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The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.

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 $K/(i\hslash )$.

Insert a complete resolution of identity in energy eigenstates on either side of the time-evolution operator. Each state evolves by its scalar energy phase, yielding the displayed quantum propagator. Its spectral sum includes an integral over any continuous spectrum, with the measure fixed by the continuum normalization. Omitting the scattering states when bound states are also present destroys completeness and the delta-function short-time limit.