According to Landau's symmetry breaking theory, there is a symmetry breaking when phase transition occurs.

  1. What is the symmetry breaking of superfluid-Mott insulator transition in Bose-Hubbard model?

  2. Why metallic state to Mott insulator state transition in Fermi-Hubbard model is not a phase transition, but a crossover.


1 Answer 1


The Mott transition in the Bose-Hubbard model is a quantum phase transition. From the point of view of field theory, that does not change much compare to standard (finite-temperature) phase transitions. The main difference is that you now have to take into account the quantum fluctuations which correspond to the "imaginary time" direction in addition to the d dimensions of space. It also means that there are at least two control parameters (i.e. parameters that have to be fine-tuned to have the transition), a non-thermal control parameter (such as the hopping amplitude or the density) and the temperature (which must be zero by definition).

Other than that, you can use Landau theory to understand the transition (which is second-order) at zero-temperature. The disordered phase is the Mott insulator, and the ordered one is the superfluid, where the non-zero order parameter is the condensate density (I will only talk about the 3D case, which is the simplest, as I won't have to deal with BKT phases). The broken symmetry is the usual one for Bose-Einstein condensate : the U(1) symmetry. One can then show that there is two universality classes, depending on the way the transition is made (at constant density or with a change of density at the transition).

Now, at finite temperature, things are different. First, the Mott insulator does not exist anymore, as a finite temperature can excite particles and one gets a finite compressibility (or conductivity). That might correspond to the cross-over you're talking about in the fermionic case. On the other hand, the superfluid exists at least up to a critical temperature.

  • $\begingroup$ thanks a lot. The superfulid oder parameter $\left<b^{\dagger}\right>$ breaks the U(1) symmetry:$b\rightarrow b e^{i\phi}$, when MI-SF transition happens. But I still cannot understand why there is no phase transition in Fermi Hubbard model? What is the essential difference between cross-over and phase transition? $\endgroup$
    – Timothy
    Sep 23, 2013 at 20:52
  • $\begingroup$ Previous answer by @Adam is mostly correct except for the statement that there are no Mott insulators at finite temperatures. See for example: prl.aps.org/abstract/PRL/v99/i12/e120405. Indeed optical lattice experiments have to conduct measurements at low enough temperatures that the gapped quasiparticles are well defined. And in a cross-over, there are no divergences associated with thermodynamic quantities or their derivatives. $\endgroup$
    – MaviPranav
    Sep 25, 2013 at 13:52
  • $\begingroup$ @MaviPranav : Strictly speaking, there is no Mott insulator at finite temperature, because the conductivity is non-zero. Of course, at very low temperature (much smaller than the gap), the system looks like an insulator (exponentially small compressibility), but you won't find a real transition, which exists only at $T=0$. Of course, experimentally one is happy with an exponentially vanishing compressibility (in the same way that insulators "experimentally exists" at finite temperature in solids) but I don't see how what I said in my answer is false. $\endgroup$
    – Adam
    Oct 1, 2013 at 3:57
  • $\begingroup$ @Jeremy : I know less about the fermionic case, but I think that in that case, the order parameter is less easy to define. For instance, you can say that the conductivity is the order parameter (zero in the insulator, finite otherwise), but there is no real symmetry breaking (but as I said, I might be wrong on that). Nevertheless, you don't need a symmetry breaking to get a phase transition. It's just that when there is a symmetry, it makes everything much easier. $\endgroup$
    – Adam
    Oct 1, 2013 at 4:03
  • $\begingroup$ @MaviPranav : it might be a matter of taste, but I think people usually describe a phase by its thermodynamical properties, and the hallmark of the Mott phase is its vanishing compressibility, hence the $T=0$. Just to say that I don't think my answer was wrong, as you suggested in your first comment. $\endgroup$
    – Adam
    Oct 5, 2013 at 19:59

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