In the standard quantum field theory we always take the vacuum to be a invariant under Lorentz transformation. For simple cases, at least for free fields, is very simple to actually prove this.

Now consider the thermal state at a given inverse temperature $\beta$ in a QFT, namely the one given by the density operator $\rho = \frac{e^{-\beta H}}{Z(\beta)}$. There is an old heuristic argument by which we loose Lorentz covariance at finite temperature: because our system is coupled to a heat bath we do have a preferred frame of reference, viz. the one in which the heat bath is static, so to ensure thermodynamical equilibrium.

Although I find the argument very reasonable I have yet to see a detailed proof of this fact. None of the usual textbooks (Kapusta, Le Bellac, etc...) furnish even a hint, nor did a keyword search for papers.

Does anyone know a reference for this, or the proof itself?

To be very clear, the proof should be able to show this: given a quantum field $\phi(t)$ (I'm suppressing space coordinates for simplicity), one can define the thermal state as the one that satisfies the KMS condition

$\langle \phi(t)\phi(t')\rangle_\beta= G(t-t')=G(t'-t-i\beta)$

or in words it is the state such that the Greens function is periodic (or anti-periodic for fermionic fields) in imaginary time with period $\beta$. Now perform a Lorentz transformation to go to new coordinates. Then the Green function in the new frame is not periodical in imaginary time. Therefore the state given by the density operator above only is a thermal state with inverse temperature $\beta$ in one frame.

Now I would ideally be interested in an "elementary proof", that is one using the usual tools of QFT. If you happen to know a proof in a more sophisticated framework, like Algebraic QFT, I would appreciate if along with the reference you could give a brief idea behind the proof.

  • 1
    $\begingroup$ See an additional reference here. $\endgroup$
    – Dilaton
    Commented Sep 26, 2014 at 11:20

2 Answers 2


Here is a proof following Ojima, "Lorentz Invariance vs. Temperature in QFT", Letters in Mathematical Physics (1986) Vol. 11, Issue 1 (1986) 73-80. The first two pages of the paper are available for free here, but the website wants money for more of the paper. (Click the orange "Look Inside" button if the paper doesn't open automatically.) Fortunately, the proof is on the second page.

Define $$w(A) = tr\bigl(e^{-\beta H} A\bigr)/tr\bigl(e^{-\beta H}\bigr).$$ The KMS condition can be written $$w(\phi(x)\phi(y)) = w(\phi(y)\phi({\tilde x}))$$ where $\tilde x$ is $x$ with the time component shifted by $i\beta$.

Now consider the Fourier transform $$\langle\phi_k\phi_{-k}\rangle = \int d^4x d^4y\ e^{i k \cdot (x-y)} w(\phi(x)\phi(y)).$$ By the KMS condition this is $$\langle\phi_k\phi_{-k}\rangle = \int d^4x d^4y\ e^{i k \cdot (x-y)} w(\phi( y)\phi({\tilde x})).$$ Shifting the time in the $x$ integral, this becomes $$e^{\beta k_0}\int d^4{\tilde x} d^4 y\ e^{i k \cdot ({\tilde x}- y)} w(\phi(y)\phi({\tilde x})) = e^{\beta k_0}\langle\phi_{-k}\phi_{k}\rangle,$$ so we have $$\langle\phi_k\phi_{-k}\rangle = e^{\beta k_0}\langle\phi_{-k}\phi_{k}\rangle.$$ Starting with the right hand side of this equation, if Lorentz invariance holds, $\langle\phi_{-k}\phi_{k}\rangle$ is a scalar under Lorentz transformations for a scalar field $\phi$. However, $e^{\beta k_0}$ is manifestly not a Lorentz scalar since it depends non-covariantly on $k_0$. This implies that the left hand side of the above equation is not a Lorentz scalar, contradicting the Lorentz invariance of $\langle\phi_k\phi_{-k}\rangle$.

This proves that Lorentz covariance cannot hold for finite $\beta$.

Another way to see that Lorentz covariance is broken at finite temperature is to Wick rotate to Euclidean spacetime. The KMS condition then implies periodicity of the Green's function in the time direction. Lorentz transformations in real spacetime are mapped to rotations in Euclidean spacetime by the Wick rotation. Since periodic boundary conditions are imposed in one direction and not the other three directions, rotation symmetry is broken in Euclidean spacetime and therefore Lorentz symmetry is broken in real spacetime.


Found a sketch of a proof on a referee's report on a paper RELATIVISTIC INVARIANCE OF THE VACUUM by Adam Bednorz.

The referee's sketch is:


Hundreds of calculations in Fnite temperature Feld theory have been published. To my knowledge, none of these calculations have ever conflicted with Lorentz invariance in the limit $\beta \to \infty$

Simple Proof

Instead of using the Keldysh contour in Sec V, it is far better to use the "symmetric" contour in which the time runs along the real axis and then returns anti-parallel to the real axis but shifted down by $\beta / 2$.

I believe this form of the propagator is used in Refs.13 and 14 (and probably in many of the other references). It is described in the the book Thermal Field Theory by Michel Le Bellac, Cambridge Univ Press, 1996. From page 55 of Le Bellac the matrix form the propagator is

enter image description here

When $\beta \to \infty$ the second contour decouples from the frst contour. The vertices do not couple the two contours. This is all that is necessary for the proof.

i think the simplest answer (apart the reference i posted above), is that both (inverse) temperature $\beta$ and (real) time $t$ are treated as imaginary times (through the usual wick-rotation), so effectively temperature $\beta$ is part of the (total, imaginary) time coordinate, plus the temperature cannot be negative (zeroth/third laws of thermodynamics). The rest follows (with some physical arguments like you already made) so

$$\langle \phi(t)\phi(t')\rangle_\beta= G(t-t')=G(t'-t-i\beta)$$

  • $\begingroup$ While I appreaciate the reference, I think the referee's report is insufficient. The sketch proves that at the limit of zero temperature it is obvious that the propagator is equivalent with the usual one and Lorentz invariance is preserved. But it does not hint at how to perform a Lorentz transformation and then show the loss of KMS property. $\endgroup$ Commented Aug 18, 2014 at 20:14
  • $\begingroup$ @cesaruliana, correct it proves the opposite of what is asked, however i referenced it for 2 reasons, 1) it provides a point of departure for the opposite direction, 2) i could not find sth better unfortunately. i think my last paragraph hints at the main reason this is so. $\endgroup$
    – Nikos M.
    Commented Aug 18, 2014 at 20:16

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