Download 1. Wave Packet and Heisenberg Uncertainty Relations En

1. Wave Packet and Heisenberg Uncertainty Relations
En electron is described via a wave-packet
Z
∞
dkφ(k)ei[kx−ω(k)t] ,
Ψ(x, t) =
(1)
−∞
where the frequency is ω(k) = h̄k 2 /(2m). The expansion coefficients φ(k) have a large
and narrow peak at k = k0 . The width of the peak is finite. We assume that the
distribution of the wave vectors is Gaussian:
r
φ(k) =
σ
σ2
√ exp − (k − k0 )2 ,
2
2π π
(2)
where σ is a finite and positive constant. What are the expectation values of the
variance of the position ∆x = (hx̂2 i − (hx̂i)2 )1/2 and the momentum hp̂x i?
A: ∆x = 0 and hp̂x i = h̄k0 .
B: ∆x = 0 and hp̂x i = 0.
√
C: ∆x = σ/ 2 and hp̂x i = h̄k0 .
√
D: ∆x = σ 2 / 2 and hp̂x i = h̄k0 .
E: ∆ = 0 and hp̂x i = h̄k0 − h̄σ.
Solution: C
From dimensional analysis of the distribution function of Eq. (2), we see that σ has dimensions of length since the dimension of the wave vector is inverse length. Therefore,
options A, B, C, and E have the correct dimensions for the variance of the position.
Furthermore, the dimension of the momentum must be equal to the dimensions of h̄k0 .
This rules out option E. Since the Gaussian is assumed to be narrow and is peaked
around k = k0 then the expectation of the momentum operator must be h̄k0 . This
rules out option B. Finally, it is stated this is a wave-packet with a finite distribution
width, represented by the length σ. It is therefore unlikely that the variance of the
position should vanish as in option A. The correct solution must therefore be option
C.
1
Explicitly, we
value of the momentum:
Z ∞can computethe expectation
h̄
∂
dxΨ∗ (x, t)
Ψ(x, t)
(3)
hp̂x i =
i ∂x
−∞
Z ∞
Z ∞
Z ∞
=
dx
dk
dk 0 h̄kφ∗ (k 0 )φ(k) exp i(k − k 0 )x exp i(ω(k) − ω(k 0 ))t (4)
−∞
−∞
Z ∞ −∞
dk |φ(k)|2 h̄k
(5)
= 2π
−∞
Z ∞
σ
=√
dkh̄k exp −σ 2 (k − k0 )2
(6)
π −∞
Similarly, we
compute the expectation value of the position to the n-the power:
Z can
∞
hx̂n i =
dxΨ∗ (x, t)xn Ψ(x, t)
(7)
−∞
Z ∞
Z ∞
Z ∞
=
dx
dk
dk 0 xn φ∗ (k 0 )φ(k) exp i(k − k 0 )x exp i(ω(k) − ω(k 0 ))t (8)
−∞
−∞
−∞
n
Z ∞
Z ∞
Z ∞
∂
0 ∗ 0
exp i(k − k 0 )x
(9)
=
dx
dk
dk φ (k )φ(k) −i
∂k
−∞
−∞
−∞
n
Z ∞
Z ∞
∂
0 ∗ 0
dk φ (k )φ(k) −i
dk
δ(k − k 0 )
(10)
= 2π
∂k
−∞
−∞
n
Z ∞
∂
dkφ(k) −i
= 2π
φ∗ (k)
(11)
∂k
−∞
We then find that hx̂i = 0 and
hx2 i = σ 2 /2
(12)
√
so that ∆x = σ/ 2.
2. Planck’s Radiation Law
Electromagnetic waves in a cavity satisfy the same boundary conditions as matter de
Broglie waves. It can then be shown that the number of states dN in phase space
V d3 p is
dN = 2
1
V d3 p ,
(2πh̄)3
(13)
where d3 p = 4πp2 dp, p is the momentum and V is the volume of the cavity. Photons
are bosons and the probability to occupy a state at energy E at temperature T is
fB (E, T ) =
1
eE/(kB T )
−1
.
(14)
In this case, the energy density (energy per frequency and volume) of the electromagnetic waves can be expressed in terms of the frequency f , the temperature T , the
speed of light c and Planck’s constant h as
2
A:
u(f ) = hf
1
ehf /(kB T )
−1
,
(15)
B:
u(f ) =
8πh
f3
,
c3 ehf /(kB T ) − 1
(16)
u(f ) =
f
c2
,
hf
/(k
BT ) − 1
8πh e
(17)
u(f ) =
c
f5
,
8πh ehf /(kB T ) − 1
(18)
u(f ) =
8πh
1
.
3
hf
/(k
T) − 1
B
c e
(19)
C:
D:
E:
Solution: B
The relation between the energy and the momentum of a massless particle is E = pc,
where c is the speed of light. The energy of a photon is E = hf , where h is Planck’s
constant and f is the frequency. We then find that dp/df = hf /c. The number of
states in the frequency interval f to f + df is then
1
h
V 4π( )3 f 2 df ,
3
(2πh̄)
c
8π
= 3 V f 2 df .
c
dN = 2
(20)
(21)
Since the energy of one photon per volume is hf /V , we then find that the energy
density of the photons (energy per frequency and volume) is
hf dN
fB (hf, T ) ,
V df
8πh
f3
= 3 hf /(k T )
.
B
c e
−1
u(f ) =
(22)
(23)
3. Scattering by a Step Potential
We consider a one-dimensional system, where there is no potential when x < 0. There
is a finite and positive potential V0 when x > 0. A particle is incident from the left
with energy E < V0 . The wave function can be written as

