Symmetry breaking in Bose-Hubbard model According to Landau's symmetry breaking theory, there is a symmetry breaking when phase transition occurs. 


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

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