Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 3rd-order Energy Correction Derivation

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Time-independent, non-degenerate perturbation theory:

    - Let |k> represent k-th order wave function correction.
    - Let [itex]E^{(k)}[/itex] represent k-th order energy correction
    - H' is the perturbed Hamiltonian.

    Want the third order energy correction:
    [tex]E^{(3)} = <1|H' - E^{(1)}|1> - 2E^{(2)}<0|1>[/tex]

    2. Relevant equations

    Equating coefficients of equal powers of the parameter:
    [tex]H_0|3> + H'|2> = E^{(0)}|3> + E^{(1)}|2> + E^{(2)}|1> + E^{(3)}|0>[/tex]


    3. The attempt at a solution

    Tried multiplying <0| on the left like when deriving the first order correction, [itex]E^{(1)}[/itex].

    I get,
    [tex]E^{(3)} = <0|H' - E^{(1)}|2>[/tex]

    I don't know what I'm missing to proceed... I'm in the same situation with a few other similar problems. I don't see how multiplying by anything else on the left would make sense either.

    Hint on what I'm missing? Thanks
     
    Last edited: Sep 16, 2010
  2. jcsd
  3. Sep 16, 2010 #2
    did I state my question poorly? :\

    This is problem 8.1 from B&J's book.
     
  4. Sep 20, 2010 #3
    One last bump... haven't gotten anywhere with this.
     
  5. May 5, 2011 #4
    Recently had to solve the same problem, and since noone has commented, I figured I'd leave a hint or two for the next person who comes along.

    Hint 1:
    The attempted solution in the opening post is the correct way to proceed, except for a term which has been forgotten:
    [tex]E^{(3)}=<0|H'-E^{(1)}\mathbb{I}|2>-E^{(2)}<0|1>[/tex]
    (0)​
    The secret now is to look at the equations of the coefficients for the lower orders
    [tex]H_0|1>+H'|0>=E^{(0)}|1>+E^{(1)}|0>[/tex]
    (1)​
    and
    [tex]H_0|2>+H'|1>=E^{(0)}|2>+E^{(1)}|1>+E^{(2)}|0>.[/tex]
    (2)​
    Using these, the third order energy correction may be expressed in the form
    [tex]E^{(3)}=<1|H'-E^{(1)}\mathbb{I}|1>-2E^{(2)}\textt{Re}<0|1>.[/tex]
    (3)​

    Hint 2:
    The "dual equation" to equation (1) may be written as
    [tex]<1|H_0=E^{(0)}<1|+E^{(1)}<0|-<0|H'.[/tex]
    (4)​

    Multiply <1| with equation (2).
    Insert (4) in the first term of this and rearrange.
    Insert this in (0).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook