A simple problem that I have given myself is to show the following identity for a generic quantum mechanical system:

$$ \langle x| U(\infty,t_{b})\hat{O}_{2}U(t_{b},t_{a})\hat{O}_{1}U(t_{a},-\infty)|x\rangle = \int \mathcal{D}[x]e^{iS[x]}\hat{O}_{2}(t_{b})\hat{O}_{1}(t_{a})$$

Here, $U(t_{f},t_{i})$ is the time evolution operator that carries us from $t_{i}$ to $t_{f}$, $\hat{O}_{i}(t)$ is the Heisenberg operator associated with $\hat{O}_{i}$, and the path integral on the right hand side runs over all paths connecting $x = 0$ at $t=-\infty$ and $x=0$ at $t=+\infty$. I'm assuming that convergence factors for the path integral are inserted appropriately. This identity seems intuitively correct, but I'm finding it difficult to show.

My attempt so far is as follows: write the operators $\hat{O}_{i}$ as $\hat{O}_{i} = \int dx |x\rangle\langle x| O_{i}(x)$, so that the LHS from equation (1) becomes

$$\int dx_{a}dx_{b}\langle x|U(\infty,t_{b})|x_{b}\rangle O_{2}(x_{b})\langle x_{b}|U(t_{b},t_{a})|x_{a}\rangle O_{1}(x_{a})\langle x_{a}|U(t_{a},-\infty)|x\rangle$$ $$=\int dx_{a}dx_{b}\mathcal{D}[x_{ia}]\mathcal{D}[x_{ab}]\mathcal{D}[x_{bf}]O_{2}(x_{b})O_{1}(x_{a})e^{i(S[x_{ia}]+S[x_{ab}]+S[x_{bf}])}$$ $$ = \int\mathcal{D}[x_{if}]e^{iS[x_{if}]}O_{2}(t_{b})O_{1}(t_{a})$$

In the second line, I've broken the path integral into three parts; by integrating over the point where they're concatenated (here $x_{a}$ and $x_{b}$) we turn it back into a single path. But when we do that, the arguments of $O_{i}$ changes from a position to a time (due to the boundary conditions). But this isn't what I set out to derive, because in the last equation the $O_{i}$ are just c-numbers, not operators. Am I doing something wrong or was my initial claim incorrect? I really thought the objects that went into the path integral when computing a correlator were operators....

  • $\begingroup$ In your first equation, the lhs is a matrix element and the rhs an operator, if I see it correctly. $\endgroup$ – Sanya Dec 23 '16 at 23:47
  • $\begingroup$ Yes it is... I can see that this can't be right but I'm confused by what I see in the literature. What does $O_{i}(t)$ mean on the RHS if it's not a Heisenberg operator...? $\endgroup$ – miggle Dec 23 '16 at 23:51
  • $\begingroup$ I am new to operator insertions (and late to your question) but if I understand correctly, in your very first equation, in the LHS, you have Schrodinger operators (because you have taken care of the time-dependence in the time evolution operators) and then the RHS is supposed to be the the matrix element of the product of the Heisenberg operators $\hat{O_2}(t_2)\hat{O_1}(t_1)$. So, what you derived (the expression without hats) seems correct but what you set out to derive looks wrong (the hats drop off when you break the braket into path-integrals). $\endgroup$ – Dvij Mankad Nov 15 '18 at 15:11

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