 eikx + re−ikx , x ≤ 0
ψ(x) =
,

Ce−κx
, x≥0
3
(24)
where E = h̄2 k 2 /(2m) and V0 − E = h̄2 κ2 /(2m). What is the correct form of the
reflection amplitude r in this case?
A:
r=
ik + κ
,
ik − κ
(25)
r=
k−κ
,
k+κ
(26)
r=
k+κ
,
k−κ
(27)
B:
C:
D:
r = 1,
(28)
E:
r=
ik
.
κ + ik
(29)
Solution: A
The energy of the particle is less than the potential barrier it enters, E < V0 . It
is therefore certain that the particle cannot continue as a propagating wave into the
barrier. The reflection probability must therefore be equal to one, R = |r|2 = 1. From
their definitions, k and κ are real numbers. Therefore, only solutions A and D satisfy
the condition that R = 1. For a barrier of a finite width, the particle can always
penetrate slightly into the barrier causing a phase shift of the reflected wave. Hence,
option D cannot be the correct solution and we are left with option A as the only
possible choice.
We can also explicitly calculate the result. The potential is finite. In this case, the
wave function and the derivative of the wave function must be continous everywhere,
also at x = 0. At x = 0, we then find the continuity conditions
1+r =C,
−κC
1−r =
.
ik
4
(30a)
(30b)
By subtracing (30b) from (30a), we can solve for the coefficient C and find
C=
2ik
.
ik − κ
(31)
In turn, by inserting the result for the coefficient C into (30a), we find
r=
ik + κ
.
ik − κ
(32)
4. Planck’s constant
Planck’s constant has the same units as
A: frequency,
B: the Hamiltonian,
C: angular momentum,
D: de Broglie wavelength,
E: momentum.
Solution: C
Recall that e.g. the eigenvalue of the orbital angular momentum is Lz = mh̄, where
m is a dimensionless quantum number.
5. The Dipole Approximation
The dipole approximation for the light-matter interaction uses the fact that
A all interactions on the atomic scale are suppressed by a factor α (the fine structure
constant),
B: the charge distribution of all matter on the atomic scale can be very well approximated by its dipole moment,
C: the vector potential of all relevant field modes is approximately stationary on the
atomic scale,
D: there is relativistic invariance,
E: we can use the Coulomb gauge.
5
Solution: B
In the dipole approximation, we utilize the fact that the wavelength of the electromagnetic field is typically much larger than the atomic size. It is then sufficient to
approximate the charge distribution on the atomic scale by its dipole moment.
6. The Lifetime of Excited States in the Hydrogen Atom
The spontaneous transition rate associated with the light-matter interaction is
(sp)
wif
3
4ωif
= α 2 |df i |2 ,
3c
(33)
where α ≈ 1/137 is the fine structure constant. The dipole moment is
Z
df i = drψf∗ (r)rψi (r)
(34)
in terms of the initial state wave function ψi (r) and the final state wave function ψf (r).
We consider the lifetime of the low-excited states in the hydrogen atom. The energy
difference between the levels is of the order of 10eV . The Bohr radius is a0 = 0.5 ×
10−10 m. The electron charge is e ≈ 1.6 × 10−19 C. Planck’s constant is roughly
h̄ ∼ 10−34 J · s. The speed of light is around c ≈ 3 · 108 m/s.
What is a rough estimate of the lifetime τ of the low-lying excited states?
A τ ∼ 10−15 s ,
B: τ ∼ 10−12 s ,
C: τ ∼ 10−3 s ,
D: τ ∼ 10−6 s ,
E: τ ∼ 10−9 s .
Solution: E
The dipole moment must be on the scale of the atom, d ∼ a0 . The lifetime is proportional to the inverse spontaneous transition rate. We then find that
ωf3i
∼ α 2 |a0 |2
c
τ ∼ 1/wi→f
6
−1
∼ 10−9 s .
(35)
7. The Position Representation in the General Formulation
In the general formulation of quantum mechanics, we consider the position representation. In this case, |xi is an eigenvector to the position operator x̂ with an eigenvalue
x:
x̂|xi = x|xi .
(36)
There is also a momentum operator p̂. By using the canonical commutation relation,
we can then find that
A
hx2 |p̂|x1 i = 0 ,
(37)
B:
hx2 |p̂|x1 i =
h̄ ∂ 2
δ(x2 − x1 ) ,
i ∂x22
(38)
C:
hx2 |p̂|x1 i = δ(x2 − x1 ) ,
(39)
D:
hx2 |p̂|x1 i =
h̄ ∂
δ(x2 − x1 ) ,
i ∂x2
(40)
h̄
δ(x2 − x1 ) .
i
(41)
E:
hx2 |p̂|x1 i =
In this problem it may or may not be useful to know the following property of the
Dirac’s delta-function (distribution), xδ 0 (x) = −δ(x).
Solution: D
The canonical commutation relation is
[p̂, x̂] =
7
h
.
i
(42)
We can use the canonical commutation relation to find
h̄
hx2 | [p̂, x̂] x1 i = hx2 | |x1 i ,
i
h̄
hx2 |(p̂x̂ − x̂p̂|x1 i = δ(x2 − x1 ) ,
i
h̄
hx2 |p̂|x1 i(x1 − x2 ) = δ(x2 − x1 ) ,
i
1
h̄
δ(x2 − x1 ) ,
hx2 |p̂|x1 i =
i (x1 − x2 )
h̄ ∂
hx2 |p̂|x1 i =
δ(x2 − x1 ) .
i ∂x2
(43)
(44)
(45)
(46)
(47)
8. The Momentum Representation in the General Formulation
In the general formulation of quantum mechanics, we consider the momentum representation. In this case, |pi is an eigenvector to the momentum operator p̂ with an
eigenvalue p:
p̂|pi = p|pi .
(48)
We then have that
A
hp2 |p̂|p1 i = 0 ,
(49)
B:
hp2 |p̂|p1 i =
h̄ ∂ 2
δ(p2 − p1 ) ,
i ∂p22
(50)
C:
hp2 |p̂|p1 i = δ(p2 − p1 ) ,
(51)
D:
hp2 |p̂|p1 i =
h̄ ∂
δ(p2 − p1 ) ,
i ∂p2
(52)
E:
hp2 |p̂|p1 i = p2 δ(p2 − p1 ) ,
(53)
where p̂ is the momentum operator. In this problem it may or may not be useful
