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.
