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3rd order energy perturbation correction

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Derive the general expression of 3rd-order perturbation energy for a

    non-degenerate quantum system.


    2. Relevant equations

    for nth order we have

    (Ho-Eo)|n>+(H'-E1)|n-1> -E2\n-2>-En|0>=0 (given)

    also,

    <0|0>=1,

    <1|0> = <0|1>=0,

    <0|2>=<2|0>=-1/2<1|1>

    <0|n>=<n|0>=-1/2*(<n-1|1>+<n-2|2>+....+<2|n-2>+<1|n-1>),

    3. The attempt at a solution

    for n=3 we get

    (Ho-Eo)|3>+(H'-E1)|2>-E2|1>-E3|0>=0

    now, multiply by <0| (this

    we get <0|(H0-E0)|3> +<0|(H'-E1)|2>-<0|E2|1>-<0|E3|0>=0

    by the rules above, the third term is = 0, the fourth is just =-E3 and the 2nd
    is -1/2<1|1>

    this gives E3=<0|(Ho-Eo)|3>-1/2*<1|(H'-E1)|1>

    for <0|n>=<n|0>=-1/2(<n-1|1>+<n-2|2>+....+<2|n-2>+<1|n-1>)

    for n=3

    i get

    E3=-1/2(<2|(H0-E0)|1>+<1|(H0-E0)|2>+<0|(H0-E0)|3>+<2|(H0-E0)|1>

    +<1|(H0-E0)|2>)-1/2*(<1|(H'-E1)|1>) (phew!)

    I am not sure what do do next

    what is <1|2>=<2|1> = to?

    i can see all the terms only have states 1 and 2 involved in them except the

    3rd term <0|3> and the last term <1|1>, are these important?

    does <1|1>=0?
     
  2. jcsd
  3. May 2, 2007 #2

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    you have to find the third order perturbation expansion for the energy correction, & not just the interaction between the n = 0,1,2 & 3 states.
     
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