1 2x2 unitary matrix, s&n, p. 256, problem 3

P6574 HW #3 - solutions

Due March 1, 2013

1 2X2 unitary matrix, S&N, p. 256, Problem 3

Consider the 2X2 matrix defined by

U =a0 + iσ · aa0 − iσ · a

,

where a0 is a real number and a is a three-dimensional vector with real components.

1. Prove that U is unitary and unimodular.

[Solution: Define A ≡ a0 + iσ · a. Then

A =

(a0 + ia3 a2 + ia1

−a2 + ia1 a0 − ia3

)

so detA = |a0 + ia3|2 + |a2 + ia1|2 = a20 + |a|2 = detA† where we use the well know properties

σ† = σ and (σ · n)2 = 1 for any unit vector n. Thus, A−1 = 1detAA

†. We find

U =A

A†=

AA

A†A=

AA

detA

Thus

UU † =1

(detA)2A2(A†)2 =

1

detAAA† = 1 = U †U,

so U is unitary. Furthermore, detU = 1(detA)2

detA2 = 1, so U is unimodular. ]

2. In general, a 2X2 matrix represents a rotation in three dimensions. Find the axis and angle

of rotation appropriate for U in terms of a0, a1, a2, and a3.

[Solution: We now write

U =1

detAA2 =

1

detA(a0+iσ · a)2 =

1

detA(a20+2ia0σ · a−|a|2) =

a20 − |a|2a20 + |a|2 +

2a0|a|a20 + |a|2 (iσ · a)

where a ≡ a/|a|. We recognize the form exp(−iσ·nφ

2

)= cos φ2 − i sin φ

2σ · n, where

cosφ

2=

a20 − |a|2a20 + |a|2

sinφ

2= − 2a0|a|

a20 + |a|2n = a]

1

2 Generators

1. Show that in any representation where Jx and Jz are real matrices (therefore symmetrical),

Jy is a pure imaginary matrix (therefore antisymmetrical).

[Solution: Recall the angular momentum algebra:

[Jx, Jy] = i~Jz, [Jy, Jz] = i~Jx, [Jz, Jx] = i~Jy

Given a basis | i〉, the matrix representations for the Jx is [Jx]ij ≡ 〈i | Jx | j〉, and likewise for

Jy and Jz. Thus, since Jx and Jz have real matrix elements, we find:

〈i | Jx | j〉 = 〈j | Jx | i〉 , 〈i | Jz | j〉 = 〈j | Jz | i〉

where we use the Hermiticty of Jx and Jz. Taking the matrix elements of the third commutator

given above, we find:

〈i | [Jx, Jz] | j〉 = −i~ 〈i | Jy | j〉

where

〈i | [Jx, Jz] | j〉 =∑

k

(〈i | Jx | k〉〈k | Jz | j〉 − 〈i | Jz | k〉〈k | Jx | j〉)

=∑

k

(〈j | Jz | k〉〈k | Jx | i〉 − 〈j | Jx | k〉〈k | Jz | i〉)

= −〈j | [Jx, Jz] | i〉

Therefore,

〈i | Jy | j〉 = −〈j | Jy | i〉

so, since Jy is Hermitian, Jy has imaginary matrix elements.]

2. Show that if any operator commutes with two components of an angular momentum vector,

it commutes with the third.

[Solution: Suppose that [O, Jx] = [O, Jy] = 0 for some operator O. Therefore, we find:

[O, [Jx, Jy]] = [Jy, [O, Jx]][Jx, [Jy,O]] = 0

where we apply the Jacobi identity, [A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0. Applying the

above result to the angular momentum algebra [Jx, Jy] = iJz, we conclude that

[O, Jz] = 0

so the result is proven. ]

3. Let u,v,w be three unit vectors forming a right-handed Cartesian system. Show that the

infinitesimal rotation

R ≡ R−1v (ε)R−1u (ε)Rv(ε)Ru(ε)

2

differs from Rw(−ε2) only by terms of higher order than ε2.

