Consider a two-dimensional infinite potential square well
of width L,

(U = 0 for 0 < x, y < L, U = infinite everywhere else) with an added perturbation

H' = g sin(2πx/L)sin(2πy/L).

(a) Calculate the first order perturbation to the ground
state energy eigenvalue.

(b) Calculate the first order perturbation to the first excited state energy
eigenvalue.

Solution:

- Concepts:

Stationary perturbation theory for non-degenerate and for degenerate states - Reasoning:

The ground state of two-dimensional infinite potential square well is not degenerate. The first excited state is two-fold degenerate. - Details of the calculation:

H = H_{0}+ H’.

The eigenfunctions of H_{0}are Φ_{nl}(x,y) = (2/L)sin(nπx/L)sin(lπy/L), with eigenvalues

E = (n^{2}+ l^{2})π^{2}ħ^{2}/(2mL^{2}).

(a) For the ground state n = l = 1. The ground state is not degenerate.

The unperturbed ground state energy is E_{0}^{0}= ħ^{2}/(mL^{2}). The first order perturbation correction is

E_{1}^{0}= <Φ_{11}|H’|Φ_{11}> = (g4/L^{2})∫_{0}^{L}sin^{2}(πx/L)^{ }sin^{2}(πy/ L)sin(2πx/L)^{ }sin(2πy/L)dxdy.

= (4g/L^{2})∫_{0}^{L}sin^{2}(πx/L)^{ }sin(2πx/L)dx)∫_{0}^{L}sin^{2}(πy/L)^{ }sin(2πy/L)dy = 0.(b) The first excited state is two-fold degenerate.

We can have n = 2, l = 1, or n = 1, l = 2.

We have to diagonalize the matrix of H’ in the subspace spanned by these two degenerate states.

<Φ

_{12}|H’|Φ_{12}> = (4g/L^{2})∫_{0}^{L}sin^{2}(πx/L)sin(2πx/L)dx)∫_{0}^{L}sin^{2}(2πy/L)^{ }sin(2πy/L)dy = 0.

<Φ_{21}|H’|Φ_{21}> = (4g/L^{2})∫_{0}^{L}sin^{2}(2πx/L)sin(2πx/L)dx)∫_{0}^{L}sin^{2}(πy/L)^{ }sin(2πy/L)dy = 0.

<Φ_{12}|H’|Φ_{21}> = (4g/L^{2})∫_{0}^{L}sin(πx/L)^{ }sin^{2}(2πx/L)dx)∫_{0}^{L}sin^{2}(2πy/L)^{ }sin(πy/L)dy = C

C = (4g/π^{2})(∫_{ 0}^{π}sin(x)^{ }sin^{2}(2x)dx)^{2}= (4g/π^{2})*64/225.

<Φ_{12}|H’|Φ_{21}> = (4g/L^{2})∫_{0}^{L}sin(πx/L)^{ }sin^{2}(2πx/L)dx)∫_{0}^{L}sin^{2}(2πy/L)^{ }sin(πy/L)dy = C

E^{2}– C^{2}= 0. E_{1}^{1}= ±C. The perturbation removes the degeneracy.

A particle of mass m is
trapped in a 3-dimensional, infinite square well.

V(x,y,z) = 0,
if x, y, and z are less than a and not negative,

V(x,y,z) = ∞, otherwise.

(a) What are the two lowest energy eigenvalues? Are these energies
degenerate?

(b) The box is placed into a uniform gravitational field with
gravitational acceleration **g** in
the negative z-direction. Use
perturbation theory to calculate first-order corrections to the energy eigenvalues
from part (a).

Solution:

- Concepts:

Three-dimensional square well potentials, perturbation theory - Reasoning:

We are supposed to find first order energy corrections to the two lowest energy eigenvalues of the 3D square well. - Details of the calculation:

(a) The eigenfunctions and eigenvalues of the 3D infinite square well are

Φ_{nml}(x,y,z) = (2/a)^{3/2}sin(k_{x}x)sin(k_{y}y)sin(k_{z}z), with k_{x}= πn/a, k_{y}= πq/a, k_{z}= πl/a, n, q, l = 1, 2, ... .

