L02-182

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IF/UFRJ
Graduate Quantum Mechanics I
2018/2 – Raimundo
Problem Set #2
15th August, 2018
1. Let |ψ〉 represent an arbitrary state for a particle in three-dimensional motion.
(a) Use the closure relation for the basis set {|vp〉} to show that
ψ(r) =
1
(2π~)3/2
∫ ∞
−∞
d3p ψ̃(p) eip·r/~.
(b) Use the closure relation for the basis set {|ξr〉} to show that
ψ̃(p) =
1
(2π~)3/2
∫ ∞
−∞
d3r ψ(r) e−ip·r/~.
2. LetH be the Hamiltonian operator, whose eigenvectors for a particular physical
system are denoted by |ϕn〉, supposed to form a discrete orthonormal basis.
Define an operator
U(m,n) ≡ |ϕm〉〈ϕn|.
(a) Obtain the adjoint U †(m,n) of U(m,n).
(b) Obtain [H, U ].
(c) Show that
U(m,n)U †(p, q) = δn,qU(m, p).
(d) Obtain TrU(m,n).
(e) The matrix elements of an operator A in this basis are Amn ≡ 〈ϕm|A|ϕn 〉.
Show that
A =
∑
m,n
AmnU(m,n).
1
(f) Show that
Apq = Tr [AU
†(p, q)].
3. The Pauli matrices, σ ≡ (σx, σy, σz), play a very important role in the de-
scription of spin-1/2 particles, as we will see throughout this course. In the
representation in which σz is diagonal, they are given by
σx =
(
0 1
1 0
)
, σy =
(
0 −i
i 0
)
, and σz =
(
1 0
0 −1
)
.
(a) Show that these matrices are Hermitian.
(b) Show that these matrices satisfy the commutation relations
[σk, σ`] = 2iεk`mσm,
where the convention of summing over repeated indices is implied.
(c) Show that these matrices anticommute, i.e.
{σk, σ`} ≡ σkσ` + σ`σk = 0, k 6= `.
(d) Show that σ2k = 1, k = x, y, z, where 1 is the 2× 2 identity matrix.
(e) Verify that Trσk = 0, k = x, y, z.
(f) Define σ± ≡ σx ± iσy, and show that
[σz, σ±] = ±2σ±
and
[σ+, σ−] = 4σz.
(g) Determine the eigenvalues and eigenvectors of σx. Comment on the influ-
ence the fact that σ2x = 1 has on the eigenvalues you found.
(h) Calculate the matrices representing the projectors onto the eigenvectors of
σx. Check that the relations of orthogonality and completeness are satisfied.
(i) Express σy and σz on the basis of eigenvectors of σz. You should convince
yourself that the results (a)–(f) do not depend on the basis used to represent
the Pauli matrices.
(j) Show that
eiασk = cosα + i σk sinα, k = x, y, z,
where α is expressed in radians.
2
(k) Show that the previous result can be generalised to
eiαn̂·σ = cosα + i n̂·σ sinα,
where n̂ is a unit vector pointing in an arbitrary direction.
4. The state space of a certain physical system is three-dimensional, and let
{|ui〉}, i = 1, 3, be an orthonormal basis in this space. Define the kets
|ψ0〉 =
1√
2
|u1〉+
i
2
|u2〉+
1
2
|u3〉,
and
|ψ1〉 =
1√
3
|u1〉+
i√
3
|u3〉.
(a) Are these kets normalized?
(b) Determine the matrices ρ0 and ρ1 which represent the projection operators
onto |ψ0〉 and |ψ1〉, respectively. Are these matrices hermitian?
5. The Hamiltonian for a particle of mass m in a one-dimensional motion is given
by
H = P
2
2m
+ V (X), (1)
where X and P are the usual position and momentum operators, respectively,
satisfying [X,P ] = i~, and V (X) is the potential. Let the eigenvectors of H be
denoted by |φn〉, with eigenvalues En, where n is a discrete index.
(a) Show that
〈φn|P |φn′ 〉 = α 〈φn|X|φn′ 〉,
where α depends on the difference between En and En′ . Determine α.
[Hint: consider the commutator [X,H].]
(b) Using the closure relation for this basis, show that∑
n′
(En − En′)2 |〈φn|X|φn′ 〉|2 =
~2
m2
〈φn|P 2|φn 〉.
6. Let H be the Hamiltonian operator for a physical system, with eigenvectors
denoted by |φn〉, with eigenvalues En, where n is a discrete index.
(a) Show that for an arbitrary operator A, one has
〈φn|[A,H]|φn 〉 = 0.
3
(b) Specializing to one dimension, assume H is given by Eq. (1). Then,
(i) Express the commutators [H, P ], [H, X], and [H, XP ] in terms of X
and P .
