# How does the $U(1)$ global symmetry break in the gauged $XY$ model?

I'm studying the particle vortex duality, and I'm confused how we're able to say that in the Coulomb phase, the "hidden" $$U(1)$$ global magnetic symmetry spontaneously breaks.

gauged XY model: $$\mathcal{L} = \frac{1}{2g^2}f_{\mu\nu} f^{\mu\nu} + |\mathcal{D} _\mu \phi|^2 - \mu |\phi|^2 - \lambda |\phi|^4$$

This theory has $$U(1)$$ gauge symmetry and a less obvious $$U(1)$$ global symmetry corresponding to the conservation of the magnetic current $$j^\mu = \frac{g^2}{\pi} \partial^\mu \sigma$$, where $$\sigma$$ is the dual photon ($$d \sigma = \frac{-g^2}{\pi}\star da$$).

In the coulomb phase phase corresponding to $$\lambda > 0, \mu > 0$$ we have an obvious unbroken $$U(1)$$ gauge symmetry. However, we say that the $$U(1)$$ global symmetry is broken, and thus, this phase corresponds to the SSB phase of the global XY model.

I don't see why this is the case, as when I change the variables from $$\phi$$ to $$\sigma$$ the Lagrangian still retains the shift symmetry $$\sigma \rightarrow \sigma + const.$$ modulo 2$$\pi$$. Thus there's no change in the magnetic current. Where am I going wrong in this reasoning?

• – SuperCiocia Sep 19 at 4:52
• It's related, and I'm aware of the answer given. However what I'm asking is not about the conservation of the current, but how it spontaneously breaks. – pyroscepter Sep 19 at 6:16

Sponteneous symmetry breaking occurs when the ground state has a different (reduced) symmetry from the Hamiltonian/Lagrangian. So $$\mathcal{L}$$ retains the $$U(1)$$ or shift symmetry, but you have to look at whether or not the ground state also does.

The "gauged $$XY$$" model is also called the Abelian-Higgs model.

The first is the usual $$U(1)$$ gauge symmetry associated with $$\phi \rightarrow e^{\mathrm{i}\varphi}\phi$$. Making this a local symmetry introduces the coupling to the gauge field $$\alpha_\mu$$ (in your gauge covariant derivative $$\mathcal{D}_\mu$$).

Another $$U(1)$$ symmetry is identified (only because we are in $$2+1$$ dimensions) that does not depend on the metric $$g_{\mu\nu}$$ and is hence distinctly topological, of the Chern-Simons type of topological field theories.

Your photon $$\alpha_\mu$$ has $$3$$ degrees of freedom, because we are in $$2+1$$ dimensions. Because it's a gauge field, one d.o.f. is taken out, and we are left with $$2$$. Depending on the sign of $$\mu$$, this will remain $$2$$ (Higgs phase, topological $$U(1)$$ symmetry unbroken) or will become $$1$$ (Coulomb phase, topological $$U(1)$$ symmetry broken).

$$\lambda > 0$$ is taken everywhere.

• Let us first look at the $$\mu<0$$ case.

The potential $$V(\phi) = \mu |\phi|^2 + \lambda |\phi|^4$$ now looks like this (with $$\lambda = -\mu = 1$$):

The potential has a ring of minima at $$|\phi|=1$$, but the ground state corresponds to a specific choice of a point on this ring. So the Lagrangian (pre-SSB) still has $$U(1)$$ symmetry, but the ground state does not. You could of course re-write the Lagrangian post-SSB in terms of real fields $$\phi$$ so that the $$U(1)$$ would not be there any longer.

The gauge $$U(1)$$ symmetry is spontaneously broken so that $$\phi$$ gains an expectation value $$\langle \phi \rangle \neq 0$$, and you have to re-write your Lagrangian (density) as per the Higgs mechanism $$\phi \rightarrow \phi_0 + \tilde{\phi}$$. This results in the "would-be" Goldstone boson to be eaten up by the gauge field $$\alpha_\mu$$ giving mass $$m$$ to the photon. Because the photon is massive, excitations (creating photons) are now gapped as they cost an energy $$\propto m$$.

The massive photon $$\alpha_\mu$$ has $$2$$ degrees of freedom (two polarisations), end of story.

• $$\mu >0$$:

In this case, the potential $$V(\phi)$$ has a global minimum at $$\phi=0$$ meaning the gauge $$U(1)$$ symmetry is not broken.

Hence, the photon $$\alpha_\mu$$ remains massless, resulting in an additional redundancy (on top of the gauge condition) which reduces the degrees of freedom from $$2$$ to $$1$$.

So your photon $$\alpha_\mu$$ can be described by a scalar field $$\sigma$$, known as the dual photon.

Let's treat this as a scalar field like we treated $$\phi$$ earlier. Its "potential" $$V(\sigma) = 0$$, and again the ground state of $$\sigma$$ corresponds to a specific choice on this potential.

The original topological $$U(1)$$ symmetry has become the freedom to translate $$\sigma \rightarrow \sigma + \mathrm{const}$$, technically $$\mathrm{mod}\,2\pi$$ as per the Dirac quantisation condition.

Breaking this topological $$U(1)$$ symmetry corresponds to choosing a specific value for $$\sigma$$.