PHY4605–Spring 2005
Problem Set 1, Solutions
Due: Jan. 14, 2005
1. Inhomogeneous field B = −αxî + (B0 + αz)k̂
(a)
∂
∂
x+0+α z
∂x
∂z
= −α + α = 0
∇ · B = −α
(b) The Hamiltonian is H = µ0~σ · B = µ0 [−αxσx + (B0 + αz)σz ]. We can then
write the time evolution operator as
¶
µ
iHt
U = exp −
h̄
= exp [i(aσx + bσz )] ≡ exp (iθn̂ · ~σ )
= cos θ + in̂ · ~σ sin θ
where
µ0 αxt
h̄
µ0 (B0 + αz)t
θnz = b = −
,
h̄
and n̂ is a unit vector (i.e., n2x + n2z = 1). Then it follows that,
θnx = a =
µ20 t2 2 2
2
2 (α x + (B0 + αz) )
h̄
µ0 t p 2 2
µ0 t √
θ=
α x + (B0 + αz)2 ≡
· · ·.
h̄
h̄
θ2 (n2x + n2z ) =
Hence,
µ
U = cos θ + i
At time t,
µ
χ+
χ−
¶
αx
B0 + αz
√ σx − √
σz sin θ.
···
···
¶
|χ(t)i = U (t)
¶
¶
¶µ
µ
µ
B0 + αz
αx
χ+
σz sin θ
= cos θ + i √ σx − √
χ−
···
···
¶
µ
¶
¶
µ
µ
iαx sin θ χ−
i(B0 + αz) sin θ
χ+
χ+
√
+ √
= cos θ
−
χ−
χ+
−χ−
···
···


iαx sin θχ− i(B0 + αz) sin θχ+
√
√
−
 cos θχ+ +

·
·
·
···

.
=
iαx sin θχ+ i(B0 + αz) sin θχ− 
√
√
+
cos θχ− +
···
···
This shows the full time-dependence of |χi. Note that this time-dependence
is found in θ.
1
(c) Assuming B0 À αx, αz,
µ0 B 0 t
=⇒ θ ≈
h̄
√
· · · ≈ B0
(These become independent of position.) Hence, for χ+ = χ− = 1,


iαx sin θ
iαz sin θ
cos θ +
− i sin θ −


B0
B0
|χ(t)i ≈ 
iαx sin θ
iαz sin θ  .
cos θ +
+ i sin θ +
B0
B0
µ ¶
1
At t = 0, or θ = 0, we have |χi =
.
1
Now we compute the momenta of the spin-up and spin-down components.
For spin-up:
µ
¶
iαx sin θ iαz sin θ
p̂χ+ (t) = (−ih̄∇)
−
B0
B0
h̄α sin θ
=
(x̂ − ẑ).
B0
For spin-down:
µ
p̂χ− (t) = (−ih̄∇)
=
iαx sin θ iαz sin θ
+
B0
B0
¶
h̄α sin θ
(x̂ + ẑ)
B0
So the spin-up and spin-down parts of the initial state are diverging in the
±ẑ directions. The “extra” αx component of the gradient required to have
∇ · B = 0 does not separate the beams, as it gives the same deflection to
both.
2. Stern-Gerlach Effect.
(a) To find the the probability amplitudes parallel and anti-parallel to ŷ, we
project the initial state onto eigenstates of Ŝy :
µ ¶
µ ¶
1
1
1
i
|χ1 i = √
, |χ2 i = √
2 i
2 1
with eigenvalues + h̄2 and − h̄2 respectively. (To check this recall that
µ
¶
0 −i
ˆ
Sy ∝
≡ σy .) So the probability of a particle ending up with a
i 0
spin antiparallel to ŷ is
r µ ¶
¢
1 ¡
1
4
−i 1 ·
P−y = | √
|2
i
17
2
1
9
= | − 4i + i|2 =
34
34
2
Now these spins are passed through a second SG apparatus ⊥ to the first.
The probability of getting a spin parallel to ẑ from a spin-down state in ŷ
is just 12 . Thus, the total fraction of particles which make it through is
1 9
9
·
= .
2 34
68
(b) For S = 1, Sz = 1, 0, or −1. With the eigenstates of Ŝz as basis, Ŝz has
the matrix form:


