PHYS 6572 - Fall 2011
PS7 Solutions
PHYS 6572 - Quantum Mechanics I - Fall 2011
Problem Set 7 — Solutions
Joe P. Chen / joe.p.chen@gmail.com
For your reference, here are some useful identities invoked frequently on this problem set:
J 2 | j, mi = j(j + 1)~2 | j, mi
Jz | j, mi = m~| j, mi
J± = Jx ± iJy
p
J± | j, mi = ~ (j ∓ m)(j ± m + 1)| j, m ± 1i
1
Angular momentum
(a). There are at least two ways to approach this problem:
1
• Using the ladder operators, Jx = 21 (J+ + J− ) and Jy = 2i
(J+ − J− ). When Jx (resp.
Jy ) acts on | j, mi, it produces a linear combination of two ket states | j, m + 1i and
| j, m − 1i, both of which are orthogonal to | j, mi. So if we put the bra hj, m | with the
ket Jx | j, mi (resp. Jy | j, mi) together, the bracket must vanish: that is, the expectation
value of Jx (resp. Jy ) in the state | j, mi 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, mi = m~| j, mi, we can compute
the expectation value of Jx in | j, mi as follows:
1
hj, m | [Jy , Jz ] | j, mi
i~
1
=
(hj, m | Jy Jz | j, mi − hj, m | Jz Jy | j, mi)
i~
1
(m~ hj, m | Jy | j, mi − m~ hj, m | Jy | j, mi)
=
i~
= 0.
hj, m | Jx | j, mi =
Essentially the same arguments go to show that hj, m | Jy | j, mi = 0.
(b). For this part we use:
• The commutation relation [Ji , Jj ] = i~ǫijk Jk (i, j, k = 1, 2.3).
• 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 ] =
ǫijk
ǫijk
]]} = 0.
J
[O,
[J
[O, [Ji , Jj ]] = −
{[Ji , i
j , O]] + [Jj , i~
i~
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PHYS 6572 - Fall 2011
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PS7 Solutions
Spin precession
We’re given a spin-1/2 particle | ψi in a B field
B(t) = B cos(ωt)î + B sin(ωt)ĵ + B0 k̂
which consists of a static z-component and an oscillating component in the xy-plane. In this lab
frame, | ψi evolves according to the Schrödinger equation
i~
d
| ψ(t)i = H(t)| ψ(t)i = −γ(S · B(t))| ψ(t)i.
dt
Due to the time-dependence of B (or H), it is more complicated to write down the solution | ψ(t)i
in the lab frame. So instead we shifts to a frame which co-rotates with the oscillating field, and
introduce | ψr (t)i = e−iωSz t/~ | ψ(t)i, which should see a time-independent B field. What is the
effective Hamiltonian Hr in the rotating frame? A direct calculation shows that
d
d −iωSz t/~
i~ | ψr (t)i = i~
e
| ψ(t)i
dt
dt
d
= i~ (−iωSz /~) e−iωSz t/~ | ψ(t)i + i~e−iωSz t/~ | ψ(t)i
dt
−iωSz t/~
= ωSz | ψr (t)i + e
H(t)| ψ(t)i
h
i
−iωSz t/~
iωSz t/~
= ωSz + e
H(t)e
| ψr (t)i.
|
{z
}
=Hr
To go on we must unravel the operator e−iωSz t/~ H(t)eiωSz t/~ . This can be done by considering its
matrix representation in the eigenbasis of Sz , {| +i, | −i}: Clearly
−iωt/2
†
e
0
iωSz t/~
−iωSz t/~
−iωSz t/~
and e
= e
.
e
=
0
eiωt/2
Meanwhile,
3
~γ X
σ j Bj
2
j=1
~γ
B0
B[cos(ωt) + i sin(ωt)]
= −
B[cos(ωt) − i sin(ωt)]
−B0
2
iωt
~γ
B0
Be
= −
.
