Consider a spontaneously broken scalar field theory with a global ${\rm U(1)}$ symmetry described by the Lagrangian $$\mathscr{L}=(\partial_\mu\phi^*)(\partial^\mu\phi)-\mathcal{V}(\phi),\\ \mathcal{V}(\phi)=\mu^2\phi^*\phi+\lambda(\phi^*\phi)^2$$ with $\lambda,-\mu^2>0$. The determination of classical vacua, in this case, amounts to the minimization of the potential. One finds that the vacua of this theory are infinitely degenerate.

Now consider a local ${\rm U(1)}$ gauge theory described by $$\mathscr{L}=(D_\mu\phi^*)(D^\mu\phi)-\mathcal{V}(\phi)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ also degenerate or unique? Here, $A_\mu$ is the gauge field and $D_\mu=\partial_\mu-ieA_\mu$. This Lagrangian is invariant under $$\phi(x)\to e^{-ie\theta(x)}\phi(x),\\A_\mu\to A_\mu+\partial_\mu\theta(x).$$

Is the vacuum of the local theory described above unique both classically and quantum mechanically? If so, what is the best and/or simplest way to understand it?

I must also say why am I interested in this question. Elitzur's theorem points out that local symmetries cannot be spontaneously broken. Also we know that the necessary requirement of spontaneous symmetry breaking is the existence of a degenerate vacua. So I wonder whether local theories have unique vacuum (or is it that despite the vacuum being degenerate, local symmetries cannot be broken.). @DominicElse answer here gave me an impression that the the vacuum may be unique.


The presence of the gauge field changes nothing - you still have a global U(1) symmetry that acts solely on the scalar field (only spacetime-dependent U(1) symmetries affect the gauge field), so the vacuum is still degenerate under that symmetry.

  • $\begingroup$ Doesn't the fifth paragraph of the answer here by Dominic Else gives an impression that the ground state is unique? physics.stackexchange.com/questions/190416/… @ACuriousMind $\endgroup$ – SRS Jun 29 at 15:33
  • $\begingroup$ I am also confused. I thought that in the case where the symmetry is gauged the vacua is essentially not physically different because it is just a gauge transformation of the original vacua, unlike in the case of global symmetry where the states are physically distinct states related by U(1) symmetry. $\endgroup$ – nGlacTOwnS Jun 29 at 15:53
  • $\begingroup$ @SRS The global U(1) transformations are not gauge transformations (since they do nothing to the gauge field - if you completely fix the gauge, there is still this global U(1)), the global U(1) here is precisely analogous to the global $\mathbb{Z}_2$ in Dominic Else's answer. $\endgroup$ – ACuriousMind Jun 29 at 16:16
  • $\begingroup$ @nGlacTOwnS See my comment above - when I say that you still have a global symmetry that acts solely on the scalar field, I mean it. It's not a gauge symmetry. Note that this is special for U(1) - if you gauge a non-Abelian group, you do not retain the full non-Abelian group as a global symmetry group, but only its center. $\endgroup$ – ACuriousMind Jun 29 at 16:17
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    $\begingroup$ @SRS A valid vacuum configuration is not just $A=0$, but any $A$ that is pure gauge, i.e. $A=\mathrm{d}f$ for any function $f$, since all these $A$ have $F=0$. $\endgroup$ – ACuriousMind Oct 16 at 13:25

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