Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 35B (Applications of Quantum Mechanics)

PutAway from $x=x^{\prime }$, this satisfies $(d^2/d{x}^2+k^2)G_k=0$. Its first derivative has jump $2ik$, so, distributionally,Applying $d^2/d{x}^2+k^2$ to the proposed integral equation therefore givesSince $k^2=2mE/{\hslash }^2$, this is precisely the stated Schrodinger equation. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x^{\prime }|}$ gives outgoing waves on both sides of each source point.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.