The associated eigenvalues are

E_{nql}= (n^{2}+ q^{2}+ l^{2})π^{2}ħ^{2}/(2ma^{2}) = (k_{x}^{2}+ k_{y}^{2}+ k_{z}^{2})ħ^{2}/(2m).

(b) The two lowest energy eigenvalues are

E_{0}^{1}= E_{111}= 3π^{2}ħ^{2}/(2ma^{2}), E_{0}^{2}= E_{112}= E_{121}= E_{211 }= 6π^{2}ħ^{2}/(2ma^{2}).

The eigenvalue E_{0}^{2}is three-fold degenerate.

(b) The perturbation is W = mgz.

E_{1}^{1}= <Φ_{111}|W|Φ_{111}> = (2/a)^{3}mg∫_{0}^{a}sin^{2}(πx/a)dx ∫_{0}^{a}sin^{2}(πy/a)dy ∫_{0}^{a}z sin^{2}(πz/a)dz

= (2/a)^{3}mg(a/2)^{4}= mga/2 = energy correction for the ground state.

To find the corrections to E_{0}^{2}and to find out if the degeneracy is removed by the perturbation, we have to diagonalize W in the subspace spanned by Φ_{211}, Φ_{121}, and Φ_{112}.

<Φ_{211}|W|Φ_{211}> = (2/a)^{3}mg∫_{0}^{a}sin^{2}(2πx/a)dx ∫_{0}^{a}sin^{2}(πy/a)dy ∫_{0}^{a}z sin^{2}(πz/a)dz = mga/2.

<Φ_{121}|W|Φ_{121}> = mga/2.

<Φ_{112}|W|Φ_{112}> = (2/a)^{3}mg∫_{0}^{a}sin^{2}(πx/a)dx ∫_{0}^{a}sin^{2}(πy/a)dy ∫_{0}^{a}z sin^{2}(2πz/a)dz = mga/2.

All the off-diagonal matrix elements of W are zero. The perturbation does not remove the degeneracy.

E_{0}^{2}= mga/2.

Consider a charged particle on a ring of unit radius with
flux Φ/Φ_{0 }= a
passing through the ring, where Φ_{0 }= h/(2e)
is the flux quantum. The Hamiltonian operator can be written as H = H_{0 }+
U,
where

H_{0} =
(i∂/∂θ
+ α)^{2}
and U = U_{0} cosθ.

θ is the angular coordinate.
We have chosen units with ħ = 2m = 1.

(a) Find the complete set of eigenvalues and
eigenfunctions of H_{0}.

(b) Use perturbation theory to find the first and second-order corrections
to the ground state energy E^{0} of H_{0} due to
the perturbation U for 0 < α < ½ .

(c) For α = ½ the ground state energy of H_{0}
is degenerate. Find the first-order correction to E^{0} for
this case.

Solution:

- Concepts:

First and second order perturbation theory for non-degenerate states, first order perturbation theory for degenerate states. - Reasoning:

We find the eigenvalues of H_{0}, which are non-degenerate if α ≠ ½. Perturbation theory for nondegenerate states therefore gives the energy corrections. For α = ½ the ground state is degenerate and perturbation theory for degenerate states must be used. -
Details of the calculation:

(a) H_{0}ψ(θ) = Eψ(θ).

Try ψ(θ) = Nexp(ibθ), H_{0}exp(ibθ) = (-b + α)^{2}exp(ibθ).

ψ(θ) = ψ(θ + 2π) --> exp(ibθ) = exp(ib(θ + 2π)) --> b = n, n = 0, ±1, ±2, ... .

ψ(θ) = Nexp(inθ), E^{n}= (-n + α)^{2},

∫_{0}^{2π}|ψ|^{2}dθ = N^{2}∫_{0}^{2π}dθ = 1 --> N = (2π)^{-½}.

