Sakurai Solutions

Solutions to Problems in

Modern Quantum Mechanics, 2

nd

edition

by J.J. Sakurai and Jim J. Napolitano

Peter Gyu Young Chang Last Modified: June 25, 2016

C

1 F

C

1. First, note that [, ] and {, } refers to the commutator and anti-commutators of the operators, respectively. Hence, using the property: [A, BC ] = B [A,C ] + [A, B ]C and [AB,C ] = A[B,C ] + [A,C ]B :

[AB,C D] = AC [B, D] + A[B,C ]D + C [A, D]B + [A,C ]DB

= AC (B D − DB) + A(BC −C B)D +C (AD − D A)B + (AC −C A)DB = AC (−BD − DB) + A(C B + BC )D +C (−AD − D A)B + (C A + AC )DB = −AC {D,B} + A{C ,B}D −C {D, A}B + {C , A}DB

2. a) The matrix X is represented in terms of the Pauli matrices and constants:

X = a0~I+ 3 X

i =1

aiσi

where ~I is the identity matrix. Since t r (σi) = 0 for i = 1, 2, 3, we have:

t r (X ) = t rµa0 0 0 a0 ¶ = 2a0 Also, we have: t r (σkX ) = t r à a0σk+ 3 X x=1 axσkσx ! = t r à ₃ X x=1 ax¡i²kx yσy+δkxI¢ ! = t r (akI ) = 2ak

sinceσaσb= i²abcσc+δabI in general and t r (axσy) = 0 for any x, y = 1, 2, 3.

b) We have the matrix elements:

X =µ a0+ a3 a1− i a2 a1+ i a2 a0− a3 ¶ Hence we have: a0= 1 2(X11+ X22) a1= 1 2(X12+ X21) a2= 1 2i(X21− X12) a3= 1 2(X11− X 22) Another way to see this is by noting that

ak=

1

2t r (σkX ) 3. The determinant of~σ ·~a is:

d et (~σ ·~a) = det µ _a 3 a1− i a2 a1+ i a2 −a3 ¶ = −a32− (a1− i a2)(a1+ i a2) = −a12− a22− a23

Without loss of generality, choose ˆn = ˆz. Then, Taylor expanding the exponentials, we have: expµ ±i σzφ 2 ¶ = ∞ X k=0 1 k! µ ±i σzφ 2 ¶k = I X k=even (−1)k2 k! µφ 2 ¶ ± σz X k=od d (−1)k−12 k! i µφ 2 ¶k = I cos µφ 2 ¶ ± i σzsin µφ 2 ¶ = Ã cosφ₂± i sinφ₂ 0 0 cosφ₂∓ i sinφ₂ ! = Ã e±iφ2 0 0 e∓i2φ !

Therefore, the transformed matrix is: ~σ ·~a0₌ à eiφ2 0 0 e−i2φ ! µ _a 3 a1− i a2 a1+ i a2 −a3 ¶Ã_e−iφ 2 0 0 ei2φ ! = à a3e iφ 2 (a_{1− i a2})e iφ 2 (a1+ i a2)e− iφ 2 −a₃e− iφ 2 ! à e−iφ2 0 0 eiφ2 ! = µ a3 (a1− i a2)eiφ (a1+ i a2)e−i φ −a3 ¶

Therefore, its determinant is:

d et (~σ ·~a0) = −a12− a22− a23= −|~a|2 = d et (~σ ·~a)

Also, noting that:

~σ ·~a0₌µ a03 a01− i a02 a₁0+ i a0₂ −a0₃ ¶ = µ a3 (a1− i a2)eiφ (a1+ i a2)e−i φ −a3 ¶ Therefore, we have: a0₁=1 2 ³ (a1− i a2)eiφ+ (a1+ i a2)e−i φ ´ = a1cosφ + a2sinφ a0₂= 1 2i ³ (a1+ i a2)e−i φ− (a1− i a2)eiφ ´ = a2cosφ − a1sinφ a0₃= a3

4. a) Expressing the trace of the operator X Y in terms of bra-ket algebra: t r (X Y ) =X a0 〈a0| X Y |a0〉 =X a0 X a00 〈a0| X |a00〉 〈a00| Y |a0〉 =X a0 X a00 〈a00| Y |a0〉 〈a0| X |a00〉 =X a00 〈a00| Y X |a00〉 = t r (Y X )

b) Looking at the matrix elements:

〈(X Y )†a0|a00〉 = 〈a0| X Y |a00〉 = 〈X†a0| Y |a00〉 = 〈Y†X†a0|a00〉

for any a0and a00. Therefore, we have (X Y )†= Y†X†. c) Suppose that A has eigenvalues:

A |ai〉 = ai|ai〉

Taylor expanding the exponential, we have:

exp[i f (A)] |a0〉 = ∞ X k=0 1 k!¡i f (A)¢ k |a0〉 = ∞ X k=0 X i 1 k!(i f (A)) k |ai〉 〈ai|a0〉 = ∞ X k=0 X i 1 k!(i f (ai)) k |ai〉 〈ai|a0〉 =X i exp[i f (ai)] |ai〉 〈ai|a0〉

for arbitrary ket |a0〉. Therefore, we have:

exp[i f (A)] =X

i

exp[i f (ai)] |ai〉 〈ai|

d) The sum given, in bra-ket algebra is: X a0 ψ ∗ a0(x0)ψa0(x00) =X a0 〈a0|x0〉 〈x00|a0〉 =X a0 〈x00|a0〉 〈a0|x0〉 = 〈x00|x0〉

5. a) The matrix element of the given operator is:

〈a(i )|α〉〈β|a( j )〉 = 〈a(i )|α〉〈a( j )|β〉∗ b) First, note that the ketβ is:

Therefore, matrix elements for the operator is: 〈+|α〉〈β|+〉 = 〈sx= ×/2|+〉 = 1 p 2 〈+|α〉〈β|−〉 =p1 2 〈−|α〉〈β|+〉 = 0 〈−|α〉〈β|−〉 = 0 Therefore, the matrix for the operator is:

|α〉〈β| =p1 2 µ1 1 0 0 ¶ 6. Suppose that: A |i 〉 =λi|i 〉 A | j 〉 =λj| j 〉 Then, we have: A(|i 〉 + | j 〉) =λi|i 〉 +λj| j 〉

Thus, we need the states to be degenerate:λi=λjfor |i 〉 + | j 〉 to also be an eigenket for A:

A(|i 〉 + | j 〉) =λi(|i 〉 + | j 〉)

7. a) We can operate the operator on an arbitrary ket |α〉: Y a0 (A − a0) |α〉 = X a00 Y a0 〈a00|α〉(A − a0) |a00〉 =X a00 Y a0 〈a00|α〉(a00− a0) |a00〉 = 0

since for each a00

Y

a0

(a00− a0) = 0

Thus the operator is the null operator.

b) If we operate the opeartor on an arbitrary ket in the same way: Y a00_6=a0 (A − a00) (a0_{− a}00₎|α〉 = X a000 Y a00_6=a0 1 (a0_{− a}00₎〈a 000_{|α〉(A − a}00_{) |a}000_〉 =X a000 Y a00_6=a0 (a000− a00) (a0_{− a}00₎ 〈a 000_|α〉|a000_〉 =X a000 δa000_,a0〈a000|α〉|a000〉 = 〈a0|α〉|a0〉 Hence it projects the |a0〉 component of an arbitrary ket.

c) For A = Sz, and a ket space spanned by {|+〉,|−〉}, we can construct an arbitrary ket |α〉:

|α〉 = a+|+〉 + a−|−〉 for some complex numbers a+and a−.

Then, operating on it by the first operator:

(Sz+× 2)(Sz− × 2)s |α〉 = (S 2 z− ×2 4 ) |α〉 = (S2z− ×2 4 )(a+|+〉 + a−|−〉) = µ a+× 2 4 |+〉 + a− ×2 4 |−〉 ¶ − µ a+× 2 4 |+〉 + a− ×2 4 |−〉 ¶ = 0

Operating on it by the second operator: first, for a0= ×/2,

Sz+×₂ × 2+×2 |α〉 =1 × µ Sz+× 2 ¶ (a+|+〉 + a−|−〉) =1 ×(a+× |+〉) = a+|+〉

and the completely analogous process leads to the a0= −×/2 case, in which it results in a−|−〉. 8. First, the commutator is trivially zero if i = j . For i 6= j :

[Sx, Sy] =i × 2 4 (|+〉〈+| − |−〉〈−| − (−|+〉〈+| + |−〉〈−|)) =i × 2 4 (2 |+〉〈+| − 2|−〉〈−|) = i ×× 2(|+〉〈+| − |−〉〈−|) = i ×Sz [Sx, Sz] =× 2 4 (−|+〉〈−| + |−〉〈+| − (|+〉〈−| − |−〉〈+|)) = i × µ −i × 2 (−|+〉〈−| + |−〉〈+|) ¶ = −i ×Sy [Sy, Sz] =i × 2 4 (|+〉〈−| + |−〉〈+| − (−|+〉〈−| − |−〉〈+|)) = i ×× 2(|+〉〈−| + |−〉〈+|) = i ×Sx Hence, we have: [Si, Sj] = i²i j k×Sk.

