__Rigid rotator__

The equation below defines the
eigenvalue problem for a two-dimensional, quantum-mechanical, rigid rotator:

-(ħ^{2}/2I)(∂^{2}/∂θ^{2})ψ(θ)
= Eψ(θ),
where ψ(θ
+ 2π) = ψ(θ).

(a)
What are the energies E_{j} and wave functions ψ_{j}(θ) of
its lowest three energy eigenstates?

(b) A perturbation U = β cosθ
is added to its Hamiltonian. Find
the effect on the ground state energy, to second order in β.

Solution:

- Concepts:

The rigid rotator in two dimensions, first and second order perturbation theory. - Reasoning:

We are asked to solve the unperturbed problem and find the energy correction for the ground state due to the perturbation to second order. -
Details of the calculation:

(a) (∂^{2}/∂θ^{2})ψ(θ) + (2I/ħ^{2})Eψ(θ) = 0.

ψ(θ) = A exp(imθ), E = ħ^{2}m^{2}/(2I).

where ψ(θ + 2π) = ψ(θ) --> m = 0, ±1, ±2, ... .

For m ≠ 0 each energy eigenvalue is two-fold degenerate. The normalized eigenfunctions corresponding to E = ħ^{2}m^{2}/(2I) are ψ_{±m}(θ) = (2π)^{-½}exp(±i|m|θ) and linear combinations. ψ_{±m}(θ) are eigenfunctions of L_{z}= (ħ/i)(∂/∂θ) with eigenvalues ±|m|ħ.

(b) H_{0 }+ U. U = β cosθ.

The ground state ψ_{0}(θ) is not degenerate, E_{0}^{0}= 0.

E_{1}^{0}= <ψ_{0}|U|ψ_{0}> = (2π)^{-1}∫_{0}^{2π}β cosθ dθ = 0.

The first order energy correction is zero.

The second order energy correction for the ground state is

E_{2}^{0}= ∑_{m≠0 }|<ψ_{m}|U|ψ_{0}>|^{2}/(E_{0}^{0 }- E_{0}^{m}).

∫_{0}^{2π}exp(imθ) cosθ dθ = 0 unless |m| = 1.

E_{2}^{0}= |<ψ_{1}|U|ψ_{0}>|^{2}/(E_{0}^{0 }- E_{0}^{1}) + |<ψ_{-1}|U|ψ_{0}>|^{2}/(E_{0}^{0 }- E_{0}^{1})

= |<ψ_{1}|U|ψ_{0}>|^{2}/(-ħ^{2}/(2I)) + |<ψ_{-1}|U|ψ_{0}>|^{2}/(-ħ^{2}/(2I))

= -(2Iβ^{2}/ħ^{2})2|(2π)^{-1}∫_{0}^{2π}cos^{2}θ dθ|^{2}= -(Iβ^{2}/ħ^{2}).

Suppose the Hamiltonian of a rigid rotator in a magnetic field
is of the form

H = AL^{2 }+ BL_{z }+ CL_{y},
if terms quadratic in the field are neglected.

Assuming B >> C, use perturbation theory to
lowest non-vanishing order to get appropriate energy eigenvalues.

Solution:

- Concepts:

Stationary perturbation theory - Reasoning:

A small correction term is added to a Hamiltonian H_{0}, whose eigenstates and eigenvalues we can solve for exactly. We are asked to find the lowest order non-vanishing corrections to the energy eigenvalues. -
Details of the calculation:

H = AL^{2 }+ BL_{z }+ CL_{y}, B >> C.

H = H_{0}+ H', H_{0}= AL^{2 }+ BL_{z}.

The eigenstates of H_{0}are {|lm>}. H_{0}|lm> = (Al(l+1)ħ^{2}+ Bmħ)|lm>

The eigenvalues are not degenerate.

The first order corrections are E_{1}^{lm}= C<lm|L_{y}|lm> = -(iC/2)<lm|L_{+}- L_{-}|lm> = 0.

We have to find the second order corrections E_{2}^{lm}.