to know the following property of the Dirac’s delta-function (distribution), xδ 0 (x) =
−δ(x).
8
Solution: E
We use that
hp2 |p̂|p1 i = hp2 |p1 |p1 i ,
(54)
= p1 δ(p1 − p2 ) ,
(55)
= p2 δ(p1 − p2 ) .
(56)
9. The Matrix Representation of Operators
We assume that the state vector |ψi can be expanded in an orthonormal basis {|ki}:
|ψi =
X
ψk |ki .
(57)
k
The state vector |ψi can also be expanded in another orthonormal basis {|k 0 i}:
|ψi =
X
ψk0 |k 0 i .
(58)
k0
We consider an operator  that can act on the state vector |ψi. We define the matrix
elements Aik = hi|Â|ki and A0i0 k0 = hi0 |Â|k 0 i.
The relation between the matrix representations of the operator  in the two different
representations is
A
A0 = SAS ,
(59)
A0 = SAS −1 ,
(60)
A0 = S −1 AS −1 ,
(61)
A0 = A ,
(62)
A0 = SS −1 A ,
(63)
B:
C:
D:
E:
9
where the elements of the matrix S are Si0 i = hi0 |ii and (S −1 )ii0 = hi|i0 i.
Solution: B
In the second basis, the matrix elements of the operator  are
A0i0 k0 = hi0 |Â|k 0 i
(64)
We use the completeness relation
X
|kihk| = 1
(65)
k
twice in Eq. (64) to find
A0i0 k0 =
X
hi0 |iihi|Â|kihk|k 0 i ,
(66)
Si0 i hi|Â|ki(S −1 )kk0 .
(67)
ik
A0i0 k0 =
X
ik
Since the matrix element of the operator  in the first basis is
Aik = hi|Â|ki
(68)
we find from Eq. (67) that the relations between the matrix representations of the
operator  in the two different basis representations is
A0 = SAS −1 .
(69)
10. The Ladder Operators for a 1D Harmonic Oscillator
We consider a 1D harmonic oscilator with the Hamilton operator
Ĥ =
1
p̂2
+ mω 2 q̂ 2 ,
2m 2
(70)
where p̂ is the momentum operator and q̂ is the position operator. We introduce new
operators
r
mω
i
q̂ + √
p̂ ,
2h̄
2mh̄ω
r
mω
i
†
â =
q̂ − √
p̂ .
2h̄
2mh̄ω
â =
The commutation relations between â and ↠are then
10
(71)
(72)
A
†
â, â = 1 ,
(73)
↠, â = 1 ,
(74)
† h̄
â, â = ,
i
(75)
†
i
â, â = ,
h̄
(76)
†
â, â = 0 .
(77)
B:
C:
D:
E:
Solution: A
The commutator between â and ↠is
r
mω
i
mω
i
p̂,
p̂ ,
q̂ + √
q̂ − √
2h̄
sh̄
2mh̄ω
2mh̄ω
r
†
mω
1
√
â, â = −2i
[q̂, p̂] ,
2h̄ 2mh̄ω
†
h̄
1
â, â = −2i
−
,
2h̄
i
†
â, â = 1 .
†
â, â =
r
(78)
(79)
(80)
(81)
11. The Number Representation for a 1D Harmonic Oscillator
We consider a 1D harmonic oscilator with the Hamilton operator
Ĥ =
1
p̂2
+ mω 2 q̂ 2 ,
2m 2
(82)
where p̂ is the momentum operator and q̂ is the position operator. We introduce new
operators
r
mω
i
q̂ + √
p̂ ,
2h̄
2mh̄ω
r
mω
i
†
â =
q̂ − √
p̂ .
2h̄
2mh̄ω
â =
11
(83)
(84)
In terms of these new operators, the Hamiltonian of Eq. (82) can be written as
1
†
.
(85)
Ĥ = h̄ω â â +
2
In this case, |ni is an eigenstate for Ĥ with eigenvalue En = h̄ω(n + 1/2):
1
Ĥ|ni = h̄ω n +
|ni .
2
(86)
We assume that all the eigenstates are orthonormal. It can then be shown that
bn |n − 1i = â|ni ,
(87)
where bn is a coefficient to be determined from the normalization conditions. Carrying
out this task and choosing bn to be real and positive, we find
A
√
n,
(88)
p
n + 1/2 ,
(89)
bn =
B:
bn =
C:
bn =
√
n + 1,
(90)
D:
bn =
p
n + 3/2 ,
E:
bn =
√
n + 2.
(91)
(92)
Solution: A
The state |n − 1i should be normalized:
hn − 1|b∗n bn |n − 1i = |bn |2 ,
(93)
hn|↠â|ni = |bn |2 ,
!
Ĥ
1
hn|
−
|ni = |bn |2 ,
h̄ω 2
1 1
|ni = |bn |2 .
hn| n + −
2 2
(94)
12
(95)
(96)
Since bn is chosen to be real and positive, we find that
bn =
√
n.
(97)
12. Angular Momentum
Quantum systems can have an angular momentum J that cannot be expressed in terms
of the position r and the momentum p as the orbital angular momentum L = r × p.
In general, angular momentum operators are defined in terms of the following commutation rules.
A
i h̄
Jˆx , Jˆy = Jˆz ,
i
h
i h̄
Jˆy , Jˆz = Jˆx ,
i
h
i h̄
Jˆz , Jˆx = Jˆy ,
i
h
(98)
(99)
(100)
B:
h
i
ˆ
ˆ
Jx , Jy = ih̄Jˆz ,
h
i
Jˆy , Jˆz = ih̄Jˆx ,
h
i
ˆ
ˆ
Jz , Jx = ih̄Jˆy ,
(101)
(102)
(103)
C:
h
i
ˆ
ˆ
Jx , Jy = 0 ,
h
i
Jˆy , Jˆz = 0 ,
h
i
Jˆz , Jˆx = 0 ,
(104)
i
ˆ
ˆ
Jx , Jy = ih̄Jˆx ,
h
i
Jˆy , Jˆz = ih̄Jˆy ,
h
i
Jˆz , Jˆx = ih̄Jˆz ,
(107)
(105)
(106)
D:
h
13
(108)
(109)
E:
i h̄
Jˆx , Jˆy = Jˆx ,
i
h
i h̄
Jˆy , Jˆz = Jˆy ,
i
h
i h̄
Jˆz , Jˆx = Jˆz .
i
h
(110)
(111)
(112)
Solution: B
The general angular momentum operators must satisfy the same commutation relations as the orbital angular momentum operator. In the position representation, we
have
L̂ = r × p̂ = r ×
h̄ ∂
.
i ∂r
(113)
For the various components, we then get
L̂x = y p̂z − z p̂y ,
(114)
L̂y = z p̂x − xp̂z ,
(115)
L̂z = xp̂y − y p̂x .
(116)
The commutation relations between L̂x and L̂y are then
h
i
L̂x , L̂y = [y p̂z − z p̂y , z p̂x − xp̂z ] ,
(117)
= [y p̂z , z p̂x ] + [y p̂z , −xp̂z ]
(118)
+ [−z p̂y , z p̂x ] + [−z p̂y , −xp̂z ]
= [y p̂z , z p̂x ] + [−z p̂y , −xp̂z ]
h̄
= (y p̂x − xp̂y ) ,
i
= ih̄L̂z .
(119)
(120)
(121)
(122)
We can find the two other commutation relations by cyclic permutations:
h
i
L̂y , L̂z = ih̄L̂x ,
h
i
L̂z , L̂x = ih̄L̂y .
14
(123)
(124)
13. The Electron Spin
The electron spin can be represented in terms of the two eigenvectors corresponding
to the angular momentum operators, |1/2, 1/2i and |1/2, −1/2i. The matrix representation of the basis vectors can then be represented by


