Sakurai Solutions

H ARVARD U NIVERSITY

Solutions to Problems in Modern Quantum Mechanics, 2nd edition by J.J. Sakurai and Jim J. Napolitano

Peter Gyu Young Chang Last Modified: June 25, 2016

C ONTENTS 1 Fundamental Concepts

2

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

1 F UNDAMENTAL C ONCEPTS 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 (−B D − 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 = a 0~ I+

3 X

a i σi

i =1

where ~ I is the identity matrix. Since t r (σi ) = 0 for i = 1, 2, 3, we have: ¶ 0 = 2a 0 a0

µ

a0 t r (X ) = t r 0 Also, we have: Ã

t r (σk X ) = t r a 0 σk +

3 X

!

a x σk σx

x=1

Ã

3 X

= tr

! ¡ ¢ a x i ²kx y σ y + δkx I

x=1

= t r (a k I ) = 2a k since σa σb = i ²abc σc + δab I in general and t r (a x σ y ) = 0 for any x, y = 1, 2, 3. b) We have the matrix elements: µ

a0 + a3 X= a1 + i a2

a1 − i a2 a0 − a3

¶

Hence we have: 1 a 0 = (X 11 + X 22 ) 2 1 a 2 = (X 21 − X 12 ) 2i

1 a 1 = (X 12 + X 21 ) 2 1 a 3 = (X 11 − X 22) 2

Another way to see this is by noting that 1 a k = t r (σk X ) 2 3. The determinant of ~ σ·~ a is: d et (~ σ·~ a ) = d et

µ

a3 a1 + i a2

a1 − i a2 −a 3

¶

= −a 32 − (a 1 − i a 2 )(a 1 + i a 2 ) = −a 12 − a 22 − a 32 = −|~ a |2

2

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

ˆ Then, Taylor expanding the exponentials, we have: Without loss of generality, choose nˆ = z. ¶ ∞ µ ¶ X 1 ±i σz φ k ±i σz φ = exp 2 2 k=0 k! k µ k−1 ¶ µ ¶k X (−1) 2 φ X (−1) 2 φ =I ± σz i 2 k! 2 k=even k! k=od d µ ¶ µ ¶ φ φ = I cos ± i σz sin 2 2 Ã ! φ φ 0 cos 2 ± i sin 2 = φ φ 0 cos 2 ∓ i sin 2 Ã iφ ! e± 2 0 = iφ 0 e∓ 2 µ

Therefore, the transformed matrix is: Ã iφ e2 0 ~ σ·~ a = 0 Ã =

0 iφ

e− 2 a3 e

!µ

¶ Ã iφ a1 − i a2 e − 2 −a 3 0 iφ ! Ã iφ (a 1 − i a 2 )e 2 e − 2

a3 a1 + i a2

iφ 2 iφ

iφ

(a 1 + i a 2 )e − 2

−a 3 e − 2

(a 1 − i a 2 )e i φ −a 3

µ

a3 = (a 1 + i a 2 )e −i φ

¶

d et (~ σ·~ a 0 ) = −a 12 − a 22 − a 32 = −|~ a |2 = d et (~ σ·~ a) Also, noting that: µ

a 30 0 a 1 + i a 20

µ

=

a 10 − i a 20 −a 30

a3 (a 1 + i a 2 )e −i φ

¶

(a 1 − i a 2 )e i φ −a 3

¶

Therefore, we have: ´ 1³ (a 1 − i a 2 )e i φ + (a 1 + i a 2 )e −i φ 2 = a 1 cos φ + a 2 sin φ ´ 1 ³ a 20 = (a 1 + i a 2 )e −i φ − (a 1 − i a 2 )e i φ 2i = a 2 cos φ − a 1 sin φ

a 10 =

a 30 = a 3 This is a counter-clockwise rotation about the z-axis of angle φ.

3

e

iφ 2

0 e

0

Therefore, its determinant is:

~ σ·~ a0 =

0

iφ 2

!

!

Peter Gyu Young Chang

4.

1 FUNDAMENTAL CONCEPTS

a) Expressing the trace of the operator X Y in terms of bra-ket algebra: t r (X Y ) =

