Past exam of the theoretical physics II course of the University of Cambridge / 2017 / 3

For a time-independent Hamiltonian operator, the time-evolution operator is $U(T)=e^{-iHT/\hslash }$, with $T=t-t^{\prime }$. Insert its resolution of identity in normalized energy eigenstates:Since $⟨x|\alpha ⟩={\phi }_{\alpha }(x)$, the retarded quantum propagator in the question isThe Heaviside step function imposes the chosen retarded convention. If the Hamiltonian operator has a continuous spectrum, the spectral sum includes its continuum integral; it must not be restricted to bound states. The spectral resolution of a quantum propagator also shows that $K$ tends to $\delta (x-x^{\prime })$ as $T\downarrow 0$. With the step factor it obeys $(i\hslash {\partial }_t-H)K=i\hslash \delta (T)\delta (x-x^{\prime })$; the corresponding Green function with a unit delta source is $K/(i\hslash )$.

Away from the delta potential, the decaying bound state obeys $E_0=-{\hslash }^2{\kappa }^2/(2m)$. Integrating the Schrödinger equation across the attractive interaction givesFor an even exponential, the left side is $-2\kappa \phi (0)$. ThereforeThe normalized bound state in a delta potential is ${\phi }_0(x)=\sqrt{\kappa }{e}^{-\kappa |x|}$, since $2\kappa {\int }_0^{\infty }{e}^{-2\kappa x}dx=1$. The single positive solution for $\kappa$ gives one bound state; an odd decaying wave cannot be both continuous and nonzero with this interaction.

Choose the even-channel scattering phase shift convention ${\phi }_{k,e}(x)={\pi }^{-1/2}\cos (k|x|+{\delta }_e)$ for $k>0$. The delta potential derivative jump givesso, choosing the branch that tends to zero as $k\to \infty$,An odd scattering eigenfunction is ${\phi }_{k,o}(x)={\pi }^{-1/2}\sin kx$: it vanishes at the interaction and its derivative is continuous, so it acquires no scattering phase shift. These parity channels, with the bound state, give the continuum part of the spectral resolution of a quantum propagator.

At the origin, the normalized bound state in a delta potential contributes $|{\phi }_0(0){|}^2=\kappa$, and the odd scattering eigenfunctions contribute zero. The even continuum functions from the preceding part haveTheir energy is $E_k={\hslash }^2{k}^2/(2m)$. Inserting these weights into the spectral resolution of a quantum propagator, and using evenness to extend the integral to the whole real axis, yields for $T>0$This is the quantum propagator of an attractive delta potential. The printed continuum exponent has an extra factor of $\hslash$: its time phase must be $-i{E}_{k}T/\hslash =-i\hslash k^{2}T/(2m)$. This correction also makes the exponent dimensionless. Oscillatory integrals are understood with an infinitesimal positive convergence factor before taking its limit.

For fixed $g>0$, write $A=\hslash T/(2m)$. The bound contribution to the quantum propagator of an attractive delta potential remains $\kappa e^{i\hslash {\kappa }^{2}T/(2m)}$. For its continuum, the stationary phase is at $k=0$, where $k^2/(k^2+{\kappa }^2)=k^2/{\kappa }^2+O(k^4)$. The Gaussian oscillatory integral and its derivative giveConsequentlyThe continuum decays as $T^{-3/2}$, but the full kernel has a persistent oscillating bound state contribution and no ordinary complex limit. This expansion requires $\hslash {\kappa }^{2}T/(2m)\gg 1$; it is not uniform in the zero-coupling limit, where the bound state disappears and the free kernel instead decays as $T^{-1/2}$.