎜⎝ 1 − c cosθ ⎞⎟⎠ 2 0

2

⎣

2 0

2 f

vf

⎦

2 0

2 f

vf

Therefore, dσ  c 3  25 α a0 dΩ mc ω

k 2fx k f v a08 k 8f (1 − f cosθ )4 c (10) 2  c 1 sin θ cos 2 φ 5 =2 α v mc ω a05 k 5f (1 − f cosθ )4 c To calculate the total cross-section we must integrate (10) over the solid angle. Here we can make the approximation: vf 1  1 + 4 cosθ 4 c vf ⎛ ⎞ ⎜⎝ 1 − c cosθ ⎟⎠ since v f / c << 1. Then,

σ  25 α

π  c 1 2π 2 cos φ d φ sin 3 θ dθ 5 5 ∫0 ∫ 0 mc ω a0 k f

27  c 1 = πα 3 mc ω a05 k 5f

(11)

However, we have already shown that: 1/2 1/2 ⎛ 2mc ⎞ ⎛ 2mc ω ⎞ kf  ⎜ k =⎜ ⎝  0 ⎟⎠ ⎝  c ⎟⎠ Inserting this in (11) and making use of a0 =

4π  2 we find after a few short steps of me2

algebra that:

⎡ e2 1 ⎤ 28 π σ α a02 ⎢ ⎥ 3 ⎣ 8π a0 ω ⎦

7/2

=

α e7 1 5/2 3/2 7/2 3( 2π ) a0 ( ω )

(12)

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18.6 a) In this problem, the geometry for classical scattering is shown in the following figure.

Fig.18.1aS Repulsive force. Force center at C.

Fig. 18.1bS Attractive force. Force center at C. In the figures above: P=projectile r0 = distance of closest approach C = force center θ = scattering angle φ = orbit angle b = impact parameter The Lagrangian is: m 2 L = T −V = r + r 2φ 2 − V (r) 2

(

)

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Since L does not contain φ , Lagrange’s equations give: d ∂L =0 dt ∂φ Hence,

mr 2φ = const ≡ J

Thus J φ = mr 2

and so: m⎛ 2 J2 ⎞ E = T + V = ⎜ r + 2 2 ⎟ + V (r) 2⎝ mr ⎠

(1)

Also,

dr  J dr φ= 2 dφ mr dφ Thus, (1) can be written: 2 J 2 ⎡ 1 ⎛ dr ⎞ 1⎤ E= ⎢ 4 ⎜ ⎟ + 2 ⎥+V 2m ⎢⎣ r ⎝ dφ ⎠ r ⎥⎦ so we have: r =

J 2 J 2 ⎛ dr ⎞ 2m(E − V ) − 2 = 4 ⎜ ⎟ r r ⎝ dφ ⎠

2

or:

∂φ = ∂r

J r 2 2m(E − V ) −

=−

J2 r2

∂ J2 2m(E − V ) − 2 ∂J r

Hence, ∂ J2 φ = −∫ 2m(E − V ) − 2 dr ' r0 ∂J r' At r = r0 we have: r

(2)

∂r r 2 J2 = 2m(E − V ) − 2 = 0 ∂φ J r Since in the repulsive (attractive) case, we have θ = π − 2φ∞ , (θ = 2φ∞ − π ) , (2) yields: Repulsive case:

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∂ 2m(E − V ) − J 2 /r '2 dr ' r0 ∂J

θ = π + 2∫ Attractive case:

∞

(3)

∂ (4) 2m(E − V ) − J 2 /r '2 dr ' r0 ∂J The impact parameter is b and the projectile has speed v0 at very large distances. Thus: (5) J = mv0b = p0b The number of classical particles scattered into solid angle dΩ = 2π sin θ dθ is proportional to the number of particles with impact parameters between b and b + db: dσ (θ ) 2π bdb b db (6) = =− dΩ −2π sin θ dθ sin θ dθ where we employ a minus sign because db/dθ is negative. From (5), JdJ bdb = 2 p0 Thus, classically: dσ JdJ (7) =− 2 dΩ p0 sin θ dθ −θ = π + 2 ∫

∞

b) Now we use the WKB approximation to calculate the scattering cross section according to quantum mechanics. We start with the following expression for the phase shift δ  of the 'th partial wave in the WKB approximation:

⎡ r r  (  + 1) 2 µV  (  + 1) ⎤ 2 δ  = lim ⎢ ∫ k 2 − − dr '− k − dr ' ⎥ ∫r0 ' r→∞ r '2 2 r '2 ⎢⎣ r0 ⎥⎦

(8)

1 ⎛ 1⎞ ( + 1) . When  >> 1, ( + 1) ≈ 2 +  + = ⎜  + ⎟ . Thus (8) can be 4 ⎝ 2⎠ k 2

where r0 ' = written:

2 2 ⎡ ⎤ 1⎞ 1⎞ ⎛ ⎛ ⎢ ⎥ + ⎟ + ⎟ ⎜ ⎜ r r ⎝ ⎝ 2 µV 2⎠ 2⎠ 2 2 ⎢ (9) δ  ≅ lim ∫ k − − 2 dr '− ∫ k − dr ' ⎥ 2 2 ⎥ r ' r→∞ ⎢ r0 0 r'  r' ⎢ ⎥ ⎢⎣ ⎥⎦  + 1/2 where to a good enough approximation we can write: r0 ' = from now on. k Let the second integral on the right hand side of this expression be called I2. Defining: 1 + 2 u= kr

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we have: 1⎞ 1 1 − u 2 du ⎛ I 2 = ⎜  + ⎟ ∫ +1/2 ⎝ 2 ⎠ kr u2 1 ⎞ π /2 ⎛ 2 = ⎜+ ⎟ ∫ +1/2 cot θ dθ ⎝ 2 ⎠ Arc sin kr

Now,

∫

π /2

cot 2 θ dθ = [ −θ − ctnθ ]

π /2

+1/2 Arc sin kr

→ Thus for large r, I 2 ≅ kr −

Arc sin

( +1/2 ) kr

kr π − for large r  + 1/2 2

(  + 1/2 ) π

2 and (9) becomes: 2 ⎡ ⎤ 1⎞ ⎛ ⎢ ⎥  + ⎜⎝ ⎟ r 2 µV 1⎞ π ⎥ 2⎠ ⎛ 2 ⎢ (10) δ  ≅ lim ∫ k − − 2 dr '− kr + ⎜  + ⎟ r→∞ ⎢ r0 ⎝ r '2  2⎠ 2 ⎥ ⎢ ⎥ ⎢⎣ ⎥⎦ c) The partial wave expansion of the scattering amplitude is: 1 ∞ f (θ ) = ( 2 + 1) e2iδ − 1 P ( cosθ ) ∑ 2ik = 0 Now consider the function: 1 ∞ F(θ ) = ∑ ( 2 + 1)P ( cosθ ) 2 = 0 Since the Legendre polynomials form a complete orthogonal set, any arbitrary function g(θ ) can be expanded in terms of them:

(

)

g(θ ) = ∑ A ' P ' ( cosθ ) '

where the A ' are expansion coefficients. Then, π π 1 I ≡ ∫ g(θ )F(θ )sin θ dθ = ∑ ∫ A ' P ' ( cosθ ) P (cosθ ) ( 2 + 1) sin θ dθ 0 2 , ' 0

=

2δ , ' 1 A ' (2 + 1) ∑ 2 , ' ( 2 + 1)

= ∑ A 

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Thus F(θ ) = δ (1 − cosθ ) which is infinite at θ = 0, but vanishes when θ ≠ 0 . Hence for θ ≠ 0 we can write: 1 ∞ f (θ ) = (11) ( 2 + 1)e2iδ P ( cosθ ) ∑ 2ik = 0 Next we consider a useful formula for P (cosθ ) when  >> 1 and θ ≠ 0. The Legendre differential equation is: d 2 P dP + cot θ  + ( + 1)P = 0 (12) 2 dθ dθ In (12) we make the substitution: (13) χ (θ ) = sin θ P (cosθ ) which yields: 2 ⎡⎛ 1⎞ 1 ⎤ (14) χ "+ ⎢⎜  + ⎟ + ⎥χ = 0 2⎠ 4 sin 2 θ ⎥⎦ ⎢⎣⎝ For large  the first term in brackets is much larger than the second if θ  >> 1, (π − θ )  >> 1. These conditions hold for all θ except for a narrow range of angles θ about 0 and π if  >>> 1 . In this case we can ignore the second term in brackets on the left side of (14), and that equation reduces to: 2 χ "+ (  + 1/2 ) χ = 0 which has the solution: χ = Asin ⎡⎣(  + 1/2 )θ + α ⎤⎦ where A, α are constants to be determined momentarily. From (13) we have: Asin ⎣⎡(  + 1/2 )θ + α ⎦⎤ P ( cosθ ) →  >>> 1, θ >>1,  (π − θ ) >> 1 (15) sin θ To determine A, α we consider (12) once again, this time for very small θ and  >> 1 : d 2 P 1 dP + + ( + 1/2)2 P = 0 2 dθ θ dθ The solution of this well known equation is: P = BJ 0 ⎡⎣(  + 1/2 )θ ⎤⎦ where J 0 is the Bessel function of order zero, and B is a constant. Now B = 1, since P (cos 0) = 1 and J 0 ( 0 ) = 1. Hence: P = J 0 ⎡⎣(  + 1/2 )θ ⎤⎦

In the limit where θ >> 1 (but θ << 1) we now use the asymptotic form of J 0 to determine A, α . We have: Lπ sin(x − ) π 2 jL ( x ) = J L +1/2 (x) → 2x x

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Thus

2 sin(x − Lπ /2) x→∞ πx In the present case, L = −1/2 and x = ( + 1/2)θ . Therefore: lim J L +1/2 (x) =

lim P ( cosθ ) =

θ →∞

2 sin ⎡⎣(  + 1/2 )θ + π /4 ⎤⎦ π ( + 1/2)θ

Therefore, (15) becomes: 2 P ( cosθ ) → sin ⎡⎣(  + 1/2 )θ + π /4 ⎤⎦ π sin θ We now substitute (16) into (11) and replace the sum by an integral to obtain: ∞ 1 f (θ ) = − 21/2 d ei [ 2δ  +(+1/2)θ + π / 4 ] − ei [ 2δ  −(+1/2)θ − π / 4 ] ∫ 0 2k 2π sin θ

{

}

(16)

(17)

The exponentials oscillate very rapidly as  varies, except near the extreme values of the exponents. These occur for:

∂ ∂δ 2δ  + ( + 1/2)θ + π /4 ] = 2  + θ = 0 [ ∂ ∂ ∂ ∂δ 2δ  − ( + 1/2)θ − π /4 ] = 2  − θ = 0 [ ∂ ∂ or, since J =  , ∂δ 2  = ±θ ∂J Now recall that δ  is given by equation (10). Therefore:

(18)

2 ⎧⎪ ∞ ∂  + 1/2 ) ∂δ  2 µV π ⎫⎪ ( 2 2 = 2 ⎨ ∫ k − − dr '+ ⎬ r ∂J r '2 2 2 ⎪ ⎪⎩ 0 ∂J ⎭ ∞ ∂ = 2∫ 2 µ (E − V ) − J 2 /r '2 dr '+ π = ±θ r0 ∂J

(19)

This is the same result for the scattering angle that we obtained with the classical theory (recall equations 3, 4). Now we want to compare the classical and quantum mechanical cross sections. For this purpose we recall (17):

f (θ ) = −

{

∞ 1 1/2 d ei [ 2δ  +(+1/2)θ + π / 4 ] − ei [ 2δ  −(+1/2)θ − π / 4 ] ∫ 0 k 2π sin θ

}

(20)

Suppose that in the first exponential, the exponent has an extreme value at  =  0 :

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∂δ  + θ = 0 for  =  0 ∂ We expand the factor in square brackets in the first exponent in (20) in a Taylor series, about  =  0 . We carry out this expansion to second order.

2

2δ  + (  + 1/2 )θ + π /4 ≅ 2δ  0 + (  0 + 1/2 )θ + π /4 +

1 ∂ 2δ  ( −  0 )2 2 2! ∂ = 

(21)

0

The second derivative in (21) is: 2 2 ∞ ∂ ∂ 2δ  J2 2 ∂ δ = = ∫ 2 µ (E − V ) − 2 dr ' r0 ∂J 2 ∂2 ∂J 2 r'

∴

∂ δ ∂2 = 2

0

⎡ ⎤ ⎢∂ ∞ ⎥ J ⎢ = − dr ' ⎥ 2 ⎢ ∂J ∫r0 2 ⎥ J r ' 2 µ (E − V ) − 2 ⎢ ⎥ r' ⎣ ⎦ =  0 = −

∂θ ≡q ∂J =  0

Thus, from (20) we have: ∞ 2 1 f (θ ) ≈ −  01/2 eiA (  0 )eiq(−  0 ) d ∫ k 2π sin θ 0 where

(22)

A (  0 ) = ei [ 2δ  0 +( 0 +1/2)θ + π / 4 ]

The integral in (22) can be written: ∞

∫  0

1/2 iA (  0 ) iq(−  0 )2 0

e

e

∞

d =  01/2 eiA (  0 ) ∫ eiq(−  0 ) d 2

0

∞

≅  01/2 eiA (  0 ) ∫ eiqx dx =  01/2 eiA (  0 ) −∞

2

π iq

Thus (22) becomes:

f (θ ) = e (

0 2qk 2 sin θ

=e(

−JdJ p sin θ dθ

i A (  0 ) − 3π / 4 )

i A (  0 ) − 3π / 4 )

2 0

Hence,

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J ∂J (23) p sin θ ∂θ which is the same result obtained in classical mechanics (recall equation 7).

