Take the BCS model of superconductivity. We have creation and annihilation operators for the electron, $\Psi/\Psi^\dagger$, and the order parameter for the phase transition from the metallic phase into the superconducting phase is given by the VEV $\chi=g\langle\Psi\Psi\rangle$, where $g$ is the electron-phonon coupling constant.

We can create a Landau-Ginzburg model for this transition, with the dominant terms of the free energy expansion given by $$ F=\int\frac{1}{4m_e}|(\nabla+2ie\mathbf A)\chi|^2+u|\chi|^2+\frac v2|\chi|^4+\frac12\mathbf B^2+... $$

When we obtain the ground state via the variational principle, we get something like $$ (u+v|\chi|^2)\chi=0 $$ which implies that either $\chi=0$ (metallic phase), or $|\chi|^2=-\frac uv$ (superconducting phase). But in the latter phase, it looks like the vacuum is degenerate, with a Goldstone degree of freedom in the phase angle! Why then is there said to be only a single ground state in this theory?

  • $\begingroup$ In what sense do they say that there is a single ground state? Josephson effect is based on the difference in these phases just as the attraction/repulsion between magnets is dependent on the orientation of their polarization. $\endgroup$
    – Roger V.
    Commented Jul 22, 2021 at 8:51
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    $\begingroup$ This is a nice question - the answer is that you have discounted the electromagnetic field in your analysis. If you take this into account, you will see that all of these ostensibly different ground states are really all gauge equivalent to each other. $\endgroup$ Commented Jul 22, 2021 at 11:32
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    $\begingroup$ @NiharKarve This is not a comment but an answer. $\endgroup$
    – my2cts
    Commented Jul 22, 2021 at 12:54

1 Answer 1


Symmetry is spontaneously broken in superconductors in essentially all of the usual ways, and is (almost) identical to the symmetry breaking in superfluids.

The gauge freedom that exists between the phase of the scalar field ($\chi$ in your question) and the longitudinal component of the photon field $\bf{A}$ is simply irrelevant for this question, because breaking of a global symmetry is about, well, the global part, and the global part of $\chi(x)$, so the uniform part at zero wavenumber $\chi(k=0)$, does not change under gauge transformations.

The main difference between superconductors and superfluids, is that, due to the coupling to the photon field, one cannot define an order parameter (like $\chi(x)$) that is both local and gauge-invariant. The solutions to this are:

  1. Use a local but gauge non-invariant order parameter. This is what BCS did.
  2. Use a gauge-invariant but non-local order parameter.
  3. Employ a gauge-fix explicitly, which can lead to 1.

While the problems with 1. were quite some issue in the early years after BCS, the question is entirely settled by the Josephson effect, which could be characterized as the hallmark for (spontaneously) broken symmetry. Nevertheless, there have been some recent published claims that this is not the case, in particular:

These claims are, respectfully and in my opinion, simply wrong.

Note that, at finite but large volume, almost all systems with spontaneous symmetry breaking have a unique groundstate, which can be pictured as a superposition of all the symmetry-broken states. This state is however extremely unstable, while the states with broken symmetry are stable, even though they are not energy eigenstates. These do become degenerate at infinite volume.

  • $\begingroup$ Something that seems commonly overlooked: the global part can be obtained a a sum of local parts, and hence if the latter is an unphysical gauge transformation, so is the former! Accordingly, I am always surprised and confused when people say that the global (sub)group is physical and not 'gauge'. I do not see how that can be a mathematically self-consistent statement. $\endgroup$ Commented Jul 23, 2021 at 3:45
  • $\begingroup$ @Ruben Verresen Please show me the gauge transformation of the global part $\chi(k=0)$ and I will delete this answer. Note that the global part of $\chi(x) \to \chi(x) + \lambda(x)$ is not a gauge transformation since it doesn't affect ${\bf A}(x)$. This is a global symmetry transformation just as it is for a neutral superfluid, related to the very physical and very gauge-invariant Noether current. $\endgroup$ Commented Jul 23, 2021 at 3:54

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