E_{2}^{lm}= ∑_{l'm'¹lm}C^{2}|<l'm'|L_{y}|lm>|^{2}/(E_{0}^{lm}- E_{0}^{l'm'})

L_{y}= -(i/2)(L_{+}- L_{-})

L_{±}|l,m> = [l(l+1) - m(m±1)]^{½}ħ|l,m±1>.

<l'm'|L_{±}|l,m> = [l(l+1) - m(m±1)]^{½}ħ δ_{l,l'}δ_{m',m±1}.

E_{2}^{lm}= (C^{2}/4)[l(l+1) - m(m+1)]ħ^{2}/(Bħ(m - (m+1))) + (C^{2}/4)[l(l+1) - m(m-1)]ħ^{2}/(Bħ(m - (m-1)))

= +C^{2}mħ/(2B).

The energy eigenvalues to second order are Al(l+1)ħ^{2}+ Bmħ + C^{2}mħ/(2B).

__Infinite well__

A particle of mass m moves in the potential energy function

U(x) = ∞, x < 0, x > a,

U(x) = εx/a, 0 < x < a

where ε is a small perturbation.

Use first-order perturbation theory to find the ground state energy of the particle.

Solution:

- Concepts:

First order perturbation theory for non-degenerate states - Reasoning:

We are asked to solve the unperturbed problem and find the energy correction for the ground state due to the perturbation. - Details of the calculation:

H = H_{0}+ H'. H_{0}is the Hamiltonian of the infinite 1D well, H' = εx/a, 0 < x < a.

The ground state eigenfunction of H_{0}is Φ_{0}= (2/a)^{½}sin(πx/a), the associated eigenvalue is

E_{0}= π^{2}ħ^{2}/(2ma^{2}).

The first order correction to the ground state energy is

ΔE_{0}= < Φ_{0}|H'| Φ_{0}> = (2/a)∫_{0}^{a}sin^{2}(πx/a)(εx/a)dx = (2ε/π^{2}) )∫_{0}^{π}sin^{2}(y) y dy = ε/2.

Ground state energy to first order: E = π^{2}ħ^{2}/(2ma^{2}) + ε/2.

A
particle of mass m is in an infinite potential well perturbed as
shown in the figure.

(a) Calculate the first-order energy shift of
the nth eigenvalue due to the perturbation.

(b) Write out the first three non vanishing terms
for the first-order perturbation expansion of the ground state in terms of the
unperturbed eigenfunctions of the infinite well.

(c) Calculate the second-order energy shift for
the ground state.

Solution:

- Concepts:

First order perturbation theory for non-degenerate states - Reasoning:

The ground state of the infinite square well is non-degenerate. -
Details of the calculation:

(a) H = H_{0 }+ H’.

H_{0}|Φ_{n}> = E_{0}^{n}|Φ_{n}>. Φ_{n}(x) = (2/a)^{½}sin(nπx/a), E_{0}^{n}= n^{2}π^{2}ħ^{2}/(2ma^{2}), n = 1, 2, 3, ... .

The eigenvalues of H_{0}are not degenerate.

H'(x) = 0, 0 < x < a/2, H'(x) = U_{0}, a/2 < x < a. U_{0}= h^{2}/(40ma^{2}) = π^{2}ħ^{2}/(10ma^{2}).

E_{1}^{n}= <Φ_{n}|H'|Φ_{n}> = (2U_{0}/a)∫_{a/2}^{a }dx sin^{2}(nπx/a) = U_{0}/2.

E_{1}^{n}= π^{2}ħ^{2}/(20ma^{2}).

To first order all energies are shifted up by U_{0}/2.

(b) First-order perturbation theory yields for the ket

|ψ_{1}> = |Φ_{1}> + ∑_{n>1}<Φ_{n}|H'|Φ_{1}>/(E_{0}^{1 }- E_{0}^{n}) |Φ_{n}> ,

<Φ_{n}|H'|Φ_{1}> = (2U_{0}/a)∫_{a/2}^{a }dx sin(πx/a)sin(nπx/a)

= (2U_{0}/π)∫_{π/2}^{π }dx sin(x)sin(nx)

= (2U_{0}/π)[sin((n-1)π/2)/(2n-2) - sin((n+1)π/2)/(2n+2)].