(ii) Show that the expectation value of the momentum vanishes, i.e.,
〈φn|P |φn 〉 = 0.
(iii) Establish a relation between the expectation value of the kinetic en-
ergy,
〈K〉 = 1
2m
〈φn|P 2|φn 〉,
and the expectation value
〈W 〉 ≡ 〈φn|X
dV
dX
|φn 〉.
(iv) Given that 〈V 〉 = 〈φn|V (X)|φn 〉, what is the relation between 〈V 〉
and 〈K〉, when V (X) = V0Xλ, with λ = 2, 4, 6, . . . ;V0 > 0?
4
Solutions
1. Assume there is a degeneracy, so that ψ1(x) and ψ2(x) correspond to the same
eigenvalue E. The time-independent Schrödinger equation (TISE) can be writ-
ten for both functions as:
− ~
2
2m
d2ψ1,2(x)
dx2
+ V (x)ψ1,2(x) = Eψ1,2(x), (2)
which, due to the single-variable (x) dependence, can be cast in the form
ψ′′1
ψ1
=
ψ′′2
ψ2
=
2m
~2
[V (x)− E]. (3)
The first equality yields
ψ′′1ψ2 − ψ1ψ′′2 = [ψ′1ψ2 − ψ1ψ′2] ′ = 0, (4)
whose solution is
ψ′1ψ2 − ψ1ψ′2 = C. (5)
The constant C is determined from the generic asymptotic behaviour of bound
states:
lim
x→∞
ψi(x) = 0, i = 1, 2 ⇒ C = 0. (6)
With this, Eq. (5) yields
ψ′1
ψ1
=
ψ′2
ψ2
, (7)
which can be integrated ∫
dx
ψ′1
ψ1
=
∫
dx
ψ′2
ψ2
(8)
⇒ lnψ1 = lnψ2 (9)
⇒ ψ1 = λψ2, λ = constant. (10)
Therefore ψ1 and ψ2 can’t be linear independent functions, so that there is no
degeneracy in one-dimensional bound states.
2. (a) By definition,
U †(m,n) = |ϕn〉〈ϕm| = U(n,m). (1)
5
(b) The commutator is given by
[H, U(m,n)] = H|ϕm〉〈ϕn| − |ϕm〉〈ϕn|H
= (Em − En)U(m,n). (2)
(c)
U(m,n)U †(p, q) = |ϕm〉 〈ϕn|ϕq〉︸ ︷︷ ︸
δnq
〈ϕp|
= δnq U(m, p). (3)
(d)
TrU(m,n) = Tr |ϕm〉〈ϕn| =
∑
`
〈ϕ`|ϕm〉〈ϕn|ϕ`〉
=
∑
`
〈ϕn|ϕ`〉〈ϕ`|ϕm〉 = 〈ϕn|
[∑
`
|ϕ`〉〈ϕ`|
]
︸ ︷︷ ︸
=1
|ϕm〉
= 〈ϕn|ϕm〉 = δnm. (4)
(e)
A = 1A 1 =
∑
m,n
|ϕm〉〈ϕm|A|ϕn〉〈ϕn|
=
∑
m,n
〈ϕm|A|ϕn 〉︸ ︷︷ ︸
Amn
|ϕm〉〈ϕn|︸ ︷︷ ︸
U(m,n)
=
∑
m,n
Amn U(m,n). (5)
Note that while Amn is a c-number, U(m,n) is an operator; hence the above
sums do not lead to a c-number, but to an operator.
6
(f)
Apq = 〈ϕp|A|ϕq 〉 = 〈ϕp|1A |ϕq 〉
=
∑
m
〈ϕp|ϕm〉〈ϕm|A|ϕq 〉
=
∑
m
〈ϕm|A|ϕq 〉〈ϕp|ϕm〉
=
∑
m
〈ϕm|A |ϕq〉〈ϕp|︸ ︷︷ ︸
=U†(p,q)
|ϕm〉
=
∑
m
〈ϕm|AU †(p, q)|ϕm 〉
= TrAU †(p, q). (6)
3. (a) Trivial.
(b)
σxσy =
(
0 1
1 0
)(
0 −i
i 0
)
=
(
i 0
0 −i
)
= iσz (1)
σyσx =
(
0 −i
i 0
)(
0 1
1 0
)
=
(
−i 0
0 i
)
= −iσz, (2)
or, subtracting the equations,
[σx, σy] = 2iσz, (3)
which can be easily generalised to the other components,
[σk, σ`] = 2iεk`mσm, (4)
(c) Adding Eqs. (1) and (2), and generalising to other components, we may
write
{σk, σ`} = 0, k 6= `. (5)
(d) By inspection, σ2k = 1,∀k.