h 21 , 21 | 12 , 12 i

| ↑i = 
1
1 1 1
h2, −2|2, 2i
 
1
= 
0
(125)
(126)
and

h 12 , 12 | 21 , − 12 i
| ↓i = 
h 21 , − 21 | 12 , − 12 i
 
0
= .
1


(127)
(128)
In the matrix representation, the electron spin operator can be expressed in terms of
the Pauli matrices σ = (σx , σy , and σz ):
S=
where

σx = 

σy = 
and

σz = 
h̄
2
σ
0 1
1 0
(129)

,
0 −i
i 0
1 0
0 −1
(130)

,
(131)

.
(132)
We assume that the spin state is an eigenstate of the electron spin operator along the
x-direction. The normalized 2-component eigenstate ψx is then
A
 
1
ψx =   ,
0
15
(133)
B:
 
0
ψx =   ,
1
C:
D:
E:

1
(134)

1
,
ψx = √ 
2 −i
(135)
 
1
1
ψx = √   ,
2 1
(136)
 
1
1
ψx = √   ,
2 i
(137)
Solution: D
The system is in an eigenstate ψx corresponding to the electron spin along the xdirection. Then, we must have
h̄
σx ψx = h̄ms ψx ,
2
(138)
where ms is the spin quantum number associated with the projection along the xdirection. To find the eigenstate

ψx = 
ψx↑
ψx↓

,
we must then find the eigenvalues of the equation



1
−ms 2
ψ
  x↑  = 0 .
h̄ 
1
−ms
ψx↓
2
(139)
(140)
By requiring that the determinant of the matrix vanishes, we determine the eigenvalues:
m2s
2
1
−
=0
2
(141)
so that mS = ±1/2. Since the electron is in a state that corresponds to an eigenstate
of the spin along the x-direction, we choose ms = 1/2 and find for the components of
the state that
−ψx↑ + ψx↓ = 0 .
16
(142)
A normalized solution is then
 
1
1
ψx = √   .
2 1
(143)
14. The Electron Spin in a Magnetic Field
We consider the electron spin in an external magnetic field along the x-direction. In
the matrix representation, the Hamiltonian can be represented by


0 1
,
H = h̄ω 
1 0
(144)
where ω is the Larmor frequency.
We assume that the initial state (at t = 0) is an eigenstate of the electron spin operator
along the z-direction,
 
1
ψ(t = 0) =   .
0
(145)
At later times (t > 0), the state will then evolve into
A:
 
1
ψ(t) =   ,
0
B:

eiωt
(146)

1
,
ψ(t) = √ 
2 e−iωt
C:

ψ(t) = 
D:
−i sin ωt
ψ(t) = 
E:

ψ(t) = 
17

0

cos ωt
0
(147)
,
(148)

,
cos ωt
−i sin ωt
(149)

.
(150)
Solution: E
The evolution of the eigenstates is determined by the Schrödinger equation:
Hψ(t) = ih̄
∂
ψ(t)
∂t
(151)
We express the 2-component state as

ψ(t) = 
a(t)
b(t)

.
The time-dependent equation we must solve is then





0 1
a(t)
a(t)

 = ih̄ ∂ 

h̄ω 
∂t
1 0
b(t)
b(t)
(152)
(153)
In other words, we need to solve the time-dependent coupled equations
∂
a(t) ,
∂t
∂
h̄ωa(t) = ih̄ b(t) .
∂t
h̄ωb(t) = ih̄
(154a)
(154b)
A general solution of Eq. (154) is
a(t) = A+ eiωt + A− e−iωt ,
(155a)
b(t) = −A+ eiωt + A− e−iωt .
(155b)
At t = 0, we must have a(t = 0) = 1 and b(t = 0) = 0, which gives the following
equations for the coefficients A+ and A− :
A+ + A− = 1 ,
(156a)
−A+ + A− = 0 ,
(156b)
with the solutions A+ = 1/2 and A− = −1/2 so that the state is


