Real and Imaginary time Green's Functions

Question

In real time, one can calculate the two point function of a given theory using

\begin{equation} G(\vec{x},t)=\langle \Omega | \phi(\vec{x},t)\phi^\dagger (0,0)|\Omega\rangle =\int_{\phi(0,0)}^{\phi(\vec{x},t)} \mathcal{D}\phi\ e^{-\frac{i}{\hbar}S[\phi]}\phi(\vec{x}',t')\phi(\vec{x},t) \end{equation}

where the limits of the path integral should match the initial and final state.

On the other hand, I know that the generating functional $\mathcal{Z}$

\begin{equation} \mathcal{Z}=\langle\phi'|e^{-iHT}|\phi\rangle=\int_\phi^{\phi'}e^{-\frac{i}{\hbar}S[\phi]} \end{equation}

can be identified with the quantum partition function $Z$ if we evaluate on imaginary time $t=-i\tau$ and we trace over the initial and final state

\begin{equation} Z=\sum_{\phi}\langle \phi | e^{-\beta H}|\phi\rangle =\int_{\phi(0)=\phi(\beta)}e^{-\beta S_E[\phi]} \end{equation}

So the relation between quantum mechanical and thermodynamic expectation values is: analytically continuate $t\rightarrow -i\tau$ with period $\tau \in [0,\beta]$, set inital and final states equal and sum over them. Now, in every book I see, the real time Green's function

\begin{equation} G(\vec{x},t)= \langle \Omega | \phi(\vec{x},t)\phi^\dagger (0,0)|\Omega\rangle \end{equation}

and the imaginary time Green's function

\begin{equation} \mathcal{G}(\vec{x},\tau)= \frac{1}{Z}\text{Tr}\Big[e^{-\beta H} \phi(\vec{x},\tau)\phi^\dagger (0,0)\Big] \end{equation}

are related by

\begin{equation} G(\vec{x},t)=\mathcal{G}(\vec{x},i\tau) \end{equation}

Meaning that we could basically define just one function $G(\vec{x},z)$ with $z\in\mathcal{C}$ that is equal to the Green Function of QM for real $z$ and equal to the thermodynamic average for imaginary $z$.

My question is the following

In the Path integral formalism, there were to things we needed to do to go from one average to the other; we need to go to imaginary time and we need to do something about the trace. In the Green's functions, however, it seems that going to imaginary time is enough, as if the trace gets automatically taken care of. How is that so?

Answer 1

@levitt has almost provided the correct answer in his comment. Although, I think he should also emphasize something that he probably implicitly implied in his comment above: that the equality $G(\vec x,t) = \mathcal G(\vec x,i t)$ as written in the original question is incorrect (apart from the typo where the argument of $\mathcal G$ is $\tau$ and not $t$).

$\mathcal G(\vec x,i t)$ computes real time correlation function in a field theory at finite temperature while $G(\vec x,t)$ (as written in the first equation of the question) computes the real time correlation function at zero temperature. These two correlation functions are different. You can obtain the zero temperature correlation function by taking $\lim\limits_{\beta\to\infty}\mathcal G(\vec x,i t)$. It should be true that $G(\vec x,t) = \lim\limits_{\beta\to\infty}\mathcal G(\vec x,i t)$.

Note: All the above statements are made assuming that the operators are consistently ordered.