(e) By inspection, Trσk = 0,∀k.
(f) With
σ± ≡ σx ± iσy, (6)
we calculate
[σz, σ±] = [σz, σx]± i[σz, σy] = 2iσy ± i(−2iσx)
= ±2σ±, (7)
7
and
[σ+, σ−] = [σx + iσy, σx − iσy] = −2i[σx, σy].
= 4σz, (8)
(g)
det(σx − λ1) = det
(
−λ 1
1 λ
)
= λ2 − 1 (9)
⇒ λ = ±1. (10)
The eigenvectors are obtained solving the following equation for b (a is
determined by normalisation),(
∓1 1
1 ∓1
)(
a
b
)
=
(
0
0
)
⇒ b = ±a, (11)
or
|±〉x =
1√
2
[|+〉 ± |−〉] . (12)
Since σ2k = 1, it follows that λ
2 = 1, so that λ = ±1 was already expected.
4. With the potential being V (x) = λ δ(x), we refer to x < 0 as Region I, and to
x > 0 as Region II.
(a) For the repulsive case, λ > 0, the positive energy solutions are plane waves
both for x < 0 and x > 0. We consider plane waves with unit amplitude
incident on the potential from Region I,
ψI(x) = e
ikx +B e−ikx (1)
ψII(x) = C e
ikx, (2)
subject to the boundary condition,
ψI(0) = ψII(0) ⇒ 1 +B = C. (3)
Note that the presence of a singular potential precludes the use of the
usual boundary condition ψ′I(0) = ψ
′
II(0). Nonetheless, the TISE can be
integrated across the potential,∫ ε
−ε
dx
[
d
dx
ψ′
]
=
2mλ
~2
∫ ε
−ε
dx δ(x)ψ − 2mE
~2
∫ ε
−ε
dxψ (4)
⇒ ψ′II(0)− ψ′I(0) =
2mλ
~2
ψ(0) (5)
⇒ ik(1−B − C) = 2mλ
~2
(1 +B). (6)
8
Solving Eqs.(3) and (6) for the two unknowns B and C yields
B = − mλ
mλ+ i k~2
(7)
C =
i k~2
mλ+ i k~2
, (8)
with k =
√
2mE/~2. The transmission and reflection coefficients are
T = |C|2 = 2E~
2
mλ2 + 2E~2
, (9)
R = |B|2 = mλ
2
mλ2 + 2E~2
, (10)
We note that T is a monotonically increasing function of E, growing from
T = 0 to T = 1 as E increases; this should be contrasted with the total
transmission found in the case of a finite potential barrier, at energiescor-
responding to wave-vectors satisfying the interference condition, ka = nπ,
a being the width of the barrier, and n an integer.
(b) For the attractive case, λ < 0, we need to distinguish two situations, E < 0
and E > 0:
(i) E < 0. The wave function must decay exponentially away from x = 0,
so we may write
ψI(x) = A e
x/ξ (11)
ψII(x) = B e
−x/ξ, (12)
with the boundary conditions,
ψI(0) = ψII(0) ⇒ A = B, (13)
and another one resulting from Eq. (5),
ψ′II(0)− ψ′I(0) =
2mλ
~2
ψ(0) (14)
⇒ ξ = − ~
2
mλ
. (15)
A, in turn, is determined from wave function normalisation:
2 |A|2
∫ ∞
0
dx e−2x/ξ = 1 (16)
⇒ A = 1√
ξ
(17)
9
The energy is determined when we insert the wave function in, say
Region I into the TISE:
− ~
2
2m
d2
dx2
ψI = −
~2
2m
1
ξ2
A ex/ξ = EA ex/ξ (18)
⇒ E = − ~
2
2mξ2
= −mλ
2
2~2
, (19)
so that there is only one bound state. This is in contrast with the
square potential well (depth V0, width a), for which the number of
bound states increases as the parameter γ ≡
√
2mV0a2/~2 increases.
(ii) E > 0. The situation is equivalent to the repulsive case: a wave
incident from the left is transmitted and reflected, with the same
coefficients as in (a) above.
5. We need to evaluate 〈x2〉 in the momentum representation (we omit the tilde
in ψ),
〈x2〉 = −~2
∫ ∞
−∞
dp ψ∗(p, t)
∂2ψ(p, t)
∂p2
. (1)
To this end we use
ψ∗
∂2ψ
∂p2
=
∂
∂p
[
ψ∗
∂ψ
∂p
]
− ∂ψ
∗
∂p
∂ψ
∂p
, (2)
which, when integrated in dp, yields
− ~2
∫
dp ψ∗
∂2ψ
∂p2
= ~2
∫
dp
∂ψ∗
∂p
∂ψ
∂p
, (3)
given that ∫
dp
∂
∂p
[
ψ∗
∂ψ
∂p
]
= 0, (4)
since ψ(p) is well behaved as p→ ±∞.
