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.

This post imported from StackExchange Physics at 2014-09-25 20:21 (UTC), posted by SE-user cesaruliana