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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

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    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

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    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

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    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).
     
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