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L.J. Sham May 4, 2009
Physics 130A Mid-Term Examination
Name:
Problem 1(a) 1(b) 1(c) 1(d) 1(e) 2 or 3 Total
Grade
Out of 10 10 10 10 10 50 100
Do problem 1 and choose either problem 2 or problem 3.
- Short questions – Answer with brief reasoning:
(a) A quantum state of a particle at an instant is given by the wave function ψ(x), where x
is the position. How would you obtain information on the particle’s (i) position and (ii)
energy?
Solution –
i. |ψ(x)|
2 gives me the probability density for the position and therefore, all the prob-
ability information I can have on position in that instant, such as the probability of
the particle being in a space interval and the position expectation value and uncer-
tainty. If I also have its Hamiltonian (which I need for the next part), then I can
also solve the time-dependent Schr¨odinger equation with a given wave function as
the initial boundary condition and obtain the time dependence of this information.
ii. If I have the Hamiltonian Hˆ, I can find its eigenvalues and eigenstates. By the ex-
pansion of ψ(x) as a linear combination of the energy eigenstates, I obtain the co-
efficients as the energy probability (or probability density) amplitude. I can obtain
the probability distribution on energy and its expectation value and uncertainty,
etc.
(b) A particle of mass m moving in the x, y plane has a potential given by,
V (x, y) = V 1 (x) + V 2 (y). (1)
Find the equations for the separate functions of x, y, t of a separable solution of the
time-dependent Schr¨odinger equation.
Solution – Let the wave function be Ψ(x, y, t) = X(x)Y (y)T (t). Substitution into
the time-dependent Schr¨odinger equation and division by the wave function yields,
X(x)
2
2 m
∂x
X(x) +
Y (y)
2
2 m
∂y
Y (y)
T (t)
−iℏ
∂t
T (t)
Since each term in the equation is a function of one variable, it must be a constant.
Thus, we have the separate equations,
(
2
2 m
∂x
X(x) = E 1 X(x), (3)
2
2 m
∂y
Y (y) = E 2 Y (y), (4)
−iℏ
∂t
T (t) = ET (t), (5)
E 1 + E 2 = E. (6)
(c) A wave function ψ(x) is given in terms of φ(k) by the relation,
ψ(x) =
dk
2 π
e
ikx φ(k). (7)
What is the momentum expectation value in terms of φ(k)?
Solution –
〈pˆ〉 =
dx; ψ
∗ (x)ˆpψ(x) =
dk
2 π
φ
∗ (k)ℏkφ(k). (8)
(d) A particle of mass m moves in one dimension confined by the potential
V (x) = λx
4 , (9)
where λ is a constant. By using the uncertainty principle or otherwise, give your best
order of magnitude estimate of the ground-state energy in terms of the given constants
and universal physical constants.
Solution – Use the uncertainties to estimate the energy expectation value,
E = 〈 Hˆ〉 =
(∆p)
2
2 m
4
2
2 m(∆x)
2
4 , (10)
(a) The state at time t is,
Ψ(x, t) =
ψ 0 (x)e
iα−iE 0 t/ℏ
−iE 1 t/ℏ
The probability at time t of finding the n-th eigenstate or the energy value En, denoted
by Pn(t) is given by the square of the coefficient in the series expansion for Ψ(x, t).