Next, for i 6= j , the anti-commutators evaluate:

and for i = j , they evaluate: {Sx, Sx} = 2S2x= 2 · ×2 4(|+〉〈+| + |−〉〈−|) = ×2 2~I {Sy, Sy} = −× 2 2 (−|+〉〈+| − |−〉〈−|) = ×2 2~I {Sz, Sz} = × 2 2 (|+〉〈+| + |−〉〈−|) = ×2 2~I Putting all these together, we have:

{Si, Sj} =µ ×

2 2

¶ ~Iδi j

9. Since the eigenket is some linear combination:

|~S · ˆn; +〉 = a+|+〉 + a−|−〉 First, check the result of the following:

Sx|+〉 =× 2|−〉 Sx|−〉 = × 2|+〉 Sy|+〉 =i × 2 |−〉 Sy|−〉 = − i × 2 |+〉 Then, the left side of the eigenvalue equation is:

  Sx Sy Sz  ·   cosαsinβ sinαsinβ cosβ 

(a+|+〉 + a−|−〉) =¡Sxcosαsinβ + Sysinαsinβ + Szcosβ¢(a+|+〉 + a−|−〉)

= cosαsinβ× 2(a+|−〉 + a−|+〉) + sinαsinβ i × 2(a+|−〉 − a−|+〉) + cosβ× 2(a+|+〉 − a−|−〉) = µ a+× 2cosβ + a− ×

2¡cosαsinβ − i sinαsinβ¢ ¶

|+〉 +

µ

a+×

2¡cosαsinβ + i sinαsinβ¢ − a− × 2cosβ ¶ |−〉 =× 2 ³

a+cosβ + a−e−i αsinβ ´ |+〉 +× 2 ³ a+eiαsinβ − a−cosβ ´ |−〉 and the right side of the equation is:

µ × 2 ¶

(a+|+〉 + a−|−〉) Since |+〉 and |−〉 are linearly independent, we have:

a+cosβ + a−e−i αsinβ = a+

a+eiαsinβ − a−cosβ = a− In other words:

µ _cosβ _e−i α_sinβ

eiαsinβ − cos β ¶ µ a+ a− ¶ =µ a+ a− ¶

From the first equation, we have:

a+=

e−i α_sinβ 1 − cosβa− Since the coefficients are normalized: |a+|2+|a−|2= 1,

µ _sin2β (1 − cosβ)2+ 1 ¶ |a−|2= 1 |a−|2= 1 2(1 − cosβ) = sin 2β 2 |a+|2= 1 2(1 + cosβ) = cos 2β 2 Therefore, we have: a+= cosβ 2 a−= sinβ 2e iθ

for some relative phaseθ.

Plugging these in to the equation relating a+and a−: cosβ 2= sinβ 1 − cosβsin β 2e i (θ−α) ei (θ−α)=1 − cosβ sinβ cot β 2 =1 − cosβ sinβ sinβ 1 − cosβ= 1 Thus, we needθ = α.

Therefore, the eigenket is:

|~S · ˆn; +〉 = cosβ

2|+〉 + sin

β

2e

iα_|−〉

10. Let |a0〉 be the energy eigenkets:

|a0〉 = a1|1〉 + a2|2〉 The Hamiltonian can be represented as a matrix:

H = aµ1 1

1 −1 ¶

Thus, the eigenvalue equation is:

aµ1 1 1 −1 ¶ µa1 a2 ¶ = a0µa1 a2 ¶

Thus, the characteristic polynomial becomes:

(a0)2− a2− a2= 0 Therefore, the eigenvalues are:

And hence: a1+ a2= ± p 2a1 a1− a2= ± p 2a2 In other words: a2= (± p 2 − 1)a1 Hence: |a0〉 = C µ ₁ ±p2 − 1 ¶

for some constant C . Normalizing:

〈a0|a0〉 = C2³1 + (±p2 − 1)2´= C2³4 ∓ 2p2´= 1

C =_q 1

2(2 ∓p2) Thus, the two eigenkets are:

|a01〉 = 1 q 2(2 −p2) ³ |1〉 + (p2 − 1)|2〉´ a₁0 =p2a |a02〉 = 1 q 2(2 +p2) ³ |1〉 − (p2 + 1) |2〉´ a₁0 = −p2a

11. We can rewrite the Hamiltonian:

H =H11+ H22

2 (|1〉〈1| + |2〉〈2|) +

H11− H22

2 (|1〉〈1| − |2〉〈2|) + H12(|1〉〈2| + |2〉〈1|) Note that we have the isomorphisms:

(|1〉〈1| + |2〉〈2|) ↔ I (|1〉〈1| − |2〉〈2|) ↔2

×Sz (|1〉〈2| + |2〉〈1|) ↔2