Phys 500, Quantum Mechanics

Homework 4 • Reference Solution

November 17, 2014

Solution to Problem 1: (a). First, Sz ∼= ~2

∑nk=1 σ

(k)z . Thus,

[H,Sz] =g~3

8

n−1∑i=1

[σ(i)x σ(i+1)

x + σ(i)y σ(i+1)y + σ(i)z σ(i+1)

z ,n∑k=1

σ(k)z

]

=g~3

8

n−1∑i=1

[σ(i)x σ(i+1)

x + σ(i)y σ(i+1)y , σ(i)z + σ(i+1)

z

]=

g~3

8

n−1∑i=1

−2iσ(i)y σ(i+1)x + 2iσ(i)x σ(i+1)

y + 2iσ(i)y σ(i+1)x − 2iσ(i)x σ(i+1)

y

= 0.

Analogously, [H,Sy] = [H,Sx] = 0. Hence also [H,S2] = 0.

(b) [H,Sz] = 0 implies exp(iαSz/~)H exp(−iαSz/~) = H, for all α ∈ R. Likewise, [H,Sx] = 0implies exp(iβSx/~)H exp(−iβSx/~) = H, for all β ∈ R. By Euler’s construction, any U ∈ SU(2)can be written as

U = exp(iγSz/~) exp(iβSx/~) exp(iαSz/~),

for some angles α, β, γ ∈ R. Combined with the above, this yields UHU † = H, for all U ∈ SU(2).

(c) We know, or find by working out the Clebsch-Gordan coefficients, that the joint eigenstatesof S2 and Sz for two spin 1/2-particles are expressed in the tensor product basis as follows

|s = 1,m = 1〉 = | ↑↑〉,|s = 1,m = 0〉 = |↑↓〉+|↓↑〉√

2,

|s = 1,m = −1〉 = | ↓↓〉,|s = 0,m = 0〉 = |↑↓〉−|↓↑〉√

2.

Using this, we can now work out that H|s = 0,m = 0〉 = −3g~24 |s = 0,m = 0〉 and H|s = 1,m〉 =

g~24 |s = 1,m〉 for all m = 0,±1. Thus, the energy eigenvalues are g~2/4 and −3g~2/4 .

Solution to Problem 2: We denote the Clebsch-Gordan coefficents (CGC) as follows

c(j1,j2)(j,m)(m1,m2)

= 〈j1j2;m1m2|j1j2; jm〉.

The coefficents c(j1,j2)(j,m)(m1,m2)

are zero unless m = m1 + m2, and |j1 − j2| ≤ j ≤ j1 + j2. Using

these two properties and the normalization (unitarity) constraint, we can obtain the CGC usingthe following recursion relations:

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• J+- move:√(j −m)(j +m+ 1) c

(j1,j2)(j,m+1)(m1,m2)

=√

(j1 −m1 + 1)(j1 +m1) c(j1,j2)(j,m)(m1−1,m2)

+

+√

(j2 −m2 + 1)(j2 +m2) c(j1,j2)(j,m)(m1,m2−1),

(1)

where the CGC on both sides of the equation are zero unless m = m1 +m2 − 1.

• J−- move:√(j +m)(j −m+ 1) c

(j1,j2)(j,m−1)(m1,m2)

=√

(j1 +m1 + 1)(j1 −m1) c(j1,j2)(j,m)(m1+1,m2)

+

+√

(j2 +m2 + 1)(j2 −m2) c(j1,j2)(j,m)(m1,m2+1),

(2)

where the CGC on both sides of the equation are zero unless m = m1 +m2 + 1.

The total angular momentum of a spin 1 and a spin 2 can be j = 1, j = 2 or j = 3. For anyfixed j the remaining parameters of the CGC that are left to vary are m1 and m2. In the following,we suppress the upper labels (j1 = 1, j2 = 2) in all CGC since they never change.

Case I. j = 3. The quantum numbers m1 and m2 are constrained by −1 ≤ m − 1,≤ 1,−2 ≤ m2 ≤ 2 (this time the additional constraint for m1 +m2 is dependent).

In this special case of highest combined angular momentum,

|1, 2; j = 3,m = 3〉 = c(3,3),(1,2)|1, 2;m1 = 1,m2 = 2〉,

and hence |c(3,3),(1,2)|2 = 1 by the normalization constraint. We choose

c(3,3)(1,2) =: +1

.We now apply the J−-move to m1 = 1, m2 = 1. Thus,

√6 c(3,2)(1,1) = 2 c(3,3),(1,2). Hence

c(3,2)(1,1) =

√2

3.

We continuue to apply J−-moves to find all CGC with m1 = 1, for m2 descending from 2 to -2. Inthis way we obtain

∗ ∗ 1

∗ ∗√

23

∗ ∗√

25

∗ ∗ 1√5

∗ ∗ 1√15

.

Therein, the vertical axis labels m2, −2 ≤ m2 ≤ 2, and the horizontal axis labels m1, −1 ≤ m1 ≤ 1.The symbols “∗” stand for Clebsch-Gordan coefficients which have not yet been evaluated.

We may now use the J±-moves to fill in the complete table for j = 3,

1√15

1√3

1

1√5

2√

215

√23√

25

√35

√25√

23 2

√215

1√5

1 1√3

1√15

. (3)

The top-left entry is for (m1 = −1,m2 = 2), and the bottom-right entry is for (m1 = 1,m2 = −2).

