Quantum Mechanics PHYS 212B
Problem Set 1
Due Tuesday, January 12, 2016
Exercise 1.1 Consider two identical linear oscillators each with spring constant k. Additionally, those
oscillators are coupled by an interaction term a x1 · x2 , where x1 and x2 are the oscillator (displacement)
variables, and a is a constant. Find the energy levels for this system. Hint: transform to new coordinates.
Solution 1.1
The Hamiltonian of the system is
H=
p21
p2
1
+ 2 + k(x21 + x22 ) + ax1 x2
2m 2m 2
Perform coordinate transformation to get rid of the cross term ax1 x2 with
√
√
2 0
2 0
x1 =
(x1 + x02 ), x2 =
(x − x02 ),
2
2 1
and p’s also satisfy the same transformation
√
2 0
(p + p02 ),
p1 =
2 1
√
p2 =
2 0
(p − p02 ).
2 1
The Hamiltonian can be expressed as
H=
p02
p02
1
1
1
02
+ 2 + (k + a)x02
1 + (k − a)x2
2m 2m 2
2
which is composed by two independent oscillators with different spring constants k ± a, respectively, with
angular frequencies
r
r
k+a
k−a
ω1 =
and ω2 =
m
m
where |a| ≤ k. Therefore the energy level will be
1
1
En1 +n2 =
+ n1 ~ω1 +
+ n2 ~ω2
2
2
where n1 = 0, 1, 2, . . . , n1 = 0, 1, 2, . . .
Exercise 1.2
Consider an operator A obeying the commutation relations,
[A, Jz ] =
1
A
2
1
3
(AJ2 + J2 A) + A
2
16
where J is the angular momentum of a system. Operators such as A arise in the decay of systems which
emit particles of half-integral spin. By employing these commutation relations find “selection rule” for
the operator A in a matrix representation which makes the z-component of angular momentum Jz and J2
diagonal (corresponding eigenvlues of m and j(j + 1), respectively). In other words, which matrix elements
hj 0 m0 |A|jmi can be non-zero? (To simplify things let Xj = j(j + 1).)
[[A, J2 ], J2 ] =
1
Solution 1.2
For Jz and J2 , we have
Jz |jmi = m|jmi,
J2 |jmi = j(j + 1)|jmi.
Using the first commutation relation,
hj 0 m0 |[A, Jz ]|jmi =
1 0 0
hj m |A|jmi
2
with Jz |jmi = m|jmi, the equation above can be simplified as
1
(m − m0 − )hj 0 m0 |A|jmi = 0
2
Therefore, the selection rule for m is ∆m = 1/2.
Using the second commutation relation, we have
3
1
hj 0 m0 |[[A, J2 ], J2 ]|jmi = hj 0 m0 | (AJ2 + J2 A) + A|jmi
2
16
with J2 |jmi = Xj |jmi, the equation above can simplified as
3
1
2
hj 0 m0 |A|jmi = 0
(Xj − Xj 0 ) − (Xj + Xj 0 ) −
2
16
Then, the selection rule is
1
3
(Xj − Xj 0 )2 − (Xj + Xj 0 ) −
=0
2
16
Exercise 1.3 An electron is initially spin up along the p
ẑ-axis. This state is then rotated by an angle θ
~
about an axis θ̂, where θ~ = θθ̂ = θx x̂ + θy ŷ + θz ẑ and θ = θ~ · θ.
(a) What is the probability (not amplitude) that the electron is still spin up along the original ẑ-axis?
Spin down along this axis?
(b) Now suppose that an electron initially spin up along the ẑ − axis is sequentially rotated first by θz
about ẑ, then by θy about ŷ, and finally by θx about x̂. What is the probability that the electron is spin up
along the original ẑ-axis?
Solution 1.3
(a)
Consider rotation operator
"
θ~ · ~σ
R(θ) = exp −
2
#
= cos(θ/2)I − i sin(θ/2)θ̂ · σ̂
!
θ
cos θ2 − i sin θ2 θθz −i sin θ2 ( θθx − i θy )
=
θ
−i sin θ2 ( θθx + i θy ) cos θ2 + i sin θ2 θθz
The amplitude that find electron spinning up
h↑ |R(θ)| ↑i =
1
0
T
cos θ2 − i sin θ2 θθz
θ
−i sin θ2 ( θθx + i θy )
!
θ
θ
θ θz
−i sin θ2 ( θθx − i θy )
1
= cos
− i sin
,
θ
θ θz
0
2
2
θ
cos 2 + i sin 2 θ
2
and then
2
θ θz
θ
2
P (↑) = |h↑ |R(θ)| ↑i| = cos
+ sin
2
2 θ2
2
Similar for
2
2
θ θx + θy2
P (↓) = |h↓ |R(θ)| ↑i| = sin
2
θ2
2
(b)
2
The rotation operator becomes
R = eiσx θx eiσy θy eiσz θz
Let us calculate eiσx θx eiσy θy first,
θx
θx
θy
θy
iσx θx iσy θy
= cos
e
e
+ i sin
σx cos
+ i sin
σy
2
2
2
2
θx
θy
θx
θy
θx
θy
θx
θy
= cos
cos
I + i sin
cos
σx + i cos
sin
σy − i sin
sin
σz
2
2
2
2
2
2
2
2


θ
θ
θ
θ
cos θ2x cos 2y − i sin θ2x sin 2y
i sin θ2x cos 2y + cos θ2x sin 2y

=
θ
θ
θ
θ
cos θ2x cos 2y + i sin θ2x sin 2y
i sin θ2x cos 2y − cos θ2x sin 2y
where σx σy = iσz is used. We know | ↑i is the eigenstate of z rotation,
θz
θz
iσz θz
e
| ↑i = cos
− i sin
| ↑i
2
2
Then
θz
θx
θy
θx
θy
θz
h↑ |e
e
e
− i sin
cos
cos
− i sin
sin
| ↑i = cos
2
2
2
2
2
2
θ
θ
θ
θ
θ
θ
z
z
x
y
x
y
h↓ |eiσx θx eiσy θy eiσz θz | ↑i = cos
− i sin
i sin
cos
− cos
sin
2
2
2
2
2
2
iσx θx iσy θy iσz θz
The final probability
θx
θy
θx
θy
2
2
2
e
e
| ↑i| = cos
P (↑) =|h↑ |e
cos
+ sin
sin
2
2
2
2
θx
θy
θx
θy
P (↓) =|h↓ |eiσx θx eiσy θy eiσz θz | ↑i|2 = sin2
cos2
+ cos2
sin2
2
2
2
2
iσx θx iσy θy iσz θz
2
2
3