Inserting a position operator in the path integral in QFT

Question

With the usual path integral description, we have the formula $$\langle q''t''|q't'\rangle =\int\mathcal{D}q \exp{(iS)}$$ where $S=\int_{t'}^{t''}L(q,\dot{q})$ is the action evaluated for $t\in (t',t'').$ My question is about the insertion of a position operator $Q(t_1)$ for the propagator, i.e.

$$\langle q''t''|Q(t_1)|q't'\rangle=\int\mathcal{D}q \ q(t_1)\exp{(iS)}.$$

What does it meane to have a path integral with an integrand $q(t_1)$? After all, one way to interpret the path integral measure $\mathcal{D}q$ is to imagine an infinite number of slice of time intervals (and therefore slices of paths). In this case, should I understand $q(t_1)$ as the position operator at time $t_1$, or in the slice-view-point, position operator living in the time interval that contains $t_1$?

Or, is there a better way to understand this?

Answer 1

The path integral itself is as far as I know not well defined as an "integration over all functions" (i.e. a corresponding measure has not been constructed yet). So let us stay in the "discrete" case where we have finitely many times $t_i$ and for each time we have a coordinate $q_i=q(t_i)$ and you have an integral over each $q_i$. So in the most extreme case with only two times $t_1, t_2$ we only have coordinates $q_1,q_2$, meaning that our integral would be of the form $\int dq_1 dq_2 q_1 \exp(iS(q_1,q_2))$. When this concept is clear you can easily just do the same for more than two different times.