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Simple perturbation theory

  1. Sep 14, 2004 #1
    we have a particle in an infinite one-dimensional square well potential

    [V(x)=0 for 0<x<L and V(x) is infinite otherwise]

    and introduce a small potential (perturbation) in the middle of the
    square well potential. Then the first order energy correction
    for the ground state is 100 times lower than the first order correction
    to the first excited state. I dont understand why...

    The wave function without the small potential step in the middle
    is zero in the middle for the first excited state and maximum in the middle
    for the ground state. Has the diffrence in energy corrections something
    to do with this?
  2. jcsd
  3. Sep 14, 2004 #2
    Welcome JohanL !

    I don't know exactly the parameters, so it is difficult to answer. Maybe the reason is that the lower energy state is especially stable, whereas excited states already have dynamics included.

    I hope better answers will come.
  4. Sep 14, 2004 #3
    Thank you!

    Im just a beginner in perturbation theory...
    The wavefunctions are just the usual standing waves in
    [V(x)=0 for 0<x<L and V(x) is infinite otherwise]

    and then a small step of width a=L/10 and height h is introduced in the
    middle of the square well potential, i.e.
    V(x)=h for 0.5(L-a)<x<0.5(L+a)

    Then with perturbation theory I calculate the first orde energy corrections
    to the unperturbed eigenvalues and I get that the corrections
    is 100 times larger for the first excited state than for the ground state.
    and i dont understand why...
  5. Sep 14, 2004 #4
    OK, now all the parameters are here.

    I don't know if my interpretation is correct. If I figure out something better, I'll let you know. The calculations barely tell you any deep reason. Do they ?
  6. Sep 14, 2004 #5


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    Do you mean <psi|V|psi> where V is your potential and psi the unperturbed state (sin(pi nx/L) ?

    So I guess you calculate something like Integral( h sin^2(pi n x/L) dx) between x = 0.5(L-a) and 0.5(L+a) with appropriate normalisations I don't know by heart ?

  7. Sep 14, 2004 #6
    Yes, thats exactly what i have done.
    and when i have calculated the integral i set n=1 for
    the energy correction for the ground state and then n=2
    for the correction for the first excited states.
    and the corrections i get is
    E(n=1)=0.0016 h
    E(n=2)=0.19 h

    and then i need a qualitative interpretation for why they differ
    so much...
  8. Sep 14, 2004 #7


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    I get something else.

    psi_n(x) = sqrt(2/L) Sin[ n Pi x/L ]

    I then find (using Mathematica) that the integral of psi_n(x)^2 from (L-a)/2 to (L+a)/2 equals:

    a/L - Cos[n Pi] Sin[ a n Pi/L] / (n Pi)

    this gives for the first energy correction:

    a/L + Sin[a Pi/L]

    for the second:

    a/L - Sin[2 a Pi/L] / (2 Pi)


    (all this times h of course).
    Filling in L -> 1, a -> 1/10, I find numerically:
    {0.198363, 0.00645107, 0.185839, 0.0243173, 0.163662, 0.0495449}

    for n = 1, 2, 3, 4, 5 and 6.

    Now maybe I'm wrong ! I can send you the (simple) mathematica sheet if you want.

  9. Sep 14, 2004 #8


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    Hi there.

    I can tell without doing the calculation that your results can't be right.

    Your perturbation is in the center of the well. Around that point, the ground state probability density has a maximum whereas the second excited state has a node. Therefore, the ground state will be much more affected by the perturbation than the first excited state.

  10. Sep 14, 2004 #9
    Thank you Patrick and Pat!
    Now I understand!
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