degenerate state perturbation theory
∑
i∈N〈φ(j)|H 0 +H 1 |φ(i)〉αi=Eαj,This is an eigenvalue equation with as many solutions as there are degnerate states in our set. audio
We recognize this as simply the (matrix) energy eigenvalue equation limited the list of degenerate
states. We solve the equation to get the energy eigenvalues and energy eigenstates, correct to first
order.
Written as a matrix, the equation is
H 11 H 12 ... H 1 n
H 21 H 22 ... H 2 nHn 1 Hn 2 ... Hnn
α 1
α 2αn
=E
α 1
α 2αn
whereHji=〈φ(j)|H 0 +H 1 |φ(i)〉is the matrix element of the full Hamiltonian. If there are n nearly
degenerate states, there are n solutions to this equation.
The Stark effect for the (principle quantum number) n=2 states ofhydrogen requires the use of
degenerate state perturbation theory since there are four states with (nearly) the same energies. For
our first calculation, we will ignore the hydrogen fine structure andassume that the four states are
exactly degenerate, each with unperturbed energy ofE 0. That isH 0 φ 2 ℓm=E 0 φ 2 ℓm. The degenerate
statesφ 200 ,φ 211 ,φ 210 , andφ21(−1).
- See Example 22.3.3:The Stark Effect for n=2 States.*
The perturbation due to an electric field in the z direction isH 1 = +eEz. The linear combinations
that are found to diagonalize the full Hamiltonian in the subspace of degenerate states are: φ 211 ,
φ21(−1)and√^12 (φ 200 ±φ 210 ) with energies ofE 2 ,E 2 , andE 2 ∓ 3 eEa 0.
22.3 Examples
22.3.1 H.O. with anharmonic perturbation (ax^4 ).
We add an anharmonic perturbation to the Harmonic Oscillator problem.
H 1 =ax^4Since this is a symmetric perturbation we expect that it will give a nonzero result in first order
perturbation theory. First, write x in terms ofAandA†and compute the expectation value as we
have done before.
∆En(1) = a〈n|x^4 |n〉=
a ̄h^2
4 m^2 ω^2〈n|(A+A†)^4 |n〉=
a ̄h^2
4 m^2 ω^2〈n|(AAA†A†+AA†AA†+AA†A†A+A†AAA†+A†AA†A+A†A†AA)|n〉=
a ̄h^2
4 m^2 ω^2[
(n+ 1)(n+ 2) + (n+ 1)^2 +n(n+ 1) +n(n+ 1) +n^2 +n(n−1)]
=
3 a ̄h^2
4 m^2 ω^2(2n^2 + 2n+ 1)