[Solution: We have

Rn(θ) ≡ e−iθn·J/~

where n is a unit vector and Rn(θ) represents a (right-handed) rotation by an angle θ about

the axis defined by n. We wish to compute

R = R−1v (ε)R−1u (ε)Rv(ε)Ru(ε) = eiεv·J/~eiεu·J/~e−iεv·J/~e−iεu·J/~

where u, v, and w form a right-handed coordinate system (u× v = w, etc), and ε is an

infinitesimal angle. To solve this problem, we will use the identity

eεAeεB = eεA+εB+ 12ε2[A,B]+O(ε3)

Multiplying on the left by e−εB and applying the original identity to the RHS, we find:

e−εBeεAeεB = e−εBeεA+εB+ 12ε2[A,B]+O(ε3) = eεA+ε

2[A,B]+O(ε3)

Multiplying on the left by e−εA and applying the original identity once more, we find:

e−εAe−εBeεAeεB = eε2[A,B]+O(ε3)

This can also be written as:

eεAeεB = eεBeεAeε2[A,B]+O(ε3)

where the higher-order terms are not present if [A,B] commutes with A and B. For the

problem at hand, we put in A = −iv · J/~ and B = −iu · J/~, so that

[A,B] = − 1

~2viuj [Ji, Jj ] =

i

~εijkujvjJk =

i

~w · J

where the last step follows from u× v = w, or εijkuivj = wk, and we use the commutator

[Ji, Jj ] = iεijkJk. Therefore, applying the identity we derived above, we find

R = exp

[i

~ε2w · J +O(ε3)

]= Rw(−ε2) +O(ε3)

which is the desired result.]

3 Electron-Positron system, S&N, p. 256, Problem 4

The spin dependent Hamiltonian of an electron-positron system in the presence of a uniform mag-

netic field in the z-direction can be written as

H = AS(e−) · S(e+) +

(eB

mc

)(S(e−)z − S(e+)

z ).

Suppose the spin function of the system is given by χ(e−)+ χ

(e+)− .

3

1. Is this an eigenfunction of H in the limit A → 0, eB/mc 6= 0? If it is, what is the energy

eigenvalue? If it is not, what is the expectation value of H?

[Solution: In the limit A→ 0,

H =

(eB

mc

)(S1z − S2

z )

and

Hχ1+χ

2− =

eB

mc

~2

(1− (−1))χ1+χ

2− =

eB

mc~χ1

+χ2−]

2. Same problem when eB/mc→ 0, A 6= 0.

[Solution: Write

Hχ1+χ

2− =

1

2A(

(S1 + S2)2 − S12 − S22)χ1+χ

2−

=1

2A~2

((2 or 0)− 3

4− 3

4

)χ1+χ

2−

According to a table of CG coefficients

χ1+χ

2− =

√1

2| 11〉+

√1

2| 00〉

Therefore

S2Tχ

1+χ

2− = S2

T

(√1

2| 11〉+

√1

2| 00〉

)

=~2√

22| 11〉

= ~2√

2| 11〉

So

〈H〉 =1

2

⟨√1

2(〈11 |+ 〈00 |) | S2

T − S21 − S2

2 |√

1

2(| 11〉+ | 00〉)

⟩

=1

4

⟨(〈11 |+ 〈00 |) | S2

T − S21 − S2

2 | (| 11〉+ | 00〉)⟩

=~2

4(2− 3

4− 3

4+ 0− 3

4− 3

4)

= −~2

4

4

4 Rigid body, S&N, p. 256, Problem 6

Let the Hamiltonian of a rigid body be

H =1

2

(K2

1

I1+K2

2

I2+K2

3

I3

)

where K is the angular momentum in the body frame. From this expression obtain the Heisenberg

equation of motion for K and then find Euler’s equation of motion in the correspondence limit.

[Solution:

∂K

∂t=−i~

[K,H]

∂K1

∂t= − i

~1

2

([K1x,

K2j

Ij]

)

=1

2

(ε1jk

(KjKk +KkKj)

Ij

)i

=1

2

((K2K3 +K3K2)

I2− (K3K2 +K2K3)

I3

)

=1

2

((2K2K3 + [K3,K2])

I2− ([K2,K3] + 2K3K2)

I3

)

=1

2

((2K2K3 − i~K1)

I2− (i~K1 + 2K3K2)

I3

)

=

((ω ×K)1 + i~K1(

1

I3− 1

I2)

)

→cassical limit = (ω ×K)1∂K

∂t= (ω ×K)]

5 Euler rotations, S&N, p. 256, problem 9

Consider a sequence of Euler rotations represented by

D(1/2)(α, β, γ) = exp

(−iσ3α2

)exp

(−iσ2β2

)exp

(−iσ3γ2

)

=

e−i(α+γ)/2 cos β2 −e−i(α−γ)/2 sin β

2

ei(α−γ)/2 sin β2 ei(α+γ)/2 cos β2

Because of the group properties of rotations, we expect that this sequence of operations is equivalent

to a single rotation abut some axis by an angle θ. Find θ.