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Case II: j = 2. We now have three constraints, namely |m1| ≤ 1, |m2 ≤ 2 and |m1+m2| ≤ 2. First,we set c(2,2)(1,1) = c. Then, we fill the diagonal above-left of c(2,2)(1,1) = c by applying J+-moves,and the column below c(2,2)(1,1) = c by applying J−-moves. The result is

∗√

2c∗ ∗ c

∗ ∗√

32c

∗ ∗√

32c

∗ c

.

From the completed diagonal we can infer the modulus of c by the normalization condition,(√22

+ 12)|c|2 = 1. Hence, with the additional convention that c be real and positive,

c =1√3.

We now use the J±-moves to fill in the complete table for j = 2, and obtain

− 1√3−√

23

− 1√2− 1√

61√3

− 1√2

0 1√2

− 1√3

1√6

1√2√

23

1√3

. (4)

We may check the normalization condition on every diagonal.

Case III, j = 1. In the same manner as in the previous case we obtain√35√310 −

√310

1√10

−√

25

1√10

−√

310

√310√35

. (5)

Marking: 1/2 point for recognizing that the combined spin j is j = 1, 2, 3. 1 1/2 points for theCGC table in each case.

Solution to Problem 3: (a) We choose the time-reversal operator Θ such that

Θ|x〉, for all position eigenstates |x〉. (6)

Then, Θ|ψ〉 = Θ∑

x |x〉〈x|ψ〉 =∑

x〈x|ψ〉∗|x〉 (by anti-linearity). Thus, Θ acts by conjugation inthe position eigenbasis.

Consider an eigenstate of a time-reversal invariant Hamiltonian H, |n〉, with energy En. Then,Θ|n〉 is also an eigenstate of H with eigenvalue En. If H is non-degenerate, then

Θ|n〉 = eiφn |n〉. (7)

An important point to note here is that the phases {φn} are not fixed by Eq. (6). Rather, theycan be chosen by convention.

Combining Eqs. (6) and (7), we find

n∗(x) = Θn(x) := 〈x|Θ|n〉 = eiφnn(x),

3

with n(x) the wave function of the energy eigenstate |n〉 in the position basis. Thus,

n(x) = nR(x)e−iφn/2,

where nR(x) is real for all x.Now consider a general state |ψ〉 =

∑n cn|n〉, with cn = 〈n|ψ〉. We may write the expansion

coefficients ascn = |cn|eiχn .

Thus,

ψ(x) =∑n

nR(x)ei(χn−φn/2).

Again, the angles φn can be chosen by convention. The choice φn = 2χn makes ψ(x) real, ∀x.

Remark: What is the physical meaning of the choice of the phases φn in Eq. (7)? – Considerthe expansion of |ψ〉 into energy eigenkets, |ψ〉 =

∑n cn|n〉. If {|n〉} is a set of eigenkets of H, so

is {eiαn |n〉}, and neither set is preferable over the other. Thus, the phase part of the expansioncoefficients cn is convention, too. (This phases can be chosen once, and then we must stick to thischoice in all subsequent calculations.) When specifying a set of phases {αn} we are specifying aphase reference for our system. Without such a reference, the phases in the expansion coefficientscn are not meaningful. Changing {αn} changes the phases {φn} in Eq. (6). Thus, choosing a phasereference amounts to choosing the phases in Eq. (6).

(b) The Hamiltonian H0 = p2

2m of which the wave functions eipx/~ are eigenstates is invariantunder time-reversal, but its energy spectrum is degenerate (all states |p〉 with the same |p| = phave the same energy). Thus, the above argument does not apply.

(c) Yes, the Hamiltonian

H = AS2z +B

(S2x − S2

y

)is invariant under time-reversal. The spin operator transforms Θ : S −→ −S, and the discussedHamiltonian is quadratic in the spin.

To obtain the energy eigenstates, we may first obtain a matrix representation of the spinoperators Sx, Sy and Sz for spin 1. By now we know the matrix representations of those operatorsfor spin 1/2, and obtain the spin 1 representation by noting that |s = 1,m = 1〉 = | ↑↑〉, |1, 0〉 =(| ↑↓〉+ | ↓↑〉)/

√2 and |1,−1〉 = | ↓↓〉, and S = S1 + S2. In this way, we find

Sz ∼= ~

1 0 00 0 00 0 −1

, Sx ∼= ~

0 1√2

01√2

0 1√2

0 1√2

0

, Sy ∼= ~

0 −i 1√2

0

i 1√2

0 −i 1√2

0 i 1√2

0

.

From this we find that

H ∼= ~2 A 0 B

0 0 0B 0 A

,

The energy eigenstates and eigenvalues are thus

|E±〉 = |1,1〉±|1,−1〉√2

, E± = ~2(A±B),

|E0〉 = |1, 0〉, E0 = 0. (8)

As for the behaviour of the states in Eq. (8) under time reversal,

Θ|1, 1〉 = eiφ+ |1,−1〉, Θ|1, 0〉 = eiφ0 |1, 0〉, Θ|1,−1〉 = eiφ− |1, 1〉.

Since Θ2 = I for integer spin, φ+ = φ−. Hence, all the energy eigenstates in Eq. (8) are invariantunder time reversal (up to acquiring a phase).

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