Thus,
P 0 (t) =
e
−iE 0 t/ℏ
2
=
P 1 (t) =
e
−iE 1 t/ℏ
2
=
P 2 (t) = 0. (18)
(b) The mean energy is given by,
〈 Hˆ(t)〉 = 〈Ψ(t)| Hˆ|Ψ(t)〉 =
n
Pn(t)En =
(E 0 + E 1 ) = ℏω. (19)
The uncertainty in energy ∆E is,
(∆E)
2 = 〈[ Hˆ(t) − 〈 Hˆ(t)〉]
2 〉 =
n
Pn(t) (En − ℏω)
2
[
1 2
2
3 2
2 )
]
(ℏω)
2
1 4
(ℏω)
2
. (20)
As a check, the uncertainty of equal probabilities between two values should be half
the distance between the two values, i.e.,
∆E =
1 2
(E 1 − E 0 ) =
ℏω. (21)
(c)
〈x(t)〉 =
ψ 0 (x)e
iα−iE 0 t/ℏ
−iE 1 t/ℏ
|x|
ψ 0 (x)e
iα−iE 0 t/ℏ
−iE 1 t/ℏ
The diagonal terms 〈ψn|x|ψn〉 = 0 from mirror symmetry about x = 0. The off-
diagonal terms give,
〈x(t)〉 = e
−iα−iωt 〈ψ 0 |x|ψ 1 〉 + e
iα+iωt 〈ψ 1 |x|ψ 0 〉. (23)
The second term is the complex conjugate of the first. So we need only evaluate one,
〈ψ 1 |x|ψ 0 〉 =
2 mω
〈ψ 1 |
c + c
†
|ψ 0 〉 =
2 mω
Hence,
〈x(t)〉 =
2 mω
e
−iα−iωt
iα+iωt
2 mω
cos(α + ωt) (25)
2 mω
sin(ωt), (26)
since e
iα = i is given.
(d) (a) and (b) are unchanged since the probabilities for the energy states are phase inde-
pendent. (c) is changed to, by Eq. (25)
〈x(t)〉 =
2 mω
cos(α + ωt) (27)
2 mω
cos(ωt), (28)
- A particle of mass m is confined to move along the x-axis in the region 0 < x < L.
(a) Find the eigenenergies and the corresponding normalized eigen-wave functions of the
particle.
(b) At time t = 0, the wave function is
Ψ(x, t = 0) =
L
[
sin
πx
L
2 πx
L
)]
Find the wave function at time t.
(c) Find the mean and uncertainty in energy at time t.
(d) Find the mean momentum at time t.
Useful integrals:
For any nonzero integer n,
0
dx sin
2 (nπx) =
1 2
0
dx sin(2πx) cos(πx) =
4 3 π
Solution –
(a) Let the eigenenergy be E = (ℏk)
2 / 2 m. Then the time-independent Schr¨odinger equa-
tion in the zero potential region may be written in the form,
d
2 ψ(x)
dx
2
= −k
2 ψ(x). (32)
As a check, the uncertainty between only two values should be half the distance be-
tween the two values, i.e.,
∆E =
1 2
(E 2 − E 1 ) =
E 1 =
2 π
2
mL
2
(d) The momentum expectation value is of the form,
〈pˆ(t)〉 = 〈Ψ(t)|pˆ|Ψ(t)〉 = 〈[ψ 1 c 1 (t) + ψ 2 c 2 (t)] |pˆ| [ψ 1 c 1 (t) + ψ 2 c 2 (t)]〉. (42)
The diagonal terms 〈ψn|pˆ|ψn〉 = 0 from mirror symmetry about L/ 2. The off-diagonal
terms give,
〈pˆ(t)〉 = c
∗ 2 (t)c^1 (t)〈ψ^2 |pˆ|ψ^1 〉^ +^ c
∗ 1 (t)c^2 (t)〈ψ^1 |pˆ|ψ^2 〉.^ (43)
The second term is the complex conjugate of the first. So we need only evaluate one,
〈ψ 1 |pˆ|ψ 2 〉 =
L
∫ L
0
dx sin
2 πx
L
i
∂x
sin
πx
L
2 ℏπ
iL
2
∫ L
0
dx sin
2 πx
L
cos
πx
L
2 ℏπ
iL
0
dy sin(2πy) cos(πy)
2 ℏπ
iL
3 π
3 iL
The product of the coefficients from Eq. (36) is,
c
∗ 2 (t)c^1 (t)^ =
ie
−iE 2 t/ℏ
e
−iE 1 t/ℏ
= −i
e
iE 2 t/ℏ
e
−iE 1 t/ℏ
= −i
e
i(E 2 −E 1 )t/ℏ = −i
e
i 3 ωt , (45)
where, for the last line,
E 2 − E 1
ℏπ
2
2 mL
2
≡ 3 ω. (46)
Hence,
〈pˆ(t)〉 = −i
e
i 3 ωt 8 ℏ
3 iL
3 L
cos(3ωt). (47)