Finally, to consider what happens when θ is very small and a large number of partial waves  up to a maximum  = 1 contribute, we once again employ the formula: P = J 0 ⎡⎣(  + 1/2 )θ ⎤⎦

σ (θ ) = f (θ ) = − 2

2 0

Then we can write: 1 ∞ f (θ ) = ∑ ( 2 + 1) e2iδ − 1 P ( cosθ ) 2ik = 0 1 1  2 e2iδ  − 1 J 0 ⎡⎣(  + 1/2 )θ ⎤⎦ d ∫ 0 2ik

(

(

)

)

In the integrand of the last integral, J 0 varies only slowly with  for small θ , but e2iδ  oscillates very rapidly. Thus in this last integrand we ignore the term in e2iδ  .

Consequently: i 1 i (24) f (θ )  ∫ J 0 [ θ ] d = 1 J1 ( 1θ ) 0 k kθ Now 1 = pa = ka where a is the radius of the scattering sphere. Thus we can write (24) as: ia f (θ )  J1 ( kaθ ) θ Therefore: dσ a2 2 = f (θ ) = 2 J12 ( kaθ ) (25) dΩ θ This can be compared with the result for Fraunhofer diffraction from a sphere of radius a, a problem that is solved in Jackson, JD: Classical Electrodynamics, 3rd Edition, Wiley, New York, 1998, Chapter 10. The result obtained there is the same as in (25).

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19.1A general solution to the Klein-Gordon equation is obtained by superposing plane wave solutions. There are 2 classes of solutions: those with momentum p and energy

E = + p 2 + m 2 , and those with momentum p and energy E = − p 2 + m 2 . Thus a general solution is: ψ ( r,t ) = ∫ A+ ( p ) exp ⎡⎣i ( pir − Et ) ⎤⎦ d 3 p + ∫ A− ( p ) exp ⎡⎣i ( pir + Et ) ⎤⎦ d 3 p (1) = ψ+ +ψ− where A± ( p ) are arbitrary functions of p. We want to evaluate the probability density function: i ρ= [ψ * ψ − ψψ *] 2mc 2 (2) i = [ψ * ψ − ψψ *] (in units where =c=1) 2m Now,

iψ ± = ± ∫ p 2 + m 2 A± ( p ) exp ⎡⎣i ( pir  Et ) ⎤⎦ d 3 p

(3)

However if the functions A± ( p ) vanish unless p << m , (3) is well approximated by:

iψ ± = ±m ∫ A± ( p ) exp ⎡⎣i ( pir  Et ) ⎤⎦ d 3 p = ±mψ ±

Thus,

iψ = m (ψ + − ψ − )

−iψ * = m (ψ + * −ψ − *)

and

i [ψ * ψ − ψψ *] 2m = ψ + *ψ + − ψ − *ψ −

ρ=

In the non-relativistic limit we not only assume that p << m , but we also ignore the negative energy solutions. Thus we obtain:

ρ = ψ + *ψ + 19.2 For  = 0 the effective one-dimensional radial Klein-Gordon equation for the function χ = rR is:

χ "+ ⎡⎣(E + V0 )2 − m 2 ⎤⎦ χ = 0

r<a

(1)

χ "− (m 2 − E 2 ) χ = 0

r>a

(2)

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Let k = (E + V0 )2 − m 2 , κ = m 2 − E 2 . Then (1), (2) become:

χ i "+ k 2 χ i = 0 χ e "− κ 2 χ e = 0

(3) (4)

where i, e denote internal, external respectively. These equations have the solutions:

χ i = Asin kr χ e = Be−κ r

(5) (6)

where A,B are constants. The wave function and its derivative must be continuous at r = a . Thus we have the conditions:

Asin ka = Be−κ a kA cos ka = −κ Be−κ a Division of (7) by (8) yields: k tan ka = − κ The first bound state just appears when: π = ka 2

(7) (8)

(9)

= a (E + V0 )2 − m 2 and when E = m. Thus we have π2 (m + V0 )2 = 2 + m 2 4a or:

π2 (10) 4a 2 Since V0 > 0, we must choose the + sign in front of the second term on the right hand side of (22.10). Thus the first bound state just appears when: ⎡ ⎤ π2 V0 = m ⎢ 1 + − 1 ⎥ (11) 4m 2 a 2 ⎢⎣ ⎥⎦ As we increase the depth of the well, (increase V0 ), the energy of the first bound state decreases ( E1 falls below m ), and eventually a second, then a third bound state appear at E = m, and so forth. Finally the energy of the first bound state reaches –m. Here, V0 = −m ± m 2 +

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k = (V0 − m)2 − m 2 and,

κ =0 Thus (9) requires

(V0 − m)2 − m 2 =

π 2a

(12)

Equation (12) implies:

π2 V0 = m ± m + 2 4a Once again we must choose the + sign, which yields the well depth condition for the first bound state energy to be E = -m: 2

⎡ ⎤ π2 V0 = m ⎢ 1 + + 1 ⎥ (13) 4m 2 a 2 ⎢⎣ ⎥⎦ We cannot have a bound state with E < −m because in that case κ would be imaginary.

19.3 In natural units, the Klein Gordon equation for a particle in one spatial dimension, with definite energy E and potential energy V is: 2 ⎡( E − V )2 − m 2 ⎤ψ + ∂ ψ2 = 0 ⎣ ⎦ ∂x Consider the potential step:

V = 0, V = V0 ,

(1)

x<0 x>0

a) E − m > V0 > 0 Here, k 2 ≡ E 2 − m2 > 0 q 2 ≡ (E − V0 )2 − m 2 > 0

Thus: x<0

ψ "+ k 2ψ = 0

x>0

ψ " + q 2ψ = 0

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As in the case of the Schroedinger equation we equate the functions and equate their derivatives at the boundary. Writing: x<0

ψ = eikx + Be−ikx

x>0

ψ = Ceiqx

we have:

2k k−q , B= q+k k+q The reflection and transmission coefficients are: C=

R=

( k − q )2 , ( k + q )2

T=

4kq ( k + q )2

b) E − m < V0 < E + m Here, (E − V0 )2 < m 2 so we have: x<0

ψ "+ k 2ψ = 0

x>0

ψ " − q 2ψ = 0

x<0

ψ = eikx + B'e−ikx

x>0

ψ = C 'e− qx

Thus:

and:

2k k − iq , B' = k + iq k + iq Thus, as in the non-relativistic case, the transmission coefficient is zero and the wave function is exponentially damped for x > 0. Also, note that since: C' =

i V (ψ *ψ − ψψ *) − ψ *ψ 2m m E −V = ψ *ψ m the probability density is negative for E < V0 .

ρ=

c) E = m < V0 Here (E − V0 )2 > m 2 once again, so the wave function is oscillatory in the region x > 0 , (but the probability density is negative there).

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20.1 For two spatial dimensions, and with units where  = c = 1 , the Dirac equation is: ∂tψ + (α1∂1 + α 2 ∂ 2 )ψ + imβψ = 0 (1 with α iα j + α jα i = 2δ ij I

α i β + βα i = 0

(2)

β =I and i,j =1,2. It is clear by inspection that relations (1) are satisfied by the 2x2 matrices: (3) α1 = σ 1 , α 2 = σ 2 , β = σ 3 (or permutations thereof), and thus that ψ is a two-component spinor. However, as we shall see in what follows, these two components do not refer to spin. (In two spatial dimensions the concept of spin angular momentum does not exist). Instead, the two components are necessary to accommodate positive and negative energies.

Consider a state of definite 2-momentum p and energy E. Defining ε = E we have: (4) ψ = χ exp ( ipir − iε t ) 2

⎛ χ1 ⎞ where χ = ⎜ ⎟ is a two-component spinor. Assuming E > 0, and inserting (3) and (4) in ⎝ χ2 ⎠ (1) we obtain: p1 − ip2 ⎞ ⎛ χ1 ⎞ ⎛ χ1 ⎞ ⎛ m E⎜ ⎟ = ⎜ −m ⎟⎠ ⎜⎝ χ 2 ⎟⎠ ⎝ χ 2 ⎠ ⎝ p1 + ip2 which gives: ( E − m ) χ1 = ( p1 − ip2 ) χ 2 (5)

( E + m ) χ 2 = ( p1 + ip2 ) χ1

Equation (6) yields:

χ2 =

p1 + ip2 χ1 E+m

(6) (7)

Thus,

⎛ 1 ⎞ χ = const ⎜ p1 + ip2 ⎟ E>0 ⎜ ⎟ ⎝ E+m ⎠ If E < 0, we have a solution of the form: ⎛ p1 − ip2 ⎞ − χ = const ⎜ ε + m ⎟ E<0 ⎜ ⎟ 1 ⎝ ⎠ If the particle is at rest we have p1 = p2 = 0 , and (5, 6) reduce to: ( E − m ) χ1 = 0, ( E + m ) χ 2 = 0

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Thus we have the two independent solutions: ⎛ 1⎞ ψ + = ⎜ ⎟ e−imt E=m ⎝ 0⎠

⎛ 0⎞ ψ − = ⎜ ⎟ e+imt ⎝ 1⎠

E = −m

20.2 The 4-component spinor described in this problem can be expressed as a superposition of the 4 linearly independent free-particle spinors with p = pz ẑ :

⎛ a⎞ ⎛ 1⎞ ⎛ 0⎞ ⎛ −q ⎞ ⎛ 0⎞ ⎜ b⎟ ⎜ 0⎟ ⎜ 1⎟ ⎜ 0⎟ ⎜ q⎟ ⎜ ⎟ = A⎜ ⎟ + B⎜ ⎟ + C ⎜ ⎟ + D⎜ ⎟ ⎜ c⎟ ⎜ q⎟ ⎜ 0⎟ ⎜ 1⎟ ⎜ 0⎟ ⎜⎝ d ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ −q ⎟⎠ ⎜⎝ 0 ⎟⎠ ⎜⎝ 1 ⎟⎠

(1)

pz , ε = E , and A,B,C,D are coefficients to be determined. Equation (1) is ε+m really a set of 4 coupled linear equations that are readily solved to yield: a + qc b − qd c − qa d + qb A= , B= ,C= , D= 2 2 2 1+ q 1+ q 1+ q 1 + q2 The desired probabilities λ are then as follows:

where q =

E > 0, spin up:

λ = A (1+ q 2 ) = a + qc / (1+ q 2 )

E > 0, spin down:

λ = B (1+ q 2 ) = b − qd / (1+ q 2 )

E < 0, spin up:

λ = C (1+ q 2 ) = c − qa / (1+ q 2 )

E < 0, spin down:

λ = D (1+ q 2 ) = d + qb / (1+ q 2 )

2

2

2

2

2

2 2

2

Given that a + b + c + d = 1 , the sum of these 4 probabilities is equal to unity. b) Any normalized positive-energy plane-wave free-particle solution to the Dirac equation has the form: 2

2

2

2

χ ⎞ ⎛ E+m 1 ⎜ σ i p ⎟ eipi r −iEt ψ= 2m χ⎟ V⎜ ⎝E+m ⎠ where χ is a normalized 2-component spinor: χ † χ = 1 , and we use units with  = c = 1.

If ψ is an eigenstate of the helicity operator Σ i p̂ with eigenvalue λ we have:

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Σ i p̂ψ = λψ Applying the operator Σ i p̂ on the left to both sides of this equation and transposing, we have: λΣ i p̂ψ = Σ i p̂Σ i p̂ψ ∴ λ 2ψ = p̂i p̂ψ = ψ which implies that λ = ±1 . Since: 0 ⎞ ⎛ σ i p̂ Σ i p̂ = ⎜ ⎝ 0 σ i p̂ ⎟⎠ the eigenvalue equation we must solve is: σ i p̂ χ ± = ± χ ± which can be written: p̂x − ip̂y ⎞ ⎛ a± ⎞ ⎛ p̂z ⎛ a± ⎞ = ±⎜ ⎟ ⎜ ⎟ ⎜⎝ p̂ + ip̂ ⎟ − p̂z ⎠ ⎝ b± ⎠ ⎝ b± ⎠ x y ⎛ a± ⎞ where χ ± = ⎜ ⎟ . It is convenient to make the substitutions: ⎝ b± ⎠ p̂x ± ip̂y = sin θ e±iφ , p̂z = cosθ Then we have:

a± cosθ + b± sin θ e−iφ = ±a± a± sin θ eiφ − b± cosθ = ±b± It is very easy to verify that solutions to these equations satisfying the normalization 2 2 condition a± + b± = 1 are:

θ ⎞ ⎛ cos ⎛ a+ ⎞ ⎜ 2 ⎟ χ+ = ⎜ ⎟ = ⎜ ⎟, ⎝ b+ ⎠ ⎜ θ iφ ⎟ sin e ⎝ 2 ⎠

θ ⎞ ⎛ sin ⎛ a− ⎞ ⎜ 2 ⎟ χ− = ⎜ ⎟ = ⎜ θ iφ ⎟⎟ ⎝ b− ⎠ ⎜ − cos e ⎝ 2 ⎠

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Obviously

these

are

spinors

with

polarization

axis

oriented

along

±p .

