Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 31B / Solution

Away from the turning points, the Liouville–Green approximation gives in the allowed regionThe bound-state condition discards the growing exponential in each forbidden region, leavingThese approximations require variation slow compared with the local wavelength, and must be replaced near $a,b$ by Airy approximations.

Near $a$, set $z=-{\lambda }^{2/3}{q}^{\prime }(a{)}^{1/3}(x-a)$. The local differential equation becomes ${\psi }_{zz}-z\psi =0$. Decay on the left selects Airy function $\mathrm{Ai}(z)$ rather than its growing companion. Comparing the supplied asymptotics across the turning point yieldsAt $b$, use $z={\lambda }^{2/3}[-q^{\prime }(b){]}^{1/3}(x-b)$; decay on the right similarly yields $2C_R{q}^{-1/4}\cos (S_b(x)-\pi /4)$, where $S_b=\lambda {\int }_x^b\sqrt{q}ds$. Let $J=S_a+S_b=\lambda {\int }_a^b\sqrt{q}ds$. The left expression has equal cosine and sine coefficients in the variable $S_a$; the right expression has those coefficients proportional to $\cos (J-\pi /4)$ and $\sin (J-\pi /4)$. They can describe the same nonzero wave only when $J-\pi /4=\pi /4+n\pi$. Therefore the WKB quantization condition isThe two simple turning points contribute the two quarter-phase shifts, giving the half-integer correction.