−iωt
Be
−B0
2
H(t) = −γ(S · B(t)) = −
Thus
e
−iωSz t/~
H(t)e
iωSz t/~
~γ
= −
2
e−iωt/2
0
iωt/2
0
e
e−iωt/2
0
iωt/2
0
e
~γ
B0
B
= −
B −B0
2
= −γ(B0 Sz + BSx ).
~γ
= −
2
2
B0
Beiωt
Be−iωt −B0
eiωt/2
0
−iωt/2
0
e
iωt/2
B0 eiωt/2
Be
−iωt/2
Be
−B0 e−iωt/2
PHYS 6572 - Fall 2011
PS7 Solutions
Putting it together, we get
ω
where Br = B îr + B0 −
k̂.
γ
d
i~ | ψr (t)i = −γ(S · Br )| ψr (t)i
dt
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)i = e−iHr t/~ | ψr (0)i = eiγ(S·Br )t/~ | ψr (0)i.
To explicitly 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
θ
θ
iφ
| n̂; +i = cos
| +i + e sin
| −i
2
2
θ
θ
| +i + eiφ cos
| −i,
| n̂; −i = − sin
2
2
where n̂ is the unit vector in R3 with azimuthal angle θ (from +k̂) and polar angle φ. Then
~
(S · n̂)| n̂; ±i = ± | n̂; ±i.
2
In the current problem, the operator S · Br = |Br |(S · n̂), where
s
ω 2
B
−1
2
|Br | = B + B0 −
, φ = 0.
and n̂ has associated angles θ = sin
γ
|Br |
Therefore it has eigenkets | n̂; ±i with eigenvalues ±|Br |~/2. By the functional calculus of operators
(see #1, PS1), the operator eiγ(S·Br )t/~ has the same eigenkets | n̂; ±i with eigenvalues e±iγ|Br |t/2 .
This means that
| ψr (t)i = eiγ|Br |t/2 hn̂; + | ψr (0)i| n̂; +i + e−iγ|Br |t/2 hn̂; − | ψr (0)i| n̂; −i.
Now suppose the initial ket is | ψr (0)i = | ψ(0)i = | +i, per the problem. Then in the rotating
frame its time evolution is given by
| ψr (t)i = eiγ|Br |t/2 hn̂; + | +i| n̂; +i + e−iγ|Br |t/2 hn̂; − | +i| n̂; −i
θ
θ
iγ|Br |t/2
−iγ|Br |t/2
= e
cos
| n̂; +i − e
sin
| n̂; −i
2
2
θ
θ
θ
iγ|Br |t/2
= e
cos
cos
| +i + sin
| −i
2
2
2
θ
θ
θ
−iγ|Br |t/2
− sin
| +i + cos
| −i
−e
sin
2
2
2
γ|Br |t
γ|Br |t
γ|Br |t
+ i sin
cos θ | +i + i sin
sin θ| −i
= cos
2
2
2
ωr t
ωr t
ωr t
ω0 − ω
γB
sin
sin
= cos
+i
| +i + i
| −i,
2
ωr
2
ωr
2
where we have used the shorthands ω0 = B0 /γ and ωr = |Br |/γ.
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PHYS 6572 - Fall 2011
PS7 Solutions
Back in the lab frame, the state would read
| ψ(t)i = eiωSz t/~ | ψr (t)i
ω0 − ω
γB
ωr t
ωr t iωSz t/~
ωr t
iωSz t/~
+i
e
| +i + i
e
| −i
sin
sin
= cos
2
ωr
2
ωr
2
ω0 − ω
γB
ωr t
ωr t
ωr t −iωt/2
iωt/2
+i
e
| +i + i
e
| −i.
= cos
sin
sin
2
ωr
2
ωr
2
Note that | ψ(0)i = | +i, as required. It follows that hSz (0)i = hψ(0) | Sz | ψ(0)i = ~/2.
