# A Path integral computes time-ordered products

1. Oct 7, 2016

### spaghetti3451

In general,

$\displaystyle{\langle q_{f}|e^{-iHt/\hbar}|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}$

and

$\displaystyle{\langle q_{f}|\hat{Q}(t)|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}\ q(t).$

How can one switch from the above expressions to the following?

$\displaystyle{\langle q_{f}|T\{\hat{Q}(t_{1})\hat{Q}(t_{2})\}|q_{i}\rangle=\int\mathcal{D}q(t)\ e^{iS[q]/\hbar}}\ q(t_{1})q(t_{2})$

In other words, why does the path integral compute time-ordered products?

2. Oct 8, 2016

### Fightfish

The technical details are given in any standard QFT textbook (see the relevant section on path integrals), so I won't reproduce them here (I'm also lazy to type that many equations haha). But intuitively, there are two quick ways to understand why this should be the case:

(1) If you examine how the path integral really is defined (my instructor in particular always emphasized that the compact notation that you always see is often misleading and usually useless for calculations - you have to go back to the original complete form), we actually broke up the operator into little bits at different times and inserted multiple resolutions of the identity (also at different times). This operation is in fact what is responsible for "moving" the fields to the respective time-ordered locations.

(2) In the path integral expression, the fields that appear there are classical fields and the ordering doesn't matter. This means that in the canonical form of the propagator, the order in which you write the field operators should not matter as well. One way to fix this is to well, have time-ordering. Of course this doesn't explain why time-ordering arises to begin with, but it tells us that
$$\int\mathcal{D}q(t) e^{iS[q]/\hbar} q(t_{1})q(t_{2}) = \langle q_{f}|\hat{Q}(t_{1})\hat{Q}(t_{2})|q_{i}\rangle$$ cannot be correct since $\hat{Q}(t_{1})$ and $\hat{Q}(t_{2})$ do not commute whereas $q(t_{1})$ and $q(t_{2})$ do.