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.

  • $\begingroup$ 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. $\endgroup$ – 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.

This picture holds as it stands if quantizing in a box. However, the intepretation can be proposed in the general case (thermodynamical limit), for instance, exploiting the algebraic approach. See for instance the second volume of the textbook of Bratteli and Robinson.

  • $\begingroup$ So it the vev of a field is non-zero, then is it the case that the vacuum state is filled with a condensate of zero momentum particles, the quanta of the field? The actual quantum field is not a collection of particles though, right?! When one expands the field in terms of its frequency modes, can this be understood heuristically as describing the field in terms of an infinite superposition of oscillators, each oscillating at one of the possible frequencies in the (infinite) spectrum?! $\endgroup$ – Will Aug 26 '16 at 22:57

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