X

〈a 0 | X Y |a 0 〉

a0

=

XX a0

=

〈a 0 | X |a 00 〉 〈a 00 | Y |a 0 〉

a 00

XX

〈a 00 | Y |a 0 〉 〈a 0 | X |a 00 〉

a 0 a 00

=

X

〈a 00 | Y X |a 00 〉

a 00

= t r (Y X ) b) Looking at the matrix elements: 〈(X Y )† a 0 |a 00 〉 = 〈a 0 | X Y |a 00 〉 = 〈X † a 0 | Y |a 00 〉 = 〈Y † X † a 0 |a 00 〉 for any a 0 and a 00 . Therefore, we have (X Y )† = Y † X † . c) Suppose that A has eigenvalues: A |a i 〉 = a i |a i 〉 Taylor expanding the exponential, we have: ∞ 1 ¡ X ¢k i f (A) |a 0 〉 k=0 k! ∞ X 1 X (i f (A))k |a i 〉 〈a i |a 0 〉 = k! k=0 i ∞ X 1 X = (i f (a i ))k |a i 〉 〈a i |a 0 〉 k! k=0 i X = exp[i f (a i )] |a i 〉 〈a i |a 0 〉

exp[i f (A)] |a 0 〉 =

i

for arbitrary ket |a 0 〉. Therefore, we have: exp[i f (A)] =

X

exp[i f (a i )] |a i 〉 〈a i |

i

d) The sum given, in bra-ket algebra is: X a0

ψ∗a 0 (x 0 )ψa 0 (x 00 ) =

X

=

X

〈a 0 |x 0 〉 〈x 00 |a 0 〉

a0

〈x 00 |a 0 〉 〈a 0 |x 0 〉

a0

= 〈x 00 |x 0 〉 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 ket β is: 1 |β〉 = p (|+〉 + |−〉) 2

4

∗

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

Therefore, matrix elements for the operator is: 1 〈+|α〉 〈β|+〉 = 〈s x = ×/2|+〉 = p 2 1 〈+|α〉 〈β|−〉 = p 2 〈−|α〉 〈β|+〉 = 0 〈−|α〉 〈β|−〉 = 0 Therefore, the matrix for the operator is: µ ¶ 1 1 1 |α〉 〈β| = p 2 0 0

6. Suppose that: A |i 〉 = λi |i 〉 A | j 〉 = λj | j 〉 Then, we have: A(|i 〉 + | j 〉) = λi |i 〉 + λ j | j 〉 Thus, we need the states to be degenerate: λi = λ j for |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

(A − a 0 ) |α〉 =

XY

a0

〈a 00 |α〉 (A − a 0 ) |a 00 〉

a 00 a 0

=

XY a 00

〈a 00 |α〉 (a 00 − a 0 ) |a 00 〉

a0

=0 since for each a 00 Y

(a 00 − a 0 ) = 0

a0

Thus the operator is the null operator. b) If we operate the opeartor on an arbitrary ket in the same way: Y (A − a 00 ) X Y 1 |α〉 = 〈a 000 |α〉 (A − a 00 ) |a 000 〉 0 00 0 − a 00 ) (a − a ) (a 00 0 000 00 0 a 6=a a a 6=a X Y (a 000 − a 00 ) 000 〈a |α〉 |a 000 〉 0 − a 00 ) (a 000 00 0 a a 6=a X = δa 000 ,a 0 〈a 000 |α〉 |a 000 〉

=

a 000

= 〈a 0 |α〉 |a 0 〉 Hence it projects the |a 0 〉 component of an arbitrary ket.

5

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

c) For A = S z , 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: × × ×2 (S z + )(S z − )s |α〉 = (S 2z − ) |α〉 2 2 4 ×2 = (S 2z − )(a + |+〉 + a − |−〉) 4 µ ¶ µ ¶ 2 × ×2 ×2 ×2 = a + |+〉 + a − |−〉 − a + |+〉 + a − |−〉 4 4 4 4 =0 Operating on it by the second operator: first, for a 0 = ×/2, S z + ×2 × 2

+ ×2

|α〉 =

µ ¶ 1 × Sz + (a + |+〉 + a − |−〉) × 2

1 (a + × |+〉) × = a + |+〉

=

and the completely analogous process leads to the a 0 = −×/2 case, in which it results in a − |−〉. 8. First, the commutator is trivially zero if i = j . For i 6= j : i ×2 (|+〉 〈+| − |−〉 〈−| − (− |+〉 〈+| + |−〉 〈−|)) 4 i ×2 = (2 |+〉 〈+| − 2 |−〉 〈−|) 4 × = i × (|+〉 〈+| − |−〉 〈−|) 2 = i ×S z

[S x , S y ] =

×2 (− |+〉 〈−| + |−〉 〈+| − (|+〉 〈−| − |−〉 〈+|)) 4µ ¶ i× = i × − (− |+〉 〈−| + |−〉 〈+|) 2

[S x , S z ] =

= −i ×S y i ×2 (|+〉 〈−| + |−〉 〈+| − (− |+〉 〈−| − |−〉 〈+|)) 4 × = i × (|+〉 〈−| + |−〉 〈+|) 2 = i ×S x

[S y , S z ] =

Hence, we have: [S i , S j ] = i ²i j k ×S k . Next, for i 6= j , the anti-commutators evaluate: {S x , S y } = {S x , S z } = {S y , S z } = 0

