Past exam of the physics A course of the University of Cambridge / 2014 / Paper 1 / Section B / B6 / Solution

Set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(U-E)}/\hslash$. For unit incident amplitude, use $\psi =e^{ikx}+r{e}^{-ikx}$ on the left and $\psi =t_s{e}^{-\kappa x}$ on the right; the growing exponential is excluded at infinity. Continuity at zero gives $1+r=t_s$ and $ik(1-r)=-\kappa t_s$. ThusThe phase is chosen in $(-\pi ,0)$ for $0<E<U$. The evanescent wave is $\psi (x)=2k{e}^{-\kappa x}/(k+i\kappa )$ for $x>0$. Its probability density is nonzero, but its probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is zero. Penetration into a forbidden region therefore does not require transmitted flux through a semi-infinite step.

For a finite rectangular barrier, matching both interfaces givesThe approximation requires $\kappa a\gg 1$, or a width much greater than the evanescent penetration length. The tunnelling exponent is $-2a\sqrt{2m(U-E)}/\hslash$; its factor of two converts an amplitude decay to a probability decay. For the electron parameters, $\kappa =1.984\times 10^9{\mathrm{m}}^{-1}$ and $\kappa a=19.84$. ThereforeKeeping only the exponential gives $5.83\times 10^{-18}$; the matched-interface prefactor accounts for the factor of three.