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Proof of Loss of Lorentz Invariance in Finite Temperature Quantum Field Theory

+ 7 like - 0 dislike

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
asked Aug 15, 2014 in Theoretical Physics by cesaruliana (215 points) [ no revision ]

3 Answers

+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-09-25 20:21 (UTC), posted by SE-user Pointless
answered Sep 25, 2014 by Pointless (40 points) [ no revision ]
+ 3 like - 0 dislike

There is a relatively recent paper by Geoffrey L. Sewell (Statistical Thermodynamics of Moving Bodies arXiv:0902.3881v1 [math-ph] 23 Feb 2009) where he elementary proves that the KMS condition is not Lorentz invariant, concluding that the notion of temperature does not transform as expected under Poincaré transformations. Here is the abstract of the paper

"We resolve the long standing question of temperature transformations of uniformly
moving bodies by means of a quantum statistical treatment centred on the zeroth law of
thermodynamics. The key to our treatment is the result, established by Kossakowski et
al, that a macroscopic body behaves as a thermal reservoir with well-defined temperature, in the sense of the zeroth law, if and only if its state satisfies the Kubo-Martin-Schwinger (KMS) condition. In order to relate this result to the relativistic thermodynamics of moving bodies, we employ the Tomita-Takesakimodular theory to prove that a state cannot satisfy the KMS condition with respect to two different inertial frames whose relative velocity is non-zero. This implies that the concept of temperature stemming from the zeroth law is restricted to states of bodies in their rest frames and thus that there is no law of temperature transformations under Lorentz boosts. The corresponding results for nonrelativistic Galilean systems have also been established."

answered Sep 26, 2014 by Valter Moretti (2,025 points) [ revision history ]
+ 1 like - 0 dislike

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.

(please scroll down to the bottom of the page and click "Collapse Sidebar" to view above equation correctly)

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

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

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

answered Aug 16, 2014 by Nikos M. (80 points) [ revision history ]
edited Sep 27, 2014 by dimension10
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.

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

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

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