1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Perturbation theory?

  1. Dec 15, 2004 #1
    I have a problem where I should calculate the ground state eigenfunction of a particle in the box where the potential V(x)=0 when 0<x<L and infinite everywhere else with the perturbation [itex]V'(x)=\epsilon[/itex] when L/3<x<2L/3.

    I get that the total ground state eigenfunction with the first order perturbation contribution is
    [tex]
    u_{1}=u_{01}+{\int_{L/3}^{2L/3} {u_{02}\hat H' u_{01}dx} \over (E_{01}-E_{02})}u_{02}+{\int_{L/3}^{2L/3} {u_{03}\hat H' u_{01}dx} \over (E_{01}-E_{03})}u_{03}
    [/tex]
    where
    [itex]
    \hat H'=\epsilon}[/itex] and [itex]u_{0n}/E_{0n}=[/itex] eigenfunctions/energies of the unperturbed system.
    I only need to use [itex]\{u_{01},u_{02},u_{03}\}[/itex] instead of all [itex]\{u_{0n}\}[/itex] when expressing the first order perturbation contribution
    [tex]u_{11}=\sum_k a_{nk}u_{0k}[/tex]

    Is this correct?
     
    Last edited: Dec 15, 2004
  2. jcsd
  3. Dec 16, 2004 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Why do you use only the three vectors [itex] \{u_{01},u_{02},u_{03}\}[/itex],when the theory states u should be using all the eigenvectors associated to eigenvalues different from the one (nondegenerate) u want to compute first orde corrections???

    Daniel.
     
  4. Dec 16, 2004 #3
    It is just an approximation and the only way I can think of an explanation to why the problem i´m supposed to solve says i only need to use these three is that they contribute most to the correction...

    Daniel
     
    Last edited: Dec 16, 2004
  5. Dec 16, 2004 #4
    On the other hand... how should i do if i needed to use all [itex]\{u_{0n}\}[/itex] How would i get [itex]a_{nk}[/itex] in that case? Please tell me if what i have got for [itex]u_1[/itex] above is correct because then i know if its me that cant integrate because i get that the integrals are zero or if i have set it all up wrong!?

    Daniel
     
  6. Dec 16, 2004 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes,the theory states that those vectors (corresponding to different energy levels) are orthogonal,and since the perturbation is a constant,then all its matrix elements between orthogonal states should annulate.
    On the other hand,are u sure with the integrations??There are products of sine/cosine functions.Only in certain conditions they annulate.And besides,the eigenfunctions are orthonormalized on the domain [0,L],and yet you're integrating them on a simmetric domain wrt to the middle of the interval L/2.So if that product of eigenfunctions is an odd function,then u shouldn't be surprised the result is zero.
    Check whether the products of eigenfunctions are odd.If so,the result is that the corrections to the eigenvector are identically null.
     
  7. Dec 16, 2004 #6
    I calculated the integrations again and only one of them was equal to zero.
    That looks alot better since it would be strange if the eigenfunction didn´t change when adding the disturbance!

    Thanks!

    /Daniel
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?