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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 $K/(i\hslash )$.

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 $H=p^2/(2m)-g\delta (x)$ with $g>0$, put $\kappa =mg/{\hslash }^2$ and $T>0$. The bound state in a delta potential has squared amplitude $\kappa$ at the origin. The even normalized continuum states have squared amplitude $k^2/[\pi (k^2+{\kappa }^2)]$ for $k>0$, and the odd states vanish there. The spectral resolution of a quantum propagator gives the displayed result. At long times, the continuum stationary point at $k=0$ is suppressed by its quadratic spectral weight: expanding that weight gives a continuum term $e^{-3i\pi /4}(2m/\hslash T{)}^{3/2}/(4\sqrt{\pi }{\kappa }^2)+O(T^{-5/2})$. The bound state term persists with constant modulus. This stationary phase expansion requires $\hslash {\kappa }^{2}T/(2m)\gg 1$.