I've been thinking about this, and it might sound like a stupid question, but I can't seem to find an answer anywhere, here goes:
Whenever we calculate expecation-values between two position eigenvectors taken at different times we need to apply the time-evolution operator $\hat{U}(t_b,t_a)$ to get the position-vectors at the same time.
So in stead of saying that:
$\langle x_b,t_b|x_a,t_a\rangle=\delta(x_b-x_a)$,
we say that: $\langle x_b,t_b|x_a,t_a\rangle=\langle x_b|\hat{U}(t_b,t_a)|x_a\rangle$.
Is this because of the fact that generally the position operator $\hat{x}$ doesn't commute with the generator of the time-evolution $\hat{H}$?
Or said otherwise, is this because of the fact that according to Heisenberg's equation of motion that we have (in the Heisenberg-image of course) that:
$i\hbar\dot{\hat{x}}=[\hat{x},\hat{H}]\neq0$ in general.
If we would have $[\hat{x},\hat{H}]=0$ then we would have that
$\langle x_b,t_b|x_a,t_a\rangle=U(t_b,t_a)\langle x_b|x_a\rangle=U(t_b,t_a)\delta(x_b-x_a)$, right?
Extra explanation on the last formula (@joshphysics):
I denote an operator with a hat and the eigenvalues without hats, so in my last formula I basically took the operator $\hat{U}(t_b,t_a)$ and took it's eigenvalue by letting it work on the ket $|x_a\rangle$ which gives me the eigenvalue $U(t_b,t_a)$. Which looks weird since an time-independant base yields a time-dependant factor this way.