5

[Solution: In terms of a rotation by θ about an axis n,

D = exp (−in · Jθ/~) = I cos θ/2− in · σ sin(θ/2)

= I cos θ/2− i sin(θ/2)

(cosκ sinκe−iφ

sinκeiφ − cosκ

)

where n = nx sinκ cosφ+ ny sinκ sinφ+ nz cosκ. Then

D =

(cos(θ/2)− i sin(θ/2) cosκ −i sin(θ/2) sinκe−iφ

−i sin(θ/2) sinκeiφ cos(θ/2) + i sin(θ/2) cosκ

)

Compare the trace of the two forms. We find

cos(β/2) cos(α+ γ)/2 = cos(θ/2)]

6 Angular momentum

A simultaneous eigenstate of J2 and Jz is denoted | j,m〉.

1. Show that the expectation values of the operators Jx and Jy for this state are zero.

[Solution: There are at least two ways to approach this problem:

• Using the ladder operators, Jx = 12(J+ + J−) and Jy = 1

2i(J+ − J−). When Jx (resp.

Jy) acts on | j,m〉, it produces a linear combination of two ket states | j,m+ 1〉 and

| j,m− 1〉, both of which are orthogonal to | j,m〉. So if we put the bra 〈j,m | with the

ket Jx| j,m〉 (resp. Jy| j,m〉) together, the bracket must vanish: that is, the expectation

value of Jx (resp. Jy) in the state | j,m〉 is 0.

• It is also possible to exploit the commutation relations of the Ji alone without invoking

the ladder operators. Since [Jy, Jz] = i~Jx, and Jz| j,m〉 = m~| j,m〉, we can compute

the expectation value of Jx in | j,m〉 as follows:

〈j,m | Jx | j,m〉 =1

i~〈j,m | [Jy, Jz] | j,m〉

=1

i~(〈j,m | JyJz | j,m〉 − 〈j,m | JzJy | j,m〉)

=1

i~(m~ 〈j,m | Jy | j,m〉 −m~ 〈j,m | Jy | j,m〉)

= 0.

Essentially the same arguments go to show that 〈j,m | Jy | j,m〉 = 0.]

2. Show that if any operator commutes with 2 components of an angular momentum operator,

it must commute with the third component.

[Solution: For this part we use:

• The commutation relation [Ji, Jj ] = i~εijkJk (i, j, k = 1, 2.3).

6

• The Jacobi identity of commutators (Lie brackets): [A, [B,C]] = −([B, [C,A]]+[C, [A,B]]).

Thus if an operator O commutes with both Ji and Jj (i 6= j), then it must commute with the

third component:

[O, Jk] =εijki~

[O, [Ji, Jj ]] = −εijki~[Ji,[Jj ,O]] + [Jj ,[O, Ji]] = 0.]

7 Spin precession

We’re given a spin-1/2 particle | ψ〉 in a B field

B(t) = B cos(ωt)i +B sin(ωt)j +B0k

which consists of a static z-component and an oscillating component in the xy-plane. In this lab

frame, | ψ〉 evolves according to the Schrodinger equation

i~d

dt| ψ(t)〉 = H(t)| ψ(t)〉 = −γ(S ·B(t))| ψ(t)〉.

Due to the time-dependence of B (or H), it is more complicated to write down the solution | ψ(t)〉in the lab frame. So instead we shifts to a frame which co-rotates with the oscillating field, and

introduce | ψr(t)〉 = e−iωSzt/~| ψ(t)〉, which should see a time-independent B field. What is the

effective Hamiltonian Hr in the rotating frame? A direct calculation shows that

i~d

dt| ψr(t)〉 = i~

d

dt

(e−iωSzt/~| ψ(t)〉

)

= i~ (−iωSz/~) e−iωSzt/~| ψ(t)〉+ i~e−iωSzt/~ d

dt| ψ(t)〉

= ωSz| ψr(t)〉+ e−iωSzt/~H(t)| ψ(t)〉=

[ωSz + e−iωSzt/~H(t)eiωSzt/~

]

︸︷︷︸=Hr

| ψr(t)〉.

