I have recently been made aware of the concept of thermal field theory, in which the introductory statement for its motivation is that "ordinary" quantum field theory (QFT) is formulated at zero temperature. Now, I have read through the introductory sections of Peskin & Schroeder, Zee and Srednicki, and none of them mention this assumption explicitly (or give a definition of what is meant by this statement). My google searches so far have been unfruitful, the only sources I have found simply state is done at zero temperature and that thermal field theory is required for systems at finite temperature.

By, "at zero temperature", is it simply meant that the vacuum state of the theory is defined as having zero energy and zero temperature. In particular one calculates vacuum to vacuum transition probabilities in which one assumes that "in" and "out" states in particle interactions exist. That is, if you go far enough from the interacting region you can have an initial unperturbed state (an "in" state), and a final state (an "out" state) that has been perturbed by its interaction with the initial state, but is now far enough away that it is no longer perturbed. Would it be correct to say that at zero temperature this notion of in/out states is possible since the vacuum contains no particles so a particle state can exist unperturbed far enough from any other interacting particle state, however, at finite temperature, there are particles everywhere (since there is enough energy available to each quantum field to create many particles), and so the notion of in/out states is no longer tenable?!

If someone could enlighten me on the subject it'd be much appreciated.

  • 1
    $\begingroup$ The ground state of the theory doesn't have zero energy at $T=0$, it's just assumed that, without external excitation, the field is in the ground state. With $T>0$ that's not true any longer and there are excited states (i.e. "real particles") around. For the electromagnetic field (at sufficiently low energies) this is basically just Planck radiation. $\endgroup$
    – CuriousOne
    Commented Apr 10, 2016 at 20:11
  • $\begingroup$ @CuriousOne But I thought with the normal ordering prescription that the ground state of the theory is redefined such that its energy is zero? So is this all that there is to the statement that QFT is done at zero temperature? Is it simply the assumption (as you put it) that without any external excitation any given quantum field will be in its ground state? So, at finite temperature, this assumption is no longer valid since each quantum field is in a constantly excited state, even in the absence of external excitation, due to the ambient energy (provided by the ambient finite temperature)? $\endgroup$
    – Will
    Commented Apr 10, 2016 at 20:44
  • $\begingroup$ I don't know what the theoretical support for normal ordering is, other than that it makes the cow spherical. Phenomenologically a non-zero ground state energy seems to be acknowledged as necessary and observable (there is some discussion about that). One could also have a philosophical discussion about what the temperature of the vacuum would be in total absence of matter, but in presence of matter and in thermodynamic equilibrium one certainly has to make $T>0$ because of the third law of thermodynamics. $\endgroup$
    – CuriousOne
    Commented Apr 10, 2016 at 21:51
  • $\begingroup$ @CuriousOne That's what I thought (in reference to the very last part of your comment). If any particles are created at all, they will have some definite energy and necessarily, due to the laws of thermodynamics, there will be a non-zero temperature. This is what confuses me about the whole statement of QFT being at zero temperature. To a certain extent I get that in "ordinary" QFT the quantum fields stay in there ground states in the absence of any external interaction (as this external influence is what provides the energy to create a particle). What about quantum fluctuations though? $\endgroup$
    – Will
    Commented Apr 11, 2016 at 8:08
  • $\begingroup$ My understanding was always that quantum fluctuations and finite temperature are orthogonal issues. One exists independently of the other. Structurally, I believe, we are mostly interested in the $T=0$ case for the exploration of models, but then, when we calculate actual physics, e.g. cosmological predictions, $T>>0$ is the driver. I would be more than happy to find myself corrected by a theoretician on this one. $\endgroup$
    – CuriousOne
    Commented Apr 11, 2016 at 8:19

1 Answer 1


First of all, note that one cannot associate a temperature to a single quantum state (cf "vacuum state of the theory is defined as having zero energy and zero temperature"), and having a zero energy vacuum state is just a convention (as it is cut-off dependent, and thus renormalized).

Furthermore, the OP is confused. Standard (i.e. zero-temperature) QFTs compute observables in the interacting ground-state of the theory, call it $|\Omega\rangle$, which is different from the non-interacting ground-state $|0\rangle$. After that, all the discussion about normal ordering and in and out state in the OP's question is "just" related to technicalities to construct $|\Omega\rangle$ from $|0\rangle$.

Thermal field theory assumes that the observables are obtained from a system the state of which is a thermal distribution, i.e. described by a density matrix $\rho= \exp(-\beta H)/Z$ with $\beta$ the inverse temperature. In the eigenbasis $|n\rangle$ of the interacting hamiltonian (with $|n=0\rangle =|\Omega\rangle$ the same ground-state as above), a single time observable is given by $$\langle A\rangle = \sum_n \frac{e^{-\beta E_n}}{Z}\langle n|A|n\rangle .$$ In the limit $\beta\to\infty$, one recovers the zero temperature QFT result $\langle A\rangle=\langle \Omega|A|\Omega\rangle$.

Of course, one need to devise a perturbation scheme to relate the interacting eigenstate to the non-interacting eigenstate, more or less copied to that of the zero-temperature QFTs.

  • $\begingroup$ Thanks for your answer. However, I still don't understand what is meant by QFT being at zero temperature? Is it simply that the ground state of the theory is one that is devoid of any particles and that in the absence of any external influence the quantum fields stay in their ground states?! Does thermal field theory differ from this in the fact that the theory is embedded in a heat bath am so the fields will never remain in their ground states (since the ambient temperature will cause them to be in a constantly excited state)?! $\endgroup$
    – Will
    Commented Apr 11, 2016 at 7:33
  • $\begingroup$ @Will: ground-state just means the lowest energy state. For a relativistic QFT (say QED), it is the state without any "real" photons and electrons but including all vacuum fluctuations. At finite temperature, you have a superposition of all states (with an arbitrary number of photons and electrons) with a Boltzmann weight. $\endgroup$
    – Adam
    Commented Apr 11, 2016 at 8:16
  • $\begingroup$ So is the point that in a relativistic QFT the assumption of zero temperature allows one to define a ground state in which there are no real particles, whereas at finite temperature it is not possible to do this since the lowest energy state will contain an arbitrary number of real particles in it?! (Would this be due to the ambient temperature forcing the lowest energy state to be one in which the field has enough energy to create particles) $\endgroup$
    – Will
    Commented Apr 11, 2016 at 9:39
  • $\begingroup$ @Will: Not really, the point of T=0 is to just look at properties of the GS. Then the ground-state is, whatever it is depending on the parameters (one could add a chemical potential, or an external classical field, etc). It happens that in QED (for example), the GS is the (interacting) vacuum. Finite temperature implies that all state have a finite probability, but if the temperature is small compare to the mass of the electron, the states with "real" electrons will be subdominant. $\endgroup$
    – Adam
    Commented Apr 11, 2016 at 10:03
  • $\begingroup$ Maybe you should read some condensed matter textbook, where finite T QFTs are discussed in length. That might help you understand what is going on here. $\endgroup$
    – Adam
    Commented Apr 11, 2016 at 10:04

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