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I have tried to think about this for some time but could not really go anywhere. Sorry for the sloppy question and thanks for any pointer.

My question is about CFT at finite temperature and nonconformal theories. I have sometimes read that CFTs are not conformal at finite temperature but I could not make clear to myself what happened precisely. What theory does a CFT at finite temperature correspond to? To begin the "temperature state" is not unique but rather an equilibrium/Gibbs distribution over several quantum states -a canonical ensemble. So correlation functions are integrals over 2 measures (vs. vev in the CFT), an average over the correlation functions between each pair of states in the ensemble. Now many theories flow to a given CFT so which one should we pick at finite temperature? For instance what is the finite temperature of a gaussian/free massless field, say in 2D, i.e. the basic free boson c=1 CFT? I could imagine this is a massive boson with $m=1/T$ but am not sure. In general one would consider the reduced temperature, centered at the critical temperature. So is the finite temperature Ising model in the "thermodynamic limit" a sort of massive Wilson-Fisher fixed point?

There are often mentions of finite temperature CFT in the AdS/CFT correspondence. The dual of a schwarzschild black hole in AdS should be a finite temperature CFT. Is this correspondence conjectured to be exact at all couplings? QCD at high temperature is free, what does it mean for the dual strongly coupled string theory on AdS with no black hole? (I would guess this means the closed strings propagate freely in some appropriate

Well sorry for the fuzzy question. But I'll be very grateful if you try to empathize.

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  • $\begingroup$ Just look at the state of finite temperature - it's not Lorentz invariant because "temperature" requires a singled-out time direction, so it spontaneously breaks Lorentz symmetry and hence also conformal symmetry. It's not the theory that loses the invariance, it's SSB. $\endgroup$ – ACuriousMind Jan 20 '16 at 13:39
  • $\begingroup$ I am sorry but I don't understand what you mean. Temperature is a distribution on various states, classical or quantum. In a quantum theory we can take weighted sums of states but I guess there's the question of phase choice/normalization. But more importantly what do you mean by ""temperature" requires a singled-out time direction"? Isn't a QFT (say QED) plasma Lorentz invariant. Aren't the classical plasma equations (I think that would be Maxwell-Vlasov) Lorentz invariant? $\endgroup$ – plm Jan 22 '16 at 5:57
  • $\begingroup$ @ACuriousMind I am really interested in your explaining your comment. Thanks. $\endgroup$ – plm Jan 24 '16 at 6:55
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A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian $H$ and compute $Z=\mathrm{tr}\,e^{-\beta H}$, where the trace is over the Hilbert space of states living on $\mathbb{R}^{d-1}$ if your CFT is in $d$ dimensions. The thermal correlators are computed in a similar way, $Z^{-1}\mathrm{tr}\,\phi(x)\phi(y)e^{-\beta H}$. Physically that means that you put your system described by the CFT in contact with a thermal bath at temperature $T=1/\beta$, and consider thermal expectation values. That could be just a box full of massless scalars sitting in a box on your stovetop, roughly speaking.

Now, there is a path integral picture to it, where you work in Euclidean signature and make your time periodic. I.e. you compute path integral on $S^1\times\mathbb{R}^{d-1}$, with circumference of circle $S^1$ being $\beta$. This is equivalent to computing the above traces. It is the same path integral you are computing in a Euclidean CFT, except it is now on a different manifold. So in particular you have the same set of local operators. You still have Weyl invariance and you can work in any metric Weyl-equivalent to the flat metric on $S^1\times\mathbb{R}^{d-1}$, e.g. in 2 dimensions you can conformally map this to a plane; you should, however, be careful about the standard transformation of correlation function under Weyl. In many aspects the theory is still the same as it was in flat space.

Now, some things are different. For starters, $S^1$ is a circle, and circle has circumference. This is a dimensionful parameter, and so you do not have scale invariance (however, for a CFT, you can use scale invariance to relate results on circles of different radius -- i.e. at different temperatures). This means that some operators of non-zero scaling dimensions can get vev's (it was the scale invariance which prevents them to have vev's on $\mathbb{R}^d$). Also, you break part of rotation invariance, since you have a preferred direction of the periodic time. This means that not only scalars can get vev's, but also operators with spin. However, you still have $SO(d-1)$ rotation invariance, so it restricts the form of vev's of tensor operators.

For example, stress energy tensor is going to get a vev. If the periodic time direction is the $0$-th coordinate, then $T^{00}$ will be equal to energy density at this temperature. Rotation invariance in other components means that the only other vev's of T are going to be $T^{11}=T^{22}=\ldots=T^{dd}$ and that is going to be proportional to pressure or something. Tracelessness of $T$ then implies a relation between pressure and energy density.

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    $\begingroup$ @plm, I am not adding 1 dimension. Original CFT is in d dimensions and I consider its Euclidean version on $S^1\times\mathbb{R}^{d-1}$, which is still d-dimensional. An example of a vev, as I said is the stress-energy tensor, which gives energy density. $\endgroup$ – Peter Kravchuk Feb 7 '16 at 7:58
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    $\begingroup$ @plm, again, this is a general procedure -- if you have a Lorentzian QFT in d dimensions, one of which is time, your physical states are defined on $\mathbb{R}^{d-1}$ spatial slices. You can either consider the time evolution of such states, which is studying this QFT in Lorentzian signature, or consider the thermodynamics of these states, which is when you study dynamics in contact with thermal bath, and this can be formally described using Euclidean periodic time, which gives you the $S^1$. $\endgroup$ – Peter Kravchuk Feb 7 '16 at 8:03
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    $\begingroup$ @plm, the finite-temperature Ising model is a different temperature. The Ising CFT does not come from a Lorentzian QFT. Instead, Ising CFT is the IR fixed point of Ising statistical model considered at the critical temperature. Ising CFT is inherently Euclidean, and it is computing the classical partition function and classical correlations in a statistical model. There is no straightforward way to get off-critical Ising model from the CFT, you can usually only do perturbation theory around the critical point. In this case the difference $T-T_c$ is a coupling in Lagrangian, and you still work $\endgroup$ – Peter Kravchuk Feb 7 '16 at 8:07
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    $\begingroup$ @plm, There are two ways d-dimensional euclidean CFT appear. 1) they describe statistical physics of (d+1)-dimensional real-world systems at critical temperature. e.g. consider a 3d Ising model -- that is nearest-neighbor interaction of "spins" $\pm 1$, at finite temperature T. NOT a CFT, but a model to describe real-world (anti-)ferromagnets in real-world conditions. Like, stuff you can hold in your hand. But it is not a real-time, dynamical, description -- it is a statistical description in thermodynamic equilibrium. Now, you can tune your temperature T to a certain critical value... $\endgroup$ – Peter Kravchuk Feb 10 '16 at 20:30
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    $\begingroup$ ...and in general you will need to specify infinitely many coupling constants to get back to off-critical Ising model. This is the same issue as UV-completing effective field theories. 2) Euclidean $d$-dimensional CFT arise as Wick rotations of Lorentzian $(d-1)+1$-dimensional CFT, which are understood either as low-energy limits of generic QFT's, or just as standalone QFT's with extra symmetry. These Lorentzian QFT's describe temporal dynamics of $(d-1)$-(space)dimensional systems. E.g. a QFT describing dynamics of a ferromagnet which you can hold in your hand is $3+1$ dimensional... $\endgroup$ – Peter Kravchuk Feb 10 '16 at 20:41

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