To go on we must unravel the operator e−iωSzt/~H(t)eiωSzt/~. This can be done by considering its

matrix representation in the eigenbasis of Sz, | +〉, | −〉: Clearly

e−iωSzt/~ =

(e−iωt/2 0

0 eiωt/2

)and eiωSzt/~ =

(e−iωSzt/~

)†.

Meanwhile,

H(t) = −γ(S ·B(t)) = −~γ2

3∑

j=1

σjBj

= −~γ2

(B0 B[cos(ωt) + i sin(ωt)]

B[cos(ωt)− i sin(ωt)] −B0

)

= −~γ2

(B0 Beiωt

Be−iωt −B0

).

7

PHYS 6572 - Fall 2010 PS7 Solutions

Back in the lab frame, the state would read

| ψ(t)〉 = eiωSzt/!| ψr(t)〉

=

[cos

(ωrt

2

)+ i

(ω0 − ω

ωr

)sin

(ωrt

2

)]eiωSzt/!| +〉 + i

(γB

ωr

)sin

(ωrt

2

)eiωSzt/!| −〉

=

[cos

(ωrt

2

)+ i

(ω0 − ω

ωr

)sin

(ωrt

2

)]eiωt/2| +〉 + i

(γB

ωr

)sin

(ωrt

2

)e−iωt/2| −〉.

Note that | ψ(0)〉 = | +〉, as required. It follows that 〈Sz(0)〉 = 〈ψ(0) | Sz | ψ(0)〉 = !/2.If ω = ω0 (”on resonance”), the effective field Br in the rotating frame consists of the transverse

component B ir only, so the spin state precesses about the ir axis at angular frequency at ωr =ω0 = γB. From the perspective of the lab frame, the state evolves as

| ψ(t)〉 = cos

(ω0t

2

)eiω0t/2| +〉 + i sin

(ω0t

2

)e−iω0t/2| −〉

= eiω0t/2

[cos

(ω0t

2

)| +〉 + eiπ/2 sin

(ω0t

2

)e−iω0t| −〉

]

= eiω0t/2| n(t); +〉,

where n(t) is the unit vector with associated angles θ(t) = ω0t and φ(t) = (π/2) − ω0t. Theorientations of the spin state form a figure-8 on the Bloch sphere (Fig. 1). Note that the up state| +〉 flips into the down state | −〉 in a duration T = π/ω0, and vice versa.

Figure 1: Time evolution of a spin-1/2 state in a field B(t) = B cos(ωt)i + B sin(ωt)j + B0k whenon resonance (ω = γB0). Initial state is the ”up” state | +〉 = | z; +〉. Dots on the Bloch sphereindicate the successive spin orientations in the lab frame.

For general ω, the z-magnetization of the state | ψ(t)〉 at time t is

〈Sz(t)〉= 〈ψ(t) | Sz | ψ(t)〉

4

Figure 1: Time evolution of a spin-1/2 state in a field B(t) = B cos(ωt)i + B sin(ωt)j + B0k when

on resonance (ω = γB0). Initial state is the ”up” state | +〉 = | z; +〉. Dots on the Bloch sphere

indicate the successive spin orientations in the lab frame.

×Sz([

cos

(ωrt

2

)+ i

(ω0 − ωωr

)sin

(ωrt

2

)]eiωt/2| +〉+ i

(γB

ωr

)sin

(ωrt

2

)e−iωt/2| −〉

)

=~2

[∣∣∣∣cos

(ωrt

2

)+ i

(ω0 − ωωr

)sin

(ωrt

2

)∣∣∣∣2

−(γB

ωr

)2

sin2

(ωrt

2

)]

=~2

[cos2

(ωrt

2

)+

(ω0 − ω)2 − (γB)2

ω2r

sin2

(ωrt

2

)]

=~2

[1 + cos(ωrt)

2+

(ω0 − ω)2 − (γB)2

ω2r

(1− cos(ωrt)

2

)]

=~2

[2(ω0 − ω)2

2ω2r

+2(γB)2

2ω2r

cos(ωrt)

]

= 〈Sz(0)〉[

(ω0 − ω)2

(ω0 − ω)2 + (γB)2+

(γB)2

(ω0 − ω)2 + (γB)2cos(ωrt)

].

