Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 16A / Solution

The one-dimensional probability current isAt a finite jump of the potential, both $\psi$ and ${\psi }^{\prime }$ are continuous. A jump in $\psi$ would create a delta derivative in the stationary Schrödinger equation, and integrating that equation across an increasingly small interval then shows continuity of ${\psi }^{\prime }$. Accordingly impose continuity of both quantities at zero and at $a$.

For an incident wave of positive energy, define $k=\sqrt{2mE}/\hslash$. With no wave incident from the right, the left region hasThe interference terms cancel from the imaginary part in the current formula. If $E>V_0$, let $k_0=\sqrt{2m(E-V_0)}/\hslash$. In the right region,For $0<E<V_0$, let ${\kappa }_0=\sqrt{2m(V_0-E)}/\hslash$. Boundedness at positive infinity excludes the growing exponential, soThe decaying solution has zero current even if its amplitude is nonzero. These exterior forms do not require $E$ to lie on a particular side of $V_1$: in the finite middle layer the solution is oscillatory for $E>V_1$, exponential for $E<V_1$, and linear at $E=V_1$; matching determines the amplitudes.

Define the reflection coefficient as reflected current magnitude divided by incident current, and the transmission coefficient as outward right current divided by incident current. ThenThe denominator is $j_{\mathrm{incident}}=\hslash k|A{|}^2/m$. For a real potential, differentiation givesby the stationary Schrödinger equation. Continuity at each interface preserves this conserved current, so $j_{\mathrm{left}}=j_{\mathrm{right}}$. Dividing by incident current yieldsThere can be penetration into the barrier without transmission to infinity, because the entire right half-line remains classically forbidden.