cos ωt
.
ψ(t) = 
−i sin ωt
(157)
15. Addition of Angular Momenta of Two Spin 1/2 Particles
We consider the total spin angular momentum Ŝ of two spin 1/2 particles, Ŝ = Ŝ1 + Ŝ2 .
Ŝ1 and Ŝ2 are the spin operators associated with particle 1 and 2, respectively. The
eigenvalue of Ŝ2 is h̄2 s(s + 1) and the eigenvalue of Ŝz is h̄m. The possible values of s
and m are
18
A: s = 0 and m = 0, ±1 or s = 1 and m = 0 ,
B: s = 0 and m = 0 or s = 1 and m = 0, ±1 ,
C: s = 1/2 and m = ±1/2 or s = −1/2 and m = ±1/2 ,
D: s = 0 and m = 0 or s = 1 and m = 1 ,
E: s = 0 and m = 0 or s = 1/2 and m = ±1/2 .
Solution: B
There are two spin 1/2 particles. The total spin quantum number s is then s =
1/2 − 1/2 = 0 or s = 1/2 + 1/2 = 1. When s = 0, we must have m = 0. When s = 1,
we can have s = 0, ±1. There are two possibles states for each particle, so in total for
the two spin 1/2 particle there are four possible states.
16. Time-dependent Perturbation Theory
We consider a system that is subject to a weak time-dependent perturbation described
by the Hamiltonian
Ĥ(t) = Ĥ0 + λV̂ (t) ,
(158)
where Ĥ0 is time-independent and λV̂ (t) is the perturbation. We assume that we
know the eigenstates of Ĥ0 ,
Ĥ0 |ψn(0) (t)i = ih̄
∂ (0)
|ψ (t)i
∂t n
(159)
so that
|ψn(0) (t)i = e−iEn t/h̄ |ψn i ,
(160)
where {|ψn i} are stationary states with associated eigenenergies {En }:
Ĥ0 |ψn i = En |ψn i .
(161)
(0)
We expand the exact state |ψi in terms of |ψn (t)i that form complete set
|ψ(t)i =
X
ak (t)e−iEk t/h̄ |ψk i ,
(162)
k
where {ak (t)} are time-dependent expansion coefficients. We can insert the expansion
of Eq. (162) in the Schrödinger equation and then find the time-dependent equation
for the expansion coefficients ak (t):
19
A:
ih̄
X
d
|an (t)|2 =
|ak (t)|2 eiωnk t |λVnk (t)|2 ,
dt
k
(163)
X
d
an (t) =
ak (t)λVnk (t) ,
dt
k
(164)
X
d
an (t) =
ak (t)eiωnk t λVnk (t) ,
dt
k
(165)
X
d
an (t) =
ak (t)eiωnk t ,
dt
k
(166)
d
an (t) = an (t)eiωnn t λVnn (t) ,
dt
(167)
B:
ih̄
C:
ih̄
D:
ih̄
E:
ih̄
where h̄ωnk = En − Ek and λVnk (t) = hψn |λV̂ (t)|ψk i.
Solution: C
The time-dependent Schrödinger equation is
ih̄
∂
|ψi = Ĥ0 + λV̂ (t) |ψi .
∂t
(168)
By inserting the expansion of Eq. (162) and using the properties of the unperturbed
eigenstates of Eq. (159), we find the time-dependent equation
X
k
ih̄
X
dak (t) −iEk t/h̄
e
|ψk i =
λV̂ (t)ak (t)e−iEk t/h̄ |ψk i .
dt
k
(169)
(0)
We take the inner product of Eq. (169) with respect to hψn (t)| = eiEn t/h̄ hψn | and find
ih̄
X
d
an (t) =
ak (t)eiωnk t λVnk (t) ,
dt
k
where h̄ωnk = En − Ek and λVnk (t) = hψn |λV̂ (t)|ψk i.
17. Fermi’s Golden Rule
Fermi’s Golden Rule describes
A: the transition probability,
20
(170)
B: the transition probability per unit time,
C: the transition amplitude ,
D: the transition amplitude per unit time ,
E: the Fermi exclusion principle.
Solution: B
Fermi’s golden rule describes the transition probability per unit time.
18. Scattering Theory - Differential Cross Section
The differential cross section dσ/dΩ is defined as
A:
# particles scattered into dΩ per unit time
,
dΩ · (# incoming particles per area and time)
(171)
# particles scattered into dΩ per unit time
,
dΩ · (# incoming particles)
(172)
# particles scattered into dΩ per unit time
,
dΩ · (# incoming particles per time)
(173)
# particles scattered into dΩ per unit time
,
(# incoming particles per time)
(174)
# particles scattered into dΩ
,