1 0 0
Sz = h̄  0 0 0 
0 0 −1
since h1, m|Sˆz |1, m0 i = mh̄δm,m0 . We also need the eigenstates of Ŝy . Construct first the matrix:
1
Sˆy |1, m0 i = (Sˆ+ − Sˆ− )|1, m0 i
2i
p
h̄ p
= ( 1 · 2 − m0 (m0 + 1)|1, m0 + 1i − 1 · 2 − m0 (m0 − 1)|1, m0 − 1i).
2i
With this the matrix elements are given by
p
h̄ p
h1, m|Sˆy |1, m0 i = [ 2 − m0 (m0 + 1)δm,m0 +1 − 2 − m0 (m0 − 1)δm,m0 −1 ].
2i
Explicitly, we have:
√




0
2
0
0
1
0
√
√
h̄ 
h̄
= √
−1 0 1  .
Sy =  − 2
0
2
√
2i
i
2
0 −1 0
0
− 2 0
The eigenvectors (which you can get with Maple or by hand) are then:
 




−1
−1
1
√
√
1
1
1
|χ1 i = √  0  ; |χ2 i =  i 2  ; |χ3 i =  −i 2  ,
2
2
2
1
1
1
corresponding to λ1 = 0, λ2 = −h̄, and λ3 = +h̄.
need to express this state in terms
Case 1 (initial state is |1, 1i, m = 1). WeP
of the eigenstates of Sˆy . Recall |1, mi = n hχn |1, mi|χn i by completeness.
The projection hχn |1, mi onto the n-th Sˆy eigenstate gives the probability
amplitude that the spin will emerge in the n-th beam:
   1
1
 √2 , n = 1
hχn |1, 1i = hχn |  0  =
−1, n = 2
 21
0
−2, n = 3
The intensity is proportional to |hχn |1, 1i|2 , and so the ratios of the intensities are 14 : 12 : 41 for Sy = 1 : 0 : −1.
3
Case 2 (initial state is |1, 0i, m = 0).
 
n=1
0
 0,
i
√ ,
n=2
hχn |1, 0i = hχn |  1  =
 √2i
0
− 2, n = 3

The intensities are related according to
1
2
:0:
Case 3 (initial state is |1, −1i, m = −1).
  
0

hχn |1, −1i = hχn |  0  =

1
The intensities are related according to
1
4
:
1
2
:
1
2
for Sy = 1 : 0 : −1.
√1 ,
2
1
,
2
1
,
2
1
4
n=1
n=2
n=3
for Sy = 1 : 0 : −1.
3. Oscillating B-field.
(a)
γB0 h̄
H = −γB · S = −γB0 cos ωt Sz = −
cos ωt
2
µ
1 0
0 −1
¶
(b)
µ
χ(t) =
α(t)
β(t)
¶
1
, with α(0) = β(0) = √
2
µ
¶
µ
¶µ
¶
∂χ
γB0 h̄
α̇(t)
1 0
α(t)
ih̄
= ih̄
= Hχ = −
cos ωt
0 −1
β(t)
β̇(t)
∂t
2
µ
¶
γB0 h̄
α(t)
= −
cos ωt
−β(t)
2
µ
α̇ =
⇒
⇒
⇒
µ
¶
γB0
dα
i
cos ωt α ⇒
=i
cos ωt
α
2
iγB0 sin ωt
+ const.
log α =
2ω
√
α = Aei(γB0 /2ω) sin ωt ; α(0) = A = 1/ 2
1
α(t) = √ ei(γB0 /2ω) sin ωt
2
γB0
2
¶
Similarly
1
β(t) = √ e−i(γB0 /2ω) sin ωt
2
4
(c) Call the probability amplitude of getting −h̄/2 with a measurement of Sx
cx− . Then we have
cx−
!
Ã
√1 ei(γB0 /2ω) sin ωt
1
2
= χx− · χ = √ (1 − 1) ·
1 −i(γB0 /2ω) sin ωt
√
e
2
2
µ
¶
µ
¶
1
1 i(γB0 /2ω) sin ωt
1 −i(γB0 /2ω) sin ωt
γB0
√ e
= √
−√ e
= i sin
sin ωt
2ω
2
2
2
Probability is |cx− |2 = sin2
¡ γB0
2ω
¢
sin ωt .
2
(d) Argument of sin must reach π/2, so we need γB0 /2ω = π/2, or B0 =
πω/γ.
5