# Symmetry breaking in Bose-Hubbard model

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.

• 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? – Timothy Sep 23 '13 at 20:52
• @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. – Adam Oct 1 '13 at 3:57
• @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. – Adam Oct 5 '13 at 19:59