Why is the path integral Lorentz invariant?

Question

It is often said one of the benefits of the path integral,
$$int Dphi ; e^{iS[phi]}$$
is that it is manifestly Lorentz covariant if $S[phi]$ is Lorentz covariant. However, this is not clear to me. When calculating the path integral (as in Peskin and Schroeder Chapter 9), we must specify the boundary conditions of the integral, i.e. fixing $phi(vec{x})$ at some initial time $-T$ and some final time $T$, before sending $Trightarrowinfty$. This to me says that space and time are on different footing, so the action is not obviously Lorentz covariant. So why is the path integral Lorentz covariant?

Dwagg · Answer

The action is Lorentz invariant. And the path integral is Lorentz invariant because the integrand is Lorentz invariant. Space and time were perhaps on an unequal footing when a finite time interval $[-T,T]$ was chosen, but then you sent $Tto infty$.
Perhaps you are wondering why you can't do everything in a Lorentz invariant way, and why an explicit spatial slices at times $pm T$ were chosen. Consider this a small detour away from manifest Lorentz invariance which is quickly remedied by sending $Tto infty$. Indeed, in the Hamiltonian formulation you start with $H$, which is completely non-Lorentz invariant, so at least comparatively the path integral formulation does a better job at maintaining manifest Lorentz invariance.

Why is the path integral Lorentz invariant?

One Answer

Add your own answers!

Ask a Question