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 / 1

Assume a real, time-independent potential with the same zero asymptote on both sides, so $k=\sqrt{2mE}/\hslash$ is the same incident and transmitted wave number. For unit incoming amplitudes,Multiplying the two Schrödinger equations crosswise and subtracting shows that the Wronskian $W={\psi }_L{\psi }_R^{\prime }-{\psi }_L^{\prime }{\psi }_R$ is constant. Its two asymptotic values are $-2ik{t}_R$ and $-2ik{t}_L$, hence $t_L=t_R$.

For a real potential the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is conserved, giving $|r_L{|}^2+|t_L{|}^2=|r_R{|}^2+|t_R{|}^2=1$. ThereforeThe reflection amplitudes can have different phases. Reality or self-adjointness is essential for the reflection-probability conclusion: a complex absorbing potential need not have equal left and right reflectances. Equal asymptotic speeds are also used in identifying transmission with the squared amplitude.

Away from $x=x^{\prime }$, the free equation is $G^{\prime \prime }+k^{2}G=0$. The outgoing boundary condition selects $G=A{e}^{ik|x-x^{\prime }|}$, continuous at its source. Integrating the differential equation across the source gives the derivative jumpThe jump for the proposed form is $2ikA$, so $A=im/(k{\hslash }^2)$. ThusFor $x$ far to either side, the $x^{\prime }$-dependent phase is absorbed into the outgoing amplitude. The outgoing prescription is equivalently the boundary value of the resolvent with $E\to E+i0$; changing that prescription would give a different Green function.

Write $L_0\psi =-U\psi$ and use the outgoing inverse of $L_0$. Adding the homogeneous incoming wave yieldsFor the attractive delta potential, let $\eta =m\alpha /(k{\hslash }^2)$. The equation reduces to $\psi (x)=e^{ikx}+i\eta e^{ik|x|}\psi (0)$ and $\psi (0)=1/(1-i\eta )$. Separating the two half-lines givesBoth sides have the same group velocity, soThe same amplitudes satisfy continuity and the exact delta-potential derivative jump ${\psi }^{\prime }(0^{+})-{\psi }^{\prime }(0^{-})=-2m\alpha \psi (0)/{\hslash }^2$, providing an independent local check on the sign.

Let $\lambda =m{U}_0{a}^2/{\hslash }^2\ll 1$ and $ka\ll 1$. Inside a region of size $a$, the Schrödinger equation estimates the curvature as $a^2{\psi }^{\prime \prime }=O(\lambda +(ka{)}^2)\psi$. Matching to the incoming and reflected waves constrains the slope as well. To leading order the interior wave has nearly constant value $C$, while its small accumulated derivative change encodes the potential's integrated strength.

This is not a Born assumption that $C$ is close to one. At very low $k$ the wave can be strongly reflected even by a weak potential, and $C$ can be small. The matching solution in the next part gives $C\sim k/(k+im\gamma /{\hslash }^2)$. It implies a fractional variation across the range of order $ka+m|\gamma |a/{\hslash }^2$, which is small under the two specified conditions. ThusThe approximation requires a finite short-range potential with a controlled width and integral; it does not describe arbitrary long-range tails.

Replace the slowly varying interior wave by $C$ and $e^{ik|x-x^{\prime }|}$ by one when both arguments lie within the range. The integral equation givesHenceThe integrated strength has dimensions of energy times length. For $U=-\alpha \delta (x)$, $\gamma =-\alpha$ and this is exactly the result in part c. For a regular weak potential the approximation sums the important ratio $m\gamma /(k{\hslash }^2)$ without requiring it to be small, which is why it remains useful beyond the naive first Born expansion at small momentum.

Write $\nu =m\gamma /{\hslash }^2$, so $t(k)\simeq k/(k+i\nu )$. Its nontrivial pole isFor negative $\gamma$ this is on the positive imaginary $k$ axis, the physical bound-state sheet. For positive $\gamma$ it lies on the negative imaginary axis and is a virtual-state pole reached by analytic continuation through the continuum cut. These different sheets can have the same negative numerical energy; specifying only $E_{\ast }$ would conceal the distinction. As a function of $E$, the momentum square root also introduces the continuum threshold branch point at zero. If $\gamma =0$, numerator and denominator cancel at this order and no pole is predicted by this leading approximation.

For a nonzero everywhere attractive weak potential, $\gamma <0$, the pole has $k=i\kappa$ with $\kappa =m|\gamma |/{\hslash }^2>0$. The exterior waves decay as $e^{-\kappa |x|}$, giving one shallow bound state to leading weak-coupling order:Its size ${\kappa }^{-1}$ is much larger than the potential range because $\kappa a=O(\lambda )\ll 1$. For an everywhere repulsive potential there is no negative-energy normalizable bound state: the kinetic and potential quadratic forms are both nonnegative. Its lower-half-plane pole is a virtual state, not a stable bound state. More complicated sign-changing potentials with a vanishing integral require higher-order analysis and are not covered by this simple leading result.