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 using Cohen-Tannoudji

  1. Sep 22, 2005 #1
    I'm reading the Cohen-Tannoudji book and I found somthing I dont understand
    in stationary perturbation theory.
    the problem the Hamiltonian is split in the known part an the perturbation:
    [tex]
    H=H_{o}+\lambda \hat{W}
    [/tex]
    [tex]
    H_{o}|\varphi_{p}^{i}\rangle=E_{p}^{o}|\varphi_{p}^{i}\rangle
    [/tex] (1)
    and we want to solve the problem:
    [tex]
    H(\lambda)|\Psi(\lambda)\rangle=E(\lambda)\Psi(\lambda)\rangle[/tex] (2)
    Expanding in [tex]\lambda[/tex] series equation (2) I get after equating each term:
    [tex]
    zeroth: (H_{o}-E_{o})|0\rangle=0
    [/tex] (3)
    [tex]
    first: (H_{o}-E_{o})|1\rangle+(\hat{W}-E_{1})|0\rangle=0
    [/tex] (4)
    [tex]
    second: (H_{o}-E_{o})|3\rangle+(\hat{W}-E_{1})|2\rangle
    E_{2}|1\rangle-E_{3}|0\rangle=0
    [/tex] (5)

    from normalizing the wave fuction order by order I get:
    [tex]
    zeroth: \langle0|0\rangle=1
    [/tex] (6)
    [tex]
    first: \langle0|1\rangle= \langle1|0\rangle=0
    [/tex] (7)
    [tex]
    second: \langle0|2\rangle=
    \langle2|0\rangle=-\frac{1}{2}\langle1|1\rangle
    [/tex] (8)

    Solution for the non-degenerated level[tex]
    H_{o}|\varphi_{n}^{o}\rangle=E_{n}^{o}|\varphi_{n}^{o}\rangle
    [/tex]
    zeroth order:
    [tex]
    E_{o}=E_{n}^{o}
    [/tex]
    [tex]
    |0\rangle=|\varphi_{n}\rangle
    [/tex]
    first order projecting (4) onto the vector [tex]|\varphi_{n}\rangle[/tex]
    [tex]
    E_{n}(\lambda)=E_{n}^{o}+\langle\varphi_{n}|W|\varphi_{n}\rangle
    [/tex]
    now this is the part that I dont understand:
    when finding the eigenvector |1> the project equation (4) onto [tex]|\varphi_{p}^{i}\rangle[/tex] why the putting the supscript i if it is non-degenerated?????? :confused:
    [tex]|\Psi_{n}(\lambda)\rangle=|\varphi_{n}\rangle+\sum_{p\neq
    n}\sum_{i}\frac{\langle\varphi_{p}^{i}|W|\varphi_{n}\rangle}{E_{n}^{o}-E_{p}^{o}}|\varphi_{p}^{i}\rangle[/tex]
    see the book page 1101
     
  2. jcsd
  3. Sep 22, 2005 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The eigenvalues are not necessarily degenerate. Read the line after equation (B-6) on page 1101.

    Regards,
    George
     
  4. Sep 22, 2005 #3
    the book is talking about the non-degenerated case
    [tex]H_{o}|\varphi_{n}\rangle=E_{n}^{o}|\varphi_{n} \rangle[/tex]
    why projecting in the degenerate subspace [tex]\{|\varphi_{p}^{i} \rangle\}[/tex]eq B-6 ?????
    degenerace refers to the new perturbed energy???? :confused: :confused:
     
  5. Sep 22, 2005 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The idea is to expand the first-order eigenvector correction |1> in terms of a complete set of states that consists of unperturbed energy eigenvectors, i.e., to arrive at equation (B-10). Since the only eigenvalue that is known to be non-degenerate is [itex]E_{0}^{n}[/itex], the index [itex]i[/itex] has to used in the labelling of this complete set of states.

    Regards,
    George
     
  6. Sep 22, 2005 #5
    I got it, thanks but the book is misleading why not to project directly in the entire basis (degenerated in general) and get a matrix in B-5 and B-10 and B-11 all make sense, then the non-degenare case the matrix shrink to one element ...
    :biggrin:
     
  7. Sep 22, 2005 #6

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Of course the non-degenerate case is a special case of the degenerate case. But the degenerate case is slightly more subtle: you cannot just take ANY eigenvector of the unperturbed system and "perturbe it": the perturbation could lift the degeneracy (partly or entirely). So you first have to find out WHICH eigenvector you can perturbe in the first place (in fact, the original, say, 5-dimensional, set of eigenvectors with identical E0 will split in, say, a 2 dimensional set with one Ea, and 3 non-degenerate vectors with Eb, Ec and Ed respectively in first order ; the Ea can still potentially split at higher order).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Perturbation theory using Cohen-Tannoudji
  1. Perturbation theory (Replies: 16)

  2. Perturbation Theory (Replies: 5)

  3. Perturbation theory. (Replies: 0)

  4. Perturbation Theory (Replies: 3)

Loading...