Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 20D / Solution

The scattering interpretation assumes $0<ϵ<U$, so $k=\sqrt{ϵ}$ is real and $K=\sqrt{U-ϵ}>0$. The incoming amplitudes are $a$ from the left and $b$ from the right; the outgoing amplitudes are $c,d$. The conserved probability current iswhich gives $j=\hslash k(|a{|}^2-|c{|}^2)/m$ on the left and $j=\hslash k(|d{|}^2-|b{|}^2)/m$ on the right. Equality proves $|c{|}^2+|d{|}^2=|a{|}^2+|b{|}^2$. Equal exterior wavenumbers make the flux weights equal.

Continuity of the wavefunction and its derivative at the finite potential steps yields the four relations, with $C=\cosh KL$ and $S=\sinh KL$,Set $\lambda =K/(ik)$. Substituting $e=a+c$, $f=(a-c)/\lambda$ and subtracting the last two relations after dividing the fourth by $ik$ giveswhere $D=C-(\lambda +{\lambda }^{-1})S/2$. HenceReflection about the barrier midpoint interchanges the left and right incidence problems and, with the chosen phase origins, exchanges $a\leftrightarrow b$ and $c\leftrightarrow d$. ThereforeFor incidence from the left alone, the transmission coefficient is the transmitted-to-incident probability flux ratio, $T=|d/a{|}^2=|D{|}^{-2}$. Since $\lambda =-iK/k$, direct calculation givesFor fixed energy inside the barrier range and $KL\gg 1$, ${\sinh }^{2}KL\sim e^{2KL}/4$, soThis is the thick-barrier quantum tunnelling limit, not a uniform approximation as energy approaches the barrier top while $KL$ remains small.