statistical mechanics - Once a quantum partition function is in path integral form, does it contain any operators?

Once a quantum partition function is in path integral form, does it contain any operators?

I.e. The quantum partition function is $Z=tr(e^{-\beta H})$ where $H$ is an operator, the Hamiltonian of the system.

But if I put this into the path integral formalism so that we have something like $Z= \int D(\bar{\gamma},\gamma) e^{-\int_0^\beta d\tau\,[\frac{i\hbar}{2}(\gamma\partial_t\bar{\gamma}-\bar{\gamma}\partial_t \gamma) + H(\bar{\gamma},\gamma)]}$, is the $H(\bar{\gamma},\gamma)$ an operator?

Thanks!

Answer

Nope, Feynman's path integral formulation of quantum mechanics is a method to directly calculate the complex probability amplitudes and all objects that appear in its formalism - not counting proofs of equivalence with other approaches to quantum mechanics - are $c$-numbers representing classical observables.

In particular, the exponent in the path integral - which should be $iS$ ($i$ times the action i.e. $i$ times the integrated Lagrangian), not the Hamiltonian - is a $c$-number-valued function of the "classical observables", the same function that is relevant for the classical (non-quantum) theory. So the path integral is an infinite-dimensional integral over otherwise "ordinary classical variables" that produces some probability amplitudes - the same ones that may be (but don't have to be) obtained from the operator formalism.

Uncertainty principle in path integral

By the way, some sorts of the path integral include integration over both positions and momenta, $\int Dx(t)\, Dp(t)$. How it is possible that both of them are treated "classically" as $c$-numbers? Doesn't it violate the uncertainty principle?

The answer is that it doesn't violate the uncertainty principle. One may still deduce that $xp-px=i\hbar$ from the path integral as long as she is careful about putting the right values of the time $t$ (the argument). The quantities $x(t)p(t-\epsilon)$ and $p(t)x(t-\epsilon)$ differ. A necessary condition for this difference to exist is the fact that "most" trajectories contributing to the path integral are discontinuous.