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Proof of vailidity of WKB approximation ! Please help !

  1. Mar 31, 2013 #1
    Hi,

    I have been looking for rigorous mathematical conditions for when the WKB approximation may be applied.

    Here is my understanding of the topic.

    We start with the most general form that the wavefunction could take, i.e. exp[if(x)/h] ,
    Where "i" stands for square root of -1, f(x) is some real function of x, h stands for "h bar",that is the original Planck's constant divide by 2pi.

    Any complex function of x can be written this way.

    Now we express f(x) as a series in powers of "h bar", i.e.

    f(x) = f0 + hf1 + h2f2 + ...

    Where again "h" actually stands for "h bar" .

    We now put this wave function into schrodinger's one dimensional equation to find various relations.

    Now in this entire process, here comes the WKB "approximation" , the approximation being, to neglect h2 dependant and all higher terms in the expansion of f(x) i.e. to take f(x) to be
    f(x) = f0 + hf1

    My question: when can we do so ? That is, mathematically, when can we ignore h2 and higher order terms ?
     
  2. jcsd
  3. Apr 1, 2013 #2

    Bill_K

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    The time-independent Schrodinger Equation in one dimension is (-ħ2/2m) ψ'' + V(x)ψ = Eψ. More simply, let k(x) = (2m(E - V(x))½/ħ, then ψ'' + k2(x)ψ'' = 0.

    The physical significance of k(x) is that it is a "local wavenumber". If V(x) ≈ const, ψ ~ eikx, and k(x) tells us how rapidly ψ is oscillating at that point.

    If V(x) is not const, it also has a characteristic length, V'(x)/V(x) ~ 1/ℓ. The WKB approximation is valid when ψ oscillates much more rapidly than V varies, that is, kℓ >> 1.
     
  4. Apr 2, 2013 #3

    DrDu

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    The tricky point about the WKB approximation is that the series in h does not converge, not even in the limit h to 0.
    This is the characteristic of a so called "asymptotic series": http://en.wikipedia.org/wiki/Asymptotic_series
    Nevertheless these series, when truncated, are often very accurate.
     
  5. Apr 2, 2013 #4

    bcrowell

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    Another issue is that V has to be sufficiently smooth in the classically allowed region. If this isn't true, then you can get reflections at discontinuities in the classically allowed region. I don't know the exact condition on V or how to prove rigorously that such a condition is necessary or sufficient. We had a thread on this recently: https://www.physicsforums.com/showthread.php?t=681988 Although I got some helpful responses, I was never able to resolve this particular technical issue to my own satisfaction.
     
  6. Apr 2, 2013 #5

    Bill_K

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    That's correct. The WKB approximation is only useful for slowly varying potentials.
     
  7. Apr 3, 2013 #6

    bcrowell

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    @Bill_K: Proof?
     
  8. Apr 3, 2013 #7

    DrDu

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  9. Apr 3, 2013 #8

    vanhees71

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  10. Apr 3, 2013 #9
    wkb approximation goes to the boundary of classical and quantum mechanics.It holds in the realm of when the action integral can be approximated by the classical integral.Also the condition kl>>1 implies that wavelengths are small compared to size of object i.e. just as in optics we go with ray diagram when the condition holds.
     
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