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

Let $x_{<}=\min (x,x^{\prime })$ and $x_{>}=\max (x,x^{\prime })$. A homogeneous solution vanishing at the left wall is $u(x)=\sin [k(x+a/2)]$; one vanishing at the right wall is $v(x)=\sin [k(a/2-x)]$. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-k\sin ka$. Continuity and the source derivative jump $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$ therefore giveThis is exactly the printed form because $x_{<}=(x+x^{\prime }-|x-x^{\prime }|)/2$ and $x_{>}=(x+x^{\prime }+|x-x^{\prime }|)/2$, with a minus sign from reversing the argument of the second sine.

The true poles occur at $k_n=n\pi /a$, $n=1,2,\dots$, corresponding toThe apparent zero at $k=0$ is removable: the numerator and $k\sin ka$ both vanish quadratically. Negative values of $k_n$ duplicate the same energies, rather than describing extra levels. The boundary conditions and source jump uniquely determine the inverse whenever $E$ is not an eigenvalue.

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 $E_n-E$.

Factorizing the projector found in part b gives the normalized eigenfunctionswith zero continuation outside. Their squared integrals are one and different positive integers give orthogonal functions. These are the familiar infinite-square-well states shifted so that the well centre is at zero. Each eigenfunction has an arbitrary overall constant phase, which cancels from the projector.

For $E<0$, inversion of the free operator gives $G_0(x,y)=m{e}^{-\kappa |x-y|}/(\kappa {\hslash }^2)$, with $\kappa =\sqrt{-2mE}/\hslash$. The bound-state equation is $\psi =-G_{0}U\psi$, henceFor the separable kernel $V(y,z)=-\alpha F(y)F^{\ast }(z)$, defineThenAn overall normalization still has to be chosen after finding a permitted energy. A nontrivial bound state must have $C\ne 0$: if $C=0$, the potential acts as zero and a square-integrable negative-energy free solution on the whole line cannot exist. Decay or square-integrability assumptions on $F$ are needed to make these integrals well defined; smoothness alone would not ensure convergence.

Multiply the expression for $\psi$ by $F^{\ast }(x)$ and integrate. Cancelling the nonzero $C$ givesThe kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:This is the scalar consistency condition for the rank-one potential, rather than an independent normalization equation. The double integral is real and nonnegative because the Fourier transform of $e^{-\kappa |x|}$ is $2\kappa /(q^2+{\kappa }^2)>0$. Thus attractive $\alpha >0$ is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free resolvent expectation decreases with $\kappa$, giving at most one bound state for this rank-one attraction.