= Solution
For a time-independent <Hamiltonian operator>, the <time-evolution operator> is $U(T)=e^{-iHT/\hbar}$, with $T=t-t'$. Insert its resolution of identity in normalized <energy eigenstates>:
$$
\langle x|U(T)|x'\rangle
=\sum_\alpha\langle x|\alpha\rangle e^{-iE_\alpha T/\hbar}\langle\alpha|x'\rangle.
$$
Since $\langle x|\alpha\rangle=\varphi_\alpha(x)$, the retarded <quantum propagator> in the question is
$$
\boxed{K(x,t|x',t')=\Theta(T)\sum_\alpha\varphi_\alpha(x)\varphi_\alpha^*(x')e^{-iE_\alpha T/\hbar}.}
$$
The <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 $\delta(x-x')$ as $T\downarrow0$. With the step factor it obeys $(i\hbar\partial_t-H)K=i\hbar\delta(T)\delta(x-x')$; the corresponding <Green function> with a unit delta source is $K/(i\hbar)$.
Back to article page
