In elementary statistical mechanics, one can think of temperature as arising from the average kinetic energy of particles in the ensemble. Is there a similar way to think about the temperature of a quantum field theory?

I know that we can talk about the temperature as being related to the periodicity of Euclidean time, but I am wondering if there is a more physical way to think about it.

  • 7
    $\begingroup$ In statistical physics, inverse temperature is generally introduced as the Lagrange multiplier implementing the constraint that the average value of energy is known. This is a general definition that applies to quantum field theory as well. The relation to kinetic energy is special to a gas of free particles (ideal gas). $\endgroup$ Jun 11, 2020 at 21:03
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/491179/50583 $\endgroup$
    – ACuriousMind
    Jun 13, 2020 at 13:17

2 Answers 2


Without temperature, if you want to know the expectation value of an operator $\langle A\rangle$ given a field state $|\psi\rangle$, you just compute: $$ \langle A\rangle = \langle \psi|A|\psi\rangle . $$

Thermal field theory (or just thermal quantum statistical mechanics) assumes that the observables are obtained from a system whose state is a thermal distribution, described by a density matrix $\rho= \mathrm{e}^{-\beta H}/Z$ with $\beta = \frac{1}{k_{\mathrm{B}}T}$.

In the eigenbasis $|m\rangle$ of the Hamiltonian, the same observable is therefore given by $$\langle A\rangle = \sum_m \frac{e^{-\beta E_m}}{Z}\langle m|A|m\rangle .$$

  • 2
    $\begingroup$ 1. It might be worthwhile to note that $\rho = \mathrm{e}^{-\beta H}$ isn't specific to field theory at all - this is what temperature is in all quantum statistical mechanics. 2. What do you mean by "a field $\lvert \psi\rangle$"? Quantum fields are operators, not states. $\endgroup$
    – ACuriousMind
    Jun 13, 2020 at 13:14
  • $\begingroup$ Yeah sorry, I was trying to oversimplify the problem. $|\psi\rangle$ is a state, "given by" $\psi(x)|0\rangle$ where $\psi(x)$ is the field operator. $\endgroup$
    – SuperCiocia
    Jun 13, 2020 at 19:34

Also another way to think about is the following (here I follow the 5th Volume of Landau&Lifshitz). Let $\rho$ be a density matrix of a system, and the system can be subdivided into 2 parts, described by denisty matrices $\rho_1, \rho_2$ with weak interaction, such that they can be considered independent, but at the same time there is thermodynamical equilibrium between them. Therefore the density matrix is product of density matrices of each subsystem: $$ \rho = \rho_1 \rho_2 \Rightarrow \ln \rho = \ln \rho_1 + \ln \rho_2 $$ Therefore $\ln \rho$ is an additive integral of motion. The system has additive following integrals of motion $E, P, J$ - total energy, momentum, angular momentum, respectively. So, $\ln \rho $ has to be some linear combination of them. Going to the frame of reference, which is comoving and corotating with the system concerned, we may set $P, J = 0$. $\beta$, the inverse temperature $1/T$, has the meaning of proportionality coefficient.
$$ \ln \rho = -\beta E $$


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