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I understand that when a system enters the BEC phase a sizable fraction of the total number of particles enters the ground state, until at some point almost all of your particles are in the ground state.

What happens in the case of degenerate ground states?

Do the particles simply distribute themselves amongst these lowest energy states, or do they actually all go into the same state?

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3 Answers 3

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In reality you almost always find that the particles prefer to go to the same state because of some tiny energy shifts in the system. For example, a ferromagnetic condensate can have many degenerate states, but there is an energy cost for particles that disagree. These systems will break symmetry by having all the particles choose the same (arbitrary) state.

One exception is if the number of degenerate states is large, for instance larger than the number of particles. I have heard that claim that if the system has a sufficiently large degeneracy, for instance if all momentum states has the same energy, Bose-Einstein condensation might not occur.

Other than that, if there really is no energy cost to arranging your particles however you want between two degenerate states, I would guess that the particles would on average be equally distributed. However, fluctuations should be enormous: the probability of any configuration (N atoms in state 1, M atoms in state 2) should all be equal.

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Why always take magnon case. In other systems, a "tiny" difference in energy (within a small energy range) can also have a set of non-degenerate states, but particles will prefer to go to the lowest energy state due either to interference and/or conversion. Does it look like a plausible explanation?

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Could you elaborate your answer? Right now it is far to sketchy and unclear to be considered a proper answer – Michiel Mar 6 '13 at 6:21

The particles are simply distributed evenly between the degenerate lowest energy states.

This is the case for a ideal spinor BEC without Zeeman effects, for example: "Because there are three internal states, the condensation temperature of an ideal spin-1 gas at $p = q = 0$ is reduced to $T_c^{\mathrm{spinor}} = (1/3)^{2/3}T_0$" where "$T_0$ is the condensation temperature of an ideal scalar gas with the same total number density." That is, $1/3$ of the atoms go into each internal state.

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