Q:

Expectation values in a quantum field theory

Consider a quantum field $\phi(x)$ (say, a scalar) at a spacetime location $x$. It turns out that $$\langle\phi(x)\rangle_0\equiv\langle 0|\phi(x)|0\rangle=0.\tag{1}$$ This is easily obtaied by using the Fourier exapansion $$\phi(x)=\int\frac{d^3\textbf{p}}{(2\pi)^{3/2}\sqrt{2E_{\textbf{p}}}}(a_{\textbf{p}}e^{-ip\cdot x}+a^\dagger_{\textbf{p}} e^{ip\cdot x}).\tag{2}$$ 
Now in quantum mechanics, any expectation value is computed w.r.t a probability distribution $p(x)$. For example, $$\langle x\rangle_\psi\equiv\langle \psi|x|\psi\rangle=\int\limits_{-\infty}^{\infty}xp(x)dx\tag{3}$$ where $p(x)=\psi^*(x)\psi(x)$. 
Which probability distribution $p[\phi(x)]$ is used to compute (1)? Does the path-integral quantization provide any answer?

A:

I think the path integral approach provides an answer. In terms of the normalized generating functional $\mathcal{Z}[J]=\frac{Z[J]}{Z[0]}$, the expectation value is given by $$\langle 0|\phi(x)|0\rangle=\frac{\delta \mathcal{Z}[J]}{\delta J(x)}\Bigg|_{J=0}=\int\mathcal{D}\phi\phi(x)\exp\Big[\int i(\mathcal{L}+j\phi)d^4x\Big]\Big|_{J=0}\tag{1}$$ which when compared with the average of any function w.r.t a normalized probability distribution given by $$\overline{f(u)}=\int \limits_{a}^{b}duf(u)p(u)$$ tells that the relevant probability distribution in this case (1) is $$p[\phi(x)]=\exp(iS)=\text{complex}!!!$$ But what bugs me here is that $p[\phi(x)]$ is complex, and therefore, the name probability distribution may not be justified.

