I am trying to apply the WKB approximation to the harmonic oscillator $$H=\frac{1}{2}p^2+\frac{\omega^2}{2}x^2\,.$$

In principle, the exact spectrum can be found using textbook methods: $$ E_n=\hbar\omega(n+\frac{1}{2})\,,\qquad \psi_n=\frac{1}{\sqrt{2^nn!}}\left( \frac{\omega}{\pi \hbar} \right)^{\frac{1}{4}}e^{-\frac{\omega}{2\hbar}x^2}H_n(\sqrt{\frac{\omega}{\hbar}}x)\,, $$ where $H_n(x)$ is the $n$-th Hermite polynomial. Now, I can imagine that I know nothing about the usual methods of quantum mechanics and "brute force" my way to the answer using the WKB ansatz for the wavefunction $$ \psi(x)=\exp\left(\frac{i}{\hbar}\int^x Q(x')\right)\,, $$ and expand $Q(x)$ as a series in $\hbar$ $$Q(x)=\sum_{k\geq0}Q^{(k)}(x)\hbar^k\,.$$ One can then find the coefficients recursively. The energy can be similarly expanded as a series in $\hbar$, and is related to $Q$ by integrating a cycle integration over the path defined the two turning points $p=0$: $$ E=\oint Q(x)dx=\sum_{k>0}E^{(k)}\hbar^k\,. $$ The quantisation condition is given by the Bohr-Sommerfeld condition $E_n=\hbar\pi(n+1/2)$ and can in principle be found by inverting the series.

For the harmonic oscillator, it is easy to find that $E^{0}=\oint\sqrt{E-\frac{\omega^2}{2}x^2}=\frac{E\pi}{\omega}$ and we get the correct condition at 0-th order.

My problem is that while the leading order vanishes, $E^{(1)}=0$, the next-to-next leading order doesn't $$\oint Q^{(2)}\sim\oint \frac{x^2}{(E-\frac{\omega^2x^2}{2})^{5/2}}\neq0\,,$$ so I would expect corrections to that result, which there can't be, as I already (accidentally) have the correct result at 0-th order.

I should also be able to reconstruct the eigenfunctions in terms of Hermite polynomials, but after computing the first few orders of the series I don't see how they can sum to $\psi_n$ written at the beginning. Am I missing something here in order to find them, or is the WKB series not applicable here? Any reference is welcome!

  • $\begingroup$ I think there is something wrong with the integration of $Q^{(2)}$. The result should be consistent with this. $\endgroup$ – secavara Jan 19 '18 at 19:53
  • $\begingroup$ Hi Bulkilol, is this taken from a reference? Which one are you in particular following? Which page? $\endgroup$ – Qmechanic Jan 23 '18 at 17:11
  • $\begingroup$ secavara : you were right, the integral of Q^2 over a cycle is proportional to Euler's beta function and for k>0, it is always zero. In the case of the wavefunction (and not the energy), it's not the case though. @Qmechanic : it's not taken from any reference in particular, any textbook about QM mentions the method, but all of them usually stop at order one. I wanted to see if there's any simple way to reconstruct the wave function from WKB. In the case of the harmonic oscillator, there's no non-perturbative effect, so one would expect the WKB expansion is enough. $\endgroup$ – Bulkilol Jan 25 '18 at 17:28

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