Wolfram Alpha:

```Time Independent Perturbation Theory
------------------------------------

Most of the time in quantum mechanics, exact solutions of the Schrodinger
equation are not possible. This is where perturbation theory comes into
play.  The basic technique is:

- Divide Hamiltonian into 2 parts:  H = H0 + εH1

- H0 is unperturbed Hamiltonian.  The Schrodinger equation
can be solved exactly.

- H' is small additive term that corrects H0

- ε turns perturbation on and off.

The basic idea of perturbation theory is to write out both sides of
the perturbed equation as power series in ε and set coefficients
of the same powers of ε to be equal.

Consider the Schrodinger Equation

H0|ψn0> = En0|ψn0>

Now perturb it

{H0 + εH1}|ψn> = En|ψn>

Take the LHS and RHS and expand them as Taylor series

ψn = ψn0 + εψn' + ε2ψn''/2! + ... where ψn' = dψn/dε

and

En = En0 + εEn' + ε2En''/2! + ... where En' = dEn/dε

If we substitute these back into the perturbed SE and set the
terms containing ε's of the same power to be equal we get
(the Dirac notation has been dropped for simplicity):

ε0:  H0|ψn0> = En|ψn0> ... 0 order
ε1:  H0|ψn'> + H'|ψn0> = En0|ψn'> + En'|ψn0> ... 1st order
ε2:  H0|ψn''> + H'|ψn'> = En0|ψn''> + En'|ψn'> + En''|ψn0> ... 2nd order

If we assume that ψ0' does not differ greatly from the
wavefunction for ψn0 we can expand ψ0' as:

ψn' = Σnanψn0

We can then substitute this into the ε1 equation to get

H0|Σnanψn0> + H'|ψn0> = En0|Σnanψn0> + En'|ψn0>

Now we can write H0|Σnanψn0> as ΣnanH0|ψn0> which equals ΣnanEn0|ψn0>

Substituting we get

H'|ψn0> + ΣnanEn0|ψn0> = En0Σnan|ψn0> + En'|ψn0>  ... 1.

Now if we multiply everything by ψn0* and integrate to get:

<ψn0*|H'|ψn0> + ΣnanEn0<ψn0*|ψn0> = En0Σnan<ψn0*|ψn0> + En'<ψn0*|ψn0>  ... 1.

∴ En' = <ψn0|H'|ψn0>

To obtain the first correction to the nth eigenstate, multiply
both sides of 1. by ψm0*, where m is an unperturbed orthogonal
state to n. (Recall that states are orthogonal if they correspond
to different eigenvalues of an Hermitian operator).

<ψm0|H'|ψn0> + ΣnanEn0<ψm0|ψn0> = En0Σnan<ψm0|ψn0> + En'<ψm0|ψn0>

<ψm0|H'|ψn0> + amEm0 = En0am

∴ am = <ψm0|H'|ψn0>/(En0 - Em0)

This can be interpreted as "how much of ψm0 is in the perturbed
state ψn'.

Example:  The He atom

The He atom is an example of a '3 body problem' that cannot be solved
analytically.

H0 = [-(h2/2m)∇12 - 2e2/4πε0r1] + [(-h2/2m)∇22 - 2e2/4πε0r2]

H' = e2/4πε0r12  electron-electron perturbation

From the H atom we get for each electron (1s):

ψ10 = (1/√π)(2/a0)3/2exp(-2r1/a0)  (a0 = Bohr radius)

and

ψ20 = (1/√π)(2/a0)3/2exp(-2r2/a0)

Where E10 = E20 = -Z2RH/n2 = -4RH = -4(2.18 x 10-18 J)  (RH = Rydberg constant)

∴ E = E10 + E20 = -1.744 x 10-17 J

In the ground state, the inner shell is full with 2 electrons that have opposite
spins.  Since the spins are antisymmetric, the spatial part of the wavefunction
must be symmetric.  Thus, using the Hartree product we can write:

ψn0 = ψ10ψ20

Now let's calculate the correction to the energy when we factor in the
electron-electron interaction term.

En' = <ψn0|H'|ψn0>

This results in the following double integral:

E' = ∫∫(1/√π)(2/a0)3/2(1/√π)(2/a0)3/2exp(-2r1/a0)exp(-2r2/a0)(e2/4πε0r12)
(1/√π)(2/a0)3/2(1/√π)(2/a0)3/2exp(-2r1/a0)exp(-2r2/a0)dτ1dτ2

= ∫∫(1/π)2(2/a0)3exp(-2r1/a0)exp(-2r2/a0)(e2/4πε0r12)(2/a0)3exp(-2r1/a0)exp(-2r2/a0)dτ1dτ2

This has the solution:

E' = (5/4)(e2/4πε0a0)

= 5.44 x 10-18 J

Therefore, the corrected energy is -1.744 x 10-17 + 5.44 x 10-18 = -1.265 x 10-17 J

This is in fairly close agreement with actual experimental measurements.
```