dΩ · (# incoming particles per area and time)
(175)
B:
C:
D:
E:
where # means ”number of”.
Solution: A
The differential cross section dσ/dΩ is defined as
dσ
number of particles scattered into dΩ per unit time
=
.
dΩ
dΩ · (number of incoming particles per area and time)
(176)
19. Scattering Theory - Asymptotic Form of the Wave function
We consider scattering at a potential that is localized at r = 0 and has a finite extent.
An incoming wave of the form
ψin = Ceik·r
21
(177)
travels towards the scattering center, where k is the incoming wave vector. The
coordinate can be expressed in terms of the distance from the origin r and the two
angles θ and φ, r = (r sin θ cos φ, sin θ sin φ, cos θ). The scattered wave ψscat far away
from the scattering center can be expressed as
A:
ψscat = Cf (θ, φ)eikr ,
(178)
ψscat = Cf (θ, φ)
eikr
,
r
(179)
ψscat = Cf (θ, φ)
eikr
,
r2
(180)
ψscat = Cf (θ, φ)
eikr
,
r3
(181)
B:
C:
D:
E:
1
,
(182)
r2
where f (θ, φ) characterizes the scattering strength and the angular dependence
ψscat = C
of the scattering.
Solution: B
The scattered wave must be proportional to the amplitude of the incoming wave,
ψscat ∼ C. Furthermore, the scattered wave must be inversely proportional to the
distance from the origin r to conserve the number of particle scattered out of a cross
section at distance r, |ψscat |2 r2 ∼ constant. Finally, there is also an angular and
strength dependence that is characterized by f (θ, φ). In conclusion, the scattered
wave is of the form
ψscat
eikr
= Cf (θ, φ)
.
r
(183)
20. Gauge Invariance
The electric field E(r, t) and the magnetic field B(r, t) are determined by the electromagnetic vector potential A(r, t) and the scalar potential Φ(r, t),
E(r, t) = −∇Φ(r, t) −
22
∂
A(r, t)
∂t
(184)
and
B(r, t) = ∇ × A(r, t) .
(185)
The electric and magnetic fields do not specify the electromagnetic vector potential
and the scalar potentil uniquely. We can carry out a gauge transformation
A(r, t) → A0 (r, t) = A(r, t) + ∇χ(r, t)
(186)
∂
χ(r, t) ,
∂t
(187)
and
Φ(r, t) → Φ0 (r, t) = Φ(r, t) −
where χ(r, t) is an arbitrary differentiable function. This gauge transformation leaves
E and B invariant.
In the absence of other potentials, the Schrödinger equation for a charged particle in
an electromagnetic field described by the field A and Φ is
(p̂ − qA)2
∂
+ qΦ ψ(r, t) = ih̄ ψ(r, t) .
2m
∂t
(188)
When we use another gauge described by A0 and Φ0 , the Schrödinger equation is
(p̂ − qA0 )2
∂
0
+ qΦ ψ 0 (r, t) = ih̄ ψ 0 (r, t) .
(189)
2m
∂t
The relation between the wave functions ψ and ψ 0 is
A:
ψ0 = ψ ,
(190)
ψ 0 = ψeqχ/h̄ ,
(191)
ψ 0 = ψe−qχ/h̄ ,
(192)
ψ 0 = ψeiqχ/h̄ ,
(193)
ψ 0 = ψe−iqχ/h̄ .
(194)
B:
C:
D:
E:
23
Solution: D
Let us try an ansatz
ψ 0 = ψeqη ,
(195)
where η is a differentiable spatiotemporal function. Inserting Eq. (195) into Eq. (189),
we find
∂
(p̂ − qA0 )2
0
+ qΦ ψ 0 (r, t) = ih̄ ψ 0 (r, t) .
2m
∂t
(196)
∂
∂ 0
∂η
qη
ih̄ ψ (, t) = e ih̄ ψ(, t) + ih̄ ψ(, t)
∂t
∂t
∂t
(197)
We now use that
and
0
qη
p̂ψ (, t) = e
h̄ ∂η
p̂ψ(, t) +
ψ(, t)
i ∂r
to find that Eq. (196) can be written as
"
#
h̄ ∂η
0 2
−
qA
)
(p̂
+
∂η
∂ 0
qη
0
0
qη
i ∂r
e
+ qΦ ψ (r, t) = e ih̄ ψ (r, t) + ih̄ ψ(, t) .
2m
∂t
∂t
(198)
(199)
By using Eq. (187) and Eq. (187), we then find the Schrödinger (196) provided
i
χ(r, t) .
h̄
η(r, t) =
(200)
21. The Pauli Matrices
The Pauli matrices σ = (σx , σy , and σz ) are
S=
where

