PHYS 508 (2015-1)
Homework II
Due date: Nov. 27, 2015, Friday.
1. A charged harmonic oscillator inside an electric field. If the oscillator has
~ = E x̂, the Hamiltonian is
charge q, and the applied electric field is E
H=
p2
1
+ mω 2 x2 − qEx .
2m 2
(1)
(a) Considering
the
E electric-field dependent term as a perturbation, compute En ,
(1)
(2)
En and ψn .
(b) Use the solutions to part (a) for expressing
i. the average electric dipole moment hPin for level n.
ii. The polarizability αn for level n. Does the polarizability depend on the level?
(c) This is a problem that has an exact solution. Note that the potential energy,
1
V (x) = mω 2 x2 − qEx
2
is quadratic in position in such a way
that the quadratic term has the same
coefficient. This implies that the V vs
x graph is a parabola and geometric
shape of the graph does not depend on
the electric field, it is just shifted on the
V –x plane. In other words, the oscillator is still a harmonic oscillator with
the same frequency, but the equilibrium
position is shifted to a new position b
and energy minimum is shifted to a new
value −c. In other words,
1
V (x) = mω 2 (x − b)2 − c .
2
and the problem at hand is the “shifted oscillator”. Consequently, the solution is
trivial.
i. Use this ideas to compute the exact energy En of the oscillator. After that,
(1)
(2)
expand En as a Taylor series in E, find En and En , and compare these to
your results in part (a).
ii. Since this is just a shifted oscillator, the energy eigenstates are also shifted.
Specifically, the nth eigenstate is given by
i
|ψn i = T (b) |ni = exp − bp |ni
~
Page 2 of 4
PHYS 508 - HW 2
where |ni is the unperturbed state and T (b) is the
E translation operator. Ex (1)
pand |ψn i as a Taylor series in E, work out ψn term and compare it with
your solution in part (a).
(d) Perhaps we did not need to use quantum mechanics for computing the polarizability. A harmonic oscillator is essentially a spring. Under a constant force, it
will expand by a constant amount proportional to the applied force (Hooke’s law).
Find the displacement b and the polarizability by this approach.
~ = B0 ẑ +B1 x̂. The Hamiltonian
2. Consider an electron inside a uniform magnetic field B
is
H = µB (B0 σz + B1 σx ) .
Even though this is an exactly solvable problem, we will get the approximate solutions
by perturbation theory. After that, we will compare these with the exact solutions.
We will consider H0 = µB B0 σ)z as the unperturbed Hamiltonian, B1 V = µB B1 σx as
the perturbation and take B1 as the perturbation parameter.
(I) Let us first suppose that B0 6= 0.
E
(0)
(0)
(a) Find the 0th order states and energies, ψn and En .
(1)
(2)
(b) Now, compute En and En for both levels.
(c) Solve the problem exactly and find the exact energy eigenvalues En for both
levels. After that, expand En as a Taylor series in B1 and compare them with
your results in part (b).
(II) Now suppose that B0 = 0. This is a degenerate perturbation theory problem now.
E
(0)
(d) Use the degenerate perturbation theory to find the 0th order states ψn .
(1)
(2)
(1)
(2)
(e) Compute En and En for both levels.
(f ) Compute En and En from the Taylor series expansion of exact solutions
and compare with part (e).
(III) Find the exact energy eigenstates |ψn i of H = µB (B0 σz + B1 σx ) for arbitrary B0
and B1 . Compute the limits
lim |ψn i
B1 →0
for constant B0 for the following two cases: (i) B0 6= 0 and (ii) B0 = 0. This
explains why we have two different expansions in (I) and (II).
Hint: The spin-up state along the unit vector n̂ = sin θ(cos φx̂ + sin φŷ) + cos θẑ
cos 2θ
.
φn̂↑ =
eiφ sin 2θ
3. A function is called concave or convex down if the region below the graph of the function
is a convex region. A function f (λ) of λ is concave if and only if the second-derivative
is negative (or zero) everywhere, f 00 (λ) ≤ 0.
Page 3 of 4
PHYS 508 - HW 2
Consider the Hamiltonian
Hλ = H0 + λV
which depends on the parameter λ linearly. Let E1 ≤ E2 ≤ E3 ≤ · · · be the ordered
sets of eigenvalues of H. You will show that the ground state energy E1 considered as
a function of λ is concave.
(a) Show that
∂ 2 E1 (λ)
∂λ2
≤0
λ=0
by using perturbation theory.
0
(a ) Can you see how you can use the same idea to show that
∂ 2 E1 (λ)
≤0
∂λ2
for all values of λ? (Hint: What can you use for the perturbed Hamiltonian? )
(b) Give an example which shows that E2 does not need to be concave. (Hint: One
of the problems in this homework might be an example.)
(c) Show that the graph of E1 + E2 + · · · + En vs λ is concave.
Remark: In general, if c1 , c2 , . . . is a decreasing sequence of positive coefficients, i.e.,
c1 ≥ c2 ≥ c3 ≥ · · · ≥ 0, then c1 E1 + c2 E2 + · · · is a concave function of λ.
4. Consider the particle-in-a-box problem. A particle with mass m is constrained to move
in the interval 0 < x < L with hard-wall boundaries at x = 0 and x = L. In class, we
have said that
ψtrial (x) = Ax(L − x)
is a very good trial wavefunction (of course, there is no variable parameter in this
case). Compute the estimate for the ground state energy and compare it with the
exact ground state energy by finding the percentage error between them.
5. Particle in infinite spherical well in 3D is an analog of the well-known 1D particle
in a box. In here, the particle feels a constant potential inside the sphere and it cannot
leave out the sphere. In short, the potential energy of the particle is given as
0 if x2 + y 2 + z 2 < R2 ,
V (x, y, z) = V (r) =
.
∞ otherwise
Naturally, the acceptable wavefunctions vanish at the boundary of the box (i.e., ψ(r, θ, φ) =
0 when r = R. This problem has an exact solution in terms of spherical Bessel functions. But a simple variational calculation gives fairly good values for the ground-state
energy.
Page 4 of 4
PHYS 508 - HW 2
(a) Let ψ be a trial wavefunction. Show (or argue) that
R
2
~2 |∇ψ|
hHiψ =
.
R
2m |ψ|2
(b) The simplest trial wavefunction which satisfies the boundary conditions is
ψ = ψ(r, θ, φ) = R2 − r2 .
Compute hHiψ and compare with the exact ground-state energy.
(c) If you wanted to find an approximation for the lowest energy for the p-states or
d-states, which trial wavefunctions would you use?
Note: The exact energy eigenvalues are given by
En`
~2 2
=
x
2mR2 n,`
where xn,` is the nth zero of the spherical Bessel function j` (x). For s-states
(` = 0), we have x1,0 = π.
6. Consider the Hydrogen atom Hamiltonian
H=
e2
p2
−
.
2m
r
Obtain the best estimate of the ground state energy by Gaussian trial functions
ψtrial (x, y, z) = N e−βr
2 /2
,
i.e., find the best value of β and the corresponding estimate. Is it above the exact
energy? What is the percentage error?