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
Thus
e−iωSzt/~H(t)eiωSzt/~ = −~γ2
(e−iωt/2 0
0 eiωt/2
)(B0 Beiωt
Be−iωt −B0
)(eiωt/2 0
0 e−iωt/2
)
= −~γ2
(e−iωt/2 0
0 eiωt/2
)(B0e
iωt/2 Beiωt/2
Be−iωt/2 −B0e−iωt/2
)
= −~γ2
(B0 B
B −B0
)
= −γ(B0Sz +BSx).
Putting it together, we get
i~d
dt| ψr(t)〉 = −γ(S ·Br)| ψr(t)〉 where Br = B ir +
(B0 −
ω
γ
)k.
Now that the effective Hamiltonian Hr = −γ(S ·Br) is time-independent, we may easily write down
the time evolution of any state in the rotating frame as
| ψr(t)〉 = e−iHrt/~| ψr(0)〉 = eiγ(S·Br)t/~| ψr(0)〉.
To explicit compute the action of the unitary evolution operator U(t) = eiγ(S·Br)t/~ on an arbitrary
state, it helps to exploit the following identities: Define
| n; +〉 = cos
(θ
2
)| +〉+ eiφ sin
(θ
2
)| −〉
| n;−〉 = − sin
(θ
2
)| +〉+ eiφ cos
(θ
2
)| −〉,
where n is the unit vector in R3 with azimuthal angle θ (from +k) and polar angle φ. Then
(S · n)| n;±〉 = ±~2| n;±〉.
In the current problem, the operator S ·Br = |Br|(S · n), where
|Br| =√B2 +
(B0 −
ω
γ
)2
and n has associated angles θ = sin−1(
B
|Br|
), φ = 0.
Therefore it has eigenkets | n;±〉 with eigenvalues ±|Br|~/2. By the functional calculus of operators
(see #1, PS1), the operator eiγ(S·Br)t/~ has the same eigenkets | n;±〉 with eigenvalues e±iγ|Br|t/2.
This means that
| ψr(t)〉 = eiγ|Br|t/2〈n; + | ψr(0)〉| n; +〉+ e−iγ|Br|t/2〈n;− | ψr(0)〉| n;−〉.
8
Now suppose the initial ket is | ψr(0)〉 = | ψ(0)〉 = | +〉, per the problem. Then in the rotating
frame its time evolution is given by
| ψr(t)〉 = eiγ|Br|t/2〈n; + | +〉| n; +〉+ e−iγ|Br|t/2〈n;− | +〉| n;−〉
= eiγ|Br|t/2 cos
(θ
2
)| n; +〉 − e−iγ|Br|t/2 sin
(θ
2
)| n;−〉
= eiγ|Br|t/2 cos
(θ
2
)[cos
(θ
2
)| +〉+ sin
(θ
2
)| −〉
]
−e−iγ|Br|t/2 sin
(θ
2
)[− sin
(θ
2
)| +〉+ cos
(θ
2
)| −〉
]
=
[cos
(γ|Br|t
2
)+ i sin
(γ|Br|t
2
)cos θ
]| +〉+ i sin
(γ|Br|t
2
)sin θ| −〉
=
[cos
(ωrt
2
)+ i
(ω0 − ωωr
)sin
(ωrt
2
)]| +〉+ i
(γB
ωr
)sin
(ωrt
2
)| −〉,
where we have used the shorthands ω0 = B0/γ and ωr = |Br|/γ.
Back in the lab frame, the state would read
| ψ(t)〉 = eiωSzt/~| ψr(t)〉
=
[cos
(ωrt
2
)+ i
(ω0 − ωωr
)sin
(ωrt
2
)]eiωSzt/~| +〉+ i
(γB
ωr
)sin
(ωrt
2
)eiωSzt/~| −〉
=
[cos
(ωrt
2
)+ i
(ω0 − ωωr
)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. The
orientations 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.
For general ω, the z-magnetization of the state | ψ(t)〉 at time t is
〈Sz(t)〉= 〈ψ(t) | Sz | ψ(t)〉
=
([cos
(ωrt
2
)− i(ω0 − ωωr
)sin
(ωrt
2
)]e−iωt/2〈+ | − i
(γB
ωr
)sin
(ωrt
2
)eiωt/2〈− |
)
9
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