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Perturbation theory

  1. Apr 29, 2014 #1
    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. jcsd
  3. Apr 30, 2014 #2

    Simon Bridge

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    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?
     
  4. Apr 30, 2014 #3
    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.
     
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