= Solution
For real $x=ka$, taking the <modulus> squared of the amplitude gives the <half-line delta-well amplitude enhancement>
$$
\boxed{|A(x)|^2=\frac{\lambda^2x^2}{(1-\cos2x)^2+(\lambda x-\sin2x)^2}.}
$$
The supplied <resonance poles> have real parts $x_n\simeq n\pi(1+\lambda/2+\lambda^2/4)$ and negative imaginary parts $-\gamma_n\simeq-(\lambda n\pi/2)^2$ for fixed low $n$ at $\lambda\ll1$.<Advanced Quantum Mechanics: S18 Exam 2015 Solutions>[Problem 1, PDF pp.5–6: pole expansion]{ref} Linearizing the denominator at a simple pole gives
$$
|A(x)|^2\simeq\frac{\lambda^2x_n^2}{|D'(p_n)|^2[(x-x_n)^2+\gamma_n^2]},\qquad D(x)=1-e^{2ix}+i\lambda x.
$$
Thus the sketch has narrow, tall peaks with a locally <Lorentzian line shape>, just above positive integer multiples of $\pi$. Their half-widths are $\gamma_n$, and the leading peak heights are $4/(\lambda^2n^2\pi^2)$. In this fixed-low-mode approximation, increasing $n$ broadens and lowers the peaks. Exactly at $x=n\pi\ne0$, the value is one. At zero it has the finite limit $\lambda^2/(2-\lambda)^2$ for $\lambda\ne2$.
The peaks represent long-lived <standing waves> between the wall and the attractive <delta potential>. Despite the source's transmission-coefficient terminology, $|A|^2$ is an internal amplitude factor and can exceed one. The interior <probability current> is zero, and the reflected <probability current> equals the incident magnitude; the hard wall passes no transmitted flux.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015-s18-delta-well-amplitude-sketch.png]
{title=Internal standing-wave resonances and the Lorentzian line shape of the first peak}
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