Past exam of the theoretical physics II course of the University of Cambridge / 2015 / Paper 1 / 3

Outside a finite-range scattering potential, the stationary Schrödinger equation becomes the free Helmholtz equation. The specified incident state is a plane wave along $+z$. The scattered disturbance must be outgoing: the scattering experiment has an incoming incident beam but no independently imposed spherical wave arriving from infinity. At large radius its angular pattern is encoded by a scattering amplitude $f_k(\theta ,\varphi )$:An outgoing three-dimensional wave has amplitude proportional to $1/r$ because its probability current spreads over area proportional to $r^2$. Indeed its radial current is $(\hslash k/m)|f_k{|}^2/r^2$ to leading order, while the incident current density is $\hslash k/m$. Therefore $d\sigma /d\Omega =|f_k{|}^2$. The sign $e^{+ikr}$ assumes the stationary time factor $e^{-i{E}_{k}t/\hslash }$.

Rearrange the continuum equation asThe incident plane wave contributes zero to the left side. The outgoing Green function identity gives $-4\pi f_k{\delta }^{(3)}(\mathbf{r})$ for the scattered term, henceThis fixes both the sign and the three-dimensional $4\pi$ factor. The model is an isotropic point coupling, so the scattered amplitude has no angular dependence. Take $g$ real, as required by the displayed Hermitian pair of coupling equations; an arbitrary phase of the discrete state can be chosen accordingly.

The outgoing wave is singular at the origin, so its value there cannot be inserted without regularization. At a short radial distance $\delta >0$, with $k\delta \ll 1$,The incident factor tends to one independently of the chosen approach direction; the singular part is isotropic. The discrete-state equation becomes, to the retained short-distance order,Keeping only $1/\delta$ and discarding $ik$ would lose the outgoing-wave imaginary part and produce an amplitude inconsistent with scattering unitarity.

Define $A=m{g}^2/(2\pi {\hslash }^2)$, which has units of energy times length. Substituting the result of (b) into (c) givesAbsorb the real cutoff shift into the physical detuning,It follows thatUsing $k=\sqrt{2m{E}_k}/\hslash$, identifyHere $\gamma$ has units of square root of energy. This is the small-cutoff, renormalized resonant-level amplitude; terms vanishing with $k\delta$ have been omitted. Taking $\delta \to 0$ at fixed measured detuning requires the bare ${ϵ}_0$ to depend on the cutoff.

The sign is checked by $f_k^{-1}=-(E_k-{\widetilde{ϵ}}_0)/A-ik$, which has the elastic $s$-wave unitarity form. Thus $\mathrm{Im}{f}_k=k|f_k{|}^2$, consistent with the optical theorem and ${\sigma }_{\mathrm{tot}}=4\pi |f_k{|}^2$. If the renormalized level is positive, at $E_k={\widetilde{ϵ}}_0$ the amplitude is $i/k$ and the scattering cross-section reaches $4\pi /k^2$.