# Defining a classical field corresponding to a quantum field

1. Why is the expectation value of the quantum field in the vacuum state $$\phi_c(x)=\langle0|\hat{\phi}(x)|0\rangle_J=\frac{\delta W}{\delta J}$$ referred to as the classical field?

2. Why not the expectation $\langle\psi|\hat{\phi}(x)|\psi\rangle$, calculated in some other quantum state $|\psi\rangle$ (such as a $N$-particle state or in an arbitrary superposition of Fock states) be interpreted as the classical field?

• Got a source for this claim? I ask because it it my understanding that for any theory symmetric in $\hat{\phi} \rightarrow -\hat{\phi}$ the expectation in the first part vanishes identically. – Sean E. Lake Jun 6 '17 at 6:25
• @SeanE.Lake Yes it does vanish (or becomes a constant if the symmetry is spontaneously broken). But only when you set $J=0$. Before you set $J=0$, it is in general, a function of spacetime. You can can look at Peskin and Schroeder, or Ryder's book on QFT for example. – SRS Jun 6 '17 at 6:27
• – AccidentalFourierTransform Aug 7 '17 at 15:24

Unless stated otherwise, in quantum field theory we almost always assume that the system is in thermal equilibrium at zero temperature. This is usually an excellent approximation to the real world, because the characteristic temperature scale for elementary particle physics is the Hagedorn temperature of $\sim 10^{12} \text{ K}$, and almost all of the universe is effective at zero temperature relative to that scale. The zero-temperature thermal density matrix is just $\rho = | 0 \rangle \langle 0 |$ where $| 0 \rangle$ is the ground state, so thermal expectation values $\langle O(x) \rangle := \text{Tr } (\rho\, O(x)) = \langle 0 | O(x) | 0 \rangle$ of any field $O(x)$ are just given by the ground-state expectation values.
Also, note that in the formalism of second quantization, all states are created by applying creation operators to the ground state, so an expectation value with respect to any pure state can be equivalently expresses as a ground-state expectation value. E.g. if $|\psi\rangle = a^\dagger(x)^N | 0 \rangle$ is an $N$-particle Fock state, then $\langle \psi | O(x) | \psi\rangle = \langle 0 | a(x)^N\, O(x)\, a^\dagger(x)^N | 0 \rangle$ is equivalent to the vacuum expectation value (VEV) of the field $a(x)^N\, O(x)\, a^\dagger(x)^N$. So at least for pure states, there's no loss of generality in only considering VEVs.