E^{0 }= E_{0}^{0 }+ E_{1}^{0 }+ E_{2}^{0 }+ ..., E_{0}^{0 }= α^{2}. E_{1}^{0 }= <ψ_{0}|H'|ψ_{0}> = (2π)^{-½}U_{0}∫_{0}^{2π}cosθ dθ = 0.

E_{2}^{0}= ∑_{n≠0 }|<ψ_{n}|H'|ψ_{0}>|^{2}/(E_{0}^{0 }- E_{0}^{n}).

<ψ_{0}|H'|ψ_{0}> = (2π)^{-1}U_{0}∫_{0}^{2π}cosθ exp(inθ)dθ

= (4π)^{-1}U_{0}∫_{0}^{2π}exp(i(n+1)θ)dθ + (4π)^{-1}U_{0}∫_{0}^{2π}exp(i(n-1)θ)dθ

= δ_{n,-1 }U_{0}/2 + δ_{n,+1 }U_{0}/2.

E_{2}^{0}= (U_{0}^{2}/4)/(E_{0}^{0}- E_{0}^{-1}) + (U_{0}^{2}/4)/(E_{0}^{0}- E_{0}^{1})

= -(U_{0}^{2}/4)[1/(1 + 2α) + 1/(1 - 2)α] = -U_{0}^{2}/(2 - 8α^{2})

(c) If α = ½ then all states are degenerate. The ground state is any linear combination of n = 0 and n = 1.

We have to diagonalize the matrix of U in the subspace spanned by

ψ_{0}(θ) = (2π)^{-½}and ψ_{1}(θ) = (2π)^{-½}exp(iθ).

E_{1}^{0}= ±U_{0}/2.

E_{1}^{0}= U_{0}/2, Φ = (1/√2)(ψ_{0}(θ) + ψ_{1}(θ)), E_{1}^{0}= -U_{0}/2, Φ = (1/√2)(ψ_{0}(θ) - ψ_{1}(θ)).

Calculate the splitting induced among
the degenerate n = 2 levels of a hydrogenic atom, when this atom is placed
in a uniform electric field **E** pointing in the z-direction.
This is the linear Stark effect. You may use the following explicit
hydrogenic wave functions |nlm>.

|200> = (4π)^{-½}(2a)^{-3/2}(2 - r/a)exp(-r/2a),

|211> = (8π)^{-½}(2a)^{-3/2}(r/a) exp(-r/2a) sinθ e^{iφ},

|210> = (4π)^{-½}(2a)^{-3/2}(r/a) exp(-r/2a) cosθ,

|21-1> = (8π)^{-½}(2a)^{-3/2}(r/a) exp(-r/2a) sinθ e^{-iφ}.

∫_{0}^{∞}e^{-br }r^{n }dr =
n!/b^{n+1}.

Hint: Exploit symmetries!

Solution:

- Concepts:

First order perturbation theory, degenerate states - Reasoning:

The uniform electric field changes the energy of the proton and the electron and therefore perturbs the Hamiltonian. -
Details of the calculation:

H = H_{0}+ q_{e}|**E**|z, where z = (z_{e}- z_{p}).

The energy of the proton in the field is -∫_{0}^{zp}q_{e}|**E**|dz = -q_{e}|**E**|z_{p}.

The energy of the electron in the field is ∫_{0}^{ze}q_{e}|**E**|dz = q_{e}|**E**|z_{e}.

The potential energy of both particles is -q_{e}|**E**|(z_{p}- z_{e}) = q_{e}|**E**|z = -**p∙E.**

Let us use first-order perturbation theory.

The first excited state is**4-fold**degenerate.

E_{0}^{2}= μe^{4}/(8ħ^{2}).

To find E_{1}^{2}we have to diagonalize the operator q_{e}|**E**|z = q_{e}|**E**|r cosθ in the subspace spanned by |200>, |210>, |211>, and |21-1>. The matrix elements between states with the same parity are zero. This implies that all diagonal matrix elements are zero and all elements between states with the same l are zero. The matrix elements between states with different m are zero. The only matrix elements left are <210|z|200> and <200|z|210>.

