Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S18 / 2 / b / Solution

Expand $G(x,x^{\prime })$ in a complete orthonormal eigenbasis in its first argument. Applying the operator yields ${\sum }_n(E_n-E)c_n{\phi }_n(x)=\delta (x-x^{\prime })$. Completeness expands the Dirac delta distribution as ${\sum }_n{\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$, soConsequently ${\lim }_{E\to E_n}(E_n-E)G={\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$.

For the explicit formula, use $E_n-E\simeq -({\hslash }^2{k}_n/m)(k-k_n)$ and $\sin ka\simeq (-1{)}^{n}a(k-k_n)$. Also
$\sin [k_n(a/2-x)]=(-1{)}^{n+1}\sin [k_n(x+a/2)]$. The resulting limit isThe expression is symmetric in $x,x^{\prime }$ and holds on either side of the source. The usual complex-analysis residue with respect to $E-E_n$ has the opposite sign, since the source requests the limit with $E_n-E$.