when calculate the energy from Hamiltonian. usually, we only know the exact form from a simple Hamiltonian but not the exact form of the Hamiltonian. Thus, we have to apply a perturbation to find the approximate energy.

the formula is :

if

$H_0 \left|n^0\right>= E_n^0\left|n^0\right>$

is the exactly solvable Hamiltonian, energy and eigenstate for the n-h state, then the REAL energy for the total Hamiltonian

$H= H_0 +V$

is

$E_n = E_n^0 + V_{nn} + \sum_{k\neq n}{\frac{|V_{nk}|^2}{E_n-E_k}}+....$

where

$V_{nk} = \left$

so far as I encounter, the exact state is not quite important.

Detail walk through.

Set the perturbation strength as

$H = H_0 + \lambda V$,

where $\lambda$ is the switch for the perturbation.

Suppose the solution for n-th level is

$\displaystyle \left|n \right> = \sum_{r}^{\infty} \lambda^r \left|n^{(r)}\right>$ ,

with energy

$\displaystyle E_n = \sum_{r}^{\infty} \lambda^r E_n^{(r)}$.

Put in the equation,

$\displaystyle (H_0 + \lambda V) (\sum_{r}^{\infty} \lambda^{r} \left| n^{(r)} \right>) = (\sum_r^{\infty} \lambda^r E_n^{(r)}) (\sum_r^{\infty} \lambda^r \left|n^{(r)}\right>)$

expand and collect $\lambda^r$

$\displaystyle H_0 \left| n^{(r)}\right> + V \left| n^{(r-1)}\right> = \sum_l^r E_n^{(k)} \left| n^{(r-k)} \right>$

we can see, for $r = 0$, this is the original, unperturbed solution.

For $r =1$, this is

$\displaystyle H_0 \left| n^{(1)} \right> + V \left| n^{(0)} \right> = E_n^{(0)} \left| n^{(1)} \right> + E_n^{(1)} \left| n^{(0)} \right>$

To get $E_n^{(1)}$, we apply $\left< n^{(0)} \right|$ from the left, the first terms of left and right side will cancel out.

$\left< n^{(0)} | V | n^{(0)} \right> = E_n^{(1)}$

For higher order, one can calculate using the lower orders. for example,

$E_n^{(2)} = - \left< n^{(1)} | (H_0 - E_n^{(0)} )| n^{(1)} \right>$

The state $\left| n^{(1)} \right>$ can be constructed using eigen states,

$\displaystyle \left| n^{(1)} \right> = \sum_{m} a_m \left| m \right>$

put in the equation, using orthogonal relation,

$\displaystyle a_m = \frac{ \left}{E_n - E_m}$

For Coulomb Potential, here is some common expected values:

$\left< n,l,m| \frac{a}{r} | n,l,m \right> = \frac{1}{n^2}$

$\left< n,l,m| \frac{a^2}{r^2} | n,l,m \right> = \frac{1}{(l+1)n^3}$

$\left< n,l,m| \frac{a^3}{r^3} | n,l,m \right> = \frac{1}{l(l+1/2)(l+1)n^3}$

$\left< n,l,m| \frac{r}{a} | n,l,m \right> = \frac{1}{2} \left(3n^2-l(l+1)\right)$

$\left< n,l,m| \frac{r^2}{a^2} | n,l,m \right> = \frac{1}{2} \left( n^2 ( 5n^2 -3l(l+1)+1) \right)$