Consider the free, real scalar field $\phi$ in Matsubara Finite-Temperature quantum field theory, where our system is kept in equilibrium with a heat bath at temperature $\frac{1}{\beta}$.

Then the field $\phi$ is an operator $\phi(\tau, \mathbf{x})$ where $\mathbf{x}$ is a position in $\mathbb{R}^3$ and $\tau$ is an imaginary time. Furthermore, the propagator for two points $(\tau_1, \mathbf{x}_1)$ and $(\tau_2, \mathbf{x}_2)$ is: $$ G\left(\tau_1, \mathbf{x}_{1};\tau_2, \mathbf{x}_{2}\right) \ = \ < \phi(\tau_1, \mathbf{x}_1)\phi(\tau_2, \mathbf{x}_2) >_{\beta} $$

Where the $\beta$ denotes a thermal average. Because of the KMS condition, we have the boundary condition $G\left(\tau_1, \mathbf{x}_{1};\tau_2+\beta, \mathbf{x}_{2}\right) = G\left(\tau_1, \mathbf{x}_{1};\tau_2, \mathbf{x}_{2}\right)$. Following the usual derivation, we then get a propagator that looks like: $$ G\left(\tau_1, \mathbf{x}_{1};\tau_2, \mathbf{x}_{2}\right) \propto \sum_{n=-\infty}^{\infty} \int d^{3}\mathbf{p}\ \mathcal{D}_{n}(\mathbf{p}) e^{- i \mathbf{p}\cdot\mathbf{x} - i \frac{2 \pi n}{\beta} \tau } $$

Where $\mathcal{D}_{n}(\mathbf{p}) = \frac{1}{\left(\frac{2\pi n}{\beta} \right)^{2} + \mathbf{p}^2 + m^2}$ is the Fourier transform of this propagator.

My questions is: what is the physical meaning of the $\tau_1$ and $\tau_2$ in $G\left(\tau_1, \mathbf{x}_{1};\tau_2, \mathbf{x}_{2}\right)$?

In ordinary QFT, there is a very nice explanation of what the propagator $G$ means physically - it's that $G(t_1, \mathbf{x}_1;t_2, \mathbf{x}_2)$ corresponds to the probability amplitude that a particle travels from a point $\mathbf{x}_1$ at time $t_1$ to another points $\mathbf{x}_2$ at time $t_2$.

I would like to attach a similar meaning to the propagator here. I initially thought that maybe this was something $G$ correlates two different temperatures $\tau_1$ and $\tau_2$, but I realize this doesn't make any sense.

How to understand the imaginary times $\tau$? Is it not possible to make a statement like this here? I know that we can only have $0<\tau<\beta$ due to the periodic boundary conditions.

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    $\begingroup$ In spite of what Pesking&Schroeder might say, in ordinary QFT the propagator is not a probability amplitude of propagation. How could it be? it is not bounded in modulus by 1, and it doesn't compose as a amplitudes should. You cannot construct meaningful propagation amplitudes in relativistic QM. People tried and failed; now we know it's just impossible. Why people insist in that lie is beyond me. $\endgroup$ Jan 28, 2018 at 19:03
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    $\begingroup$ @AccidentalFourierTransform it should be very obvious why the propagator is thought of as a probability for propagation. In ordinary QM that's what it is. In QFT it isn't that way only because you work on a continuum. That's like complaining about people who think about Feynman diagrams in terms of real particle processes are wrong. Sure that's true, but as long as you know there are more subtleties, who does it hurt? $\endgroup$
    – KF Gauss
    Jan 29, 2018 at 14:19
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    $\begingroup$ @user157879 I strongly disagree. In non-relativistic QM the propagator is a propagation amplitude. In relativistic QM, it is factually impossible to write down a propagation amplitude: Lorentz invariance precludes such an object from existing. We think of the propagator as a propagation amplitude by extension from the non-relativistic case, but there is no reason at all to make such an identification. It helps no one, and it causes much confusion (as is made manifest by the question in the OP). It is not just a subtlety: it is the whole picture what is wrong and misleading. $\endgroup$ Jan 29, 2018 at 15:21
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    $\begingroup$ @AccidentalFourierTransform I can't comment on a global scale if the picture of amplitudes in relativistic QM causes true problems for young physicists so you may be completely right. But I believe it isn't a big deal here. OP never stated that he is discussing a relativistic theory. In fact, most of the discussion on finite temperature field theories that I've seen is in the realm of solid state physics, where relativity is often not imposed, though that may be just my small depth of knowledge speaking. $\endgroup$
    – KF Gauss
    Jan 29, 2018 at 15:35
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    $\begingroup$ @AccidentalFourierTransform The fact that the propagator in relativistic QFT isn't bounded in modulus by 1 is not a good argument for its not being a probability amplitude; the actual probability-amplitude propagator in nonrelativistic QM isn't bounded in modulus by 1 either. $\endgroup$
    – tparker
    Jan 31, 2018 at 23:30

1 Answer 1


Because finite temperature QFT in $d$-dimensions can be translated to classical statistical mechanics in $d+1$ dimensions, I believe you can reinterpret the propagator accordingly. In statistical mechanics, you don't have a time variable, instead the imaginary time variable $\tau$ would be thought of as an additional spatial dimension, and so you work in periodic boundary conditions with $\beta$ being the system size [1].

Because there is no time axis, you shouldn't interpret the function $G$ as a propagator, but instead of as a statistical correlation function of the classical field in $d+1$ spatial dimensions, one of which is periodic. Actually, as alluded to by AccidentalFourierTransform, thinking of $G$ in terms of correlation functions is always the more sensible thing to do.

[1] http://galileo.phys.virginia.edu/~pf7a/msm2.pdf

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    $\begingroup$ Okay interesting. Your suggestion regarding thinking about propagators as correlation functions I take to heart. So I should think of $\tau$ as some additional (periodic) spatial dimension with no real meaning attached? It's a little confusing in that when you look at Keldysh-Schwinger finite temperature QFT (aka. closed-time path formalism), you get time back into the picture swapped with the $tau$ variable. I will think about this some more thank you. $\endgroup$ Jan 29, 2018 at 19:03

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