From this we deduce that the up state reverses orientation in a duration

T =π

ωr=

π√(ω0 − ω)2 + (γB)2

.

10

8 Angular momentum of an unknown particle, S&N p.258, prob-

lem 17)

1. It helps to rewrite ψ(x) in spherical coordinates:

ψ(x) = rf(r)(sin θ cosφ+ sin θ sinφ+ 3 cos θ).

To check whether ψ is an eigenfunction of L2, we may carry out a direct computation. First

note that the operator L2 can be explicitly written in spherical coordinates [e.g. Sakurai Eq.

(3.6.15)]:

L2ψ(x) = −~2[

1

sin2 θ∂φφ +

1

sin θ∂θ[(sin θ)∂θ]

]ψ(x)

∂φφψ(x) = −rf(r)(sin θ cosφ+ sin θ sinφ)

∂θψ(x) = rf(r)(cos θ cosφ+ cos θ sinφ− 3 sin θ)

∂θ [(sin θ)∂θ]ψ(x) = rf(r)∂θ[sin θ cos θ(cosφ+ sinφ)− 3 sin2 θ

]

= rf(r)[cos(2θ)(cosφ+ sinφ)− 3 sin(2θ)]

So

L2ψ(x) = −~2[

1

sin2 θ[− sin θ(cosφ+ sinφ)] +

1

sin θ[cos(2θ)(cosφ+ sinφ)− 3 sin(2θ)]

]rf(r)

= −2~2rf(r) [− sin θ(cosφ+ sinφ)− 3 cos θ]

= 2~2ψ(x).

Thus ψ(x) is an eigenfunction of L2 with eigenvalue l(l + 1)~2 = 2~2, or l = 1.

2. Alternatively, we can re-express ψ(x) as a linear combination of spherical harmonics. Using

the normalized spherical harmonics

Y 01 =

√3

4πcos θ , Y ±11 = ∓

√3

8πsin θe±iφ,

we find

ψ(x) = rf(r)

[sin θ

(eiφ + e−iφ

2+eiφ − e−iφ

2i

)+ 3 cos θ

]

= rf(r)

[1

2(1− i) sin θeiφ +

1

2(1 + i) sin θe−iφ + 3 cos θ

]

= rf(r)

[−√

2π

3(1− i)Y 1

1 +

√2π

3(1 + i)Y −11 + 2

√3πY 0

1

].

In one fell swoop, we’ve shown that ψ(x) is an eigenfunction of L2 with eigenvalue 1(1 +

1)~2 = 2~2, and expanded ψ(x) in the eigenbasis of the j = 1 Hilbert space, i.e,. ψ(x) =

rf(r)∑1

m=−1 cmYm1 where

c1 = −√

2π

3(1− i) , c0 = 2

√3π , c−1 =

√2π

3(1 + i).

11

It ought to be clear that the probability of ψ being found in the state | 1,m〉 is given by

P (m) =|cm|2∑1

m=−1 |cm|2.

Since |c0|2 = 9|c1|2 = 9|c−1|2, we have

P (1) =1

11, P (0) =

9

11, P (−1) =

1

11.

3. Recall that the Laplacian in R3 can be written as

∆ =1

r2∂r(r2∂r

)+

1

r2

[1

sin2 θ∂φφ +

1

sin θ∂θ ((sin θ)∂θ)

]=

1

r2

[∂r(r2∂r

)− L2

~2

].

So the time-independent Schrodinger equation in R3 takes the form

− ~2

2mr2∂r(r2∂r

)+

L2

2mr2+ V (r)

Ψ(x) = EΨ(x). (1)

Now suppose the energy eigenstate Ψ(x) is the known wavefunction ψ(x) = rf(r)ζ(Ω), where

ζ(Ω) = sin θ cosφ+sin θ sinφ+3 cos θ. From Part (a) we already saw that L2ψ(x) = 2~2ψ(x).

