We can get the Bohr-Sommerfeld quantization from the WKB method as answered. Since we use approximation, there should be an error in the system, I know this is not right all the time; in some conditions approximation can yield the real result. So, is there a way to find the $ \hbar^2$ correction?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Linked, as well as this. Actually, there are striking failures of the approximation even at a level below $\hbar^2$. The best link is through deformation quantization. $\endgroup$– Cosmas ZachosCommented Nov 5, 2023 at 23:58
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
OK, from the point of WKB, if we could find $S_2$, coefficient of the term $ (\hbar/i)^2$, and plug it into the Schrödinger equation, we would have find the correction. From the WKB, we already know that $S_0 = \int p(x)dx$ and $S_1 = - \frac{1}{2} \ln p + C$. We also have the condition, $ S_1''+(S_1')^2+2S_0'S_2' = 0$, so plugging $S_1$and $S_0$ written in terms of $p$, we got $S_2$ = $\frac{1 \cdot p'}{4 \cdot p^2} + \frac{1}{8} \int \frac{p'^2}{p^3}dx$. WKB says that, $\psi = e^{\frac{i}{\hbar}s}$ and $s = s_0 + \frac{\hbar}{i}s_1 + \frac{\hbar}{i}^2 s_2$. If we plug $s$ containing the $s_2$ term to the Schrödinger equation, we can find the correction.