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  Superfluid-Mott insulator transition in Bose-Hubbard model in terms of vortex condensation

+ 5 like - 0 dislike

I have heard that there is some effective field theoretic type understanding of the superfluid-Mott insulator transition in Bose-Hubbard model. It says if the system is in a superfluid phase where the $U(1)$ symmetry is broken, then by proliferating vortices we can destroy the phase coherence and obtain a Mott insulator. 

I understand that there is no phase coherence in a Mott insulator and having many vortices will mess up the phase coherence, but I do not understand why the mechanism to destroy phase coherence has to be proliferating vortices. To be more precise, suppose we can describe the superfluid phase by the following Lagrangian:
where $r$ and $u$ are positive so that the $U(1)$ symmetry is broken. Now if you asked me how to go from this $U(1)$ symmetry broken phase to a $U(1)$ symmetric phase, I would say we need to decrease $r$ until $r$ starts to be negative. However, when $r$ is decreased, I do not expect there will be more and more vortices in the ground state since here the energy of a single vortex always diverges logarithmically with the system size. Then how is decreasing $r$ related to vortex condensation?

By the way, if that vortex condensation picture is somehow correct, should one see vortices in experiments with Bosons in an optical lattice?

asked Sep 9, 2014 in Theoretical Physics by Mr. Gentleman (270 points) [ no revision ]
recategorized Sep 9, 2014 by dimension10

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