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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.

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For the sake of completeness, let's first unravel some definitions and prove the identity \begin{align} \langle x_b, t_b|x_a, t_a\rangle = \langle x_b|\hat U(t_b,t_a)|x_a\rangle. \end{align} The definition of each position eigenstate $|x\rangle$ is that it's a normalized eigenvector of the position operator $\hat x$ corresponding to eigenvalue $x$; \begin{align} \hat x |x\rangle = x|x\rangle. \end{align} Given a time interval $t$, time evolution operator for that interval is given by an operator exponential; \begin{align} \hat U(t) = e^{-i\hat tH/\hbar}. \end{align} We can use the time-evolution operator to define a time-evolved (aka heisenberg picture) position operator; \begin{align} \hat x(t) = \hat U(t)^\dagger \hat x \hat U(t). \end{align} Then, the states $|x,t\rangle$ are defined as the normalized eigenstates of the operator $\hat x(t)$ corresponding to eigenvalue $x$; \begin{align} \hat x(t) |x,t\rangle = x|x,t\rangle. \end{align} But notice that \begin{align} \hat x(t) \big[\hat U(t)^\dagger |x\rangle\big] &= \hat U(t)^\dagger\hat x\hat U(t)\hat U(t)^\dagger |x\rangle \\ &= \hat U(t)^\dagger \hat x |x\rangle \\ &= \hat U(t)^\dagger x|x\rangle \\ &= x\big[\hat U(t)^\dagger |x\rangle\big] \end{align} which implies that \begin{align} \hat U(t)^\dagger|x\rangle = |x,t\rangle. \end{align} It follows that \begin{align} \langle x_b,t_b|x_a,t_a\rangle &= \langle x_b|\hat U(t_b)\hat U(t_a)^\dagger|x_a\rangle \\ &= \langle x_b|\hat U(t_b)U(-t_a)|x_a\rangle\\ &= \langle x_b| \hat U(t_b-t_a)|x_a\rangle. \end{align} This is exactly what we wanted to prove if we simply note the conventional definition \begin{align} \hat U(t_b,t_a) = \hat U(t_b-t_a). \end{align} Now let's address your question

Is this because of the fact that generally the position operator $\hat x$ doesn't commute with the generator of the time-evolution $\hat H$?

Basically, yes. Notice that if it did commute, then $\hat x$ would also commute with $U(t)$ and $U(t)^\dagger$, so we would have \begin{align} \hat x(t) = \hat U(t)^\dagger \hat x \hat U(t) = \hat x \hat U(t)^\dagger \hat U(t) = \hat x \end{align} in which case the eigenstates of $\hat x$ and $\hat x(t)$ can be chosen equal. In particular, the inner products between them would be the same; \begin{align} \langle x_b, t_b|x_a, t_a\rangle = \langle x_b| x_a\rangle \end{align} Notice, in particular, that your last equation is incorrect. You could have seen this immediately in fact because you are equating an inner product to an operator.

Addendum. As you remark in your comment below, your intention was for the symbol $U(t_b,t_a)$ to denote a complex number. In that case, your last expression is correct. However, when $\hat U$ commutes with $\hat x$ and we have $\hat x(t) = \hat x$, we can simply choose normalizations such that $|x,t\rangle = |x\rangle$ for all $t$, in which case we would have $U(t_b,t_a) = 1$.

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  • $\begingroup$ maybe I should have clarified my last equation a bit more (see edit). But it did indeed look weird to me in the sence that $U(t_b,t_a)$ is a time-dependant factor while I'd expect the product to be time-independant. Thanks for the great answer, I should have known that the time should be added by definition. $\endgroup$ – Nick Nov 20 '13 at 10:51
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    $\begingroup$ @Nick Ah I see yeah then your expression is fine. See the addendum. $\endgroup$ – joshphysics Nov 20 '13 at 18:27

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