# Wilson-Sommerfeld quantization to solve square-well potential

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## Summary:

Applying the Wilson-Sommerfeld quantization rule to solve the square-well potential problem

## Main Question or Discussion Point

The Wilson-Sommerfeld quantization rule claims (##\hbar=1##)
$$\frac{1}{2\pi} \oint p(x)\,dx=n,\,n=1, 2, ...$$
where integration is done in the classically allowed region. Applying this to a square-well potential with a depth of ##V_0## and width ##a##, we get $$E=\frac{\pi^2 n^2}{2a^2}$$
This only gives the correct result in the limit ##V_0 \rightarrow \infty##, and for low ##V_0## the error is quite substantial. I would like to understand why.

As I leant, the Wilson-Sommerfeld rule can be obtained from a zeroth-order WKB approximation. Let us consider a potential well described by a continuous function ##V(x)##, and pick two classical turning points ##x_1 < x < x_2##. In zeroth-order, the wave function can be given by
$$\psi_1 = exp\left( \pm i \int_{x_1}^x p\,dx\right)$$
but also as
$$\psi_2 = exp\left( \pm i \int_{x}^{x_2} p\,dx\right)$$
They have to be equal, and the real and imaginary parts yield the same result, namely, for the real part
$$\cos\left(\int_{x_1}^x p\,dx\right)=\cos\left(\int_{x}^{x_2} p\,dx\right)$$
Here comes a part that I don't understand ##(1)##: this implies that the sum of their phases have to be ##2\pi n##, from which
$$\int_{x_1}^{x_2}p\,dx=2\pi n$$
which is the desired result. The other part that I don't understand ##(2)## is that what step caused this result to only be valid for an infinitely deep square-well potential? I read about more detailed calculations that treat the regions around the turning points and those lead to the Bohr-Sommerfeld rule, which gives the correct result for a harmonic oscillator for example, but for this square potential it doesn't work at all. Can you help me figure this out?

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anuttarasammyak
Gold Member
Solutions of finite well have exponential decay part beyond the wall that the rule you mentioned does not include. Infinite well does not allow such permeation of wave function. I assume it is at least a part of reason to explain why.

Solutions of finite well have exponential decay part beyond the wall that the rule you mentioned does not include. Infinite well does not allow such permeation of wave function. I assume it is at least a part of reason to explain why.
That is true, but I assume there's a (hidden) step in this calculation that assumes that the wavefunction at the turning points is zero.

anuttarasammyak
Gold Member
If you go further to WKB approximation this exponential dumping is considered.

Summary:: Applying the Wilson-Sommerfeld quantization rule to solve the square-well potential problem
I read about more detailed calculations that treat the regions around the turning points and those lead to the Bohr-Sommerfeld rule, which gives the correct result for a harmonic oscillator for example,
I am afraid your method gives right wave function for harmonic oscillator. You say the correct result, it is of eigenvalue, wave function or any there features ?

That is true, but I assume there's a (hidden) step in this calculation that assumes that the wavefunction at the turning points is zero.
I do not think ##e^{ipx}## cannot be zero at turning point mathematically.

I am afraid your method gives right wave function for harmonic oscillator. You say the correct result, it is of eigenvalue, wave function or any there features ?
Right wave function for harmonic oscillator? Those should be Hermite-polynomials, not simple exponentials, no? By correct result I mean eigenvalues.

I do not think ##e^{ipx}##cannot be zero at turning point mathematically.
That is true, but the ##\pm## in the argument means that it is a linear combination of a ##+## and a ##-## term. My writing was a bit misunderstandable, sorry. It can be sine and cosine, so it can be zero at the turning points.

Could it be that the correctness of the quantization rule obtained from the consistency of given-order WKB wave functions does not imply the correctness of the wave function itself? It wouldn't answer my question, but I would be one step closer, I guess.

Another idea: classically, the bound states in a square-well potential are independent of the depth of the well, while in quantum mechanics for a given level the wave functions take different values at the classical turning points for different depths. In zeroth-order, no quantum correction should appear, so one way of choosing consistent boundary conditions is to zero out the wavefunctions at the turning points. However, I still don't see how this appears mathematically here.