Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B6

Let $\rho ={\psi }^{\ast }\psi$ be the probability density. For a real potential energy $V(x)$, the Schrödinger equation and its complex conjugate areDifferentiate $\rho$ and substitute both equations. The potential terms cancel, leavingComparison with the probability continuity equation ${\partial }_t\rho +{\partial }_{x}J=0$ identifiesFor a plane wave of amplitude $A$ and positive wavenumber $k$, $J=\hslash k|A{|}^2/m$, the density times the velocity. Integrating the continuity equation over an interval makes its probability change equal to the incoming current minus the outgoing current.

SetAll are real and $k_0>k_1>k_2>0$. Suppressing a common time quantum phase $e^{-iEt/\hslash }$, chooseThe rightmost region has no left-moving term because no wave is incident from the right. The shifted quantum phase convention makes $F$ the transmitted amplitude at the second interface.

For finite potential steps and equal mass, wavefunction matching at a finite potential step requires continuity of both $\psi$ and ${\partial }_x\psi$. Explicitly,A discontinuous wavefunction would introduce an inadmissible singular kinetic term, while integrating the stationary Schrödinger equation across a finite step shows that the derivative cannot jump. These conditions also keep the probability current continuous, so a finite interface neither creates nor destroys probability. The derivative rule would need modification for a delta-function potential or a different effective mass; neither occurs here.

Zero reflection means $B=0$. Matching the first interface then givesThe specified layer quantum phase is $k_{1}a=\pi /2$, so $e^{i{k}_{1}a}=i$ and $e^{-i{k}_{1}a}=-i$. The second-interface conditions becomeSince $C+D=A$, compatibility requires $\boxed{k_1^2=k_0{k}_2}$. This is a quantum quarter-wave matching layer: the intermediate wavenumber is the geometric mean, and the two interface reflections cancel after their round trip through the layer.

In terms of the physical energies,Rearranging also givesA finite positive energy satisfying the stated ordering requires $V_2/2<V_1<V_2$. Under that condition $E-V_2=(V_2-V_1{)}^2/(2V_1-V_2)>0$, consistent with the assumed propagating transmitted wave. The width must simultaneously satisfy the given quarter-wave quantum phase condition.

From the second-edge amplitude in part (c),This uses ${\psi }_{\mathrm{R}}=F{e}^{i{k}_2(x-a)}$. If the transmitted wave is instead written $T{e}^{i{k}_{2}x}$, its coefficient is $\boxed{T/A=i\sqrt{k_0/k_2}{e}^{-i{k}_{2}a}}$; the extra factor only changes the quantum phase convention.

Using the probability current of a plane wave,Thus the flux transmission probability is one. The larger transmitted density $|F{|}^2>|A{|}^2$ compensates for the smaller transmitted velocity $\hslash k_2/m$; equal currents need not have equal wave amplitudes.