If ω = ω0 (”on resonance”), the effective field Br in the rotating frame consists of the transverse
component B îr only, so the spin state precesses about the îr axis at angular frequency ωr = ω0 =
γB. From the perspective of the lab frame, the state evolves as
ω0 t iω0 t/2
ω0 t −iω0 t/2
| ψ(t)i = cos
e
| +i + i sin
e
| −i
2
2
ω0 t
ω0 t −iω0 t
iω0 t/2
iπ/2
cos
= e
| +i + e
sin
e
| −i
2
2
= eiω0 t/2 | n̂(t); +i,
where n̂(t) is the unit vector with associated angles θ(t) = ω0 t and φ(t) = (π/2) − ω0 t. The
orientations of the spin state trace out a figure-8 on the Bloch sphere (Fig. 1). Note that the up
state | +i flips into the down state | −i 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)î + B sin(ωt)ĵ + B0 k̂ when
on resonance (ω = γB0 ). Initial state is the ”up” state | +i = | ẑ; +i. Dots on the Bloch sphere
indicate the successive spin orientations in the lab frame.
For general ω, the z-magnetization of the state | ψ(t)i at time t is
hSz (t)i
= hψ(t) | Sz | ψ(t)i
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PHYS 6572 - Fall 2011
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=
=
=
=
=
PS7 Solutions
ωr t
ωr t
ωr t iωt/2
ω0 − ω
γB
−iωt/2
cos
sin
sin
−i
e
h+ | − i
e
h− |
2
ωr
2
ωr
2
ωr t
ωr t
ωr t −iωt/2
ω0 − ω
γB
iωt/2
cos
×Sz
sin
sin
+i
e
| +i + i
e
| −i
2
ωr
2
ωr
2
"
#
ω0 − ω
γB 2 2 ωr t
ωr t 2
ωr t
~ +i
−
sin
cos
sin
2 2
ωr
2 ωr
2
ωr t
ωr t
(ω0 − ω)2 − (γB)2
~
sin2
cos2
+
2
2
2
ωr
2
2
2
1 − cos(ωr t)
~ 1 + cos(ωr t) (ω0 − ω) − (γB)
+
2
2
ωr2
2
2
2
2(γB)
~ 2(ω0 − ω)
+
cos(ωr t)
2
2ωr2
2ωr2
(ω0 − ω)2
(γB)2
hSz (0)i
+
cos(ωr t) .
(ω0 − ω)2 + (γB)2 (ω0 − ω)2 + (γB)2
From this we deduce that the up state reverses orientation in a duration
T =
3
π
π
.
=p
ωr
(ω0 − ω)2 + (γB)2
Angular momentum of an unknown particle (Sakurai 3.15)
(a). It helps to rewrite ψ(x) in spherical coordinates:
ψ(x) = rf (r)(sin θ cos φ + sin θ sin φ + 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)]:
1
1
2
2
L ψ(x) = −~
∂θ [(sin θ)∂θ ] ψ(x)
∂φφ +
sin θ
sin2 θ
∂φφ ψ(x) = −rf (r)(sin θ cos φ + sin θ sin φ)
∂θ ψ(x) = rf (r)(cos θ cos φ + cos θ sin φ − 3 sin θ)
∂θ [(sin θ)∂θ ] ψ(x) = rf (r)∂θ sin θ cos θ(cos φ + sin φ) − 3 sin2 θ
= rf (r)[cos(2θ)(cos φ + sin φ) − 3 sin(2θ)]
So
1
1
L ψ(x) = −~
[cos(2θ)(cos φ + sin φ) − 3 sin(2θ)] rf (r)
[− sin θ(cos φ + sin φ)] +
sin θ
sin2 θ
= −2~2 rf (r) [− sin θ(cos φ + sin φ) − 3 cos θ]
2
2
= 2~2 ψ(x).
Thus ψ(x) is an eigenfunction of L2 with eigenvalue l(l + 1)~2 = 2~2 , or l = 1.
