Is Bose-Einstein condensate a good example of a classical massive boson field?

Question

Physically, we know that a BEC has formed if a macroscopic number of bosons occupy a single quantum state. The wave-function $\Psi(x)$ of the latter, normalized to the total number of condensed atoms $N \gg 1$, gives the macroscopic description of the condensate. $\Psi(x)$ will satisfy the ususal Schrodinger euqation in the trapping potential (which is the special case of the Gross-Pitaevskii equation with zero interactions).

On the other had, a macroscopic limit of a boson field should be described by the (classical) Klein-Gordon equation. In the relevant non-realtivistic limit, dispersion of the Klein-Gordon field is quadratic, as in the Schrodinger equation, but I struggle to derive the former from the latter.

Despite this technical difficulty of mine, is it a valid claim that BEC is a physical realization of the non-relativistic, classical-wave-limit of a boson field? I see it in the same vein as electric-field description of the laser radiation being a classic limit of the quantum coherent state.

(The difference between number states and coherents states in the limit $N \gg 1$ is, hopefully, a mere technicality here).

Answer 1

Yes, it is valid. The Schrodinger field is the nonrelativistic limit of the Klein Gordon field, but you have to take the limit slightly carefully, because you have to separate the creation and annihilation operators into separate fields, which become, appropriately mixed real/imaginary parts of the Schrodinger field.

The way to do this is to expand the dispersion to second order in the scalar Klein Gordon field, and (in the free theory) to identify $\psi(k,t)$ with $e^{imt}a(k)$ (throwing away the rest mass), where $a(k)$ is the nonrelativistically normalized creation operator for mode k. Then $\psi(x,t)$ is the Fourier transform of $\psi(k,t)$, and unlike the relativistic theory, has an immediate interpretation as an annihilation operator removing a particle at x, and from the expansion of the dispersion, it obeys the Schrodinger equation, and from complex conjugation, it's a complex field whose conjugate involves modes near -m energy in the relativistic theory.

That's the free field anyway. The interacting field is easier in the nonrelativistic limit, so it is easier to introduce the interactions by starting there directly (although you can do it by taking the limit, of course).