- #1
Robin04
- 259
- 16
- TL;DR
- Applying the Wilson-Sommerfeld quantization rule to solve the square-well potential problem
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?
$$\frac{1}{2\pi} \oint p(x)\,dx=n,\,n=1, 2, ...$$
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?