Relativistic Bose-Einstein Condensation (BEC)

Question

I wonder if there is a concept similar to the one of BEC but arising from Quantum Field Theory instead that from the usual one developed in non-relativistic many-body Quantum Mechanics.

In non-relativistic many-body QM the particles undergo condensation by occupying the single-particle ground state (if the system is non-interacting or weakly interacting). The description is then in terms of a "collective" wave function $\Psi$ that is the "order parameter" and is subject to the Gross-Pitaevskii equation (GPE).

Now, if the theory has to be relativistic, I suppose that instead of the condensate wave-function $\Psi$, we should have a Klein-Gordon scalar field $\phi$, and that instead of the GPE we should have something like a Klein-Gordon wave equation.

This is just speculative and (if correct) it is not clear to me which is the exact meaning of this relativistic Klein-Gordon field $\phi$ that should play the role of "order parameter". In particular, the scalar bosons that undergo the condensation are already described in terms of a scalar field, but is this the same scalar field $\phi$ that play the role of "order parameter"?

PS: my claim is based on the fact that if you write $\phi = e^{im t} \Psi$, then you can convert the Lagrangian of $\phi$ into the Lagrangian for the non-relativistic field $\Psi$, where $m$ is the mass of the boson. Making the variations of the the Lagrangian for $\Psi$ we obtain the time-dependent GPE equation (basically a Schrodinger equation), see e.g. this or this.

Answer 1

Bose-Einstein Condensation happens for all bosons, provided the potential and the system's dimenionsality supports it. It just depends on the statistics that the particles obey.

The GPE is just an equation that adequately describes (in the mean-field approximation), interacting Bose-condensed bosons. You cannot show that atoms condense at $T=T_{\mathrm{c}}$ from the GPE. You have to use to already condensed bosons.

The Klein-Gordon equation describes free spin-$0$ particles, that is, a particular type of (free) bosons. Hence they will indeed Bose-Condense, even though BEC in free space is kinda boring and impossible to observe (in the thermodynamic limit) as the lowest state is a flat distribution that would tend to $0$ for normalisation's sake.

While BECs are "intuitively" described in quantum mechanics (first quantisation), they should really be dealt with withint QFT (second quantisation). Only in the latter formalism does the concept of symmetry breaking really make sense (though for a non interacting BEC it turns out you don't necessarily need symmetry breaking). The order parameter is the field operator $\hat\psi(r)$: $$ \hat\psi(r) = \sum_i a_i \varphi_i(r),$$ where $a_i$ is the annihilation operator of a particle in the single particlestate $\varphi_i(r)$. You can easily see that $\langle \hat \psi(r) \rangle = 0$ whenever the state you are averaging over is in a number eigenstate. BEC occurs when $\langle \hat \psi(r) \rangle \neq 0$, which agrees with the state now being a coherent state and hence having a "more fixed" phase while an uncertain number of particles$^\dagger$. Hence $\langle \hat \psi(r) \rangle$ can be used as the order parameter.

$^\dagger$: this is a subtlety usually ignored by most literature, but the symmetry breaking and phase-selection formalism, while intuitively applicable and useful, is not required for non-interacting BECs. You could actually do without it. The coherent state picture kinda works for $0 \ll T \ll T_{\mathrm{c}} $ where the thermally depleted atoms provide a reservoir that allows the Bose-condensed number to fluctuate. But for $T=0$, baryon number conservation prohibits the particle-number fluctuations expected for a coherent state. See this reference for a nice discussion on coherent state and number-state formalisms.