σx = 

σy = 
and

σz = 
h̄
2
0 1
1 0
(201)

,
0 −i
i 0
1 0
0 −1
The commutator between σx and σy is then
24
σ
(202)

,
(203)

.
(204)
A:
[σx , σy ] = 2iσz ,
(205)
[σx , σy ] = iσx ,
(206)
[σx , σy ] = 2iσx ,
(207)
[σx , σy ] = 2iσy ,
(208)
[σx , σy ] = 0 ,
(209)
B:
C:
D:
E:
Solution: A
We multiply the matrices and find
 




0 −i
1 0
0 1
·
 = i
 = iσz .

1 0
i 0
0 −1
(210)
Similarly, we find


0 −i
i 0
 
·
0 1
1 0


 = −i 
1 0
0 −1

 = −iσz .
(211)
Hence, we have that
[σx , σy ] = 2iσz .
(212)
22. Confinement in a Delta-Function Potential
We consider a particle in a one-dimensional system described by the Hamiltonian
H=−
h̄2 ∂ 2
− βδ(x) ,
2m ∂x2
(213)
where δ(x) is the Dirac δ-function and β is a positive constant. The negative bound
state energy EB is then
A:
Eb = −
25
mβ
,
2h̄2
(214)
B:
Eb = −
mβ
,
16h̄2
(215)
Eb = −
mβ 2
,
16h̄2
(216)
Eb = −
mβ 3
,
2h̄2
(217)
Eb = −
mβ 2
.
2h̄2
(218)
C:
D:
E:
Solution: E
We consider the energy to be negative. The wave function must be normalizable. The
solution when x > 0 must then be of the form
ψ+ (x) = C+ e−κx
(219)
where
h̄2 κ2
.
2m
Similarly, the solution when x < 0 must be of the form
E=−
ψ− (x) = C− eκx .
(220)
(221)
Continuity of the wave function at x = 0 requires
C+ = C− ≡ C .
(222)
Next, we integrate the wave function from x = 0− to x = 0+
Z 0+
Z 0+
Hψ(x) =
Eψ(x) ,
0−
−
(223)
0−
h̄2 dψ(x) 0+
| − − βψ(x = 0) = 0 ,
2m dx 0
h̄2
−
(−2Cκ) − βC = 0 .
2m
(224)
(225)
So that
h̄2
κ.
m
The bound state energy can then be written as
β=
Eb = −
26
mβ 2
.
2h̄2
(226)
(227)
23. The Continuity Equation
The time-dependent Schrödinger equation is
2
p̂
∂ψ
+ V (r) ψ = ih̄
,
2m
∂t
(228)
where p̂ = h̄∇/i is the momentum operator and V (r) is the scalar potential. From
the Schrödinger equation (228), we can derive a continuity equation
∂n
+ ∇ · j = 0,
∂t
(229)
n = ψ(r, t)∗ ψ(r, t)
(230)
j = [ψ ∗ ∇ψ − (∇ψ ∗ )ψ] ,
(231)
j = ψ ∗ ∇ψ ,
(232)
where the particle density is
and the current density is
A:
B:
C:
j=
h̄
[ψ ∗ ∇ψ − (∇ψ ∗ )ψ] ,
2mi
(233)
j=
h̄
[ψ ∗ ∇ψ + (∇ψ ∗ )ψ] ,
2mi
(234)
D:
E:
j=∇·
h̄
[ψ ∗ ∇ψ − (∇ψ ∗ )ψ] .
2mi
(235)
Solution: C
We can use the Schrödinger equation (228) to evaluate the rate of change of the particle
density:
∂n
∂
∂ψ ∗
∂ψ
=
(ψ ∗ ψ) =
ψ + ψ∗
,
∂t
∂t ∂t ∂t ψ p̂2
ψ ∗ p̂2
∗
=−
+ V (r) ψ +
+ V (r) ψ ,
ih̄ 2m
ih̄ 2m
h̄ ∗ 2
=−
ψ ∇ ψ − ψ∇2 ψ ∗ ,
2mi
h̄
∗
∗
=∇ −
[ψ ∇ψ − ψ∇ψ ]
2mi
27
(236)
(237)
(238)
(239)
which we can write as the continuity equation (229) with the current density
j=
h̄
[ψ ∗ ∇ψ − (∇ψ ∗ )ψ] .
2mi
(240)
24. Scattering in One Dimension
We consider scattering in one dimension, the x-direction. An incoming electron from
the left enters a scattering region when 0 ≤ x ≤ a. The electron can be reflected
back in the negative x-direction or transmitted to propagate towards the right when
x > a. We assume the potential vanishes when x < 0 and x > a. In these areas, the
particle propagates freely and the relation between the wavevector k and the energy
E is k = (2mE/h̄2 )1/2 . We assume the energy E is positive. The correct form of the
wave function when x < 0 for this scattering problem is then
A: ψ(x) = eikx + re−ikx ,
B: ψ(x) = ekx + re−kx ,
C: ψ(x) = eikx + re−kx ,
D: ψ(x) = ekx + re−ikx ,
E: ψ(x) = re−ikx .
Solution: A
In the left region, the wave function should describe an incoming wave that propagates
to the right and a reflection wave that propagates to the left. The time-dependent
wave function is Ψ(x, t) = ψ(x) exp iEt/h̄ and we then see that only option A describes
an incoming wave that propagates to the right and a reflected wave that propagates
to the left.
25. Wave Packet - Phase and Group Velocity
A photon is described via a wave-packet
Z ∞
Ψ(x, t) =
dkφ(k) exp i(kx − ω(k)t) ,
(241)
−∞
where the frequency is ω(k) = ck. It is assumed that the expansion coefficients φ(k)
has a large and narrow peak with a finite width at k = k0 . What is the relation
between the phase velocity and the group velocity for this photon?
28
A: vf /vg = 2 .
B: vf /vg = 1/2 .
C: vf /vg = 3 .
D: vf /vg = 1 .
E: vf /vg = ω(k)/k .
Solution: D
The phase velocity is
vf =
ω(k)
= c.
k
(242)
vg =
dω(k)
= c.
dk
(243)
The group velocity is
The relation between the phase velocity and the group velocity is therefore,
vf /vg = 1 .
(244)
26. Density of States
We consider a two-dimensional system of a finite area L2 . Our focus is on the density
of states in the limit that L → ∞. We take into account that there is a factor of 2
degeneracy of each state due to the electron spin. The density of states is in this limit
A:
g(E) = 1/L2 ,
(245)
B:
g(E) =
dN
L2 2m
=
,
dE
2π h̄2
(246)
C:
g(E) = E ,
D:
L
g(E) =
π
r
2m −1/2
E
,
h̄2
(247)
(248)
E:
g(E) = E 3/2 .
29
(249)
Solution: B
Taking into account the spin-degeneracy, the number of states with a wave vector less
than k is
Z
k
0
2πk dk
N =2
0
0
L
2π
2
=
(Lk)2
L2 2mE
=
.
2π
2π h̄2
(250)
The density of states is then
g(E) =
dN
L2 2m
=
,
dE
2π h̄2
(251)
27. Quantum Probabilities
The Hilbert space of a system is spanned by two orthonormal states, |1i and |2i. The
associated eigenenergies are E1 = 1 and E2 = 16, respectively. We assume that the
system is in the state
r
|ψi =
1
|1i +
5
r
4
|2i .
5
(252)
The expectation value of the energy, Eavg , is then
A:
Eavg = 1 ,
(253)
Eavg = 16 ,
(254)
Eavg = 12 ,
(255)
Eavg = 13 ,
(256)
Eavg = 14 .
(257)
B:
C:
D:
E:
Solution: D
The expectation value of the energy is
1
4
Eavg = E1 + E2 = 13
5
5
30
(258)
28. Matrix Mechanics
The Hilbert space of a system is spanned by three orthonormal states, |1i, |2i, and
|3i. Using the matrix mechanics formulation, an operator  can be represented in this
basis as


