Past exam of the further quantum theory course of the University of Oxford / 2023 / 2

For $E>0$, put $k=\sqrt{2mE}$ and $q=k/\hslash >0$. Outside the support of the potential energy, a generalized eigenfunction of the Hamiltonian operator has the plane wave formHere $R_0$ denotes the support radius, to distinguish it from the quantum reflection probability $R$. The momenta are $\pm k$ and the velocities are $\pm k/m$. Thus $A,D$ describe incoming scattering channels from the left and right, and $B,C$ describe outgoing scattering channels towards the left and right. These energy eigenfunctions are not square-integrable functions; normalizable incoming wave packets are formed by superposition over energy, and their passage through the interaction region gives the physical quantum scattering experiment.

For incidence from the left, normalize $A=1,D=0$ and write $B=r_L,C=t_L$. The quantum reflection amplitude is $r_L$ and the transmission amplitude is $t_L$. The ratios of outgoing to incident probability currents defineNo extra velocity ratio occurs because both exterior regions have the same potential energy and mass. The scattering matrix, with a fixed choice of channel order, isIts first and second columns are the responses to unit incidence from the left and right, respectively. For a reflection-symmetric potential energy, these conventions give $r_L=r_R=r$ and $t_L=t_R=t$.

For a real potential energy and a stationary state, the probability current isThe time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$, so $j^{\prime }=\hslash \mathrm{Im}({\psi }^{\ast }{\psi }^{\prime \prime })/m=0$. In an exterior plane wave superposition, interference terms contribute no current andConsequently $|A{|}^2+|D{|}^2=|B{|}^2+|C{|}^2$ for every choice of incoming amplitudes. If $u=(A,D{)}^T$, this says $u^{†}(I-S^{†}S)u=0$ for every complex vector $u$. Since $I-S^{†}S$ is Hermitian, vanishing of its quadratic form implies that the matrix itself vanishes. ThereforeThe second identity follows by choosing $u=(1,0{)}^T$. Since $S$ is a square matrix, the first also implies $S{S}^{†}=I$, proving unitarity. A real delta potential preserves this proof: continuity of $\psi$ and the real derivative jump make the current continuous across the interaction point.

Assume $a>0$ and put $\mu =mg/{\hslash }^2$, $q=k/\hslash$, and $\gamma =\mu /q$. At a delta potential located at $x_0$, integration of the Schrödinger equation gives the matching conditionsWriting the adjacent plane wave coefficients as $(A_L,B_L)$ and $(A_R,B_R)$ givesSolving these two linear equations yields the transfer matrix of a delta potential:For the symmetric double-delta potential, use separate plane wave pairs in $x<-a$, $-a<x<a$, and $x>a$. Successive matching gives $M=D(a)D(-a)$ between the two exterior pairs. For real $q$, multiplication givesThe determinant identity $\det M=1$ reads $|d{|}^2-F^2=1$. Left incidence means $(t,0{)}^T=M(1,r{)}^T$, whenceHere the first relation also follows from $t=M_{11}-M_{12}{M}_{21}/M_{22}=\det M/M_{22}$. Therefore the requested transmission probability and quantum reflection probability, as functions of the printed momentum $k$, areThey obey $R+T=1$. As checks, $g=0$ gives $T=1$, and the coincident limit $a\to 0$ gives $T=q^2/(q^2+4{\mu }^2)$, the answer for a single delta potential of strength $2g$.

Continue the transmission amplitude to complex wavenumber $q=i\kappa$ with $\kappa >0$, equivalently complex momentum $k=i\hslash \kappa$. Then $E=-{\hslash }^2{\kappa }^2/(2m)<0$. In the left exterior region the $A$ term grows at negative infinity and the $B$ term decays; in the right exterior region the $D$ term grows and the $C$ term decays. A bound state thus requires $A=D=0$, with nonzero $B,C$. The transfer matrix relation gives $M_{22}(i\kappa )=0$. Equivalently, a zero of the incoming coefficient is a bound-state pole of $t(q)$, whose numerator is nonzero at $q=i\kappa$.

The denominator found in part (c) givesFactorization produces two branches. To identify their parity, use an even interior wavefunction $A\cosh (\kappa x)$ or an odd interior wavefunction $A\sinh (\kappa x)$ for $|x|<a$, joined to a decaying exterior wavefunction at $a$. Continuity and the derivative jump give, respectively,Rearrangement gives the implicit energy eigenvalue equationsOnly strictly positive equation roots $\kappa$ are retained. With $\nu =-\mu >0$, the even equation root lies above $\nu$ and the odd equation root, when it exists, lies below $\nu$, so the even bound state has lower energy. The formal zero equation root in the odd equation does not itself establish a normalizable wavefunction.

For $g\ge 0$, the Hamiltonian operator is nonnegative: its quadratic form is ${\hslash }^2\int |{\psi }^{\prime }{|}^{2}dx/(2m)+g(|\psi (a){|}^2+|\psi (-a){|}^2)\ge 0$. Thus it has no negative-energy bound states. A nonnegative-energy exterior wavefunction cannot be square integrable unless it vanishes identically, so there are no other bound states either.

For $g<0$, set $\nu =-mg/{\hslash }^2>0$. The even equation is the zero of $f(\kappa )=\nu (1+e^{-2a\kappa })-\kappa$. Since $f(0)=2\nu >0$, $f(\kappa )\to -\infty$ and $f^{\prime }(\kappa )=-2a\nu e^{-2a\kappa }-1<0$, the intermediate value theorem and strict monotonicity give exactly one positive equation root.

The odd equation is the zero of $h(\kappa )=\nu (1-e^{-2a\kappa })-\kappa$. Here $h(0)=0$, $h^{\prime }(0)=2a\nu -1$, $h^{\prime \prime }(\kappa )=-4a^2\nu e^{-2a\kappa }<0$, and $h(\kappa )\to -\infty$. If $2a\nu \le 1$, strict concavity makes $h(\kappa )<0$ for every $\kappa >0$. If $2a\nu >1$, $h$ initially increases, has exactly one maximum, and then decreases through zero exactly once. This proves the bound-state count of an attractive symmetric double-delta potential:At the threshold $g=-{\hslash }^2/(2ma)$, the odd zero-energy solution is proportional to $x$ inside $(-a,a)$ and to $a\mathrm{sgn}x$ outside. It is bounded but not a square-integrable function, so it is a zero-energy scattering resonance rather than a second bound state. Each negative-energy bound state is either even or odd by nondegeneracy of one-dimensional bound states; the two branches above exhaust the count.