ψ ± Σ i p̂ ψ ± = ± ψ ± ψ ± ⎛ ⎞ χ † ⎞ ⎛ ⎛ σip ⎞ ⎡ 1 −i( pir−Et ) i( pir−Et ) 3 ⎤ E + m † ⎜ ⎟ = ±⎢ ∫e e d r⎥ χ⎟ ⎟ σ i p ⎜ χ ⎜ ⎜ ⎟ ⎝ E + m ⎠ ⎟⎠ ⎣V ⎦ 2m ⎜⎝ ⎜⎝ E + m χ ⎟⎠ = ±1i =±

E + m †⎛ p2 ⎞ i χ ⎜ 1+ χ 2m ⎝ (E + m)2 ⎟⎠

E + m 2E E =± 2m E + m m

Thus expectation values of Σ i p̂ are:

Σ i p̂ =

ψ ± Σ i p̂ ψ ± = ±1 ψ± ψ±

Also,

ψ ± γ Σ i p̂ ψ ± 0

⎛ ⎞ χ † E+m⎛ † ⎛ σ i p ⎞ ⎞ 0⎜ ⎟ =± χ⎟ ⎟ γ ⎜ χ ⎜ σip ⎜ ⎜ ⎟ ⎝ ⎠ 2m ⎝ E+m χ ⎟ ⎠ ⎜ ⎝ E + m ⎟⎠

E + m †⎛ p2 ⎞ =± i χ ⎜ 1− χ 2m ⎝ (E + m)2 ⎟⎠ E + m 2m = ±1 2m E + m γ 0 Σ i p̂ m 0 Thus, γ Σ i p̂ = =± .

E ψψ =±

c)

χ ⎞ ⎛ σxχ ⎞ ⎛ ⎜ ⎟ Σ x pσ z ⎟ = ⎜ −ip ⎜ σ yχ⎟ χ⎟ ⎜ ⎠ ⎝E+m ⎠ ⎝E+m Thus,

χ ⎞ ⎛ † ⎛ σz p ⎞ †⎞ ⎛ ⎡ † ⎤ p2 † ⎜ ⎟ χ χ Σ = χ σ χ − i ⎢ x ⎜⎝ ⎟⎠ ⎟ x pσ z ⎜ 2 χ σ zσ y χ ⎥ E+m χ⎟ ⎢ (E + m) ⎝ ⎠ ⎜⎝ ⎥⎦ E+m ⎠ ⎣ ⎡ ⎤ p2 † = ⎢ χ †σ x χ − 2 χ σxχ ⎥ (E + m) ⎢⎣ ⎥⎦

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For χ we must obviously choose: 1 ⎛ 1⎞ χ= 2 ⎜⎝ 1⎟⎠ which is the eigenvector of σ x with eigenvalue +1. Then, χ ⎞ ⎛ † ⎛ σz p ⎞ †⎞ ⎛ p2 2m ⎜ ⎟ = 1− = ⎜ χ ⎜⎝ E + m χ ⎟⎠ ⎟ Σ x pσ z 2 (E + m) E+m χ⎟ ⎝ ⎠ ⎜⎝ E+m ⎠ Including the normalization factor obtain: Σ x = 1 ; hence Σ x =

E+m for the original 4 component spinor, we 2m

m .

E

However,

γ 0Σ x =

E+m⎛ p 2 ⎞ E + m 2E E 1+ = = 2⎟ ⎜ 2m ⎝ (E + m) ⎠ 2m E + m m

Thus γ 0 Σ x = 1 .

20.3a) In the standard representation,

⎛ 0 0 0 ⎛ 0 σ1 ⎞ ⎜ 0 0 1 γ1=⎜ ⎟ =⎜ ⎜⎝ −σ 1 0 ⎟⎠ ⎜ 0 −1 0 ⎜⎝ −1 0 0

1 ⎞ 0 ⎟ ⎟ = −γ1 0 ⎟ 0 ⎟⎠

⎛ 0 0 0 −i ⎞ ⎛ 0 σ2 ⎞ ⎜ 0 0 i 0 ⎟ γ2 =⎜ ⎟ = γ2 ⎟ =⎜ ⎜⎝ −σ 2 0 ⎟⎠ ⎜ 0 i 0 0 ⎟ ⎜⎝ −i 0 0 0 ⎟⎠ ⎛ 0 0 ⎛ 0 σ3 ⎞ ⎜ 0 0 γ3=⎜ ⎟ =⎜ ⎜⎝ −σ 3 0 ⎟⎠ ⎜ −1 0 ⎜⎝ 0 −1 ⎛ 1 ⎜ 0 γ0 =⎜ ⎜ 0 ⎜⎝ 0

1 0 ⎞ 0 −1 ⎟ ⎟ = −γ3 0 0 ⎟ 0 0 ⎟⎠

0 0 0 ⎞ 1 0 0 ⎟ 0 ⎟ =γ 0 −1 0 ⎟ 0 0 −1 ⎟⎠

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We require:

Sγ 1 = −γ 1S Sγ 2 = γ 2 S Sγ 3 = −γ 3S Sγ 0 = γ 0 S S = γ 1γ 3 satisfies these requirements. Also, we have S † = γ 3†γ 1† = γ 3γ 1 = S −1 = −S . Note that:

⎛ 0 σ1 ⎞ ⎛ 0 σ 3 ⎞ ⎛ σ 2 0 ⎞ S = γ 1γ 3 = ⎜ ⎟⎜ ⎟ = i⎜ ⎟ = iΣ 2 ⎜⎝ −σ 1 0 ⎟⎠ ⎜⎝ −σ 3 0 ⎟⎠ ⎜⎝ 0 σ 2 ⎟⎠ 20.4 a) In the Weyl representation the wave function is: ⎛ φ1 ⎞ 1 ⎛ψ A + ψ B ⎞ ψ'=⎜ ⎟ = ⎝ φ2 ⎠ 2 ⎜⎝ ψ A − ψ B ⎟⎠

(1)

ψ ' = Sψ

(2)

Thus,

where ⎛ψ A ⎞ ψ =⎜ ⎟ ⎝ψ B ⎠

Pauli-Dirac representation

and

S=

1 ⎛ I 2 ⎜⎝ I

1 I ⎞ = γ 0 +γ 5) ( ⎟ −I ⎠ 2

(3)

Note that S † = S −1 = S . Then: 1 γ ' = Sγ S −1 = (γ 0 + γ 5 ) γ (γ 0 + γ 5 ) 2

⎛ 0 2 1 = − γ (γ 0 + γ 5 ) = −γ = ⎜ 2 ⎝ σ

−σ ⎞ 0 ⎟⎠

(4)

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1 0 γ + γ 5 )γ 0 (γ 0 + γ 5 ) ( 2 1 = γ 0 (γ 0 − γ 5 ) (γ 0 + γ 5 ) 2 1 = γ 0 ( I + γ 0γ 5 − γ 5γ 0 − I ) 2 ⎛ 0 I ⎞ =γ 5 =⎜ ⎝ I 0 ⎟⎠

(5)

1 0 γ + γ 5 )γ 5 (γ 0 + γ 5 ) ( 2 1 = − γ 5 (γ 0 − γ 5 ) (γ 0 + γ 5 ) 2 1 = − γ 5 ( I + γ 0γ 5 − γ 5γ 0 − I ) 2 ⎛ I 0 ⎞ =γ 0 =⎜ ⎝ 0 −I ⎟⎠

(6)

γ 0'=

and:

γ 5'=

b) The free-particle Dirac equation in the standard representation is: Eψ = α i pψ + mγ 0ψ = γ 0γ i pψ + mγ 0ψ

Applying the matrix S on the left to both sides of this equation we obtain:

E ( Sψ ) = ( Sγ 0 S −1 ) ( Sγ S −1 ) i p ( Sψ ) + m ( Sγ 0 S −1 ) ( Sψ ) which yields: Eψ ' = γ 5 ( −γ i p )ψ '+ mγ 5ψ ' ⎛ 0 I ⎞⎛ 0 ⎛ 0 I ⎞ −σ i p ⎞ =⎜ ψ' ⎜ ⎟ ψ '+ m ⎜ ⎟ 0 ⎝ I 0 ⎠⎝ σ i p ⎝ I 0 ⎟⎠ ⎠ ⎛ σip m ⎞ =⎜ ⎟ψ ' −σ i p ⎠ ⎝ m Thus we have:

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( E − σ i p )φ1 = mφ2 ( E + σ i p )φ2 = mφ1

(7)

which are sometimes called the free-particle Weyl equations. In the limit m → 0 , we have E = p and the two equations (7), which are now decoupled from one-another, read:

σip φ1 = +φ1 p σip φ2 = −φ2 p Thus φ1 , φ2 are both two-component eigenstates of helicity, with helicity eigenvalues ±1 respectively. In the period 1956-1958, immediately following the discovery by T.D. Lee and C. N. Yang of parity non-conservation in weak interactions, it was determined from various experiments that a neutrino emitted or absorbed in a weak decay or scattering process is “left-handed” (has negative helicity h = -1), while an anti-neutrino is “righthanded” (has positive helicity h = +1). Also, until 1998, the possibility existed that neutrinos and anti-neutrinos have zero rest mass. (Since 1998 convincing evidence exists for neutrino oscillations, which implies that the three neutrino mass eigenstates have mass greater than zero). Thus, during the period 1958-1998, it appeared possible that ν was described by a Weyl 2-component spinor φ1 , and ν was described by a Weyl 2component spinor φ2 .

20.5 a)We first show that U is unitary.

2 ε + m + βα i p ) ( ε + m − βα i p ) ( ε + m ) + p 2 ( UU= = 2ε ( ε + m ) 2ε ( ε + m ) 2ε ( ε + m ) (ε + m + ε − m ) = 1 = †

Next,

2ε

(ε + m − βα i p )(ε + m − βα i p ) 2ε ( ε + m ) (α i p + β m )( m − βα i p ) = βε =

H ' = U † (α i p + β m )U = (α i p + β m )

ε ⎡ ∂ ⎤ 1 p b) xi = [ xi , βε ] = ⎢ , βε ⎥ = β i i ε ⎣ ∂ pi ⎦ Chapter 21

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21.1 We choose natural units (  = me = c = 1 ) and assume that the final electron state is a negative energy plane wave. The transition probability per unit time is given by the Golden rule formula: 3 3 3 3 2 L d k L d kf (1) dW = 2π H ' fi δ (Ei − E f ) ( 2π )3 ( 2π )3 We use the following notation: k = photon momentum k = ω = photon energy k f = final electron momentum Ee = − k 2f + 1 = final electron energy

ε = Ee Ei = 1− α 2 ≈ 1 E f = k − k 2f + 1 = k − ε

Now,

H ' = eα i A

(2)

⎛ 0 σ⎞ where A is the radiation field vector potential, α = ⎜ , and for spontaneous ⎝ σ 0 ⎟⎠ emission of a photon, the relevant term in A is: 1 1 −ikir A → 3/2 ε̂ e (3) L 2k where ε̂ is the photon polarization unit vector. The final electron wave function is the negative-energy plane wave: 1 m ik ir (4) ψ f = 3/2 ufe f L ε where: ⎛ −σ i k f ⎞ ⎛ −σ i k f ⎞ χf ⎟ χf ⎟ ε+m⎜ ε +1 ⎜ (5) uf = ε+m ε +1 ⎟= ⎟ 2m ⎜ 2 ⎜ χf χf ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ where χ f is a two-component spinor. Thus:

H ' fi =

σ i kf e 1 ε + 1 −i( k+k f )ir ⎛ † e ⎜ −χ f 3 ∫ L 2k 2ε E +1 ⎝

⎞⎛ 0 σ i ε̂ ⎞ ⎛ ψ A ⎞ 3 χ †f ⎟ ⎜ ⎟ d r (6) ⎟⎜ 0 ⎠ ⎜⎝ ψ B ⎟⎠ ⎠ ⎝ σ i ε̂

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⎛ψ A ⎞ where the hydrogen ground state wave function is ⎜ ⎟ . Now, ψ B << ψ A so in what ⎝ψ B ⎠ follows, we neglect ψ B . Also, to an adequate approximation:

ψ A  (π a )

3 −1/2 0

e

− r/a0

α 3/2 −α r χ i = 1/2 e χ i π

Here we have used a0 = α −1 in the present system of units, where e = 4πα . Thus, defining q = k + k f , we rewrite (6) as follows:

α 2 ( ε + 1) † χ f σ i ε̂χ i ∫ e−iqir e−α r d 3r L3 kε The integral on the right hand side is evaluated in the usual way as follows:

(

H ' fi 

∫e

−iqir − α r

e

)

(7)

π

∞

d 3r = 2π ∫ e−α r r 2 dr ∫ e−iqr cosθ sin θ dθ 0

0

∞

= 2π ∫ e−α r r 2 0

e − e−iqr dr iqr iqr

=

2π ∞ r ⎡⎣ e−(α −iq )r − e−(α +iq )r ⎤⎦dr ∫ 0 iq

=

⎤ 2π ⎡ 1 1 ⎢ 2 − 2 ⎥ iq ⎢⎣ (α − iq ) (α + iq ) ⎥⎦

=

8πα

(α + q ) 2

2 2

Thus (7) becomes:

8πα 3 ε + 1 1 † H ' fi  3 2 χ f σ i ε̂χ i 2 2 L kε (α + q )

(

)

(8)

Now, q 2 = k 2 + k 2f + 2kk f cosΘ where Θ is the angle between k and kf. Employing this in (8) we find that (1) becomes: 2

2α 6 1+ ε χ f σ i ε̂χ i dΓ  3 ( k 2dk ) dΩγ k f ε dε dΩ f δ (1+ E − k) π kε (α 2 + q 2 )4 †

(

)

2

χ †f σ i ε̂χ i 2α 6 = 3 (1+ ε ) 4 ( kdk ) dΩγ k f d ε dΩ f δ (1+ E − k) π α 2 + k 2 + k 2f + 2kk f cosΘ

(

)

