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 / 1

Away from $x=x^{\prime }$, an outgoing solution has form $G=C{e}^{ik|x-x^{\prime }|}$. Integrate the differential equation across $x^{\prime }$ to get $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$. Since the derivative jump is $2ikC$, $C=im/(k{\hslash }^2)$. ThusContinuity excludes a derivative-of-delta contribution. The outgoing condition, equivalently the appropriate limiting resolvent, removes the arbitrary incoming homogeneous solution.

Write the full equation as $L{\psi }_s=-U{\psi }_s$. Apply the outgoing inverse of $L$ and add the incident homogeneous solution with unit left-incident amplitude. This givesFor sufficiently short-range potentials this is the outgoing Lippmann-Schwinger equation. The minus sign follows from the definition $L=H_0-E$, rather than $E-H_0$.

As $x\to -\infty$, the scattered kernel is $e^{-ikx}{e}^{ik{x}^{\prime }}$; as $x\to +\infty$, it is $e^{ikx}{e}^{-ik{x}^{\prime }}$. Therefore ${\psi }_s\sim e^{ikx}+r{e}^{-ikx}$ on the left and $t{e}^{ikx}$ on the right, withEqual asymptotic wavenumbers give $R=|r{|}^2$, $T=|t{|}^2$. Squaring the first amplitude proves the requested expression. These are current ratios, which would require a velocity factor if the two asymptotic potentials differed.

Put $\beta =m\alpha /{\hslash }^2$. The delta interaction reduces the integral equation to ${\psi }_s=e^{ikx}+i\beta e^{ik|x|}{\psi }_s(0)/k$. At zero, ${\psi }_s(0)=(1-i\beta /k{)}^{-1}$. HenceTheir sum is one, as required for a real potential. The explicit wave is the indicated incident-plus-reflected form for $x<0$ and $t{e}^{ikx}$ for $x>0$; it is continuous and has derivative jump $-2\beta \psi (0)$.

The first Born approximation substitutes the incident wave for the exact wave inside the scattering integral. ThusFor the attractive delta this is ${\beta }^2/k^2$, the leading small-$\beta /k$ expansion of the exact result. Physically one neglects the distortion of the incident state by repeated interaction with the potential. At low $k$ the Born result can exceed one, while the exact reflection remains bounded and approaches one; it is not valid in that limit.

For $E=-{\hslash }^2{\kappa }^2/(2m)$, the decaying Green's function is $m{e}^{-\kappa |x-x^{\prime }|}/(\kappa {\hslash }^2)$, with the same derivative jump as before. A normalizable negative-energy state has no incident homogeneous part. Applying the inverse to $L\psi =-U\psi$ givesThe condition $\kappa >0$ is necessary for decay on both sides.

At the delta potential, the bound equation becomes $\psi (x)=\beta \psi (0)e^{-\kappa |x|}/\kappa$. A nonzero solution therefore requires $\kappa =\beta$, givingThere is one bound level. The transmission amplitude $t(k)$ has its physical-sheet pole at $k=i\beta$, which produces exactly this decaying state. The real-axis probability's rational continuation $T(k)=k^2/(k^2+{\beta }^2)$ has both $+i\beta$ and $-i\beta$ poles; the lower pole is not an additional bound state. It is essential to distinguish the analytic amplitude from an even continuation of its modulus squared.