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Background: This question is inspired by Why is a relativistic quantum theory of a finite number of particles impossible?

1 - QFT is typically used to calculate relativistic scattering. The ground state of relativistic QFT is the vacuum, which is an invariant state under Lorentz transformations. This is not the case when we are in a medium (finite density): a finite-density medium in the ground state is not Lorentz invariant (from the point of view of Lorentz transformations, the "finite-density" ground state is different from the vacuum, see the attempted answer below).

2 - On the other hand, we have non-relativistic many-body Quantum Mechanics that is used in condensed matter for finite-density systems. This can be expressed in a "second quantization" fashion, becoming a non-relativistic QFT. Moreover, we also have a sort of non-relativistic "thermal QFT" that allows us to calculate the thermodynamic equilibrium properties of many-body quantum systems (i.e., many-body non-relativistic quantum mechanics after Wick rotation).

Question: How about a "relativistic QFT for finite-density systems"? Namely, a QFT that aims at describing a dense medium of relativistic particles (not just scattering in vacuum). Do we have such a theory or the only thing we have are some "approximations"? By "approximation" I mean a theoretically problematic situation resembling that of the "early relativistic QM" (i.e. the so-called relativistic quantum mechanics), which was then replaced by relativistic QFT.

Answer: Exercise V.2.3 of Zee's QFT in a Nutshell (2nd edition) asks us to develop "QFT at finite density", which would be exactly the framework I am looking for: we have to add a chemical potential term to the Lagrangian in the path integral formalism. Moreover, Zee comments that "finite density, as well as finite temperature, breaks Lorentz invariance", see also this article. Some extra useful and related questions are Chemical potential in quantum field theories, Quantum field theory with a constraint: energy-momentum conservation?. The relation between the use of Lagrange multipliers to impose a certain density and path integral is discussed here. Interesting posts on Lagrange multipliers in QFT: QFTs which are pure constraint, Can auxiliary fields be thought of as Lagrange multipliers?, see also this.

Collection of references: Thermal/finite temperature quantum field theory: online lectures and best books.

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    $\begingroup$ Are you talking about QFT at finite temperature? $\endgroup$
    – Prof. Legolasov
    May 27, 2020 at 23:09
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    $\begingroup$ There is already a name for many-body QFT: it's just "QFT". The whole point of quantum field theory is that it can accommodate arbitrary numbers of particles, so there's no special case for many particles. $\endgroup$
    – knzhou
    May 28, 2020 at 5:06
  • $\begingroup$ For example, all the quantum systems studied in condensed matter physics contain lots of particles, and they of course can be described in the language of QFT. $\endgroup$
    – knzhou
    May 28, 2020 at 5:07
  • $\begingroup$ @knzhou , thank you. I agree that in some sense "many body QFT" is "just QFT". However, the many body formulation of QM as a QFT (usually used in condensed matter physics) and the QED or the QCD look very different to me, even though there are some "language similarities". For example, I was wondering if a theory for a "dense phase of matter made of quarks" would look like the QCD we use in high energy-physics or some modifications are needed: adding new terms in the Lagrangian? In fact, in "non-relativistic many-body" for example we add a chemical potential term to the Hamiltonian. $\endgroup$
    – Quillo
    May 28, 2020 at 11:35
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    $\begingroup$ Related, useful questions: physics.stackexchange.com/q/248704/226902 , physics.stackexchange.com/q/425459/226902 , physics.stackexchange.com/q/149463/226902 , physics.stackexchange.com/q/519802/226902 , finite density: physics.stackexchange.com/q/34616/226902 , loss of Lorentz invariance physics.stackexchange.com/q/131197/226902 , references: physics.stackexchange.com/q/353334/226902 $\endgroup$
    – Quillo
    Jan 24, 2022 at 16:12

1 Answer 1

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Constructing a macroscopic relativistic quantum field theory is done in exactly the same way all such quantum theories are. You assume the system is in equilibrium, i.e. at some temperature $T$. The partition function for a quantum system is given by $$\mathcal{Z}=\text{tr}\big(e^{-\beta H}\big)$$

You use this to calculate all sorts of statistically relevant quantities, as you would in statistical mechanics. The difference now is the Hamiltonian is that for a relativistic quantum field theory!

This can be incorporated with every quantum field theory you can think of, even the standard model.

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  • $\begingroup$ Thank you @LucashWindowWasher (+1). My question maybe is not super clear but I was thinking about a slightly different thing. In non-relativistic many-body we can (at least in principle) calculate the ground state of many particles. Then, we have the thermal theory but also the dynamical theory with time evolution. The same is true also for QED or QCD, but the ground state is "the vacuum". So I was wondering if there is a relativistic "many-body" QED or QCD that deals with "non-trivial" ground states. $\endgroup$
    – Quillo
    May 28, 2020 at 11:55
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    $\begingroup$ In that case, the same formalism I described can accommodate this, except the partition function is not enough. You can consider correlation functions on top of the finite temperature background. These correlation functions are calculated with respect to the canonical ensemble density matrix $\rho=e^{-\beta H}/\mathcal{Z}$. In principle, a QFT allows you to calculate correlation functions over ANY background, you just replace the vacuum with any state you wish. $\endgroup$
    – fewfew4
    May 28, 2020 at 13:32
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    $\begingroup$ The finite temperature scalar two point correlation function for example goes like $\langle \phi(x)\phi(y)\rangle=\text{tr}(\phi(x)\phi(y)e^{-\beta H})/\mathcal{Z}$ $\endgroup$
    – fewfew4
    May 28, 2020 at 13:39
  • $\begingroup$ Very cool thank you. May I ask you an introductory reference if you know a good one? $\endgroup$
    – Quillo
    May 28, 2020 at 15:05
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    $\begingroup$ One reference I would recommend is chapter 18 of Quantum Field Theory, A Modern Perspective, by V.P. Nair $\endgroup$
    – fewfew4
    May 29, 2020 at 7:43

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