(

)

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We now integrate over k and employ the delta function to obtain:

(

)

(1+ ε ) k f dε 2 2α 6 dΓ = 3 χ †f σ i ε̂χ i dΩγ dΩ f 4 π α 2 + (1+ ε )2 + k 2f + 2(1+ ε )k f cosΘ 2

(

)

(9)

We next integrate over the electron solid angle, keeping the direction of k fixed. Here, it is necessary to evaluate the integral: π sin ΘdΘ (10) I = 2π ∫ 0 ( A + B cosΘ )4 where A = α 2 + (1+ ε )2 + k 2f ≈ (1+ ε )2 + k 2f

B = 2(1+ ε )k f Thus

du 2π A+B dx = 4 −1 ( A + Bu ) B ∫A−B x 4

I = 2π ∫

1

=

⎤ 2π ⎡ 1 1 ⎢ 3 − 3⎥ 3B ⎢⎣ ( A − B ) ( A + B ) ⎥⎦

=

4π 3A 2 + B 2 3 ( A 2 − B 2 )3

Now,

A 2 − B 2 = k 4f + (1+ ε )4 + 2k 2f (1+ ε )2 − 4k 2f (1+ ε )2

(

= k 2f − (1+ ε )2

)

2

= 4 (1+ ε ) and simple algebraic steps yield: 3A 2 + B 2 = 4 4 ε 4 + 8ε 3 + 3ε 2 − 2ε − 1 2

(

)

Therefore, we obtain: 4 3 2 π 4 ε + 8ε + 3ε − 2ε − 1 I= 12 (1+ ε )6 Hence, 4 ε 4 + 8ε 3 + 3ε 2 − 2ε − 1 2 α6 dΓ = k dε χ †f σ i ε̂χ i dΩγ 4 2 f 6π (1+ ε ) Now evaluate the factor:

(

(

)

)

(11)

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(

)

*

χ †f σ i ε̂χ i = χ †f σ i ε̂χ i χ †f σ i ε̂χ i = χ i†σ i ε̂χ f χ †f σ i ε̂χ i

(

) (σ i ε̂ ) ( χ ) . = ( χ χ ) (σ i ε̂ ) ( χ χ ) (σ i ε̂ ) = ( χ i† )α (σ i ε̂ )αβ χ f χ †f i

† i δα

αβ

f

i δ

γδ

βγ

† f βγ

γδ

= tr ⎡⎣ χ i χ σ i ε̂χ f χ σ i ε̂ ⎤⎦ ≡ S † i

† f

Summing over final spins and averaging over initial spins we have: ∑ χ f χ †f = I

1 1 χ i χ i† = I ∑ 2 2 Thus, S=

1 1 tr [σ iε̂σ iε̂ ] = tr(I )ε̂ iε̂ = 1 2 2

Hence, the factor

χ †f σ iε̂χ i

2

is replaced by unity. Now integrate dΓ in (11) over the

photon solid angle to obtain: 4E 4 + 8E 3 + 3E 2 − 2E − 1 2α 6 dΓ = kf dE 3π (1 + E )4 To complete the calculation we integrate dΓ numerically over ε from 1 to 2: m c2 Γ  .31α 6 → .31α 6 e = 3.64i10 7 s-1 (12)  This shows that if unoccupied electron negative energy states really existed, no atoms would be stable.

(

)

21.2 In this problem it is advantageous to use the second-order equation (21.7). For an electron, q = −e , and in this example, the electric field is E = 0 . Thus that equation becomes: ⎡ m 2 + ( ∂ν − ieAν ) ( ∂ν − ieAν ) + e Σ i B ⎤ψ = 0 ⎣ ⎦ We choose B = B0 ẑ and Α =

B0 ( xŷ − yx̂ ) . Thus we obtain: 2

(

)

2 2 m 2ψ + ⎡( px − eB0 y/2 ) + py + eB0 x/2 + pz2 ⎤ ψ + eΣ z B0ψ = E 2ψ ⎣ ⎦

(1)

Choose any representation where Σ z is diagonal (e.g. the standard representation). Since its eigenvalues are ±1, we have:

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(

)

(

)

⎡( p − eB y/2 )2 + p + eB x/2 2 + p 2 ⎤ ψ = E 2 − m 2  eB ψ 0 y 0 z 0 ⎣ x ⎦ Define K ± =

(

(2)

)

1 E 2 − m 2  eB0 . Then, 2m

K ±ψ =

(

)

2 1 ⎡ ( px − eB0 y/2 )2 + py + eB0 x/2 + pz2 ⎤⎦ψ ⎣ 2m

(3)

Obviously if we find the eigenvalues of K±, we determine the eigenvalues E2 and thus the eigenvalues E. Now (3) is the same eigenvalue equation that was encountered in nonrelativistic quantum mechanics: see for example problem 6.12. We do not repeat the solution of that problem here, but just summarize the main result. Defining the operator: Q± = K ± − pz2 /2m

we find that the eigenvalues of Q± are ( n + 1/2 )ω , where ω = eB0 /m and n = 0,1, 2,... .

Hence, 1 E 2 = m 2 + pz2 ± eB0 + 2eB0 (n + ) 2 2 2 2 +: E = m + pz + 2eB0 (n + 1)

−:

E 2 = m 2 + pz2 + 2eB0 n

(4) (5)

Exhibiting ,c explicitly we have:

+:

E 2 = m 2 c 4 + pz2 c 2 + 2ecB0 (n + 1)

−:

E 2 = m 2 c 4 + pz2 c 2 + 2ecB0 n

(6)

From (5) or (6) we see that for n=1,2,3,… each eigenvalue E2 is doubly degenerate.

However the eigenvalue E2 for n=0 is not degenerate.

21.3a) Let the zeroth order neutron wave function be: ⎛ ⎞ χ ε+m⎜ ⎟ ei( pir−Et ) ψ= σip ⎜ ⎟ 2ε ⎜⎝ E + m χ ⎟⎠ The perturbation Hamiltonian for interaction of the neutron magnetic moment with the magnetic field B is: ⎛ σ iB ⎞ 0 H ' = − µγ 0 Σ i B= − µ ⎜ −σ i B ⎟⎠ ⎝ 0 213

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The energy shift to first order in this perturbation is: ψ H 'ψ ΔE = = ψ H 'ψ ψψ

ε +m⎛ † = −µ χ 2ε ⎜⎝ = −µ

⎛ ⎞ χ ⎞ ⎛ ⎞ σ i p σ i B 0 ⎜ ⎟ χ† σip ⎟ E + m ⎟⎠ ⎜⎝ 0 −σ i B ⎟⎠ ⎜ χ ⎜⎝ E + m ⎟⎠

ε + m †⎛ σ i pσ i Bσ i p ⎞ χ ⎜σ i B − χ 2ε (E + m )2 ⎟⎠ ⎝ (1)

Taking into account the fact that p, B are ordinary vectors, not operators, we have:

σ i pσ iBσ i p = ( piB + iσ i p × B ) σ i p

= σ i ppiB + ip × Bi p + σ i p × ( p × B ) However, p × Bi p = 0 . Thus, σ i pσ iBσ i p = σ i ppiB + σ i p × ( p × B )

= σ i ppiB + σ i( ppiB − Bpi p ) = 2σ i ppiB − p 2σ iB

Thus (1) becomes:

ΔE = − µ

ε + m † ⎡⎛ p2 ⎞ 2σ i pp i B ⎤ χ ⎢⎜ 1+ σ i B − ⎥χ 2⎟ 2ε (ε + m )2 ⎥⎦ ⎢⎣⎝ ( ε + m ) ⎠

⎡ σ i pp i B ⎤ = − µχ † ⎢σ i B − ⎥χ ε (ε + m ) ⎦ ⎣

(2)

Substituting ε = mγ , p = mvγ , µ = µσ in (2), we obtain:

⎡⎛ ⎞ ⎤ γ (3) ΔE = − χ † ⎢⎜ µ − µ ivv ⎟ iB ⎥ χ γ +1 ⎠ ⎦ ⎣⎝ In (3) the quantity in (…) is the Lorentz-contracted magnetic dipole moment. The magnetic dipole moment in the neutron rest frame is µ ; this quantity has dimensions of electric charge x length. Under the Lorentz boost that we make to go to the frame where the neutron is moving, the charge remains invariant but the length undergoes a Lorentz contraction.

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b. Our goal in this part of the problem is to show how to get from the equation:

1 σ i( p + i µnE ) σ i( p − i µnE )ψ A 2mn to the Schroedinger equation with the “Foldy potential”: 1 µe − ∇ 2ψ A + n δ 3 ( r )ψ A = Wψ A 2mn 2mn Wψ A =

(4)

(5)

In (4), employing units where  = c = 1 , we start with:

σ i( p + i µnE ) σ i( p − i µnE )ψ A = ( p + i µnE )i( p − i µnE )ψ A

+ iσ i ⎡⎣( p + i µnE ) × ( p − i µnE ) ⎤⎦ ψ A

(6)

Consider the first term on the right hand side of (6). It is:

( p + iµnE )i( p − iµnE )ψ A = p 2ψ A + iµn (Ei p − piE )ψ A + µn2E 2ψ A = −∇ 2ψ A − µn ( ∇iE )ψ A + µn2E 2ψ A

(7)

In what follows, we ignore the last term on the right hand side of (7), since it is much smaller than the other two terms. This is justified at the end of this solution. The second term on the right side of (6) vanishes when we average over the spins of a sample of initially unpolarized neutrons. Thus we obtain: p2 µ (9) Wψ A = ψ A − n ( ∇iE )ψ A 2mn 2mn Now, ∇ i E = ρ = −eδ 3 ( r ) . Thus, we obtain (5).

Finally, we justify neglect of the third term on the right hand side of (7). If this term had µn2 e2 been included, it would have yielded the following additional portion ΔV = 32π 2 mn r 4 eµ to the potential in (5). We now compare the contributions of ΔV and of V = n δ 3 ( r ) to 2mn the scattering amplitude in the first Born approximation. The scattering amplitude due to V (in units with  = c = 1 ) is: µ eµ fV = − eiqir n δ 3 ( r ) d 3r ∫ 2π 2mn

eµn 4 π mn where µ is the reduced mass of the neutron-electron system and q is the momentum transfer. The scattering amplitude due to ΔV is: = −µ

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fΔV = −

µe2 µn2 1 eiqir 4 d 3r 3 ∫ 64π mn r

(10)

Thus:

µe2 µn2 d 3r µe2 µn2 (11) = 64π 3mn ∫R r 4 16π 2 mn R where R is the neutron radius. (We cut off the integral at the neutron radius, since the neutron magnetic moment is not a point dipole but is spatially distributed). The ratio of the magnitudes of the two scattering amplitudes is: fΔV e µn ≤ fV 4π R e e α In natural units, R ≈ 10 −13 cm ≈ , while µn = 1.91 . Thus: ≈ 2 2mn c mn fΔV ≤

fΔV e µn 2e2 1 2 2 ≤ ≈ = = ≈ 10 −3 fV 4π R 4πα mn mn 1836

21.4 a) To discuss the numbers of nodes of F, G we start with the coupled radial equations: ∂G κ Zα ⎞ ⎛ + G = ⎜m+ E + ⎟F ⎝ ∂r r r ⎠ (1) ∂F κ Zα − F = (m − E − )G ∂r r r For very large r these equations become: ∂G = (m + E)F ∂r ∂F = (m − E)G ∂r which imply that:

∂2G = λ 2G 2 ∂r

(2)

with λ = m 2 − E 2 . The solution to (2) is:

G = ae− λ r where a is a constant. Thus, ∂G = − λG = ( m + E ) F ∂r Hence, for very large r, F and G must have opposite signs. For very small r, equations (1) become:

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∂G κ Zα + G= F ∂r r r ∂F κ Zα − F=− G ∂r r r Therefore, κ κ Zα Zα G"+ G '− 2 G = F '− 2 F r r r r Zα ⎛ κ Zα ⎞ Zα = G⎟ − F ⎜⎝ F − r r r ⎠ r2 Hence, Z 2α 2 − κ κ κ − 1⎛ κ ⎞ G"+ G '+ G= ⎜⎝ G '+ G ⎟⎠ 2 r r r r which yields: 1 κ 2 − Z 2α 2 G"+ G '− G=0 r r2 This equation has the solution: G = br γ

(

(3)

)

(4)

where b is a constant and γ = + κ − Z α . From (4) and the first of the equations (3) we see that for small r, ⎛ κ + κ 2 − Z 2α 2 ⎞ (5) F=⎜ ⎟G Zα ⎝ ⎠ Thus for small r, F and G have the same signs for κ > 0 but opposite signs for κ < 0 . We now consider separately the cases κ < 0, κ > 0 for arbitrary r. We assume E > 0 throughout. κ < 0.

Zα Since m + E + > 0 for all r, it is obvious from (1) that where G = 0 , G ' and F have r the same sign, and also F ' and F have opposite signs. Therefore, in the vicinity of a node of G the possibilities are as shown in Figs. 21.1aS, 21.1bS: 2

2

2

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Fig. 21.1aS

21.1bS

Thus between 2 successive nodes of G there must be at least one node of F, and the number n of nodes of F must be odd. Can n be greater than one? For example can the possibility shown in Fig. 21.2S occur?

