For the Bose-Hubbard model, we know that there are the Mott insulator phase with $\langle a_i \rangle = 0$ and the superfluid phase with $\langle a_i \rangle \neq 0$. However, when we are trying to solve the Bose-Hubbard model numerically, we often assume a fixed site number and a fixed total particle number. But if $\langle a_i \rangle \neq 0$, the total particle number is not conserved. Does that mean we cannot really get the true superfluid phase numerically, thus we don't really have a phase transition if we solve the Bose-Hubbard model numerically?

  • $\begingroup$ Experiments also (typically) have a fixed particle number, yet they see the superfluid phase without a problem. $\endgroup$
    – Rococo
    Oct 27, 2020 at 22:32
  • 1
    $\begingroup$ To distinguish these phases, the single-particle momentum density is often calculated. That is flat in the Mott regime, and sharply peaked in the superfluid case. $\endgroup$ Oct 27, 2020 at 23:27

1 Answer 1


The numerical calculations are usually always carried out in the grand canonical ensemble, so you assume your system of $N$ particles is connected to a large particle reservoir. The chemical potential is fixed, which ensures $\langle N \rangle$ is fixed.

This is of course not physical, but in the thermodynamic limit everything tends to the same anyway.

But what really happens is that the total number of particles (system + reservoir) is conserved because of baryon number conservation. So the fluctuations in atom number for the superfluid phase would be balanced by equal and opposite fluctuations in atom number in the reservoir. They have actually proven this experimentally for partially and fully condensed Bose-Einstein condensates.


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