This question already has an answer here:

Technically, this is a math problem but I think it is better here than in the math stacks. Consider the differential equation

\begin{equation} y'' = (x^4 - E)y \end{equation}

The boundary conditions are the usual $y(\pm\infty)=0$. I want to find the eigenvalues, $E$, using WKB. According to my textbook (Bender and Orszag), we have the guessed form

\begin{equation} y=e^{S_0(x)} \end{equation}


\begin{equation} S_0'(x) = \pm\sqrt{x^4-E} \end{equation}

One picks the sign such that the boundary conditions are obeyed.

Now, let us consider the "classical region" where $E>x^4$. After lots of work, the book reaches the eigenvalue condition \begin{equation} \int_{-E^{1/4}}^{E^{1/4}} \sqrt{E-x^4}dx =\left(n+\frac{1}{2}\right)\pi \end{equation}

Notice, that in this condition, the $\pm$ disappeared and the sign under the square root flips. The way this condition was obtained was fairly complicated, patching together solutions at infinity with an Airy function in between the turning points but can someone explain, in a simple way, how the boundary conditions forbade

\begin{equation} \int_{-E^{1/4}}^{E^{1/4}}-\sqrt{E-x^4}dx=\left(n+\frac{1}{2}\right)\pi~? \end{equation}


marked as duplicate by Qmechanic Apr 24 '15 at 5:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ I don't think there is any other answer than that, if you look at the derivation of the Bohr-Sommerfeld condition, you see that there is not actually a choice of sign. $\endgroup$ – ACuriousMind Apr 23 '15 at 11:20
  • $\begingroup$ Could I phrase the question the other way? What boundary conditions would yield the negative sign and make the last equation I wrote true? $\endgroup$ – user1936752 Apr 23 '15 at 12:07
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/168448/2451 and links therein. $\endgroup$ – Qmechanic Apr 23 '15 at 12:16