Meanwhile,

∂rψ(x) = [f(r) + rf ′(r)]ζ(Ω)

∂r(r2∂rψ(x)) = 2r[f(r) + rf ′(r)]ζ(Ω) + r2[2f ′(r) + rf ′′(r)]ζ(Ω)

= r[2f(r) + 4rf ′(r) + r2f ′′(r)]ζ(Ω).

Plugging the various terms into (1) yields

− ~2

2mr[2f(r) + 4rf ′(r) + r2f ′′(r)]ζ(Ω) +

~2

mr2[rf(r)ζ(Ω)] + V (r)[rf(r)ζ(Ω)] = E[rf(r)ζ(Ω)].

Thus

V (r) = E +1

rf(r)

~2

2mr

[4rf ′(r) + r2f ′′(r)

]= E +

~2

2mr2

[4rf ′(r) + r2f ′′(r)

f(r)

].

9 Rotated angular momentum, S&N, p. 258, problem 20

A state | ψ〉 rotated by an angle β about the y-axis becomes e−iJyβ/~| ψ〉. So the probability for

the new state to be in | 2,m′〉 (m′ = 0,±1,±2) is given by the modulus squared of the projection

of e−iJyβ/~| l = 2,m = 0〉 onto the subspace | l = 2,m′〉, i.e.,∣∣∣D(2)

m′0(α = 0, β, γ = 0)∣∣∣2

=∣∣∣⟨

2,m′ | e−iJyβ/~ | 2, 0⟩∣∣∣

2,

12

where α, β, and γ are the Euler angles. At this stage we may invoke Sakurai Eq. (3.6.52):1

D(l)m0(α, β, γ = 0) =

√4π

2l + 1Y m∗l (β, α).

Using the expressions Y m2 (θ, φ) in Appendix A, we find

D(2)2,0(α = 0, β, γ = 0) =

√4π

5Y 2∗2 (β, 0) =

√4π

5

√15

32π(sin2 β) =

√3

8sin2 β,

D(2)1,0(α = 0, β, γ = 0) =

√4π

5Y 1∗2 (β, 0) =

√4π

5

[−√

15

8π(sinβ cosβ)

]= −

√3

2(sinβ cosβ),

D(2)0,0(α = 0, β, γ = 0) =

√4π

5Y 0∗2 (β, 0) =

√4π

5

√5

16π(3 cos2 β − 1) =

1

2(3 cos2 β − 1),

D(2)−1,0(α = 0, β, γ = 0) = −D(1)

1,0(α = 0, β, γ = 0),

D(2)−2,0(α = 0, β, γ = 0) = D(1)

2,0(α = 0, β, γ = 0).

Thus∣∣∣D(2)±2,0(α = 0, β, γ = 0)

∣∣∣2

=3

8sin4 β,

∣∣∣D(2)±1,0(α = 0, β, γ = 0)

∣∣∣2

=3

2sin2 β cos2 β,

∣∣∣D(2)0,0(α = 0, β, γ = 0)

∣∣∣2

=1

4(3 cos2 β − 1)2.

It is straightforward to check that∑2

m=−2

∣∣∣D(2)m0(α = 0, β, γ = 0)

∣∣∣2

= 1.

10 Rotation matrix for j = 1 states, S&N p. 260, problem 26

1. Since Jy = 12i(J+ − J−), it is clear that the matrix element 〈j,m′ | Jy | j,m〉 vanishes for any

m, m′ where |m−m′| 6= 1. Also 〈j,m | Jy | j,m′〉 = (〈j,m′ | Jy | j,m〉)∗ by hermiticity of Jy.

So for j = 1, it is enough to compute the matrix elements 〈1, 0 | Jy | 1, 1〉 and 〈1, 0 | Jy | 1,−1〉.From

Jy| 1, 1〉 =1

2i(J+| 1, 1〉 − J−| 1, 1〉) = − 1

2i(√

2~)| 1, 0〉 =i~√

2| 1, 0〉;

Jy| 1,−1〉 =1

2i(J+| 1,−1〉 −

J−| 1,−1〉) =1

2i(√

2~)| 1, 0〉 = − i~√2| 1, 0〉,

we deduce that 〈1, 0 | Jy | 1, 1〉 = (i~)/√

2 and 〈1, 0 | Jy | 1,−1〉 = −(i~)/√

2. So the matrix

representation of Jy in the | 1, 1〉, | 1, 0〉, | 1,−1〉 basis reads

J (j=1)y =

0 − i~√2

0i~√2

0 − i~√2

0 i~√2

0

=

~2

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

.

