 Solution

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 $4\pi /k^2$.