Expectation values of the form $$\langle \psi| x(t')\, x(t) |\psi\rangle$$ are very common in quantum field theory (Heisenberg picture operator $x(t)$). Focusing on $1$-dimensional QFT (i.e. ordinary quantum mechanics with just a time dimensions) for simplicity, what is the interpretation of that expectation value?
I suppose what's throwing me is that I'm used to thinking in the Schrödinger picture, and I was always told that translating between the two was equivalent to moving the time dependence between the operators and wave functions. In this case, though, there doesn't seem to be a clean way to move the time dependence of the $x(t')\, x(t)$ operator to the wave functions. The first version of this quantity in the Schrödinger picture is (assuming $H$ commutes with itself at all times): \begin{align} \langle \psi | x(t')\, x(t)|\psi\rangle_{\mathrm{Heis}} & = \langle\psi(t')| x_{\mathrm{Sch}} \operatorname{e}^{-iH(t'-t)}x_{\mathrm{Sch}}|\psi(t)\rangle. \end{align} which doesn't isolate the time evolution into the state vectors cleanly. At best, I read this as:
- evolve $|\psi\rangle$ to time $t$,
- measure its position,
- evolve the result for time $t'-t$ (possibly backwards),
- measure position again,
- evolve state backwards by $t'$,
- calculate overlap with $|\psi\rangle$, and
- scale the the product by the results of the two position measurements.
That can't be right, though, because real "measurements" collapse superpositions, and that's not done here.