Now we consider the time-dependent momentum distribution for the free packet
as
ψ(p, t) = φ(p) e−iεpt/~ (5)
ψ(p, t)∗ = φ(p) eiεpt/~ (6)
with εp = p
2/2m, so that
∂ψ
∂p
=
(
φ′ − i pt
~m
φ
)
e−iεpt/~ (7)
∂ψ∗
∂p
=
(
φ′ + i
pt
~m
φ
)
eiεpt/~, (8)
10
with φ(p) real. Taking these into (3), we get
〈x2〉t = ~2
∫
dp
[
φ′
2
+
φ2(p)p2
~2m2
t2
]
, (9)
with
〈x2〉t=0 = ~2
∫
dp φ′
2
. (10)
Similarly,
〈x〉t = i~
∫
dp φφ′︸ ︷︷ ︸
=〈x〉t=0=0, by hypothesis
+
t
m
∫
dp φ2 p︸ ︷︷ ︸
〈p〉
=
〈p〉t
m
. (11)
We finally arrive at
(∆x)2 = (∆x)2t=0 +
(∆p)2t2
m2
, (12)
with
(∆x)2t=0 = ~2
∫
dp φ2(p), (13)
(∆p)2t=0 =
∫
dp φ2(p) p2︸ ︷︷ ︸
〈p2〉
−
[∫
dp φ2(p) p
]2
︸ ︷︷ ︸
〈p〉2
. (14)
6. (a) Given
ψ(x, 0) =
1
(σ2π)1/4
e−x
2/2σ2 eik0x, (1)
the initial momentum distribution is the Fourier transform of the wave
packet,
ψ̃(p, 0) =
1√
2π~
∫ ∞
−∞
dx ψ(x, 0) e−ipx/~
=
(
σ2
π~2
)1/4
e−(p−p0)
2σ2/2~2 , (2)
where the integral was evaluated by completing the squares.
(b) The uncertainties in x and p can be read off straight from the gaussian
form, exp[−y/2(∆y)2], for |ψ(x, 0)|2 and |ψ̃(p, 0)|2:
|ψ(x, 0)|2 = 1√
πσ2
e−x
2/σ2 ⇒ ∆x = σ√
2
(3)
|ψ̃(p, 0)|2 =
√
σ2
π~2
e−(p−p0)
2σ2/~2 ⇒ ∆p = ~√
2σ
(4)
11
Therefore, the uncertainty at t = 0 is indeed minimum,
∆x∆p = ~/2 (5)
(c) The current density is defined as
j(x) =
~
2mi
{
ψ∗(x, 0)ψ′(x, 0)− ψ(x, 0)ψ∗′(x, 0)
}
, (6)
so it is convenient to write the initial packet as
ψ(x, 0) =
1
(σ2π)1/4
ef(x), with f(x) = ik0x− x2/2σ2, (7)
so that
dψ(x, 0)
dx
= f ′(x)ψ(x, 0), with f ′(x) = ik0 − x/σ2. (8)
Similarly,
dψ∗(x, 0)
dx
= [f ′(x)]∗ ψ∗(x, 0), since f ∗(x)′ = [f ′(x)]∗ (9)
Therefore,
j(x) =
~
2mi
ρ(x) {2i Im [f ′(x)]} (10)
= ρ(x) v0, (11)
where ρ(x) ≡ |ψ(x, 0)|2 and v0 ≡ ~k0/m.
(d) The time-dependent packet is
ψ(x, t) =
1√
2π~
∫
dp ψ̃(p) ei(kx−εpt/~) (12)
=
√
σ√
π(σ2 + i~t/m)
exp
[
− (x− v0t)
2
2(σ2 + i~t/m)
]
exp[ik0(x− v0t/2)],
(13)
where the integral was evaluated by completing the squares. As before, the
width of the packet is read off |ψ(x, t)|2,
∆x(t) =
σ√
2
√
1 +
~2t2
σ4m2
, (14)
which is the same as that of Problem 5 .
We note that the centre of the packet moves with the group velocity v0,
while the phase velocity is v0/2.
12
(e) Here again we may write
ψ(x, t) = A(t) exp f(x, t), (15)
leading to
j(x, t) =
~
2mi
ρ(x, t) {2i Im [f ′(x, t)]} (16)
=
~k0
m
1− 2
(
x
k0
− ~t
m
)
~t
m
σ4 +
(~t
m
)2
 ρ(x, t). (17)
If one sits at the maximum of the packet, xm = v0t, then
j(xm, t) = ρ(xm, t) v0, (18)
the same as for t = 0.
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