<Φ_{n}|H'|Φ_{1}> = 0 if n = odd.

<Φ_{2}|H'|Φ_{1}> = -4U_{0}/(3π), <Φ_{4}|H'|Φ_{1}> = 8U_{0}/(15π).

E_{0}^{1}- E_{0}^{2}= -3π^{2}ħ^{2}/(2ma^{2}). E_{0}^{1}- E_{0}^{4}= -15π^{2}ħ^{2}/(2ma^{2}).

|ψ_{1}> = |Φ_{1}> + [4U_{0}/(3π)]/[3π^{2}ħ^{2}/(2ma^{2})] |Φ_{2}> - [8U_{0}/(15π)]/[15π^{2}ħ^{2}/(2ma^{2})] |Φ_{4}> + ...

= |Φ_{1}> + 8/(90π) |Φ_{2}> + 16/(2250π)|Φ_{4}> + ... .(c) E

_{2}^{1}= ∑_{n>1 }|<Φ_{n}|H'|Φ_{2}>|^{2}/(E_{0}^{1 }- E_{0}^{n})

= -[4U_{0}/(3π)]^{2}/[3π^{2}ħ^{2}/(2ma^{2})] - [8U_{0}/(15π)]^{2}/[15π^{2}ħ^{2}/(2ma^{2})] - ...

= -ħ^{2}/(100ma^{2})[32/27 + 128/3375 + ...] = second-order energy shift for the ground state.

The second-order energy shift of the ground state is always negative.

A particle of mass m is constrained to move in an
infinitely deep, one-dimensional square well extending from -a to +a.

If this
particle is under the influence of a perturbation H' = -Aδ(x),
where A is a constant and δ(x) is a
delta function at x = 0, calculate the first order corrections to the energy
levels.

What is the condition that A must satisfy if the corrected n* energy
level is to have a negative energy?

Solution:

- Concepts:

First order perturbation theory for non-degenerate states, the infinite square well - Reasoning:

The energy level of the 1D infinite square well are non-degenerate. - Details of the calculation:

For the unperturbed Hamiltonian we have:

H = (ħ^{2}/2m)(∂^{2}/∂x^{2}) + U(x), U(x)_{ }= 0 for |x| < a and U(x)_{ }= ∞ for |x| > a.

(∂^{2}/∂x^{2})Φ(x) + (2m/ħ^{2})EΦ(x) = 0 in the region |x| < a, Φ(x) = 0 for |x| = a.

Φ(x) = Cexp(ikx) + Dexp(-ikx), with E = ħ^{2}k^{2}/(2m).

Cexp(ika) + Cexp(-ika) = 0, Cexp(-ika) + Dexp(ika) = 0.

Two solutions:

C = D: coska = 0, ka = nπ/2, n = odd.

C = -D: sinka = 0, ka = nπ/2, n = even.

E_{n}= n^{2}π^{2}ħ^{2}/(8ma^{2}), Φ_{n}(x) = Ncos(nπx/(2a)), n = odd, Φ_{n}(x) = Nsin(nπx/(2a)), n = even.

Normalization: N^{2}= 1/a.Let us calculate the first order energy corrections due to the perturbation.

n = odd:

E_{n}^{1}= -AN^{2}∫_{-a}^{a}cos^{2}(nπx/(2a))δ(x)dx = -(A/a)

n = even:

E_{n}^{1}= -AN^{2}∫_{-a}^{a}sin^{2}(nπx/(2a))δ(x)dx = 0.

E_{n}* = n^{2}π^{2}ħ^{2}/(8ma^{2}) - A/a if n = odd. We need A > n^{2}π^{2}ħ^{2}/(8ma) for E_{n}* to be negative.

E_{n}* = n^{2}π^{2}ħ^{2}/(8ma^{2}) if n = even. The perturbation does not change this energy level to first order.