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PHYS 6572 - Fall 2011
PS7 Solutions
(b). Alternatively, we can re-express ψ(x) as a linear combination of spherical harmonics. Using
the normalized spherical harmonics
r
r
3
3
±1
0
cos θ , Y1 = ∓
sin θe±iφ ,
Y1 =
4π
8π
we find
iφ
e + e−iφ eiφ − e−iφ
+
ψ(x) = rf (r) sin θ
+ 3 cos θ
2
2i
1
1
iφ
−iφ
= rf (r) (1 − i) sin θe + (1 + i) sin θe
+ 3 cos θ
2
2
" r
#
r
√
2π
2π
= rf (r) −
(1 − i)Y11 +
(1 + i)Y1−1 + 2 3πY10 .
3
3
In one fell swoop, we’ve shown that ψ(x) is an eigenfunction of L2 with eigenvalue 1(1 +
2 , and expanded ψ(x) in the eigenbasis of the j = 1 Hilbert space, i.e,. ψ(x) =
1)~2 =
P2~
1
rf (r) m=−1 cm Y1m where
r
r
√
2π
2π
(1 − i) , c0 = 2 3π , c−1 =
(1 + i).
c1 = −
3
3
It ought to be clear that the probability of ψ being found in the state | 1, mi is given by
|cm |2
.
2
m=−1 |cm |
P (m) = P1
Since |c0 |2 = 9|c1 |2 = 9|c−1 |2 , we have
1
9
1
P (1) =
, P (0) =
, P (−1) = .
11
11
11
(c). Recall that the Laplacian in R3 can be written as
L2
1
1
1
1
1
2
2
∆ = 2 ∂r r ∂r + 2
∂φφ +
∂θ ((sin θ)∂θ ) = 2 ∂r r ∂r − 2 .
r
r sin2 θ
sin θ
r
~
So the time-independent Schrödinger equation in R3 takes the form
L2
~2
2
∂r r ∂r +
+ V (r) Ψ(x) = EΨ(x).
−
2mr2
2mr2
(1)
Now suppose the energy eigenstate Ψ(x) is the known wavefunction ψ(x) = rf (r)ζ(Ω), where
ζ(Ω) = sin θ cos φ + sin θ sin φ + 3 cos θ. From Part (a) we already saw that L2 ψ(x) = 2~2 ψ(x).
Meanwhile,
∂r ψ(x) = [f (r) + rf ′ (r)]ζ(Ω)
∂r (r2 ∂r ψ(x)) = 2r[f (r) + rf ′ (r)]ζ(Ω) + r2 [2f ′ (r) + rf ′′ (r)]ζ(Ω)
= r[2f (r) + 4rf ′ (r) + r2 f ′′ (r)]ζ(Ω).
Plugging the various terms into (1) yields
2
~2
+ 4rf ′ (r) + r 2 f ′′ (r)]ζ(Ω) + ~ [
[rf
(r)ζ(Ω)]
+ V (r)[rf (r)ζ(Ω)] = E[rf (r)ζ(Ω)].
2f
(r)
2
2mr
mr
Thus
~2 ~2
1
4rf ′ (r) + r2 f ′′ (r)
′
2 ′′
4rf (r) + r f (r) = E +
V (r) = E +
.
rf (r) 2mr
2mr2
f (r)
−
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PHYS 6572 - Fall 2011
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PS7 Solutions
Rotated angular momentum (Sakurai 3.?)
A state | ψi rotated by an angle β about the y-axis becomes e−iJy β/~ | ψi. So the probability for
the new state to be in | 2, m′ i (m′ = 0, ±1, ±2) is given by the modulus squared of the projection
of e−iJy β/~ | l = 2, m = 0i onto the subspace | l = 2, m′ i, i.e.,
2 D
E2
(2)
Dm′ 0 (α = 0, β, γ = 0) = 2, m′ | e−iJy β/~ | 2, 0 ,
where α, β, and γ are the Euler angles. At this stage we may invoke Sakurai Eq. (3.6.52):1
r
4π
(l)
Dm0 (α, β, γ = 0) =
Y m∗ (β, α).
