What is the integrand of a path integral?

Question

I have been reading the Deligne et. al notes "Quantum Fields and Strings: A Course For Mathematicians" and I am trying to reconcile what I learned with what physicists are doing. I am confused about what physicists mean by expressions of the form $$\langle 0|\phi(x_1)...\phi(x_n)|0\rangle= \int \mathcal{D}\phi \; \phi(x_1)...\phi(x_n)e^{iS[\phi]}$$ I am not asking about the measure $\mathcal{D}\phi$. This integral is supposed to be the "$n$-point correlation function of the field $\phi$" and should return a scalar, but the $\phi(x_i)$ are all supposed to be operators on a Hilbert space, so it is not clear to me how this happens.

This notation also suggests to me that this correlation function is independent of the field $\phi$ (since $\phi$ is also used for the variable of integration), but this seems unreasonable. Would it be correct to say $$\langle 0|\psi(x_1)...\psi(x_n)|0\rangle= \int \mathcal{D}\phi \; \psi(x_1)...\psi(x_n)e^{iS[\phi]}$$ instead?

Answer 1

The field $\phi$ has slightly different meanings inside the path integral and outside of it, as the Feynman path integral is relating two different interpretations.

Inside the path integral, imagine that we're summing an action of classical fields over a measure of all possible values for those classical fields. As a variable of integration, $\phi$ is thus a classical field and each value $\phi(x_i)$ is a scalar value.

Outside the path integral is where $\phi$ is a Hilbert space operator.

Answer 2

The integrand is the functional $F[\phi]:= \phi(x_1)\cdots\phi(x_n)\exp(iS[\phi])$. Your confusion probably stems from the fact that physicists like to jump around between the Hamiltonian picture where your field is some kind of operator-valued distribution, and the Feynman picture where you're working with path integrals and $\phi$ is just a function (or distribution). The link between them is precisely that the vacuum expectation value of the operator $\phi(x_1)\cdots\phi(x_n)$, i.e. $$ \langle 0|\phi(x_1)\cdots\phi(x_n)|0\rangle, $$ can be calculated as the path integral of the integrand I wrote above.