F G

a

1

2

3

b

Fig. 21.2S An impossible situation: between 2 successive nodes (a,b) of G, there are three nodes (1,2,3) of F. In Fig.21.2S we have the following at nodes 1, 2, 3: ⎛ Zα ⎞ F' = ⎜m − E − G<0 r1 ⎟⎠ ⎝ ⎛ Zα ⎞ F' = ⎜m − E − G>0 r2 ⎟⎠ ⎝ ⎛ Zα ⎞ F' = ⎜m − E − G<0 r3 ⎟⎠ ⎝ Since at each of these nodes, G > 0, this implies:

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Zα Zα >m−E− r2 r3 which is impossible, since r3 > r2 . Therefore, between any two successive nodes of G there is one and only one node of F. Thus, since for κ < 0 , F,G have opposite signs at very small r and very large r , F and G must have the same number of nodes. Note also that the first node of G must come before the first node of F. κ >0 From (1), G ' , F , and F ' all have the same sign at a node of G. The situation is shown in Fig.21.3S. m−E−

Fig. 21.3S From Fig. 21.3S it is clear that between 2 successive nodes of G, there must be an odd number of nodes of F. However the argument associated with Fig.21.2S is independent of the sign of κ so it also applies here: between any two successive nodes of G there is one and only one node of F. However, we have already shown that for κ > 0 , F and G have the same sign for very small r and opposite signs for very large r. Therefore F must have one more node than G. This happens in the following way: The first node of F occurs before the first node of G, and the last node of F occurs after the last node of G. b) What if G = qF where q is a constant? Then equations (1) can be written:

κ Zα ⎞ ⎛ G = ⎜m+ E + ⎟ qG ⎝ r r ⎠ κ 1⎛ Zα ⎞ G '− G = ⎜ m − E − ⎟G r q⎝ r ⎠ Subtracting the second of these equations from the first, we obtain for G ≠ 0 : G '+

⎛ 2κ ⎡ ⎛ 1⎞ 1 ⎞ ⎤ Zα ⎛ 1⎞ = ⎢m ⎜ q − ⎟ + E ⎜ q + ⎟ ⎥ + q+ ⎟ ⎜ ⎝ r q⎠ q⎠ ⎦ r ⎝ q⎠ ⎣ ⎝

(6)

(7)

Since (7) must hold for all values of r where G ≠ 0 , we have:

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E = mε = m

1 − q2 1 + q2

(8)

and:

κ=

Zα ⎛ 1⎞ q+ ⎟ ⎜ 2 ⎝ q⎠

(9)

1− ε . Which sign should we choose? Since F,G 1+ ε always have opposite signs for very large r, we choose the minus sign: 1− ε (10) q=− 1+ ε Employing (10) in (9), we have: Zα ⎛ 1 − ε 1+ ε ⎞ Zα (11) κ =− + =− ⎜ ⎟ 2 ⎝ 1+ ε 1− ε ⎠ 1− ε2 Thus, Z 2α 2 Z 2α 2ε 2 2 2 γ 2 ≡ κ 2 − Z 2α 2 = − Z α = 1− ε2 1− ε2 Therefore, Zαε γ = 1− ε2 and Zαε n' = −γ = 0 1− ε2 1 This last equation requires κ = − (  A + 1) and n = j + . We list the states for which 2 F = constiG :

Equation (8) implies that q = ±

n

j

Spectroscopic notation

1

1/2

12 s1/2

2

3/2

2 2 p3/2

3

5/2

32 d5 /2

4 7/2 4 2 f7 /2 .............................................

21.5 Start with the M1 hyperfine perturbation Hamiltonian in natural units:

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µp × r 4π r 3 r̂ × α = eµ p i 4π r 2

H ' = eα i A = eα i

µ p = g p µ pΙ is the proton spin magnetic moment operator, with e . Also, I is the proton spin operator. The energy shift in first g p = 5.58 and µ p = 2m p order of perturbation is: Here

ΔE = ψ H ' ψ = g p

α Ι i r̂ × α ψ† ψ d 3r ∫ 2m p r2

(1)

The hyperfine splitting is δ = ΔE(F = 1) − ΔE(F = 0) . For F=1, we can employ the wave function: ⎛ ig(r)Y 1/2 χ p ⎞ 1/2,0 + ⎟ ψ (F = 1, mF = 1) = ⎜ (2) 1/2 ⎜⎝ f (r)Y1/2,1 χ +p ⎟⎠ For F=0, we have: 1/2 −1/2 p ⎛ ⎛ Y1/2,0 χ −p −Y1/2,0 χ+ ⎞ ⎞ ⎜ ig(r) ⎜ ⎟⎠ ⎟ 2 ⎝ ⎜ ⎟ (3) ψ ( F = 0, mF = 0 ) = ⎜ ⎟ 1/2 p −1/2 p ⎜ f (r) ⎛ Y1/2,1 χ − −Y1/2,1 χ + ⎞ ⎟ ⎜⎝ ⎟⎠ ⎟ ⎜ 2 ⎝ ⎠ where the Y j,j are the usual Dirac spin-angular functions, and χ ±p are proton spin up (down) spinors.

To develop formula (1) we label the coefficient of ig(r) in (2) or (3) by the generic symbol YA , and the coefficient of f (r) in (2) or (3) by YB . In either (2) or (3), σ i r̂YA = YB σ i r̂YB = YA Thus we write (1) as: m

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ΔE = g p

α 1 ∫ 2m p r 2

α 1 = gp ∫ 2m p r 2 = −ig p

(

−igYA†

fYB†

( −igY

)

⎛ 0 ⎜ I i r̂ × σ ⎝

fY σ i r̂

† A

† A

)

I i r̂ × σ ⎞ ⎛ igYA ⎞ 3 ⎟ ⎜⎜ fY ⎟⎟ d r 0 ⎠⎝ B ⎠

⎛ f I i r̂ × σ (σ i r̂ )Y ⎞ A 3 ⎜ ⎟d r ⎜⎝ ⎟⎠ igI i r̂ × σ YA

α gf ⎡Y † I i r̂ × σ (σ i r̂ ) − σ i r̂I i r̂ × σ )YA ⎤⎦d 3r 2 ⎣ A( ∫ 2m p r (4)

Using the repeated index summation convention we write:

I i r̂ × σ (σ i r̂ ) − σ i r̂I i r̂ × σ = I i ( r̂ × σ )i σ j r̂j − σ j r̂j I i ( r̂ × σ )i = I i ⎡⎣ r̂mσ k ε mkiσ j r̂j − σ j r̂j r̂mσ k ε mki ⎤⎦ = I i r̂m r̂j ε mki ⎡⎣σ kσ j − σ jσ k ⎤⎦ = 2iI i r̂m r̂jσ nε kjnε mki = 2iI i r̂m r̂jσ nε jnk ε imk

(

= 2iI i r̂m r̂jσ n δ ijδ nm − δ inδ mj = 2i ( I i r̂σ i r̂ − I i σ )

)

Thus (4) becomes: α gf † ΔE = g p YA ( I i r̂σ i r̂ − I i σ )YA d 3r mp ∫ r2 In the integrand on the right hand side of (5), we have:

(5)

1 ⎡1 ⎤ Ii r̂σ i r̂ − Iiσ = ⎢ I + sin θ e−iφ + I − sin θ eiφ + I z cosθ ⎥ ⎡⎣ S+ sin θ e−iφ + S− sin θ eiφ + σ z cosθ ⎤⎦ 2 2 ⎣ ⎦ − ⎡⎣ I + S− + I − S+ + I zσ z ⎤⎦ (6)

Here S± , I ± are the electron spin, proton spin raising and lowering operators, respectively.

Only the following terms on the right hand side of (6) can make a contribution to the integral of (5):

⎛1 ⎞ ⎛1 ⎞ I + S− ⎜ sin 2 θ − 1⎟ + I − S+ ⎜ sin 2 θ − 1⎟ − I zσ z sin 2 θ (7) ⎝2 ⎠ ⎝2 ⎠ Now calculate ΔE(F = 1, mF = 1) and ΔE(F = 0, mF = 0). In the case F = 1, mF = 1 we have: 222

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YA = Y00 χ +e χ +p =

1 χ +e χ +p 4π

Thus, from (7), (5) reduces to: 1 α gf ΔE (1,1) = − gp χ +e† χ +p† I zσ z χ +e χ +p ) ∫ 2 sin 2 θ d 3r ( 4π m p r =−

∞ 1 α gp sin 2θ dΩ ∫ gf dr ∫ 0 8π m p

=−

1 α 8π ∞ gp gf dr 8π m p 3 ∫0

=−

g pα ∞ gf dr 3m p ∫0

(8) 1 ( χ +e χ −p − χ −e χ +p ) . Applying the operator in 8π (7) to Y A, multiplying on the left by YA† and carrying out simple algebra analogous to the steps leading to (8), we obtain:

In the case F = 0, mF = 0 we have YA =

g pα ∞ gf dr m p ∫0 Thus, the hyperfine splitting is: ΔE(0,0) =

δ = ΔE(1,1) − ΔE(0,0) = −

(9)

4g pα ∞ gf dr 3m p ∫0

(10)

It remains to evaluate the integral in (10). For the 12 s1/2 state, we have

g = 2Nα 1− α +1/2 r 1− α −1e−α r

(11)

1− 1−α2 g α

(12)

2

2

and

f =− with

N = 2 1− α −1 2

1+ 1−α2 Γ(1 + 2 1 − α 2 )

(13)

Thus:

223

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2 2 1− α α 2 1− α + 2

∞

0

2

0

2

− ∫ fgdr =

2

∫ r Γ(1 + 2 1 − α

= 2α

2 1− α 2 − 2 −2 α r

e

dr

2 3 Γ(−1 + 2 1 − α )

(14)

Γ(1 + 2 1 − α 2 ) Expanding the ratio of gamma functions in powers of α and saving only the leading term we obtain:

Γ(−1 + 2 1 − α 2 )

Γ(1) 1 = Γ(1 + 2 1 − α ) Γ(3) 2 Hence, we find that to lowest order in α the hyperfine splitting is: 4 α4 2 m (15) δ = gp = g p gsα 4 e 3 mp 3 mp where gs = 2 . In atomic units, this result is: 2 m (16) δ = g p gsα 2 e 3 mp which is the same result obtained in non-relativistic theory (Chapter 10, equation (10.32)).

2



21.6 Start with:

(

)

d γ µ ∂ µ − ieAµ ψ − i γ 5σ µν Fµνψ − meψ = 0 2

(1)

Multiply on the left by γ 0 . This yields: d Hψ = i ∂t ψ = −iα i ∇ψ + meγ 0ψ − eΦψ + eα i Aψ + i γ 0γ 5σ µν Fµνψ 2 Let H 0 = α i ( p + eA) − eΦ + mγ 0 and d H ' = i γ 0γ 5σ µν Fµν = −d (γ 0 Σ i E + iγ i B ) 2

(2) (3) (4)

b) To see that H ' as given by (4) is parity-odd, note that under space inversion, E is odd, B is even, γ 0 Σ is even, and γ is odd since γ anticommutes with γ 0 . H’ is also odd under time reversal, since under time-reversal B and Σ are odd, E is even, and iγ is even. c) Start with: 224

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H ' = −d Σ i E − d (γ 0 − I ) Σ i E = Let E = −∇Φ . Then,

H1 +

H2

d [ Σ i p,eΦ ] e d = i ⎡⎣ Σ i p, H 0 − α i p − mγ 0 ⎤⎦ e 0 Now ⎡⎣ Σ i p, γ ⎤⎦ = [ Σ i p, α i p ] = 0 . Thus, H 1 = d Σ i ∇Φ = i

id [ Σ i p, H 0 ] e Let ψ be an eigenstate of H 0 : H1 =

H0 ψ = E ψ

Then the first order energy shift in ψ due to H 1 is:

id ψ [ Σ i p, H 0 ] ψ e id = (E − E) ψ Σ i p ψ = 0 e

ψ H1 ψ =

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22.1a The time reversal transformation is given by the formula:

ψ '(t) = Tψ (−t) = ηT γ 1γ 3Kψ (−t) where ηT is a phase factor chosen for convenience, and K means complex conjugation of all quantities to the right of K. We start with the free particle Dirac wave function: ⎛ 1 ⎞ ⎜ 0 ⎟ ⎜ ⎟ 1 ε + m ⎜ p3 ⎟ i ( pi x − ε t ) ψ (t) = e 2ε ⎜ ε + m ⎟ V ⎜ ⎟ ⎜ p1 + ip2 ⎟ ⎝ ε+m ⎠ for positive energy, spin up, and linear momentum p.

