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?

  • $\begingroup$ So the question (v1) is actually just about the trace procedure, not the Wick rotation? $\endgroup$ – Qmechanic May 15 '19 at 9:11
  • $\begingroup$ The question is that in the formalism of path integrals, quantum expectation values and thermodynamic averages can be identified by analytically continuing to imaginary time AND changing the boundary conditions on path integral. However, for Green functions it seems that only going to imaginary time is enough... If I were to express the Green functions as path integrals, I don't see how the boundary conditions get fixed too. $\endgroup$ – P. C. Spaniel May 15 '19 at 16:21
  • 1
    $\begingroup$ My understanding is that, to do finite temperature field theory, one goes to euclidean signature(wick rotation) and compactifies the time direction -the period of which gives the inverse temperature $\beta$. One then calculates the partition function and wick rotate back to real time, your last equation, $\mathcal{G}(\vec{x},\tau)= \frac{1}{Z}\text{Tr}\Big[e^{-\beta H} \phi(\vec{x},\tau)\phi^\dagger (0,0)\Big]$. After you wick rotate back, you don't do anything to the periodic boundary condition you imposed while working with imaginary time. $\endgroup$ – levitt May 18 '19 at 9:01

@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.

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