# Time-independent perturbation theory

1. Dec 23, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
In each of my QM books, they always say something like "we can write the perturbed energies and wavefunctions as"

$$E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \cdots$$

$$|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots$$

without any justification. This is really not obvious to me and although it seems reasonable, it do not see why the perturbed energies and wavefunctions might not be something completely different like

$$E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 \log \left(E_n^{(2)}\right)^{-1}\sin E_n^{(2)} + \cdots$$

$$|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \cdots$$
2. Relevant equations

3. The attempt at a solution

Last edited: Dec 23, 2007
2. Dec 23, 2007

### Dr Transport

Expanding the energies and wave functions as a function of the perturbation parameter is an iterative manner which is well behaved. The function you wrote for the energy has poles and in multivalued.

3. Dec 23, 2007

### kdv

You are right that it's an assumption that is being made. The assumption is that the perturbation Hamiltonian is a small perturbation, which means that as lambda goes to zero, the energies and wavefunctions smoothly approach the unperturbed results. This assumption does fail in some cases (for example, trying to treat the Couloumb potential in Hydrogen as a perturbation would fail completely because the energy expansion would diverge in lambda which would signal the fact that the Coulomb potential has to be included to all orders so it cannot really be treated as a perturbation)