Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 33C (Applications of Quantum Mechanics)

With the incoming basis ${\mathcal{I}}_{\pm }$ and outgoing basis ${\mathcal{O}}_{\pm }$ as printed, expand any scattering solution as $\psi \sim a_{+}{\mathcal{I}}_{+}+a_{-}{\mathcal{I}}_{-}+b_{+}{\mathcal{O}}_{+}+b_{-}{\mathcal{O}}_{-}$. Linearity definesThe column indexed by a channel records outgoing amplitudes for a unit incoming wave in that channel. The extra minus sign in ${\mathcal{O}}_{-}$ is a channel phase convention; it affects the sign of the free odd-channel amplitude but not flux probabilities. For a real potential and no absorbing or additional channels, probability-current conservation makes the flux-normalized matrix unitary.

An even potential $V(x)=V(-x)$ preserves parity. Its Hamiltonian operator commutes with the parity operator, so an even incoming solution produces only an even outgoing solution, and similarly for an odd one. ThusAn asymmetric potential generically couples the parity channels; an accidental vanishing at one energy does not imply the symmetry for all energies.

For the odd channel, solve on $x>0$ with $\psi (0)=0$. Inside $0<x<a$, write $\psi =C\sin (kx)$; outside write $\psi =e^{-ikx}-S_{--}{e}^{ikx}$, respecting the outgoing odd-channel minus sign. Continuity at $a$ and the delta-potential derivative jump areEliminating $C$ yieldsThe equations remain valid when $\sin (ka)=0$ by taking the limiting value, rather than dividing blindly. For real $k,U_0$, numerator and denominator inside the fraction are complex conjugates and the prefactor has unit modulus. Thus $|S_{--}{|}^2=1$. This is conservation of flux in the odd parity channel: the real potential causes only a phase shift there, not absorption. It does not assert unit left-to-right transmission in the usual incident-from-one-side basis.

A negative-energy odd bound state has $k=i\kappa$, $\kappa >0$, with $\psi =C\sinh (\kappa x)$ inside and a decaying exponential outside. Its jump condition, equivalently the nontrivial upper-half-plane pole of the preceding S-matrix, givesSet $w=2\kappa a$ and $g=-a{U}_0>0$. The condition is $w=g(1-e^{-w})$. The right side is strictly increasing and strictly concave, has slope $g$ at zero and is bounded by $g$. Hence there is exactly one positive intersection iff $g>1$. ThereforeThe formal zero root is not a positive-decay bound state. At $a{U}_0=-1$ there is a zero-energy threshold solution, not a normalizable bound state.

Let $g=a{U}_0\gg 1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeysNear $z=\pi$, write $z=\pi +\delta$ and choose the logarithm branch with $\delta \to 0$. Expanding givesConsequentlyThe lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma /2={\hslash }^2{k}^2/(2m)$, then $\Gamma \simeq 2{\hslash }^2\pi \gamma /(m{a}^2)$ and its probability lifetime is $\hslash /\Gamma$. It is not a genuine bound state of the repulsive potential.