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.

For Coulomb Potential, here is some common expected values:

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

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

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

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

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