Quantum fields and condensates

From my fairly naive understanding of quantum field theory (QFT), a quantum field $\hat{\phi}$ is an operator field, i.e. for each spacetime point $x^{\mu}$, $\hat{\phi}(x)$ is an operator acting on the corresponding Fock space of particle states. Then, at least in the free-field case, $\hat{\phi}(x)$ acting on the vacuum state, $\lvert 0\rangle$ creates a particle in a given single particle state. As far as I understand it, one can heuristically interpret this as a quantised excitation in the quantum field, and in this sense, all particles of a given species are simply excitations (the so-called quanta) of their corresponding quantum field.

Now, here comes my question: I have read that if a given field has a non-zero vacuum expectation value (vev), that is $$\langle 0\rvert\hat{\phi}(x)\lvert 0\rangle\neq 0$$ then this is referred to as a condensate, and furthermore can be interpreted as a collection of field quanta all residing in the vacuum state.

Why is it that, when a quantum field as a non-zero (i.e. it forms a condensate), it can be interpreted as a collection of particles in the vacuum state?$^{(\ast)}$ I thought a quantum field is simply as I described early - not a collection of particles, but an operator field from which particles can be created (from the vacuum state) at each spacetime point?!

$^{(\ast)}$ I have been reading about inflation and the subsequent reheating process. According to what I've read, at the end of inflation, the inflaton field forms a condensate, oscillating about the minimum of its potential, which is interpreted as a collection of zero momentum particles in a single quantum state.

• good question, I think it is related to the coherent state but have no idea how such condensation is formed from the very beginning. Hope somebody will give a clear answer. – Wein Eld Aug 26 '16 at 20:22

If $\langle 0|\phi (x) |0\rangle = c$, then $\phi$ is related to the quantum field $\Phi$, whose $|0\rangle$ is the vacuum state, by means of $$\phi (x) = \Phi (x) + cI\:.$$ In other words, $$\phi (x) = U \Phi (x) U^*$$ where $$U = e^{b (a_0 -a^*_0)}$$ for some constant $b$ depending on $c$ and where $a_0$ and $a^*_0$ are the annihilation and creation operators referred to the zero mode of the quantum field $\phi$. The fact that only the zero mode enters the game is due to the fact that $c$ is a constant and not a function of $x$. Alternatively, we may say that $$|0\rangle = U^*\Psi_0\:,\tag{1}$$ where $\Psi_0$ is the vacuum state of $\phi (x)$.
(1) says that the left hand side is a coherent state of zero modes constructed over the vacuum $\Psi_0$: a condensate of zero modes.