6

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

and for i = j , they evaluate: ×2 ×2 (|+〉 〈+| + |−〉 〈−|) = ~ I 4 2 ×2 ×2 {S y , S y } = − (− |+〉 〈+| − |−〉 〈−|) = ~ I 2 2 ×2 ×2 (|+〉 〈+| + |−〉 〈−|) = ~ I {S z , S z } = 2 2 {S x , S x } = 2S 2x = 2 ·

Putting all these together, we have: µ

{S i , S j } =

¶ ×2 ~ I δi j 2

9. Since the eigenket is some linear combination: ˆ +〉 = a + |+〉 + a − |−〉 |~ S · n; First, check the result of the following: × |−〉 2 i× |−〉 S y |+〉 = 2

× |+〉 2 i× S y |−〉 = − |+〉 2

S x |+〉 =

S x |−〉 =

Then, the left side of the eigenvalue equation is:    Sx cos α sin β ¡ ¢ S y  ·  sin α sin β  (a + |+〉 + a − |−〉) = S x cos α sin β + S y sin α sin β + S z cos β (a + |+〉 + a − |−〉) Sz cos β 

× i× = cos α sin β (a + |−〉 + a − |+〉) + sin α sin β (a + |−〉 − a − |+〉) 2 2 × + cos β (a + |+〉 − a − |−〉) 2 µ ¶ ¢ × ×¡ = a + cos β + a − cos α sin β − i sin α sin β |+〉 2 2 µ ¶ ¢ ס × + a + cos α sin β + i sin α sin β − a − cos β |−〉 2 2 ´ ³ ´ ׳ × = a + cos β + a − e −i α sin β |+〉 + a + e i α sin β − a − cos β |−〉 2 2 and the right side of the equation is: µ ¶ × (a + |+〉 + a − |−〉) 2

Since |+〉 and |−〉 are linearly independent, we have: a + cos β + a − e −i α sin β = a + a + e i α sin β − a − cos β = a − In other words: cos β e −i α sin β e i α sin β − cos β

µ

7

¶µ

¶ µ ¶ a+ a+ = a− a−

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

From the first equation, we have: a+ =

e −i α sin β a− 1 − cos β

Since the coefficients are normalized: |a + |2 +|a − |2 = 1, µ ¶ sin2 β + 1 |a − |2 = 1 (1 − cos β)2 β 1 |a − |2 = (1 − cos β) = sin2 2 2 1 β |a + |2 = (1 + cos β) = cos2 2 2 Therefore, we have: β 2 β iθ a − = sin e 2 a + = cos

for some relative phase θ. Plugging these in to the equation relating a + and a − : sin β β β = sin e i (θ−α) 2 1 − cos β 2 1 − cos β β e i (θ−α) = cot sin β 2 1 − cos β sin β = =1 sin β 1 − cos β cos

Thus, we need θ = α. Therefore, the eigenket is: ˆ +〉 = cos |~ S · n;

β β |+〉 + sin e i α |−〉 2 2

10. Let |a 0 〉 be the energy eigenkets: |a 0 〉 = a 1 |1〉 + a 2 |2〉 The Hamiltonian can be represented as a matrix: µ

1 1 H =a 1 −1

¶

Thus, the eigenvalue equation is: µ ¶µ ¶ µ ¶ 1 1 a1 0 a1 a =a 1 −1 a 2 a2

Thus, the characteristic polynomial becomes: (a 0 )2 − a 2 − a 2 = 0 Therefore, the eigenvalues are: p a 0 = ± 2a

8

Peter Gyu Young Chang

1 FUNDAMENTAL CONCEPTS

And hence: p a 1 + a 2 = ± 2a 1 p a 1 − a 2 = ± 2a 2 In other words: p a 2 = (± 2 − 1)a 1 Hence: |a 0 〉 = C

µ

1 p ± 2−1

¶

for some constant C . Normalizing: ´ ³ ³ p ´ p 〈a 0 |a 0 〉 = C 2 1 + (± 2 − 1)2 = C 2 4 ∓ 2 2 = 1

C=q

1 p 2(2 ∓ 2)

Thus, the two eigenkets are: |a 10 〉 = q

1

³

´ p |1〉 + ( 2 − 1) |2〉

p 2(2 − 2) ³ ´ p 1 |1〉 − ( 2 + 1) |2〉 |a 20 〉 = q p 2(2 + 2)

a 10 =

p 2a

p a 10 = − 2a

11. We can rewrite the Hamiltonian: H=

H11 + H22 H11 − H22 (|1〉 〈1| − |2〉 〈2|) + H12 (|1〉 〈2| + |2〉 〈1|) (|1〉 〈1| + |2〉 〈2|) + 2 2

Note that we have the isomorphisms: (|1〉 〈1| + |2〉 〈2|) ↔ I 2 (|1〉 〈1| − |2〉 〈2|) ↔ S z × 2 (|1〉 〈2| + |2〉 〈1|) ↔ S x ×

9