<210|z|200> = q_{e}|**E**|(4π)^{-1}(2a)^{-3}∫_{0}^{2π}dφ∫_{0}^{π}cos^{2}θ sinθ dθ ∫_{0}^{∞}r^{3}dr (r/a)(2 - r/a) exp(-r/a)

= -3q_{e}|**E**|a.

We therefore have to diagonalize the matrix

in the |200>, |210>, |211>, |21-1> basis.

(E

_{1}^{2})^{2}[(E_{1}^{2})^{2}- (3q_{e}|**E**|a)^{2}] = 0.We have (E

_{1}^{2})^{2}= 0 (two fold degenerate ), or (E_{1}^{2})^{2 }= (3q_{e}|**E**|a)^{2}_{, }E_{1}^{2 }= ±(3q_{e}|**E**|a)^{.}The eigenvectors corresponding to E_{1}^{2 }= 0 are |211> and |21-1>.

The eigenvector corresponding to E_{1}^{2 }= 3q_{e}|**E**|a is (1/√2)(|210> - |200>).

The eigenvector corresponding to E_{1}^{2 }= -3q_{e}|**E**|a is (1/√2)(|210> + |200>).

The four fold degeneracy is changed into two non degenerate eigenvalues and one two fold degenerate eigenvalue.

Positronium is a bound state of an electron and a positron.

(a) What is the energy eigenvalue for the 1s state?

(b) If the spin-spin interaction is neglected, what is the degree of degeneracy
of the ground state?

(c) Assume that the spin-spin interaction is given by A**S**_{e}∙**S**_{p},
where **S**_{e} and **S**_{p} are the spin operators for
the electron and positron respectively, and A is a coupling constant. Show that
the ground state is split into two states and find the degree of degeneracy of
those states.

Solution:

- Concepts:

Hydrogenic atoms, a system of two spin ½ particles. - Reasoning:

Positronium is a hydrogenic atom. It is a bound state of two spin ½ particles. The wave function in coordinate space and the energy levels can be obtained from the wave function and energy levels of the hydrogenic atom using scaling rules. - Details of the calculation:

(a) For the hydrogen atom we have E_{n}= -E_{I}/n^{2}.

For a hydrogenic atom we have E_{n}= -E_{I}'/n^{2}, where E_{I}' = Z^{2}E_{I}(μ'/μ).

In this problem μ' = m_{e-}m_{e+}/(m_{e-}+ m_{e+}) = m_{e}/2 , Z = 1.

Therefore E_{0}= -13.6/2 eV.

E_{0}is the energy of the 1s (ground) state.

(b) The spins of the two particles can combine to form the S = 1 (triplet) and S = 0 (singlet) state. The ground state is therefore 4-fold degenerate.

(c) The energy shift due to the spin-spin interaction is given by

A <S,M_{s}|**S**_{e}∙**S**_{p}|S,M_{s}>.**S**_{e}∙**S**_{p}= ½(S^{2}- S_{e}^{2}- S_{p}^{2}).

**S**_{e}∙**S**_{p}|S,M_{s}> = (ħ^{2}/2)(S(S+1) - ½(3/2) - ½(3/2))|S,M_{s}> = (ħ^{2}/2)(S(S+1) - (3/2))|S,M_{s}>.

For the triplet state we therefore have ΔE = ¼ħ^{2}A and for the singlet state we have ΔE = -¾ħ^{2}A. The singlet state is not degenerate and the triplet state is 3-fold degenerate.

A valence electron in an alkali atom is in a p-orbital (l = 1).
Consider the simultaneous interactions of an external magnetic field **B** =
B**k** and the spin-orbit interaction. The two interactions are described by
the potential energy

U = (A/ħ^{2})**L**∙**S – (**μ_{B}/ħ)(**L** + 2**S**)∙**B**.