Then: ⎛ 1 ⎞ ⎜ 0 ⎟ ⎜ ⎟ 1 ε + m ⎜ p3 ⎟ −i ( pi x + ε t ) Kψ (−t) = e 2ε ⎜ ε + m ⎟ V ⎜ ⎟ ⎜ p1 − ip2 ⎟ ⎝ ε+m ⎠ ⎛ 1 ⎞ ⎜ 0 ⎟ ⎜ ⎟ 1 ε + m ⎜ p3 ⎟ i (-pi x − ε t ) = e 2ε ⎜ ε + m ⎟ V ⎜ ⎟ ⎜ p1 − ip2 ⎟ ⎝ ε+m ⎠ Now, ⎛ 0 σ 1 ⎞ ⎛ 0 σ 3 ⎞ ⎛ iσ 2 0 ⎞ γ 1γ 3 = ⎜ ⎟⎜ ⎟ =⎜ ⎟ ⎜⎝ −σ 1 0 ⎟⎠ ⎜⎝ −σ 3 0 ⎟⎠ ⎜⎝ 0 iσ 2 ⎟⎠ ⎛ 0 ⎜ −1 =⎜ ⎜ 0 ⎜⎝ 0

1 0 0 0 0 0 0 −1

0 ⎞ 0 ⎟ ⎟ 1 ⎟ 0 ⎟⎠

Thus, choosing ηT = 1 we have:

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⎛0 1 ε + m ⎜ −1 ⎜ ψ '(t) = V 2ε ⎜ 0 ⎜⎝ 0

1 0 0 0 0 0 0 −1

⎛ 1 ⎞ 0⎞ ⎜ 0 ⎟ ⎜ ⎟ 0 ⎟ ⎜ p ⎟ i (-pi x − ε t ) 3 ⎟ e 1⎟ ⎜ ε + m ⎟ ⎜ ⎟ 0 ⎟⎠ ⎜ p1 − ip2 ⎟ ⎝ ε+m ⎠

⎛ 0 ⎞ ⎜ −1 ⎟ ⎜ ⎟ 1 ε + m ⎜ p1 − ip2 ⎟ i (- pi x − ε t ) = e V 2ε ⎜ ε + m ⎟ ⎜ ⎟ ⎜ − p3 ⎟ ⎝ ε+m ⎠ 0 ⎛ ⎞ ⎜ ⎟ 1 ⎜ ⎟ 1 ε + m ⎜ −( p1 − ip2 ) ⎟ i (- pi x − ε t ) =− e V 2ε ⎜ ε + m ⎟ ⎜ ⎟ ⎜ −(− p3 ) ⎟ ⎝ ε+m ⎠ Apart from the overall minus sign, this last expression is the wave function for a positive energy free particle with spin down and linear momentum –p.

22.2

Q=

−e 3 m2 d x ∑ εε ' i V ∫ p,s p',s '

⎡⎣u ( p') eip'⋅x bs†' ( p') + v†s ' ( p') e−ip'⋅x ds ' ( p') ⎤⎦ ⎡⎣us ( p ) e−ip⋅x bs ( p ) + vs ( p ) eip⋅x ds† ( p ) ⎤⎦ † s'

=

−e 3 m2 † ⎡⎣us ' ( p') us ( p ) bs†' ( p') bs ( p ) eip'⋅x e−ip⋅x d x ∑ ∫ V εε ' p,s p',s '

+v†s ' ( p') vs ( p ) ds ' ( p') ds† ( p ) eip⋅x e−ip'⋅x ⎤⎦ 1 ±i( p'− p )⋅x 3 e d x = δ p', p we reduce the sum over p, p' to a sum V∫ over p and we also have ε = ε ' . Thus:

Carrying out the integrals

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m † ⎡⎣us ' ( p ) us ( p ) bs†' ( p ) bs ( p ) +v†s ' ( p ) vs ( p ) ds ' ( p ) ds† ( p ) ⎤⎦ ε p,s,s '

Q = −e ∑

mε δ ss ' ⎡⎣bs†' ( p ) bs ( p ) − ds†ds ' ( p ) + 1⎤⎦ ε m p,s,s '

= −e ∑

= e∑ ⎡⎣ N s+ ( p ) − N s− ( p ) − 1⎤⎦ p,s

22.3 Start with the Lagrangian density: 1 µ L = − F µν Fµν − jEM Aµ 4 and the Euler-Lagrange equations: ⎛ ∂L ⎞ ∂L − ∂ν ⎜ ⎟ =0 ∂Aµ ⎝ ∂ ∂ν Aµ ⎠ ∂L µ We have: . Also, = − jEM ∂Aµ

(

(1)

(2)

)

(

1 1 − F µν Fµν = − ( ∂ µ Aν − ∂ν A µ ) ∂ µ Aν − ∂ν Aµ 4 4 1 = − ∂ µ Aν F µν 2 Therefore (2) yields: ∂ µ Fµν = jνEM

)

22.4 Start with the Lagrangian density in the previous solution, written in the form: 1 µ L = ( E 2 − B2 ) − jEM Aµ 2 ∂A and recall that E = − for the radiation field in Coulomb gauge. Thus, ∂t ∂L ∂A H = i −L ⎛ ∂ A ⎞ ∂t ∂⎜ ⎝ ∂t ⎟⎠ yields 1 µ H = ( E 2 + B2 ) + jEM Aµ 2 22.5 Start with: 1 1 L D = Ψ ⎡⎣iγ µ ∂ µ − m ⎤⎦ Ψ − ⎡⎣i∂ µ Ψγ µ + mΨ ⎤⎦ Ψ 2 2 We have:

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∂L D ⎛ 1 ⎞ = ⎜ i ∂− m ⎟ Ψ ⎝ ⎠ ∂Ψ 2 ∂L D 1 − ∂µ = i ∂Ψ 2 ∂ ∂µ Ψ

(

∴

)

∂L D ∂L D − ∂µ = ( i ∂− m ) Ψ ∂Ψ ∂ ∂µ Ψ

(

)

∂L D 1 = − i ∂ µ Ψγ µ − mΨ ∂Ψ 2 ∂L D 1 − ∂µ = − i ∂ µ Ψγ µ 2 ∂ ∂µ Ψ

(

∴

)

∂L D ∂L D − ∂µ = −i ∂ µ Ψγ µ − mΨ ∂Ψ ∂ ∂µ Ψ

(

)

22.6 Start with: δΨ ≡ Ψ '− Ψ = iαΨ

δΨ ≡ Ψ '− Ψ = −iαΨ

(1)

and i i L D = Ψγ µ ∂ µ Ψ − ∂ µ Ψγ µ Ψ − mΨΨ 2 2 Now we use δ ∂ µ Ψ = ∂ µ δΨ, δ ∂ µ Ψ = ∂ µ δΨ

and write:

(

)

(

)

i i i i δ L D = δΨ γ µ ∂ µ Ψ + Ψ γ µ ∂ µ δΨ − ∂ µ δΨγ µ Ψ − ∂ µ Ψγ µδΨ 2 2 2 2 − mδΨΨ − mΨδΨ

(2)

Now using relations (1) in (2) we obtain

α α 1 α 1 α Ψ ∂ Ψ − Ψ ∂ Ψ − ∂ µ α Ψγ µ Ψ − ∂ ΨΨ − ∂ µ α Ψγ µ Ψ + ∂ ΨΨ 2 2 2 2 2 2 = − ∂ µ α Ψγ µ Ψ

(

δLD =

(

)

(

)

)

22.7 δ ∂ µ + iqAµ Ψ = ∂ µ δΨ + iqδ Aµ Ψ + iqAµδΨ

(

)

( ) = i ( ∂ α ) Ψ + iqδ A Ψ + iα ∂ Ψ − qα A Ψ = i ∂ µ α Ψ + iα ∂ µ Ψ + iqδ Aµ Ψ − qα Aµ Ψ µ

µ

µ

(1)

µ

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Now

1 δ Aµ = − ∂ µ α q Thus (1) becomes: δ Dµ Ψ = δ ∂ µ + iqAµ Ψ = iα ∂ µ Ψ − qα Aµ Ψ = iα Dµ Ψ

(

)

Similarly, δ Dµ Ψ = −iα Dµ Ψ

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23.1 If the Coulomb potential transformed as a scalar rather than the 4th component of a 4-vector, the factor M = u2γ 0u1 in the transition amplitude would be replaced by the factor: M ' = u2u1 Then: 2 M ' = tr(u2u2u1u1 ) Averaging over initial spins and summing over final spins, we obtain: 2 1 M' = tr ⎡( m − p2 ) ( m − p1 ) ⎤⎦ 8m 2 ⎣ 1 ⎡ 4m 2 + 4 p1 i p2 ⎤⎦ = 2 ⎣ 8m 1 ⎡ m 2 − p1 i p2 + E 2 ⎤⎦ = 2 ⎣ 2m 1 ⎡ E 2 − E 2 v2 − E 2 v2 cosθ + E 2 ⎤⎦ = 2 ⎣ 2m E2 ⎛ θ⎞ = 2 ⎜ 1− v2 cos 2 ⎟ m ⎝ 2⎠

θ⎞ ⎛ Thus in the formula for the Mott scattering cross section, the factor ⎜ 1 − v2 sin 2 ⎟ would ⎝ 2⎠ θ⎞ ⎛ be replaced by the factor ⎜ 1 − v2 cos 2 ⎟ with no other changes. In the non-relativistic ⎝ 2⎠ limit where v <<1, both forms yield the same result: the original Rutherford crosssection.

23.2

The initial state is described by the 4-component spinor: ⎛ 1 ⎞ ⎜ 0 ⎟ ⎜ ⎟ u1 = N ⎜ p ⎟ ⎜ E + m⎟ ⎜ ⎟ ⎝ 0 ⎠ where N is a normalization factor. A final state with spin and momentum along the z ' axis is described by the spinor u2 obtained from u1 by employing the rotation operator for rotation about y by angle θ :

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⎛ i ⎞ u2+ = exp −iJ yθ u1 = exp ⎜ − Σ yθ ⎟ u1 ⎝ 2 ⎠

(

)

θ ⎛ cos ⎜ 2 ⎜ ⎜ sin θ ⎜ 2 = N⎜ ⎜ 0 ⎜ ⎜ ⎜⎝ 0

θ 2 θ cos 2

− sin

0 0

0 0

θ 2 θ sin 2

cos

⎞ ⎟ ⎟⎛ 1 ⎞ 0 ⎟⎜ 0 ⎟ ⎟ ⎟⎜ p ⎟ ⎜ ⎟ θ − sin ⎟ ⎜ E + m ⎟ ⎟ 2⎟ ⎜ ⎝ 0 ⎠ θ ⎟ cos ⎟ 2⎠ 0

θ ⎞ ⎛ cos ⎜ 2 ⎟ ⎜ ⎟ θ ⎜ ⎟ sin ⎜ ⎟ 2 = N⎜ p θ⎟ ⎜ cos ⎟ 2⎟ ⎜E+m ⎜ p θ⎟ sin ⎟ ⎜ ⎝E+m 2⎠ Similarly, an electron moving along the z ' axis with negative helicity would be described by the spinor:

θ ⎛ ⎞ − sin ⎜ ⎟ 2 0 ⎞ ⎛ ⎜ ⎟ θ ⎜ ⎜ ⎟ 1 ⎟ cos ⎟ ⎜ ⎟ 2 ⎛ i ⎞ ⎜ u2 − = exp ⎜ − Σ 2θ ⎟ N ⎜ 0 ⎟ = N⎜ ⎝ 2 ⎠ p θ ⎟ ⎜ ⎟ ⎜ ⎟ sin p 2 ⎟ ⎜− ⎟ ⎜ E+m ⎝ E + m⎠ ⎜ p θ⎟ cos ⎟ ⎜− ⎝ E+m 2⎠ The amplitudes to obtain u2 ± are proportional to:

θ⎛ p2 ⎞ θ † 2 2E M + = u2+γ 0u1 = u2+ u1 = N 2 cos ⎜ 1+ cos 2⎟ = N 2 ⎝ (E + m ) ⎠ E+m 2

(1)

and

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θ p2 θ 2m 2 (2) M − = u u = −N sin (1 − 2 ) = −N sin 2 2E+m (E + m ) The probability to obtain positive helicity along z ' is: θ 2 E 2 cos 2 M+ 2 P+ = (3) 2 2 = θ 2 2 2 2θ M+ + M− E cos + m sin 2 2 The probability to obtain negative helicity along z ' is: θ 2 m 2 sin 2 M− 2 P− = (4) 2 2 = θ 2 2 2 2θ M+ + M− E cos + m sin 2 2 These results can be obtained in a more complicated way by using the spin projection operator: 1 1+ γ 5 s ) ( 2 in a trace calculation. † 2− 1

2

23.3 Consider a Feynman diagram with N external photon lines and no external fermion or anti fermion lines. Let the number of incoming photon lines be n. Then the number of outgoing photon lines is N-n. Also the initial state charge conjugation eigenvalue is Ci = (−1)n , while the final state eigenvalue is (−1)N − n . Since the electromagnetic interaction is charge conjugation invariant, we must have: (−1)n = (−1)N − n which implies that N − 2n is an even integer, and therefore that N is also an even integer.

23.4 From Feynman rules 6), 7) in Sec. 23.5, the first-order vertex factors for e− ,e+ scattering on a nucleus of charge Ze must be opposite in sign.

Now consider the following 4 reactions up to and including second order:

e− on Z :

A(1) fi = aZ

2 A(2) fi = bZ

2 e− on − Z : A(1) A(2) fi = −aZ fi = bZ

e+ on Z :

2 A(1) A(2) fi = −aZ fi = cZ

2 e+ on − Z : A(1) A(2) fi = aZ fi = cZ where a, b, c are complex numbers. By charge conjugation invariance, we must have: σ (e− , Z ) = σ (e+ , −Z ) to all orders Thus up to and including second order, and for any Z, we must have:

2

aZ + bZ 2 = aZ + cZ 2

2

Thus,

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a Z 2 + (a * b + b * a)Z 3 + b Z 4 = a Z 2 + (a * c + c * a)Z 3 + c Z 4 which implies: 2 2 b = c hence: b = ceiθ , θ real and: a *b + b *a = a *c + c*a hence: a * c(eiθ − 1) = ac * (1 − e−iθ ) If θ = 0, b = c. If θ ≠ 0, we have: a * c 1 − e−iθ c = iθ = e−iθ = ac * e − 1 b Thus, a* a = c* b Therefore for any θ the interference terms for e− on Z and e+ on Z have opposite signs.

