1. Limited time only! Sign up for a free 30min personal 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. Feb 18, 2012 #1
    time-independent, non-degenerate. I am referring to the following text, which I am reading:
    http://www.pa.msu.edu/~mmoore/TIPT.pdf
    On page 4, it represents the results of the 2nd order terms. In Eqs. (32), (33) and (34) I don't understand the second equality, i.e. basing on which formula he has turned the potential terms into a sum.
    For example, in (32) how he got from [itex]\langle n^{(0)}|V|n^{(1)} \rangle [/itex] to [itex]-\sum_{m\neq 0}\frac{|V_{nm}|^2}{E_{mn}}[/itex]
     
  2. jcsd
  3. Feb 18, 2012 #2
    You substitute in the expression you found for the second order coefficients in the expansion.
    The sum is something along the lines of

    [itex]E_n^2 = \sum_{m \ne n} V_{n,m} c^1_m[/itex]

    and in the first approximation you find that [itex]c^1_m = \frac{V_n,m}{E_{m,n}}[/itex] and you realise that V is hermitian and you multiply them together and get what you have, the negative sign comes from interchanging the m and n in the [itex]E_{m,n} = -E_{n,m}[/itex] term after you hermitian conjugate [itex]c^1_m[/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Perturbation theory
  1. Perturbation Theory (Replies: 1)

  2. Perturbation theory (Replies: 0)

  3. Perturbation Theory (Replies: 2)

  4. Perturbation theory (Replies: 2)

Loading...