When water freezes continuous translational symmetry is broken. When a metal becomes superconducting, what is the symmetry that gets broken?


1 Answer 1


In most of the textbooks discussing this point, you should find something like : superconductors breaks the U(1)-gauge symmetry down to $\mathbb{Z}_{2}$. Fine, but what does it mean ?

To explain it, let me be a bit outside the main stream discussion. What I'll discuss below is more a personal reflexion than something clearly stated in any book.

Clearly, the origin of superconductivity -- as explained by Bardeen, Cooper and Schrieffer (BCS) -- is the instability of the Fermi surface due to the Cooper pairing. So the first question to ask is: why is the Fermi surface stable? You can find more details in the previous link, so let me give you the ultimate answer: the Fermi surface is stable since it is a topological concept. In short, the Fermi surface can be defined as a quantity which is not perturbed by some interactions. You can add impurities in your solid and/or various interactions between your electrons, the Fermi surface will not be so much deformed. Of course, an other arrangement of atoms in the solid gives an other Fermi surface, but the stability of this new one is still verified.

Over the years, this concept of the stability of the Fermi surface has been refined, down to the work by Horava, reproduced in the book by Volovik. There you will see the topological invariant responsible for the stability of the Fermi surface (chapter 8), and the reason why it is a U(1) stability (well, it has to be Abelian for simple reasons you can guess easily, and the Fermi surface has a volume in the energy-space, so it can be reduced to a circle).

The point is: the Fermi surface is stable with respect to almost all interactions, with the exception of the Cooper pairing. The reason is simple to understand: most of the interaction conserve the number of particles, but the Cooper pairing transmute the particles. In short, the vacuum of the Cooper pair is no more a Fermi gas/liquid, but a Bose gas/liquid. Then the volume of the Fermi surface (i.e. the number of fermions) is no more conserved. In other words, the topological protection ensuring the stability of the Fermi surface is no more at work when the Cooper pairing enters the stage.

Now let us picturesquely understand why it is a $\text{U}\left(1\right)\rightarrow\mathbb{Z}_{2}$ breaking. The disappearance of the electrons at the Fermi surface creates a gap, reminiscent of the physics of semi-conductors. There, you know that there are 2-bands (conduction and valence). This is the first hint why only $\mathbb{Z}_{2}$. The second ingredient is that the Bose gas/liquid has no Fermi surface (tautology !), so it is not stable at all with respect to any interactions (re-tautology !), and so in principle the breaking should be from U(1) to nothing. But you still have two species of bosons: the hole-like and the particle-like, hence the doubled $\mathbb{Z}_{2}$ symmetry.

Of course, all the arguments above are sketchy, so a more precise definition is still warm welcome.

So, let us go back to the mainstream argument: the BCS-interaction reads $$H_{\text{BCS}}\sim c^{\dagger}\left(x\right)c^{\dagger}\left(x\right)c\left(x\right)c\left(x\right)$$ in a simplified form. To this Hamiltonian you can apply the transform $$c\left(x\right)\rightarrow e^{\mathbf{i}\varphi\left(x\right)}c\left(x\right)\;\;;\;\; c^{\dagger}\left(x\right)\rightarrow e^{-\mathbf{i}\varphi\left(x\right)}c^{\dagger}\left(x\right)$$ such that $H_{\text{int}}\rightarrow H_{\text{int}}$ and so $H_{\text{int}}$ is invariant with respect to a U(1)-gauge-transformation, since any real phase $\varphi$ is allowed and the group of multiplication by a phase $e^{\mathbf{i}\varphi\left(x\right)}$ is the group U(1).

The mean-field counterpart of $H_{\text{int}}$ in the Cooper channel reads $$\tilde{H}_{\text{BCS}}\sim\Delta\left(x\right)c^{\dagger}\left(x\right)c^{\dagger}\left(x\right)+\Delta^{\dagger}\left(x\right)c\left(x\right)c\left(x\right)$$ and so it is only invariant when we choose $\varphi\in\left\{ 0,\pi\right\} $ in the above gauge-transformation. So there is only two possibilities if one wants to change the phase of the operators and keeps the mean-field Hamiltonian invariant. The group with only two elements is called $\mathbb{Z}_{2}$. That's the microscopic origin of the $\text{U}\left(1\right)\rightarrow\mathbb{Z}_{2}$ gauge-symmetry breaking.

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    $\begingroup$ @huotuichang and FraSchelle I've moved the conversation to chat in case you'd like to continue. $\endgroup$
    – David Z
    Oct 16, 2014 at 16:26
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    $\begingroup$ @DavidZ why move to chat? The formula can not display well in the chat. $\endgroup$ Oct 17, 2014 at 13:35
  • $\begingroup$ @DavidZ Thanks, it was too long indeed :-) I think we should open a new question in case we need to continue :-) $\endgroup$
    – FraSchelle
    Oct 17, 2014 at 14:09
  • $\begingroup$ @luming There were no much equations actually, so that's perfectly fine for me :-) $\endgroup$
    – FraSchelle
    Oct 17, 2014 at 14:10
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    $\begingroup$ @FraSchelle, you argue that it is in the mean field limit, that the theory goes from $U(1)$ to $Z_2$ but if I treat the BCS interaction in full, your answer seems to indicate that there is no symmetry breaking. Shouldn't we talk about the symmetry breaking as in the ground state of the theory instead of comparing the two Hamiltonians? Is that what you mean to say by writing the exactly solvable mean field Hamiltonian?(actually the other one is also exactly solvable but let's not go into that) $\endgroup$
    – cleanplay
    Feb 8, 2017 at 4:45

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