Quantum field theory: zero vs. finite temperature

Question

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.

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.