(a) Describe the energy levels of this l = 1 electron for B = 0.

(b) Describe the energy levels of this l = 1 electron for weak magnetic
fields. Use the projection theorem. (This is the Zeeman effect.)

(c) Describe the energy levels for strong magnetic fields so that the
spin-orbit term in U can be ignored. (This is called the Paschen-Back effect.)

Projection theorem: Inside a subspace E(k,j) the matrix elements of the
component of a vector observable **V **are proportional to the matrix
elements of the corresponding component of **J**.

Therefore <k,j,m_{j}'|(**L+**2**S)**|k,j,m_{j}> = [<**(L+**2**S)**∙**J**>_{kj}/(j(j
+ 1)ħ^{2})] <k,j,m_{j}'|**J**|k,j,m_{j}>.

Solution:

- Concepts:

Stationary perturbation theory - Reasoning:

We use the basis functions {|k,l,s;j,m_{j}>} and diagonalize U in the subspace defined by a fixed k and l = 1. - Details of the calculation:

(a) B = 0, U = (A/ħ^{2})**L**∙**S**.**J**=**L**+**S**.**L**∙**S**= ½(J^{2}– L^{2}– S^{2}).

**L**∙**S**is diagonal in the subspace defined by a fixed k and l = 1 and the matrix elements

<j,m_{j}|**L**∙**S**|j,m_{j}> do not depend on m_{j}.

<U>_{klj}= ½A(j(j + 1) – l(l + 1) – s(s + 1)) = ½A(j(j + 1) – 2 – ¾). j = ½ or j = 3/2.

For the p_{½ }level we have <U> = -A, for the p_{3/2}level we have <U> = ½A.

The p_{3/2}level lies 3A/2 above the p_{½ }level.

(b) If the magnetic field is weak, we consider U' =**– (**μ_{B}/ħ)(**L**+ 2**S**)∙**B**a perturbation that further splits the p_{½ }and p_{3/2 }levels. For each value of j we diagonalize the perturbation in the subspace spanned by basis functions with different m_{j}. We use

<k,j,m|**(L+**2**S)**|k,j,m_{j}> = [<(**L+**2**S**)∙**J**>_{kj}/(j(j + 1)ħ^{2})] <k,j,m_{j}|**J**|k,j,m_{j}>. =

Define g_{j}= <(**L+**2**S**)∙**J**>_{kj}/(j(j + 1)ħ^{2}) = Landé**(L+**2**S)**∙**J**= ½(3J^{2}+ S^{2}– L^{2}), <(**L+**2**S**)∙**J**>_{kj}= ½(3j(j + 1) + s(s + 1) – l(l + 1)) ħ^{2}.

g_{j}= 1 + (j(j + 1) + s(s + 1) – l(l + 1))/(2j(j + 1)). g_{½}= ⅔, g_{3/2}= 4/3 (for l = 1, s = ½).

<U'>_{½}=**– (**μ_{B}B/ħ)⅔m_{j}ħ = -⅔μ_{B}m_{j}B, <U'>_{3/2}= -(4/3)μ_{B}m_{j}B.

The weak magnetic field completely removes the degeneracy.

(c) U = -**(**μ_{B}/ħ)(**L**+ 2**S**)∙**B**= -**(**μ_{B}Bħ)(L_{z}+ 2S_{z})∙ (We neglect the spin orbit term.)

The matrix of U is diagonal in the {|k,l,s;m,m_{s}>} basis.

<U>_{m,ms}= -**(**μ_{B}B/ħ)(m + 2m_{s})ħ = -μ_{B}B(m + 2m_{s}).

m = -1, 0, 1; 2m_{s}= -1, 1;

possible values for m + 2m_{s}are -2. -1, 0 (two-fold degenerate), 1, 2.

Each of the levels |k,l,s;m,m_{s}> can be written as a linear combination of |k,l,s;j,m_{j}>, and the spin-orbit separates the sublevels with different j.