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It seems that WKB is applicable for a given $E$ if and only if $\hbar$ is sufficiently small. Or in other words, WKB is applicable if and only if the quantum number is large enough.

Is this understanding right?

I would take the exactness of WKB for the harmonic oscillator as purely accidental.

DanielSank
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Jiang-min Zhang
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    The exactnes of HO is not accidental. If you look at the evolution of the Wigner function in phase space, you see that only the intial conditions, not the evolution of the HO are quantum, and it is thus semi-classical in a perfectly well-defined way (...and now I'm going to dig in my stuff till I find that exact argument...) – ACuriousMind Aug 05 '14 at 17:05

3 Answers3

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Comments to the question (v3):

  1. Yes, given a Hamiltonian $H$, the semiclassical WKB method can give a qualitative but not a quantitative prediction for the ground state energy $E_0$.

  2. The WKB prediction (incl. the metaplectic correction/Maslov index) for the harmonic oscillator (HO) happens to be exact due to a hidden supersymmetry, cf. this Phys.SE post.

  3. For an example where WKB is not exact, see e.g. my Phys.SE answer here.

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Qmechanic
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The WKB approximation can be derived using an expansion in powers of $\hbar$. However, that doesn't imply that it can only be used if some quantum number is big. For example, a classic application of the WKB approximation is to alpha decay, which typically occurs from the ground state.

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    I have never understood the idea of expanding in a dimensioned parameter. "Expanding in $\hbar$" must certainly be expanding in $\hbar$ divided by some action scale of the problem, but nobody ever says this and certainly nobody ever seems to discuss what that action scale is. – DanielSank Jan 21 '15 at 07:47
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@DanielSank, for your question about "expanding in h_bar", please see the following explanation from http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_approx.pdf.

semiclassical_limit

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