Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section F / 46 / a / iv

For $0<E<V_0$, put $\alpha =i\kappa$ with $\kappa =\sqrt{2m(V_0-E)}/\hslash$. Since $\sin (i\kappa a)=i\sinh (\kappa a)$, both signs in the quotient change andIt is finite and positive for nonzero incident energy and finite width: this is quantum tunnelling. In a thick barrier $P\simeq [16E(V_0-E)/V_0^2]e^{-2\kappa a}$. Inside, the scattering wave is a combination $C{e}^{-\kappa x}+D{e}^{\kappa x}$, not an isolated real decaying exponential; their complex coefficients allow a nonzero conserved current.

Let ${\kappa }_0=\sqrt{2m{V}_0}/\hslash$. As $E\to 0^{+}$, $\kappa \to {\kappa }_0$ while the large denominator term scales as $1/E$. HenceThe limit is linear in incident energy for fixed positive height and nonzero width. A below-barrier wave has nonzero penetration at finite energy, but that does not imply a finite zero-energy transmitted flux.

For $E\gg V_0$, expand the reciprocal to obtainRetain the exact $\alpha$ in the bounded oscillatory factor unless a further phase expansion is controlled. Reflection is small, with an oscillatory interference correction of order $(V_0/E{)}^2$. The wavefunction is nearly a travelling wave throughout; its wave number inside is slightly smaller than outside, so its wavelength is slightly larger there. The reflected wave adds weak standing-wave modulation on the incident side.

For fixed $V_0$ and $a$, $0\le {\sin }^2(\alpha a)\le 1$ while the multiplier of this bounded factor tends to zero as $E^{-2}$. Consequently $\boxed{\underset{E\to \infty }{\lim }P(E)=1}$, even though the sine factor itself does not have a limit. This is the high-energy classical transmission limit.