2

2

2

2

23.5 The amplitude corresponding to the bremsstrahlung diagrams in Fig. 23.1S is: −Ze2 1 A fi = 1/2 δ (E '+ ω − E) 2V 2ω

m2 1 ⋅ EE ' q 2

⎡ ⎤ i i u ' ⎢( −i ε ) −iγ 0 ) + ( −iγ 0 ) ( −i ε )⎥ u ( p '+ k − m p −k−m ⎣ ⎦

k

p'

p' k

q p-k

p'+k

q p

p

Fig. 23.1S 23.6 The spin-averaged cross section is given in the problem as: σ = π r02 / β

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Thus the cross section for 2-photon annihilation from the 1 S0 state, which has a much higher rate than 3-photon annihilation from the 3 S1 states, is 4 times larger:

σ ( 1 S0 ) = 4π r02 / β

The transition probability per unit time is: Γ = σ 1 S0 ρβ = 4π r02 ρ

( )

where ρ = ψ (0) = 2

1

3 =

α3 . Here a0 is multiplied by 2 because the reduced mass 8π

π ( 2a0 ) of the positronium system is me / 2 .

Thus, α5 Γ = σ ( 1 S0 ) ρβ = 2 To obtain Γ in s

−1

(1)

me c 2 we multiply the right hand side of (1) by and thus obtain: 

α 5 me c 2 = 8.03⋅10 9 s −1 2 which gives τ = 1.24 ⋅10 −10 s Γ=

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24.1 The rate of energy generation per unit volume due to the reaction: e+ e− = νν is: λ = σ v n+ n−Q where σ is the cross section for the reaction (which depends on the relative velocity of the incoming particles), v is the relative velocity, n± are the number densities of positrons and electrons, and Q is the energy generated per reaction. The cross section has 2 4 dimension L . However it must be proportional to GF2 which has dimension L ; hence −1

σ ≈ GF2 E 2 where E is an energy, which has dimension L in natural units. Just as in neutrino-electron scattering discussed in Sec. 24.7, E can be chosen as the relative energy of the initial reactants: in the present case E ≈ kT. Also v ≈c=1. The number densities are: 3 1.8 ( kT ) 3 n± = 2 3 3 ≈ ( kT ) in natural units. π c Also, Q ≈ kT. Thus, in natural units, 9 λ ≈ GF2 ( kT ) Meanwhile, the thermal energy per unit volume due to black-body radiation is: 4 π 2 ( kT ) 4 U = aT = 15  3c 3 π2 = ( kT )4 ≈ ( kT )4 in natural units.

15 The time scale for stellar evolution due to neutrino losses is thus: U 1 ≈ 2 λ GF ( kT )5 In natural units: m p4 1 = = (1836)4 1010 = 1.13i10 23 2 −5 2 GF (1.03i10 ) Also, since the unit of energy in natural units is mec2, and kT = me c 2 when T = 6 • 10 K , 9

5

⎛ 1010 ⎞ = 12.9 . Thus, we have ( kT ) = ⎜ ⎝ 6i10 9 ⎟⎠ 5

U 1.13i10 23  = = 8.8i10 21 λ 12.9 me c 2 = 11 s.

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24.2 If beta decay obeyed an S-P law, the leptonic part of the amplitude would take the form: 1 ue (1− γ 5 )vν = ue (1− γ 5 )(1− γ 5 )vν 2 1 = ue†γ 0 (1− γ 5 )(1− γ 5 )vν 2 1 = (1+ γ 5 )ue (1− γ 5 )vν 2 − e , ν Thus the polarization of would be +v/c, +1 respectively.

24.3 Start with the reduced amplitude:

M=

Gβ 2

u pγ µ (1− λγ 5 )un ⋅ gµσ ⋅ueγ σ (1− γ 5 )vν

(1)

Here Gβ = GF cosθ C where θ C = 13o is Cabibbo’s angle. In the non-relativistic limit, appropriate for the recoiling proton as well as the neutron in the neutron rest frame: ⎛ χp⎞ ⎛ χn ⎞ u p  ⎜ ⎟ , un = ⎜ ⎟ ⎝ 0⎠ ⎝ 0⎠ where χ p,n are two-component Pauli-Schroedinger wave functions. Thus:

u pγ u n = 0 u pγ 0un = χ †p χ n u pγγ 5un =

(

χ †p

)

⎛ 0 0 ⎜ ⎝ σ

σ ⎞ ⎛ 0 I ⎞ ⎛ χn ⎞ † ⎜ ⎟ = χ pσχ n 0 ⎟⎠ ⎜⎝ I 0 ⎟⎠ ⎝ 0 ⎠

u pγ 0γ 5un = 0 Therefore,

M=

Gβ

M* =

⎡⎣ χ †p χ nueγ 0 (1− γ 5 )vν + λχ †pσ i χ n i ueγ i (1− γ 5 )vν ⎤⎦ 2

Gβ

⎡ χ n† χ p vνγ 0 (1− γ 5 )ue + λ * χ n†σ j χ p i vνγ j (1− γ 5 )ue ⎤⎦ 2⎣

Hence,

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M = 2

{

Gβ2 tr ⎡⎣ χ p χ †p χ n χ n† ⎤⎦ tr ⎡⎣ueueγ 0 (1− γ 5 ) vν vνγ 0 (1− γ 5 ) ⎤⎦ 2 + λ tr ⎡⎣ χ p χ †pσ i χ n χ n† ⎤⎦ tr ⎡⎣ueueγ i (1− γ 5 ) vν vνγ 0 (1− γ 5 ) ⎤⎦

(1− γ ) v v γ (1− γ )⎤⎦ + λ tr ⎡⎣ χ χ σ χ χ σ ⎤⎦ tr ⎡⎣u u γ (1− γ ) v v γ (1− γ ) ⎤⎦} + λ tr ⎡⎣ χ p χ χ n χ σ j ⎤⎦ tr ⎡⎣ueueγ *

† p

2

p

† p

† n

i

n

† n

0

5

i

j

e e

ν ν

5

j

ν ν

(2)

5

j

5

In (2) consider the nucleon traces first. If the proton spin lies along +z, we have: ⎛ 1⎞ ⎛ 1 0⎞ 1 χ p χ †p = ⎜ ⎟ (1 0 ) = ⎜ = (I + σz ) ⎝ 0⎠ ⎝ 0 0 ⎟⎠ 2 while for proton spin along –z, ⎛ 0⎞ ⎛ 0 0⎞ 1 χ p χ †p = ⎜ ⎟ ( 0 1) = ⎜ = (I − σz ) ⎝ 1⎠ ⎝ 0 1 ⎟⎠ 2 In general, if the proton spin is in direction ŝ p ,

χ p χ †p =

(

1 I + σ i ŝ p 2

)

(3)

and similarly,

1 ( I + σ i ŝn ) (4) 2 In the present calculation, we assume ŝn = ẑ , but we sum over all final spins, including the proton spin. Thus 1 1 χ p χ †p → I + σ i ŝ p + I − σ i ŝ p = I 2 2 and (2) becomes: 2 Gβ2 ⎧ ⎡ I + σ z ⎤ ⎡ pe − me 0 pν 0 5 5 ⎤ M = tr γ 1− γ γ 1− γ ( ) ( )⎥ ⎨tr ⎢ ⎢ 2 ⎩ ⎣ 2 ⎥⎦ ⎣ 2me 2mν ⎦

χ n χ n† =

(

) (

)

p ⎤ ⎡ I + σ z ⎤ ⎡ pe − me i + λ tr ⎢σ i tr ⎢ γ (1− γ 5 ) ν γ 0 (1− γ 5 ) ⎥ ⎥ 2 ⎦ ⎣ 2me 2mν ⎣ ⎦ p ⎤ ⎡ I + σ z ⎤ ⎡ pe − me 0 + λ *tr ⎢ σ j ⎥ tr ⎢ γ (1− γ 5 ) ν γ j (1− γ 5 ) ⎥ 2mν ⎣ 2 ⎦ ⎣ 2me ⎦

(5)

p ⎤ ⎫⎪ 2 ⎡ I + σ z ⎤ ⎡ pe − me i + λ tr ⎢σ i σ j ⎥ tr ⎢ γ (1− γ 5 ) ν γ j (1− γ 5 ) ⎥ ⎬ 2 2mν ⎣ ⎦ ⎣ 2me ⎦ ⎭⎪ Now,

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⎛ I + σz ⎞ tr ⎜ =1 ⎝ 2 ⎟⎠

⎛ I + σz ⎞ tr ⎜ σ i ⎟ = δ iz ⎝ 2 ⎠ ⎛ I + σz ⎞ tr ⎜ σ j ⎟ = δ jz ⎝ 2 ⎠ and:

1 ⎛ I + σz ⎞ 1 tr ⎜ σ i σ j ⎟ = tr σ iσ j + tr σ jσ iσ z ⎝ ⎠ 2 2 2

(

)

(

(

)

1 = δ ij + tr δ jiσ z + iε jikσ kσ z 2 = δ ij − iε ijz

)

Also,

(1− γ 5 )γ µ = γ µ (1+ γ 5 ) (1− γ 5 )2 = 2(1− γ 5 ) Thus (5) becomes 2 Gβ2 M = tr ⎡⎣ peγ 0 pνγ 0 (1− γ 5 ) ⎤⎦ 4me mν

{

λ tr ⎡⎣ peγ 3 pνγ 0 (1− γ 5 ) ⎤⎦

(6)

+ λ tr ⎡⎣ peγ pνγ (1− γ ) ⎤⎦ *

+λ

2

0

3

5

(δ − iε ) tr ⎡⎣ p γ p γ (1− γ )⎤⎦} i

ij

ij 3

e

ν

j

5

Evaluation of the traces is tedious but straightforward. We obtain: 2 Gβ2 M = {( pe i pν + EeEν ) + 2 Re(λ )(Ee pν 3 + Eν pe3 ) − 2 Im(λ )( pe1 pν 2 − pν1 pe2 ) me mν

+ λ

2

( 3EeEν − pe i pν ) + 2 λ 2 (Ee pν 3 − Eν pe3 )}

Rearrangement of this expression yields:

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M = Gβ

2

EeEν me mν

2

(1 + 3 λ ) i

2 Re( λ ) − 2 λ 2 Re( λ ) + 2 λ 2 Im( λ ) ⎡ pe i pν 1 − λ ⎤ + v̂ ν i ŝn − ŝn i v e × v̂ ν ⎥ 2 2 ⎢1 + E E 1 + 3 λ 2 + v e i ŝn 1 + 3 λ 2 1+ 3 λ 1+ 3 λ ⎣ ⎦ e ν 2

2

2

(7) where v e = pe /Ee and v̂ ν is a unit vector in the direction of pν . (Note that throughout this −1

calculation we assume the neutrino mass to be negligible. The factor of mν appearing immediately after Gβ2 on the right hand side of (7) causes no problem because it is cancelled by another factor mν in the formula for the transition rate.)

The second term on the right hand side of (7) describes the electron-antineutrino angular correlation. Although it is not practical to observe the antineutrino directly in neutron beta decay, this correlation (which is not a parity violating effect) has been observed by detecting the electron and recoiling proton in coincidence.

If we integrate over all neutrino momenta, the terms in (7) proportional to pν or v̂ ν yield zero and we find: 2 ⎡ 2 Re ( λ ) − 2 λ ⎤ M ∝ ⎢1 + v e i ŝn (8) ⎥ 2 1+ 3 λ ⎢⎣ ⎥⎦ Expression (8) describes the angular distribution of emitted electrons with respect to the direction of neutron spin polarization. The effect is a manifestation of parity violation and has been measured accurately in a number of experiments. If we had λ = 1 instead of 1.26, this angular distribution would be isotropic.

The next to last term on the right hand side of (7), another manifestation of parity violation, describes the angular distribution of emitted antineutrinos with respect to direction of neutron spin polarization. It has been observed by correlating the directions of electron and recoil proton in coincidence, with the direction of neutron spin. The effect is large for λ = 1.26 : 2 2 Re(λ ) + 2 λ B= = .988 2 1+ 3 λ If λ were equal to unity, B would also be equal to unity.

The last term in (7) does not violate parity (it involves the scalar product of two axial vectors ) but it would violate time reversal invariance, which requires that Im ( λ ) = 0.

Finally, it is possible to gain some intuitive understanding of why the beta decay 2 2 Re(λ ) − 2 λ asymmetry coefficient A = is close to zero and why the antineutrino 2 1+ 3 λ asymmetry coefficient B is close to unity. Consider Fig.24.1S, which illustrates various 2

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features of the decay schematically, as seen in the neutron rest frame. The thin arrows refer to linear momenta, and the thick arrows refer to spin angular momenta.

