Stationary perturbation theory is concerned with finding the changes in the discrete energy levels and the changes in the corresponding energy eigenfunctions of a system, when the Hamiltonian of a system is changed by a small amount. Let
H = H_{0} + H’ = H_{0} + λW
H_{0} is the unperturbed Hamiltonian whose eigenvalues E_{0}^{p} and eigenstates φ^{i}_{p}> are known. LetH_{0}  φ^{i}_{p}> = E_{0}^{p}  φ^{i}_{p} >.
Here i denotes the degeneracy. Assume that the matrix elements of H’ in the eigenbasis of H_{0} are small compared to the matrix elements of H_{0}.<φ^{i}_{p}H ' φ^{i}_{p}> << <φ^{i}_{p}  H_{0}  φ^{i}_{p} > = E_{0}^{p}.
We write
H’ = λW
with λ << 1 and <φ^{i}_{p}W φ^{i}_{p}> ≈ E_{0}^{p}.
Example:H = H_{0} + H’
is the Hamiltonian of a perturbed onedimensional harmonic oscillator.

We are looking for the eigenvalues
E(λ)
and the eigenstates ψ(λ) >
of H(λ) = H_{0} + λW.
Since λW is small, we assume that E and ψ> can be expanded as a power series in λ.
E^{p}=E_{0}^{p}+λE_{1}^{p}+λ^{2}E_{2}^{p}+...,
ψ_{p}>=ψ_{p}^{0}>+λψ_{p}^{1}>+λ^{2}ψ_{p}^{2}>+....
We may then write
(H_{0} + λW)(ψ_{p}^{0}>+λψ_{p}^{1}> + λ^{2}ψ_{p}^{2}> + ...)
=(E_{0}^{p} + λE_{1}^{p} + λ^{2}E_{2}^{p} + ...)(ψ_{p}^{0}> + λψ_{p}^{1}> + λ^{2}ψ_{p}^{2} > +...).
This equation is must be
valid over a continuous range of λ.
Therefore we equate coefficients of equal powers of λ
on both sides to obtain a series of equations that represent successively
higher orders of the perturbation.
(H_{0}E_{0}^{p})ψ_{p}^{0}>=0  
(H_{0}E_{0}^{p})ψ_{p}^{1}>=(E_{1}^{p}W)ψ_{p}^{0}>  
(H_{0}E_{0}^{p})ψ_{p}^{2}>=(E_{1}^{p}W)ψ_{p}^{1}>+E_{2}^{p}ψ_{p}^{0}>  
(H_{0}E_{0}^{p})ψ_{p}^{3}>=(E_{1}^{p}W)ψ_{p}^{2}>+E_{2}^{p}ψ_{p}^{1}>+E_{3}^{p}ψ_{p}^{0}>  
... 
(H_{0}E_{0}^{p})ψ_{p}^{0}> = 0 implies that ψ_{p}^{0}> is a linear combination of unperturbed eigenfuctionns φ^{i}_{p}> with the corresponding eigenvalue E_{0}. We choose <ψ_{p}^{0}ψ_{p}^{0}> = 1. ψ_{p}^{s}> is not uniquely defined. We can add an arbitrary multiple of ψ_{p}^{0}> to each ψ_{p}^{s}> without affecting the left hand side of the above equations. Most often this multiple is chosen so that <ψ_{p}^{0}ψ_{p}^{s}>=0. The perturbed ket is then not normalized. We then have
0 =
To calculate the energy
to sth order, we only need to know the state vector to order s1.
E^{n}=E_{0}^{n}+<φ_{n}H’φ_{n}>+O(&lambda^{2})
Example:E^{n} = E_{0}^{n} + O(λ^{2}). To first order, the eigenvalues of H are equal to the eigenvalues of the unperturbed Hamiltonian H_{0}. 
Firstorder eigenvector corrections:
(H_{0}E_{0}^{p}) ψ_{p}^{1}> = (E_{1}^{p}W) ψ_{p}^{0} > = (E_{1}^{p}W) φ_{p} >.
ψ_{p}^{0}> is an eigenstate of the unperturbed Hamiltonian. We may expand
in terms of the basis vectors φ^{i}_{p'} >. In the expansion b_{p} = 0 because <ψ_{p}^{0} ψ_{p}^{i} > = 0.
Multiply from the left by <φ^{i}_{p''} .
Therefore
Example:

Since we have found the expression for the state vector to first order, we can now find the expression for the energy to second order.
Example:

Let H = H_{0} + H’ = H_{0} + λW. In practice, after having derived the perturbation expansion, we often set λ = 1 and let H’ = W be small.
Solution:
The ground state is not degenerate. The potential energy of an electron outside a uniformly charged sphere of radius r_{0} and total charge q_{e} is Inside the sphere the potential energy is From Gauss’ law we know that (SI units).
Let
since r_{0} << a_{0},
=4⋅10^{9}eV.

A particle of mass m is in an infinite potential well perturbed as shown in the figure.
(a) Calculate the firstorder energy shift of the nth eigenvalue due to the perturbation.
(b) Write out the first three non vanishing terms for the firstorder perturbation expansion of the ground state in terms of the unperturbed eigenfunctions of the infinite well.
(c) Calculate the secondorder energy shift for the ground state.
Solution:
H = H_{0} + H’.
The eigenvalues of H_{0} are not degenerate.

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