2l + 1 l
Using the expressions Y2m (θ, φ) in Appendix A, we find
r
r r
r
4π 2∗
4π
15
3
(2)
2
D2,0 (α = 0, β, γ = 0) =
Y2 (β, 0) =
(sin β) =
sin2 β,
5
5
32π
8
#
r " r
r
r
4π
4π
15
3
(2)
1∗
Y2 (β, 0) =
(sin β cos β) = −
(sin β cos β),
−
D1,0 (α = 0, β, γ = 0) =
5
5
8π
2
r r
r
4π 0∗
4π
5
1
(2)
Y2 (β, 0) =
(3 cos2 β − 1) = (3 cos2 β − 1),
D0,0 (α = 0, β, γ = 0) =
5
5
16π
2
(2)
(1)
D−1,0 (α = 0, β, γ = 0) = −D1,0 (α = 0, β, γ = 0),
(2)
(1)
D−2,0 (α = 0, β, γ = 0) = D2,0 (α = 0, β, γ = 0).
Thus
2
(2)
D±2,0 (α = 0, β, γ = 0) =
2
(2)
D
(α
=
0,
β,
γ
=
0)
=
±1,0
2
(2)
D0,0 (α = 0, β, γ = 0) =
3
sin4 β,
8
3
sin2 β cos2 β,
2
1
(3 cos2 β − 1)2 .
4
2
P
(2)
It is straightforward to check that 2m=−2 Dm0 (α = 0, β, γ = 0) = 1.
5
Rotation matrix for j = 1 states (Sakurai 3.22)
1
(a). Since Jy = 2i
(J+ − J− ), it is clear that the matrix element hj, m′ | Jy | j, mi vanishes for any
m, m′ where |m − m′ | =
6 1. Also hj, m | Jy | j, m′ i = (hj, m′ | Jy | j, mi)∗ by hermiticity of Jy .
So for j = 1, it is enough to compute the matrix elements h1, 0 | Jy | 1, 1i and h1, 0 | Jy | 1, −1i.
From
Jy | 1, 1i =
Jy | 1, −1i =
1
1 √
1
i~
J+
|
1, 1i
− J− | 1, 1i) = − ( 2~)| 1, 0i = √ | 1, 0i;
(
2i
2i
2
√
1
1
i~
(J+ | 1, −1i − J−
|
1, −1i) = ( 2~)| 1, 0i = − √ | 1, 0i,
2i
2i
2
Please read Sakurai Eqs. (3.6.46) through (3.6.51) and the accompanying text for the derivation.
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PHYS 6572 - Fall 2011
PS7 Solutions
√
√
we deduce that h1, 0 | Jy | 1, 1i = (i~)/ 2 and h1, 0 | Jy | 1, −1i = −(i~)/ 2. So the matrix
representation of Jy in the {| 1, 1i, | 1, 0i, | 1, −1i} basis reads


Jy(j=1) = 
− √i~2
0
0
i~
√
2
i~
√
2
0

√


0
0
0
− 2i
√
√
~
− √i~2 
− 2i  .
 =  2i √0
2
0
2i
0
0
(b). A direct computation shows that
√
√





2
2
0
0
0
− 2i
0
− 2i
2 0 −2
√
√
√
√
~ 
~ 
0 4 0 
[Jy(j=1) ]2 =
2i √0
− 2i   2i √0
− 2i  =
2
2
−2 0 2
0
2i
0
0
2i
0
and
√
√





3
2i
0
2i
0
0
−
0
−
2
0
−2
3
√
√
~  √
~  √
[Jy(j=1) ]3 =
2i √0
− 2i   0 4 0  =
2i √0
− 2i  .
