The WKB approximation and boundary conditions

Question

To estimate a quantization rule using the WKB approximation, one is usually working in the 'classically allowed region'. You apply the WKB approx in the middle of the region and use the Airy connection formulae to match the wavefunction at the turning points of the classical motion.

Consider a situation where we have a slowly-varying potential on the half line $x \geq 0$. I want to enforce mixed (Robin) boundary conditions at the origin so that $$ \psi(0) + a\psi'(0) = 0.$$ Such a boundary condition has no classical analog (the introduction of the dimensionful parameter $a$ is technically an anomaly), and hence no defined notion of a "turning point". How, then, can I use the WKB approximation to estimate a quantization rule for the energy levels?

Morally, such a mixed boundary condition is similar to having a delta function potential $V \supset 1/a \cdot\delta(x)$. The WKB approximation relies on continuity of the potential so I expect that the approximation fails at zero. However, the potential would still be continuous infinitesimally close to zero, and so the approximation should hold in its neigborhood. Furthermore, there is no way to solve the classical turning point equation $E = V(x)$ at $x = 0$.

Some approaches have you integrate Schrodinger to determine the discontinuity in the derivative, and use WKB elswhere. This is complicated as the delta function lives on the boundary of the domain.

What is the proper treatment of mixed boundary conditions in the WKB approximation?

Answer 1

1. A Dirac delta potential yields a 2-sided condition $$\psi^{\prime}(0^+)-\psi^{\prime}(0^-)~=~a\psi(0),$$ not a Robin/mixed boundary condition (BC) $$\psi^{\prime}(0^+)~=~a\psi(0)$$ per se.

2. The WKB Airy connection formula is for a linearized potential at a turning point, which do not apply to a Robin/mixed BC.

3. Given a (possibly $\hbar$-dependent) Robin/mixed BC, the WKB ansatz for the wavefunction should presumably satisfy the BC to the leading order in $\hbar$.