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Perturbation Theory - First Order Approximation

  1. Jul 21, 2014 #1
    If:

    ##\hat{H} \psi (x) = E \psi (x)##

    where E is the eigenvalue of the *disturbed* eigenfunction ##\psi (x)##

    and ##E_n## are the eigenvalues of the *undisturbed* Hamiltonian ##\hat{H_0}##

    and the *disturbed* Hamiltonian is of the form:

    ##\hat{H} = \hat{H_0} +{\epsilon} \hat{V} ##

    where ## \hat{V} ## is the potential and

    then we can expand, using a Taylor expansion, ##\psi (x) ## and ##\epsilon## is a very small (positive?) number

    ##\psi (x) = {\psi}_0 (x) + {\epsilon}{\psi}_1 (x) + {\epsilon}^2 {\psi}_2 (x) + ... ## (1)

    where ##{\psi}_k (x)## is the eigenfunction of ##\hat{H}## corresponding to the eigenvalue ##E_n## (eigenvalues are assumed to be non-degenerate)

    We also get, using a Taylor expansion:

    ##E = E_0 +{\epsilon}{E_1} + {\epsilon}^2 {E_2} + ...## (2)

    MY QUESTIONS:

    I tried deriving formulae (1) and (2) - i.e. the expansions - on my own but I was unable to do it. Could anyone please show me *in detail* how to obtain those expansions?

    Another question is:
    How do we know that, when using the Taylor expansion, the coefficient ## {\epsilon} ## applies to ##{\psi}_1 (x)## and not to some other eigenfuncion, the coefficient ## {\epsilon}^2 ## applies to ##{\psi}_2 (x)## and not to some other eigenfuncion, and so on? The same question for eigenvalues.
     
    Last edited: Jul 21, 2014
  2. jcsd
  3. Jul 21, 2014 #2
    ##E^{(n)}## is *defined* as the coefficient of ##\epsilon^n## in the power series expansion of ##E##. Same for ##\psi^{(n)}##.

    Note that the ##E^{(n)}##'s and ##\psi^{(n)}##'s are *not* eigenvalues and eigenfunctions, except for ##E^{(0)}## and ##\psi^{(0)}## which are an eigenvalue and eigenfunction of ##H_0##.
     
  4. Jul 22, 2014 #3
    Thank you The_Duck, that clarifies it. I was obviously misled by the confusing notation. You made it clear, so thank you again :)
     
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