Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S18 / 2

For $E<0$, impose decay and continuity on $G=C{e}^{-\kappa |x-x^{\prime }|}$. Its derivative jump is $-2\kappa C$, so $-{\hslash }^2(G_{+}^{\prime }-G_{-}^{\prime })/(2m)=1$ requiresThere is no nonzero free solution decaying at both infinities that can be added without spoiling the derivative condition.

The full equation rearranges to $L\psi =-U\psi$. Since a bound state has no incoming free wave, convolution with the decaying inverse gives $\boxed{\psi (x)=-m\int e^{-\kappa |x-x^{\prime }|}U(x^{\prime })\psi (x^{\prime })d{x}^{\prime }/(\kappa {\hslash }^2)}$. Applying $L$ to this identity recovers the differential equation and the decay boundary conditions.

Let $b=a/2$, $q=\sqrt{2m(U_0-|E|)}/\hslash$ and $\kappa =\sqrt{2m|E|}/\hslash$. Bound levels have $-U_0<E<0$. Up to normalization, parity givesThe respective inverse squared normalization constants are $b+\sin (2qb)/(2q)+{\cos }^2(qb)/\kappa$ and $b-\sin (2qb)/(2q)+{\sin }^2(qb)/\kappa$. These include both exponential tails.

Continuity has already been built into these forms. At $b$, derivative continuity requires $-q\sin (qb)=-\kappa \cos (qb)$ for even parity and $q\cos (qb)=-\kappa \sin (qb)$ for odd parity. ThusUse $q/\kappa =\sqrt{U_0/|E|-1}$ to obtain the two printed equations. The right side is positive, so only the corresponding allowed tangent/cotangent branches can describe a bound state; squaring without retaining this sign restriction adds spurious roots.

Matching $\psi$ and ${\psi }^{\prime }$ at both walls, or propagating the pair $(\psi ,{\psi }^{\prime })$ through the interior by the matrix $\left(\begin{array}{cc}\cos \zeta a& \sin \zeta a/\zeta \\ -\zeta \sin \zeta a& \cos \zeta a\end{array}\right)$, givesHere ${\zeta }^2=k^2+2m{U}_0/{\hslash }^2$. For real positive $k$, taking the modulus givesThe equal exterior velocities make this also the current transmission coefficient.

For the amplitude continue to $k=i\kappa$ on the decaying physical sheet. Then $\zeta =q$ and $D=\cos (qa)+({\kappa }^2-q^2)\sin (qa)/(2\kappa q)$. The double-angle identity factors its zero condition into the even and odd matching conditions $\kappa =q\tan (qb)$ and $\kappa =-q\cot (qb)$. Thus physical upper-half-plane amplitude poles are exactly the bound energies.

The literal statement for the probability requires correction. The displayed real-axis expression continued algebraically is $1/[D(k)D(-k)]$, and includes poles of the opposite-boundary-condition amplitude too. A concrete counterexample uses ${\hslash }^2/(2m)=1$, $a=2$, $q=1$, $\kappa =\cot 1$ and $U_0=1+{\cot }^{2}1$. Its continued probability denominator vanishes at $k=i\cot 1$, because $D(-k)=0$. But neither $\kappa =\tan 1$ nor $\kappa =-\cot 1$ holds, so $E=-{\cot }^{2}1$ is not a bound energy. Keeping the amplitude and the decaying boundary condition removes this spurious identification.

Rewrite the real-energy probability as $T=[1+U_0^2{\sin }^2(\zeta a)/(4E(E+U_0)){]}^{-1}$. Thus $\boxed{T_{\max }=1}$ at $\zeta a=j\pi$, namely $E_j={\hslash }^2{j}^2{\pi }^2/(2m{a}^2)-U_0\ge 0$, with a strictly positive resonance energy in the ordinary scattering problem. Between resonances it is less than one and its oscillations approach one as energy increases. Generically $T(0^{+})=0$; a well tuned to a zero-energy resonance instead has $T(0^{+})=1$. The original transmission plot in S1 illustrates these maxima for a chosen well, not a universal specified depth.

On the physical sheet the bound-state amplitude poles lie at $k=i{\kappa }_j$, ${\kappa }_j>0$, and map to real negative $E_j=-{\hslash }^2{\kappa }_j^2/(2m)$. The square-root map has two sheets meeting at the continuum threshold. Analytic continuation can also have virtual-state poles on the negative imaginary $k$ axis and resonance poles in the lower half-plane; these are not normalizable bound states. In the energy plane the resonances belong to the nonphysical sheet, rather than additional negative-axis bound levels.

For the algebraically continued probability, include the mirror poles contributed by $D(-k)$ as explained in part d; the two $k$ signs map to the same numerical energy but need not impose the same boundary condition. A sketch must label the sheet or the amplitude being plotted, rather than treating every singularity as a bound state.