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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: \begin{equation} \mathcal{L}=|\partial_\mu\phi|^2+\frac{r}{2}|\phi|^2-\frac{u}{4!}|\phi|^4 \end{equation} 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?

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Because $|\phi|^2=\rho$ is the boson density, and you expect to have finite boson density in the Mott phase (like one boson per site), so you can not just let the boson field $\phi$ goes to zero.

In fact, there are two ways to disorder a superfluid and restore the U(1) symmetry, either by the amplitude fluctuation or by the phase fluctuation. The $r$-reversal process you described is to introduce the amplitude fluctuation, which usually corresponds to a finite temperature transition to the thermal boson gas. In this case, $\phi$ describes the condensed part of the boson, and decreasing the amplitude $|\phi|$ corresponds to depleting the boson from the condensate to the thermal gas. However the superfluid-Mott transition is a zero temperature quantum phase transition. At zero temperature, there is no thermal boson to absorb the lost of $|\phi|^2$, so if you send $|\phi|$ to zero, the boson will be physically lost from the system, and the resulting U(1) symmetric state is actually the vacuum state. We know this should not be the case for the superfluid-Mott transition, as the boson density $|\phi|^2$ is fixed at a non-zero level in both phases, so the transition is not driven by the amplitude fluctuation, but actually driven by the phase fluctuation without $|\phi|$ going to zero.

The correct field theory for the superfluid-Mott transition is the XY model. In this theory, one must first separate the boson field into the amplitude and the phase modes $\phi=\sqrt{\rho}\mathrm{e}^{\mathrm{i}\theta}$. Substitute into the $\phi^4$-type of theory and integrate out the small amplitude fluctuation, we obtain the low-energy theory for $\theta$ $$\mathcal{L}=\frac{\eta}{2}(\partial_\mu\theta)^2,$$ where $\eta$ is the superfluid stiffness. Because $\theta$ is a compact variable (meaning that it has the periodicity $\theta\sim\theta+2\pi$), it is more convenient to consider its dual field theory in terms of $\psi$, such that $\epsilon^{\mu\nu}\partial_\nu\psi$ is the conjugate variable of $\partial_\mu\theta$, with $v\equiv\mathrm{e}^{\mathrm{i}\psi}$ being the annihilation operator of the vortex in the superfluid, $$\mathcal{L}=\frac{1}{2\eta}(\partial_\mu\psi)^2+\frac{g}{2}(v+v^\dagger)=\frac{1}{2\eta}(\partial_\mu\psi)^2+g\cos\psi.$$ In the large $\eta$ limit, $(\partial_\mu\psi)^2$ costs little energy, so $\psi$ field fluctuates strongly (and $g\cos\psi$ is averaged to zero and hence irrelevant), then the dual variable $\theta$ will not fluctuate, which corresponds to the U(1) breaking superfluid phase. In the small $\eta$ limit, the fluctuation of $\psi$ is suppressed, and the $g\cos\psi$ term becomes relevant, which leads to the vortex proliferation (creation and annihilation of vortices). So the long-range order of $\theta$ is destroyed, and the U(1) symmetry is restored, which corresponds to the Mott insulating phase.

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    $\begingroup$ The equation $|\phi|^2=\rho$ for (non-relativistic) bosons is not valid at all close to the Mott transition, only in the mean-field (Gross-Pitaevskii) regime. In the superfluid phase, close the Mott transition, the condensate density is finite but small $|\phi|^2\ll \rho$ ($\rho$ is of order one). The idea that $|\phi|=0$ corresponds the vacuum is wrong, because $|\langle \hat\psi\rangle|^2\neq \langle \hat\psi^\dagger\hat\psi\rangle$. Though it works in the dilute limit, where $|\phi|^2\simeq \rho$, but that's not the relevant limit here. $\endgroup$ – Adam Sep 10 '14 at 12:36
  • $\begingroup$ @Adam I see what you mean. But do you have any understanding of how to write down a field theory description close to the Mott transition? $\endgroup$ – Everett You Sep 10 '14 at 17:17
  • $\begingroup$ Yes! Close to the tip of the Mott lobes, where the density is an integer, you write exactly the same action. But the field $\phi$ is not related to the density. Away from the tip (but close to the transition line), one writes an action which looks the same as that of a dilute Bose gas, but where $|\phi|^2$ now gives the excess of particles or holes compared to the integer filling, i.e. $\rho=n\pm|\phi|^2$. The case $\phi=0$ is then the vacuum of holes or excess of particles. In fact, the true vacuum is the trivial Mott insulator with zero particles per site ! $\endgroup$ – Adam Sep 10 '14 at 19:39

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