Conservation of linear momentum and of spin angular momentum in this decay allow for 3 possibilities a priori: cases a), b), and c). However, the V-A law requires the electron and proton to be left handed (negative helicities), and the anti-neutrino to be right handed (positive helicity). This excludes case a) and leaves only cases b) and c), for which the electron linear momenta point in opposite directions, yielding A ≈ 0. However in both cases b) and c) the antineutrino linear momentum is parallel to the neutron spin, yielding B ≈1. n

p

e

!

a)

•

Forbidden

b)

•

Allowed

c)

•

Allowed

Fig.24.1S 24.4 Consider the transition: µ + + e− → ν e + ν µ in the muon rest frame and start with the following formula for the differential cross section: ( 2π )4 δ 4 pe + pµ − p1 − p2 m1m2 mµ me d 3 p1d 3 p2 2 (1) dσ = M ve m µ E e E1E2 ( 2π )6

(

)

where 1,2 refer to ν µ , ν e respectively, and:

GF vµγ λ (1− γ 5 )v1 ⋅u2γ λ (1− γ 5 )ue 2 G M * = F v1γ σ (1− γ 5 )vµ ⋅ueγ σ (1− γ 5 )u2 2 M=

Thus:

GF2 M = T1T2 2

(2)

T1 = tr ⎡⎣ vµ vµγ λ (1− γ 5 ) v1 v1γ σ (1− γ 5 ) ⎤⎦

(3)

2

where

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and

T2 = tr ⎡⎣u2u2γ λ (1− γ 5 ) ueueγ σ (1− γ 5 ) ⎤⎦

(4)

Averaging over initial spins, summing over final spins, and simplifying the traces, we obtain: 1 T1 = tr ⎡⎣ pµγ λ p1γ σ (1− γ 5 ) ⎤⎦ (5) 4mµ m1

1 tr ⎡⎣ p2γ λ peγ σ (1− γ 5 ) ⎤⎦ 4me m2 It can be shown that T1T2 is 4 pµ i p2 pe i p1 T1T2 = mµ me m1m2 T2 =

(6)

Thus, 2

M = 2GF2

pµ i p2 pe i p1 mµ me m1m2

and (1) becomes:

dσ = GF2

(

pµ i p2 pe i p1δ 4 pe + pµ − p1 − p2

)d p d p

(7) 2 2π ve mµ Ee E1E2 In the muon rest frame, pµ i p2 = mµ E2 .Also, the electron is very non-relativistic in the 3

3

1

2

muonium bound state, so we can make the replacements: pe i p1 = meE1 , me = Ee .Thus (7) becomes: δ (mµ + me − E1 − E2 )δ 3 ( p1 + p2 ) 3 3 2 dσ = GF d p1d p2 (8) 2π 2 ve Integrate over p2 , thereby eliminating the delta function δ 3 , and at the same time imposing the restriction p1 = − p2 , so E1 = E2 . Then ignoring me in comparison to mµ we obtain:

dσ = GF2

δ (mµ − 2E1 )

E12 dE1dΩ1

2π ve Now integrate over the solid angle and over the energy to find: 2 2 GF2 mµ GF2 mµ σ = 4π 2 = 2π ve 8 4 π ve The transition probability per unit time is related to the cross section as follows: Γ = σ i(incident flux) = σρve where ρ is the electron probability density at the muon (at the origin). Thus: 1 1 2 ρ = ψ 1s (0) = 3 = me3α 3 π a0 π 2

(9)

(10)

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Hence,

(

)

(

)

( (

) )

GF2 2 3 3 m µ me α 4π 2 Comparing this result to the transition probability per unit time for muon decay: GF2 mµ5 + + Γ µ → e + νe + νµ = 192π 3 we obtain the ratio: Γ µ + + e− → ν e + ν µ 48π me3α 3 = ≈ 6.3i10 −12 3 + + m Γ µ → e + νe + νµ µ Γ µ + + e− → ν e + ν µ =

(11)

(12)

This very small numerical value means that the process: µ + + e− → ν e + ν µ is of negligible significance for the lifetime of a muonium atom.

24.5 Consider the isodoublet Dirac field: ⎛ Ψ1 ⎞ Ψ=⎜ ⎟ ⎝ Ψ2 ⎠ with the Lagrangian density: 1 1 L = ∂ µ Ψ γ µ Ψ − Ψγ µ ∂ µ Ψ − mΨΨ (1) 2 2 Make a finite SU(2) gauge transformation, defined by: Ψ ' = exp ( iε it ) Ψ and

(

)

Ψ ' = Ψ exp ( −iε it ) . The resulting Lagrangian density is:

(

)

1 1 ∂ µ Ψ ' γ µ Ψ '− Ψ 'γ µ ∂ µ Ψ '− mΨ 'Ψ ' 2 2 1 1 = ∂ µ ⎡⎣ Ψ exp ( −iε i t ) ⎤⎦ γ µ exp ( iε i t ) Ψ − Ψ exp ( −iε i t )γ µ ∂ µ ⎡⎣ exp ( iε i t ) Ψ ⎤⎦ 2 2 − mΨ exp ( −iε i t ) exp ( iε i t ) Ψ

L' =

(

)

(

)

(

)

1 = L + Ψγ µ ⎡⎣ ∂ µ e−iε it eiε it − e−iε it ∂ µ eiε it ⎤⎦ Ψ 2 = L − iΨγ µ e−iε it ∂ µ ε i t eiε it Ψ

(

)

Thus

(

)

δL = −iΨγ µ e−iε i t ∂ µ ε it eiε i t Ψ

(2)

To restore local gauge invariance we have to add gauge fields to the Lagrangian density.

Their transformation properties under the finite gauge transformation are determined as follows. We first write out the covariant derivatives:

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( ) D Ψ = ( ∂ Ψ − ig A Ψ i t ) Dµ Ψ = ∂ µ + ig Aµ i t Ψ µ

µ

(3)

µ

Now require that under the finite gauge transformation, Dµ Ψ must transform in the same way as Ψ itself. This implies that: Dµ Ψ ' = eiε i t Dµ Ψ

(

)

which is:

( ∂ + ig A 'it ) e Ψ = e ( ∂ + ig A it ) Ψ µ

iε i t

µ

iε i t

µ

µ

This yields: ieiε i t ∂ µ ( ε it ) Ψ + ig Aµ 'iteiε i t Ψ = eiε i t ig Aµ itΨ Since (4) must hold for any arbitrary Ψ we have: ieiε i t ∂ µ ( ε it ) + ig Aµ 'iteiε i t = eiε i t ig Aµ it

(4)

from which we obtain the desired result:

⎛ 1 ⎞ Aµ 'it = eiε i t ⎜ − ∂ µ ( ε it ) + Aµ it ⎟ e−iε i t ⎝ g ⎠ b) Start with the equations: Dµ Dν − Dν Dµ Ψ = igEµν itΨ

(

and

(5)

)

(6)

( D D − D D ) Ψ = −igE Ψit µ

ν

ν

µ

(7)

µν

Substituting the first of equations (3) in (6) we obtain: ∂ µ + ig Aµ it ( ∂ν + ig Aν it ) Ψ − ( ∂ν + ig Aν it ) ∂ µ + ig Aµ it Ψ = igEµν itΨ

(

)

(

)

(8)

Expand the left hand side of (8) to obtain: ∂ µ + ig Aµ it ( ∂ν + ig Aν it ) Ψ − ( ∂ν + ig Aν it ) ∂ µ + ig Aµ it Ψ =

(

)

( ) ig∂ ( A itΨ ) − ig A it∂ Ψ − ig∂ ( A itΨ ) + ig A it∂ Ψ − g ( A itA it − A itA it ) Ψ = ig ( ∂ A it − ∂ A it ) Ψ − g ( A itA it − A itA it ) Ψ = i ⎡⎣ g ( ∂ A it − ∂ A it ) − g A × A it ⎤⎦ Ψ µ

ν

ν

µ

ν

µ

2

µ

ν

µ

ν

ν

2

µ

ν

µ

2

µ

µ

µ

ν

ν

ν

µ

ν

ν

µ

µ

ν

Comparison of this expression with the right hand side of (8) yields the result:

Eµν = ∂ µ Aν − ∂ν Aµ − g Aµ × Aν

(9)

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24.6a) We show that (24.174) reduces to (24.175) in the limit k 2 >> mZ2 . To begin, note that the second term in the Z propagator numerator of (24.174) makes no contribution because k = p1 + p2 and vp2 ≈ 0 ≈ p1u . Thus in the limit k 2 >> mZ2 we only have to simplify: α

−ig vγ (1 − γ )u

Mb =

5

2

4mW

k α

=

−ig vγ (1 − γ )u 5

2

4mW

k α

=

[( k − k ) k ⋅ k − k ( 2k + k ) ⋅ k + k ( 2k + k ) ⋅ k ] +

2

5

4mW

−

+α

+

−

−α

−

−

+

+

2

+

+

α

(k − k ) = −

2

+

α

( k − k ) (k ⋅ k + m ) −

2

−ig vγ (1 − γ )u 2

−

+

α

−

−ig 2

4mW

W

vq (1 − γ )u 5

Here we have used:

k 2 = 2mW2 + 2k+ ⋅ k− k− − k+ = 2 q b) In the case: + − e+ + e− → Wlong. pol + Wlong. pol

(10)

the three lowest-order diagrams are as shown in Fig.24.2S. W k

-

W k

-

+

W k

+

-

W

+

k

-

k

W k

+

Z

-

W

+

k +

-

k

!

q p 1

2

+

p

e

e a)

-

e

p2

p1

+

e b)

p 2

e

p 1

+

e

-

c)

Fig. 24.2S In the high energy limit the amplitude for Fig. 24.2Sa is:

245

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

ig 2 (11) vq (1+ γ 5 )u 4mW2 where v , u refer to the incoming positron, electron respectively, and: (12) k+ = p2 + q, k− = p1 − q Note that these equations are not the same as in part a) of this problem. For Fig. 24.2Sb, the amplitude is: ig 2 Mb = v ⎡⎣γ σ (1− 4 sin 2 θW + γ 5 ) ⎤⎦ u ⋅ ( k− − k+ )σ 2 8mW (13) −ig 2 2 = vq (1− 4 sin θW + γ 5 )u 4mW2 For Fig. 24.2Sc the amplitude is: ig 2 sin 2 θW Mc = vγ σ u ⋅ ( k− − k+ )σ 2 2mW Ma =

−ig 2 vqu mW2 where we have used e = gsin θW . Combining (13) and (14) we obtain: =

Mb + Mc =

(14)

−ig 2 vq (1− γ 5 )u 4mW2

which cancels Ma to this order of perturbation theory. In arriving at this result we used k+ − k− = 2 q which differs in sign from the analogous relation in part a) of the problem.

24.7 At time t =0 the state vector is: 1 ψ (0) = K 0 = K S0 + K L0 2

(

)

Downstream along the beam, we have: 1 ψ (t) = K S0 e− λSt + K L0 e− λLt 2

(

)

where t = d / βγ , d is the distance, and λS =

γS γ + imS , λ L = L + imL . The K 0 2 2

fraction of the beam at time t is:

I(K 0 ) = K 0 ψ (t)

2

=

1 − γ St − γ Lt ⎡⎣ e + e + 2e−(γ S +γ L )t /2 cos Δmt ⎤⎦ 4

(1)

2

=

1 − γ St − γ Lt ⎡⎣ e + e − 2e−(γ S +γ L )t /2 cos Δmt ⎤⎦ 4

(2)

Similarly,

I(K 0 ) = K 0 ψ (t)

246

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

The beam strangeness is defined as: I(K 0 ) − I ( K 0 ) S = I(K 0 ) + I ( K 0 )

Δm . Since Δm and γ ( K S0 ) are comparable in magnitude, 2π the strangeness oscillations are visible before they are damped away by decay of the K S0 component. For times t such that: 1 / γ L >> t >> 1 / γ S the K S0 component is gone, and the beam is essentially pure K L0 . Now consider what happens if a piece of matter is placed in the path of the beam. Neglecting time dependence, the state prior to entering the slab of matter is: 1 1 ψ = K L0 = K 0 + K 0 2 2 0 0 Since K , K have very different scattering amplitudes a,b respectively on matter nuclei, the state vector after passage through the slab of matter is: 1 1 ψ = cK 0 + dK 0 = ( c − d ) K S0 + ( c + d ) K L0 2 2 2 c+d 0 = K L + rK S0 2 2 where c ≠ d are two complex numbers and: c−d r= c+d is the “regeneration parameter”.

It oscillates with a frequency

24.8 Assuming the non-relativistic limit, choosing natural units, making the approximation that the deuterium nucleus is infinitely heavy, and choosing m j = +1 / 2 we have: 1/2 ψ (12 s1/2 , m j = 1 / 2) = 2α 3/2 e−α rY1/2,0

α 3/2 ⎛ α r ⎞ −α r/2 1/2 ⎜ 1− ⎟⎠ e Y1/2,0 2 2 ⎝ 5/2 α 1/2 ψ (2 2 p1/2 , m j = 1 / 2) = re−α r/2Y1/2,1 2 6 The matrix element in (24.182) is: ψ (2 2 s1/2 , m j = 1 / 2) =

M=

(

)

† GFQW ⎡ σ i pψ 2 p1/2 ψ 2 s ⎤⎥ ⎢ ⎦ r=0 4 2 ⎣

Now: 247

Solutions to problems in Quantum Mechanics by Eugene D. Commins Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014

∂ 2 ⎡ 1/2 ⎤ ⎡⎣σ i pψ 2 p1/2 ⎤⎦ = ⎢ −iσ i r̂ ⎛⎜ + ⎞⎟ R2 pY1/2,1 ⎥ 0 ⎝ ∂r r ⎠ ⎣ ⎦0 3α 5/2 1/2 ⎡ ⎛ ∂ 2⎞ 1/2 ⎤ = ⎢ −i ⎜ + ⎟ R2 p ⋅Y1/2,0 = −i Y1/2,0 ⎥ 2 6 ⎣ ⎝ ∂r r ⎠ ⎦0

Thus,

M=

−3iGFQW α 4 16 6

In natural units the energy denominator is 1059 Mhz = 8.58 ⋅10 −12 and GF = 3⋅10 −12 .

Also for deuterium, QW = (1− 4 sin 2 θW − 1) = −.996 . Thus

ψ 2 p H P(1) ψ 2 s

3⋅ 3⋅10 −12 α 4 ≅i = 7.6i ⋅10 −11 −12 E2 s − E2 p 8.58 ⋅10 16 6 The electric dipole matrix element for the 1s1/2-2p1/2 transition is:

215/2 ea0 35 Thus the parity violating electric dipole matrix element connecting the 1s and 2s states is: M0 = −

M ' = M0

2 p H P(1) 2s E2 s − E2 p

≅ −5.6i ⋅10 −11 ea0

248