2
2
−2 0 2
0
2i
0
0
2i
0
h
i3
(j=1)
(j=1)
/~, which means that for positive integers n,
In other words, Jy
/~ = Jy
h
Jy(j=1) /~
in
Therefore
e
(1)
−iJy β/~
= 1+
i
 h
 Jy(j=1) /~ , n odd
h
i2
.
=
 Jy(j=1) /~ , n even
∞
X
(−iβ)2n+1
n=0
= 1−i
(2n + 1)!
∞
X
(−1)n β 2n+1
n=0
(2n + 1)!
!
(1)
= 1−i
Jy
~
sin β +
(1)
(1)
Jy
~
!2n+1
(1)
Jy
~
(1)
Jy
~
!
!2
+
+
∞
X
(−iβ)2n
n=1
(2n)!
∞
X
(−1)n β 2n
n=1
(2n)!
(1)
!2n
(1)
!2
Jy
~
Jy
~
(cos β − 1).
(c). The matrix representation d(1) (β) of e−iJy β/~ reads
√




2
0
−
2i
0
2
0
−2
√
i sin β  √
1 
0 4 0 
d(1) (β) = 1 −
2i √0
− 2i  + (cos β − 1)
2
2
−2 0 2
0
2i
0
 1

1
√1
2 (1 + cos β) − 2 sin β 2 (1 − cos β)

cos β
− √12 sin β 
=  √12 sin β
.
1
1
1
√
sin β 2 (1 + cos β)
2 (1 − cos β)
2
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PHYS 6572 - Fall 2011
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PS7 Solutions
Neutrino oscillations
By assumption, the initial state of the neutrino is the weak eigenstate
| ψ(0)i = | νe i = cos θ| ν1 i + sin θ| ν2 i,
where | νj i (j = 1, 2) are the mass eigenstates, and θ is the mixing angle. Since the neutrinos are
assumed free, the mass eigenstates evolves in time according to
q
| νj (t)i = e−iHt/~ | νj (0)i = e−iEj t/~ | νj (0)i where Ej = (mj c2 )2 + (pc)2 .
The assumption that | νe i is a momentum eigenstate allows us to replace the operator p̂ with the
scalar p. As a result, | ψi evolves in time according to
| ψ(t)i = e−iHt/~ | ψ(0)i
= e−iE1 t/~ hν1 | ψ(0)i| ν1 i + e−iE2 t/~ hν2 | ψ(0)i| ν2 i
= e−iE1 t/~ cos θ| ν1 i + e−iE2 t/~ sin θ| ν2 i.
Thus the probability of the system being in | νµ i at time t is
2
|hνµ | ψ(t)i|2 = (− sin θhν1 | + cos θhν2 |) e−iE1 t/~ cos θ| ν1 i + e−iE2 t/~ sin θ| ν2 i 2
= (sin θ cos θ)2 −e−iE1 t/~ + e−iE2 t/~ "
#2
−i(E1 +E2 )t/(2~) −e−i(E1 −E2 )t/(2~) + ei(E1 −E2 )t/(2~) 2
= sin (2θ) ie
2i
(E1 − E2 )t
= sin2 (2θ) sin2
.
2~
Knowing that the mass of the neutrino is very small, i.e., mj c2 ≪ pc, we may approximate Ej in
the usual way:
s
!#
"
q
2 2
2 4
2 2
m
c
m
c
m
c
1
j
j
j
.
+O
= pc 1 +
Ej = (mj c2 )2 + (pc)2 = pc 1 +
pc
2
pc
pc
So
"
1 (m21 − m22 )c4
E1 − E2 = pc
+O
2
(pc)2
mj c2
pc
4 !#
=
(m21 − m22 )c4
+ (higher-order terms).
2pc
Using the shorthand ∆(m2 ) = m21 − m22 , we deduce that
2 3
2
2
2 ∆(m )c t
,
|hνµ | ψ(t)i| ≈ sin (2θ) sin
4p~
which shows that neutrino oscillation occurs at period T = (4πp~)/(∆(m2 )c3 ).
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