1Please read Sakurai Eqs. (3.6.46) through (3.6.51) and the accompanying text for the derivation.

13

2. A direct computation shows that

[J (j=1)y ]2 =

(~2

)2

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

=

(~2

)2

2 0 −2

0 4 0

−2 0 2

and

[J (j=1)y ]3 =

(~2

)3

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

2 0 −2

0 4 0

−2 0 2

=

~3

2

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

.

In other words,[J(j=1)y /~

]3= J

(j=1)y /~, which means that for positive integers n,

[J (j=1)y /~

]n=

[J(j=1)y /~

], n odd

[J(j=1)y /~

]2, n even

.

Therefore

e−iJ(1)y β/~ = 1 +

∞∑

n=0

(−iβ)2n+1

(2n+ 1)!

(J(1)y

~

)2n+1

+

∞∑

n=1

(−iβ)2n

(2n)!

(J(1)y

~

)2n

= 1− i∞∑

n=0

(−1)nβ2n+1

(2n+ 1)!

(J(1)y

~

)+

∞∑

n=1

(−1)nβ2n

(2n)!

(J(1)y

~

)2

= 1− i(J(1)y

~

)sinβ +

(J(1)y

~

)2

(cosβ − 1).

3. The matrix representation d(1)(β) of e−iJ(1)y β/~ reads

d(1)(β) = 1− i sinβ

2

0 −√

2i 0√2i 0 −

√2i

0√

2i 0

+ (cosβ − 1)

(1

2

)2

2 0 −2

0 4 0

−2 0 2

=

12(1 + cosβ) − 1√

2sinβ 1

2(1− cosβ)1√2

sinβ cosβ − 1√2

sinβ12(1− cosβ) 1√

2sinβ 1

2(1 + cosβ)

.

11 Rotations and Clebsch Gordon coefficients

Prove

Djmm′(R) =∑

m1,m′1,m2,m′2

〈j,m, j1, j2 | j1,m1, j2,m2〉Dj1m1m′1

(R)Dj2m2,m′2

(R)⟨j1,m

′1, j2,m

′2 | j,m′, j1, j2

⟩

14

Use this along with the known values of the rotation matrices d for j = 12 and j = 1 to compute

d3232, 32

(θ), d3232, 12

(θ), d3212, 12

(θ)

[Solution: Write

| j,m, j1, j2〉 =∑

m1,m2

| j1,m1, j2,m2〉〈j,m, j1, j2 | j1,m1, j2,m2〉 (2)

Rotate by R

D(R)| j,m, j1, j2〉 =∑

m1,m2

D(R)| j1,m1〉D(R)| j2,m2〉〈j,m, j1, j2 | j1,m1, j2,m2〉

Multiply from the left by the dual of 2.

〈j,m′, j1, j2 |D(R)| j,m, j1, j2〉 =∑

m′1m′2

∑

m1,m2

〈j1,m′1 |D(R)| j1,m1〉〈j2,m′2 |D(R)| j2,m2〉

×⟨j1,m

′1, j2,m

′2 | j,m′

⟩〈j,m, j1, j2 | j1,m1, j2,m2〉

Djm′,m =∑

m′1m′2

∑

m1m2

Dj1m′1m1

Dj2m′2m2

⟨j1,m

′1, j2,m

′2 | j,m′

⟩〈j,m, j1, j2 | j1,m1, j2,m2〉

Then

d3232, 32

=∑D1m′1m1

D12

m′2m2

⟨1,m′1,

1

2,m′2 |

3

2,3

2

⟩⟨1,m1,

1

2,m2 |

3

2,3

2

⟩

= D111D

1212

12

⟨1, 1,

1

2,1

2| 3

2,3

2

⟩⟨1, 1,

1

2,1

2| 3

2,3

2

⟩

= e−i(1+12)α 1

2(1 + cosβ) cos(β/2)e−i(1+

12)γ

15

1 2x2 unitary matrix, s&n, p. 256, problem 3

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