Is the propagator the same as the matrix elements of the time evolution operator?

Question

So Sakurai in their QM book defines the propagator in wave mechanics as:

$$K(x'',t;x',t_0)=\sum_{a'}\langle x''\vert a'\rangle \langle a'\vert x'\rangle \exp\left[\dfrac{-iE_{a'}(t-t_0)}{\hbar}\right].$$

Since the exponential terms are the eigenvalues of $\vert a'\rangle$, you could write this expression as:

\begin{align} K(x'',t;x',t_0)&=\sum_{a'}\langle x''\vert \exp\left[\dfrac{-iH(t-t_0)}{\hbar}\right] \vert a'\rangle \langle a'\vert x'\rangle \\ &= \sum_{a'}\langle x''\vert U(t,t_0) \vert a'\rangle \langle a'\vert x'\rangle \end{align} with $U(t,t_0)$ being the time evolution operator. Then, just eliminating the completeness relation you'd have:

$$\langle x''\vert U(t,t_0)\vert x'\rangle\,,$$

which are the matrix elements of the time evolution operator in the position basis. Is this correct or did I miss something? Are the two concepts describing the same thing?

Answer 1

Yes, you are correct. Actually a few pages later Sakurai derives the same relation:

Because of these two properties, the propagator (2.6.8), regarded as function of $\mathbf{x}''$, is simply the wave function at $t$ of a particle that was localized precisely at $\mathbf{x}'$ at some earlier time $t_0$. Indeed, this interpretation follows, perhaps more elegantly, from noting that (2.6.8) can also be written as $$K(\mathbf{x}'',t;\mathbf{x}',t_0)=\langle\mathbf{x}''|\exp\left[\frac{-iH(t-t_0)}{\hbar}\right]|\mathbf{x}'\rangle, \tag{2.6.10}$$