0 1 0




A=1 0 0


0 0 0
(259)
The possible eigenvalues of the operator  are then
A: -1,0,1 ,
B: -i, 0, i ,
C: -1,1 ,
D: -i, i ,
E: -2,0,2 .
Solution: A
The matrix is block-diagonal. We see that one eigenvalue is zero. The remaining
upper-left corner 2 × 2 matrix equals σx , the Paul matrix, that has eigenvalues -1 and
1. In total, we therefore have the eigenvalues -1,0,1.
29. Addition of Angular Momenta
We consider two non-interacting spinless particles. The first particle is in an eigenstate
corresponding to the its total angular momentum operator
2
L̂1 |l1 , m1 i = h̄2 l1 (l1 + 1)
(260)
with l1 = 10 while the second particle is in an eigenstate corresponding to its total
angular momentum operator
2
L̂! |l2 , m2 i = h̄2 l2 (l2 + 1) .
(261)
with l2 = 5.
The possible values for the total angular momentum operator
L̂2 |l, mi = h̄2 l(l + 1)
31
(262)
are then
A: l = 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 ,
B: l = 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 8.5, 10,
10.5, 11, 11.5, 12, 12.5, 13, 13.5, 14, 14.5, 15 ,
C: l = 5, 10, 15 ,
D: l = 5, 15 ,
E: l = 0, 5, 10, 15 .
Solution: A
The minimum total angular momentum quantum number is lmin = |l1 − l2 | = 5 and
the maximum total angular momentum quantum number is lmin = l1 + l2 = 15. All
possible integral numbers in between lmin and lmax are possible. Hence, the possible
values are l = 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
30. Three-dimensional Box
A particle is confined in a three-dimensional box of a volume V = Lx Ly Lz , where Lx
is the length in the x-direction, y is the length in the y-direction, and z is the length
in the z-direction. The potential vanishes inside the box and is infinitely large outside
the box. The energy levels are then
A:
Enx ,ny ,nz
Enx ,ny ,nz
π 2 h̄2
=
2m
Enx ,ny ,nz
π 2 h̄2
=
m
Enx ,ny ,nz
h̄2
=
2m
B:
C:
D:
E:
n2y
n2z
n2x
+
+
L2x L2y L2z
h̄2
=
2m
,
n2y
n2x
n2z
+
+
L2x L2y L2z
n2y
n2x
n2z
+
+
L2x L2y L2z
nx
ny
nz
+
+
Lx Ly Lz
(263)
,
(264)
,
(265)
,
(266)
π 2 h̄2 nx
ny
nz
Enx ,ny ,nz =
+
+
,
(267)
2m Lx Ly Lz
where nx = 1, 2, . . ., ny = 1, 2, . . ., and nz = 1, 2, . . . are integral numbers.
32
Solution: B
The particle is confined in a three-dimensional box. Since the potential vanishes inside
the box and is infinitely large outside the box, the wave function is separable into in
the three directions. It is then sufficient to consider the one-dimensional box first.
Inside the box, when 0 ≤ x ≤ Lx , the wave function is
ψx (x) = A cos kx + B sin kx .
(268)
The wave function must vanish at the boundary, ψ(x = 0) = 0 and ψ(x = Lx ) = 0.
This gives that A = 0 and kLx = nx π, where nx = 1, 2, . . . is an integral number. The
energy is then
Ex =
h̄2 2 h̄2 π 2 n2x
k =
.
2m
2m L2x
(269)
We get similar results for the energies associated with the motion in the y and z
direction so that the total energy is
EnX ,ny ,nz
h̄2 π 2
=
2m
n2y
n2x
n2z
+
+
L2x L2y L2z
.
(270)
31. The Average Kinetic Energy and the Fermi Energy
We consider an ideal gas of spin 1/2 fermions at zero temperature. The system is
then spin degenerate and there are 2 spin states associated with each orbital state. It
can then be shown that the relation between the number of particles and the Fermi
momentum pF is
Np = 2
V 4π 3
p .
(2πh̄)3 3 F
(271)
The average kinetic energy per fermion hEi is then in terms of the Fermi energy EF
A:
1
hEi = EF ,
2
(272)
hEi = EF ,
(273)
3
hEi = EF ,
2
(274)
B:
C:
33
D:
3
hEi = EF ,
5
(275)
2
hEi = EF .
3
(276)
E:
Solution: D
The total kinetic energy is
Z
Etot =
d3 p2
V
p2
,
(2πh̄)3 2m
(277)
where the factor of 2 is due to spin-degeneracy and the factor V /(2πh̄)3 is the density
of states in momentum space. We should integrate up to the Fermi momentum. We
then find
Etot
Z pF
V
p2
2
= 4π2
p
dp
,
(2πh̄)3 0
2m
V 4π 3 3 pF
=2
p
,
(2πh̄)3 3 3 5 2m
3
= Np EF .
5
(278)
(279)
(280)
The average kinetic energy per particle is therefore hEi = 3EF /5.
32. Classical Scattering Theory
We consider the classical motion of a particle that moves towards a hard wall potential
in the shape of a sphere with radius r. The potential is infinitely large inside the sphere
and vanishes outside the sphere.
The classical total cross section σ associated with the scattering off the hard wall
potential is then
A:
σ=
4π 2
r ,
3
(281)
σ=
4π 3
r ,
3
(282)
B:
C:
σ = r2 ,
34
(283)
D:
σ = 2πr ,
(284)
σ = πr2 .
(285)
E:
Solution: E
When a particle scatters off a hard wall potential, the classical total cross section
equals the cross section of the scattering center, e.g.
σ = πr2 .
(286)
33. One-Dimensional Harmonic Oscillator and Time-Dependence
The Hamiltonian of a one-dimensional harmonic oscillator can be written in terms of
the number operator n̂ as
1
Ĥ = h̄ω n̂ +
2
,
(287)
where ω is the classical oscillation frequency. The time-dependent Schrödinger equation is
Ĥ|ψ(t)i = ih̄
∂
|ψ(t)i .
∂t
(288)
The possible energy eigenstates of the time-dependent Schrödinger equation (288) are
|n(t)i = eiEn t/h̄ |ni ,
(289)
n̂|ni = n|ni
(290)
1
|ψ(t = 0)i = √ [|1i + |2i + |3i] .
3
(291)
where
and En = (n + 1/2)h̄ω.
A system is at t = 0 in a state
At t = t1 ≡ 2πh̄/E1 , the system is then in the state
A:
1 |ψ(t1 )i = √ |1i + ei2π/3 |2i + ei4π/3 |3i ,
3
35
(292)
B:
1 |ψ(t1 )i = √ |1i + ei4π/3 |2i + ei2π/3 |3i ,
3
(293)
1
|ψ(t1 )i = √ [|1i + |2i + |3i] ,
3
(294)
1
|ψ(t1 )i = √ [|1i − |2i + |3i] ,
3
(295)
1 |ψ(t1 )i = √ |1i + ei1π/3 |2i + ei2π/3 |3i .
3
(296)
C:
D:
E:
Solution: B
The state evolves as
|ψ(t = 0)i =
1 iE1 t/h̄
e
|1i + eiE2 t/h̄ |2i + eiE3 t/h̄ |3i .
3
(297)
At t = 2πh̄/E1 , we therefore have
1 |ψ(t = 2πh̄/E1 )i = √ ei2π |1i + ei2πE2 /E1 |2i + ei2πE3 t/E1 |3i ,
3
1 i2π
= √ e |1i + ei2π5/3 |2i + ei2π7/3 |3i ,
3
1 = √ |1i + ei4π/3 |2i + ei2π/3 |3i .
3
(298)
(299)
(300)
34. Energy Uncertainty
A system is described by a time-independent Hamiltonian Ĥ. A particle is in a state
|n(t)i = eiEn t/h̄ |ni
(301)
Ĥ|ni = En |ni .
(302)
where
The uncertainty of the energy in the state
2
∆En = hn(t)|Ĥ |n(t)i − hn(t)|Ĥ|n(t)i
is then
36
2 1/2
(303)
A:
∆En = En /5 ,
(304)
∆En = En /3 ,
(305)
∆En = En /4 ,
(306)
∆En = 0 ,
(307)
∆En = En /2 ,
(308)
B:
C:
D:
E:
Solution: D
The system is in an energy eigenstate. There is then no uncertainty of the energy,
∆En = 0.
35. Matrix Mechanics - Hamiltonian
In the matrix mechanics notation, a system is described by the Hamiltonian H = h̄ωX,
where

X=
0 −i
i 0

.
(309)
What are the possible energy eigenvalues of the system?
A: h̄ω and −h̄ω ,
B: 2h̄ω and −2h̄ω ,
C: 3h̄ω and −3h̄ω ,
D: 4h̄ω and −4h̄ω ,
E: 2ih̄ω and −2ih̄ω .
Solution: A
The eigenvalues of the matrix X are -1 and 1. The energy eigenvalues are therefore
−h̄ω and h̄ω.
37