How to show causality is preserved in path integral approach? I am struggling to find a good way to show causality is conserved using the path integral approach. In the operator approach one can show that $[\hat{\phi}(x),\hat{\phi}(y)]=0$ if $x$ and $y$ are space-like separated.
In the path integral approach we have:
$$\int \phi(x)\phi(y) e^{iS[\phi]} D\phi = \Delta_F(x-y)$$
which gives the Feynman propagator which comes from $T(\phi(x)\phi(y))$, but to show causality is conserved I feel I need to get the non-time ordered propagator somehow.
I have been examining the transition functional:
$$U(\phi_{in},\phi_{out};t) = \int \limits_{\Phi(0)=\phi_{in}}^{\Phi(t)=\phi_{out}} e^{i \int\limits_0^tL[\Phi]dt}D\Phi$$
which should give the amplitude for a field to transition from the in state to the out state over time $t$ but also don't know how to show that this satisfies causality, i.e. that a signal can't travel faster than light.
I prefer the functional approach rather than the operator approach so I would really like to find out how to show causality using this approach.
 A: Since the path-integral formulation is manifestly Lorentz symmetric, we only need to show that $\phi(\vec x,t)$ and $\phi(\vec y,t)$ commute when $\vec x\neq \vec y$. If they commute when the times are equal, then the path integral's manifest Lorentz symmetry implies that they commute for all spacelike separations.
In the canonical formulation with operators on a Hilbert space, two different kinds of ordering are involved: operator order, and time order. Nothing prevents us from considering quantities like $\phi(\vec x,t)\phi(\vec x',t')|0\rangle$ with $t'>t$, where the time order is different than the operator order.
In contrast, the path integral always gives us time-ordered correlation functions. It doesn't come with any separate notion of operator order, but we can still obtain equal-time commutation relations using the following identity. Let $X$ and $Y$ be arbitrary products of field operators whose time-arguments are all $>t$ and $<t$, respectively, with enough margin to avoid interfering with the $\epsilon\to 0$ limit shown below, where $\epsilon$ is positive. Then
\begin{align}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newcommand{\pl}{\partial}
\la 0|X[\phi(\vec x,t),\phi(\vec y,t)]Y|0\ra
&=\lim_{\epsilon\to 0}\Big(
\la 0|X\phi(\vec x,t+\epsilon)\phi(\vec y,t)Y|0\ra
-
\la 0|X\phi(\vec y,t)\phi(\vec x,t-\epsilon)Y|0\ra
\Big)
\\
&=\lim_{\epsilon\to 0}\Big(
\la 0|T\phi(\vec x,t+\epsilon)\phi(\vec y,t)XY|0\ra
-
\la 0|T\phi(\vec x,t-\epsilon)\phi(\vec y,t)XY|0\ra
\Big).
\end{align}
The operators $X$ and $Y$ are arbitrary (as long as they're not too close to the time $t$), so this gives us the commutator $[\phi(\vec x,t),\phi(\vec y,t)]$ sandwiched between arbitrary states, which determines the commutator itself as an operator relation.
When $\vec x\neq\vec y$, the two terms on the last line are clearly equal to each other in the limit $\epsilon\to 0$. That implies $[\phi(\vec x,t),\phi(\vec y,t)]=0$ for $\vec x\neq \vec y$, and the path integral's manifest Lorentz symmetry then implies that they commute for all spacelike separations.
A similar approach can be used to derive the equal-time commutation relation $[\phi(\vec x,t),\pi(\vec y,t)]\propto \delta(\vec x-\vec y)$, where $\pi$ is the canonical conjugate of $\phi$. This takes more work, but it isn't necessary for answering the question, because (if $\pi=\dot\phi$) it follows easily from the unequal-time commutator of $\phi(x)$ and $\phi(y)$ when $\vec x\neq \vec y$, which we already deduced.
