# Perturbation theory

1. Apr 29, 2014

### PeteSampras

In a text a exercice says that for the Hamiltonian

$H_0 = \frac{p^2}{2m}+V(x)$ the eigenfunction and eigen energy are $\phi_n, E_n$. If we add the perturbation $\frac{\lambda}{m}p$ ¿what is the new eigenfunction?

The solution is

$\frac{p^2}{2m} + \frac{\lambda}{m}p+V= \frac{(p+\lambda)^2}{2m}- \frac{\lambda^2}{2m}+V$

but, the solution says $\psi= exp(-i x \lambda/ \hbar) \phi_n$

I understand that $H_0 \phi_n = E \phi_n$ but,

¿why the solution $\psi$ has it form?,

2. Apr 30, 2014

### Simon Bridge

Welcome to PF;

Perhaps because that is how the maths comes out?
Did you go through the calculation? Did you come up with anything else?

3. Apr 30, 2014

### L.S.H

The form of the new psai is a phase factor mutiplied by the previous eigenfunction. You can to some extent anticipate this answer form because the moment p has undergone a certain 'shift' by lamda.