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, 2019 at 9:11
  • 1
    $\begingroup$ 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. $\endgroup$ May 15, 2019 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, 2019 at 9:01

1 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.

  • $\begingroup$ In the last equation you wrote, is it an equality? i.e. aren't they related by a Wick rotation and hence are not equal? $\endgroup$
    – Alex
    Feb 16, 2021 at 10:40
  • $\begingroup$ Do you mean $G(\vec{x},t) = \lim\limits_{\beta\to\infty}\mathcal G(\vec x,it)$? It is an exact statement. The Wick rotation enters as an imaginary argument of $\mathcal G$. $\endgroup$
    – nGlacTOwnS
    Feb 22, 2021 at 17:25
  • $\begingroup$ I see. So the Wick rotation only works for infinite temperature, is that right? This is to say $G(t+i\delta) = G(it)$ is only true when the integral is from infinity to infinity, since in this case we assume the integrand goes to zero sufficiently fast. At finite temperature, the integral on the imaginary axis is periodic, hence can't do Wick rotation. $\endgroup$
    – Alex
    Feb 27, 2021 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.