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Would it be legitimate to use the WKB approximation for a particle in a spherically symmetric Gaussian potential?

$$V(r)~=~V_0(1-e^{-r^2/a^2}).$$

I'm not sure when to use which approximation method. Variational, perturbation, or WKB. Can anyone enlighten me? What are the "rules" to using WKB? I'm given the particles $n$ and $\ell$ quantum numbers and I'm trying to approximate its energy.

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It looks like I want to use the variational principle. I'm still curious If its possible with WKB though. –  Synthetic Dec 13 '12 at 23:04
    
WKB will give you an approximate solution valid in the quasi classical regime ($n,l\gg 1$). –  Fabian Dec 15 '12 at 6:21
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By the way: shouldn't there by a minus sign in the exponent? –  Fabian Dec 15 '12 at 6:21
    
Yes there should. Good catch. Thanks Fabian. I see so only a classical approximation. Interesting. –  Synthetic Dec 15 '12 at 7:20
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2 Answers

up vote 1 down vote accepted

The WKB approximation is useful for highly excited states or scattering states that have an associated wavelength much smaller than the length scale over which the potential term in the interaction changes. In your case the product of the wave number and the length scale $a$ should be much larger than unity.

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for big $ n>>>1 $ the WKB energies are

$$ 2\pi (n+1/2)=\int_{a}^{b}dr\sqrt{E_{n}-V_{0}+e^{-r^{2}/a^{2}}} $$

here a and b are the turning points so $$ V(a)=V(b)=E_{n} $$

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