Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 30E (Asymptotic Methods)

The endpoints are the two classical turning points, determined by $V(x)=E$. Since the potential is even, they are symmetric. For $0\le E\le 1$, they lie in the linear part of the potential and satisfy $|x|/4=E$; for $E\ge 1$, they lie in the outer part and satisfy $\sqrt{|x|}-1=E$. HenceThe two-turning-point WKB quantization condition uses

By the evenness of $V$, the semiclassical action integral is twice its integral over the positive half-axis. If $0\le E\le 1$, then

If $E\ge 1$, split the integral at $x=4$:The first integral isIn the second, put $r=\sqrt{x}$ and then $u=E+1-r$. It becomesAfter simplification,This is the action integral for a linear-to-square-root potential.

The graph begins at the origin, is continuous and strictly increasing, passes through $(1,16/3)$, and tends to infinity. Therefore for every $n\ge 0$ and $\epsilon >0$, the equationhas exactly one solution $E=E_n$ by the intermediate value theorem and strict monotonicity.

For fixed $n$, the right-hand side of the quantization equation tends to zero with $\epsilon$. Ifthe unique energy satisfies $E_n<1$, so the first branch of the action integral for a linear-to-square-root potential applies exactly. SolvinggivesThus this formula is valid whenever

For $E\to \infty$, the outer part of the action integral for a linear-to-square-root potential dominates:The WKB quantization condition therefore givesTaking the power $2/5$ yields the asymptotic equivalenceHence