Suppose we have quantum mechanical plane waves of energy $E$ incident upon a one-dimensional potential barrier $V(x)$ with sloping sides.
One can compare the WKB solutions in the three relevant regions to asymptotics of the Airy function solutions to the linearized Schrödinger equation near the turning points to obtain matching formulae, relating $A, B$ (the coefficients of the right- and left-propagating waves, respectively, left of the barrier) to $F, G$ (the coefficients of the right- and left-propagating waves, respectively, right of the barrier). One can then construct the relevant scattering matrix $S$, defined by:
$$\left[\begin{array}{c} B\\ F \end{array}\right]=S\left[\begin{array}{c} A\\ G \end{array}\right] \, .$$
I have carried out the above computation and found that
$$S=\left[\begin{array}{cc} i\left(\dfrac{1-4e^{2\gamma}}{1+4e^{2\gamma}}\right) & \dfrac{4e^{\gamma}}{1+4e^{2\gamma}}\\ \dfrac{4e^{\gamma}}{1+4e^{2\gamma}} & i\left(\dfrac{1-4e^{2\gamma}}{1+4e^{2\gamma}}\right) \end{array}\right]$$
where
$$\gamma=\frac{1}{\hbar}\int_{x_{1}}^{x_{2}}\sqrt{2m\left[V\left(x\right)-E\right]}dx$$
measures the "strength" of the reflecting, classically-forbidden region.
This scattering matrix $S$ is in fact unitary, and hence this scattering process conserves the probability current across the barrier. But why should this be true?
I know that a solution of the Schrödinger equation will necessarily conserve probability current, but here we have both (1) assumed approximate WKB-type solutions and (2) carried out a complicated asymptotic matching procedure. It is not clear to me why, after this approximation and matching procedure, the resulting solution should still conserve probability current.
Said another way: is there an a priori reason one should expect the WKB-matched solutions to conserve probability current (in spite the fact that they are inexact), or is this simply a happy coincidence that falls out of the computation?
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