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In the beginning of subparagraph about superconductors (which corresponds to paragraph about spontaneously symmetry breaking) Weingberg states that in superconductors EM gauge invariance is spontaneously broken. He then says that fermion field function corresponds to electron in superconductor may be written as $$ \tag 1 \psi_{n} = e^{iq_{n}\varphi(x)/\hbar}\tilde{\psi}_{n}, $$ where Goldstone boson $\varphi (x)$ which parametrizes subspace $U(1)/Z_{2}$ is identified with $\varphi (x) + \frac{\pi \hbar}{e}$.

The question: I don't understand how do we break $U(1)$ symmetry by writing $(1)$, and how $\varphi$ paranetrizes subspace $U(1)/Z_{2}$ only because $\varphi (x) "=" \varphi + \frac{\pi \hbar }{e}$.

Edit. It seems that the answer is following. By rewriting field $\psi$ through $\psi = e^{i\kappa (x)}\tilde{\psi}$ we remove Goldstone boson degrees of freedom from it. For doing this, we must set $\kappa (x)$ as transformation which leaves $\tilde{\psi}$ orthogonal to massless states vector. Our symmetry $U(1)$ is broken to $Z_{2}$, so transformation $\kappa$ corresponds to $Z_{2}$ group.

Name YYY
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    My naive understanding is that the restriction of the transformation $\varphi\rightarrow\varphi+n\pi$ is sufficient to restrict the symmetry to $U\left(1\right)/\mathbb{Z}_{2}$. For a $U\left(1\right)$ gauge transformation, any $\chi$ such that $\varphi\rightarrow\varphi+\chi$ would be allowed. The restriction $\chi=n\pi$ is the breaking. In short, writing your eq.(1) alone is not sufficient, but the restriction of $\varphi$ is the key point. Perhaps you might find http://physics.stackexchange.com/questions/133780 of interest for you, as well as http://arxiv.org/abs/cond-mat/0503400 – FraSchelle Feb 13 '15 at 06:04
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    Ah, for your second question: the fact that, from the full allowed circle (say $U\left(1\right)$) you identified two points $n=\pm 1$ gives you the factor group $U\left(1\right)/\mathbb{Z}_{2}$. – FraSchelle Feb 13 '15 at 06:08
  • @FraSchelle : thank you. I have one more question on you first comment: what are the restrictions on $\tilde{\psi}{n}$ from $(1)$ under $U(1)$ transformations? I.e., if under $U(1)$ transformation $\psi{n}$ transforms as $$ \psi_{n} \to e^{iq_{n}\kappa (x)\hbar}, $$ which transformation law will correspond to $\tilde {\psi}_{n}$? – Name YYY Feb 13 '15 at 13:04
  • I'm not sure I understand your last point. I'd say $\psi$ is not constrained. Of course $\left|\psi\right|^{2}=\psi^{2}=1$ if it's a wave function, but otherwise I do not see any further restriction. A gauge transformation in principle does not affect the modulus of the wave-function, only its phase. – FraSchelle Feb 13 '15 at 14:29
  • Ah, ok, I opened the Weinberg's book to be sure (your point is discussed in section 21.6, I think there is no re-edition of this book). I was naively thinking that $\tilde{\psi}$ was the modulus of the wave-function. But it's not. Weinberg introduces the Goldstone mode $\phi$ as you did for $\varphi$ (his eq.(21.6.3)), then he says "All the $\tilde{\psi}$ are gauge-invariant, and so when integrated out leave the Lagrangien as a gauge-invariant functional of $\phi$ and $A_{\mu}$ alone." So $\tilde{\psi}$ is not really an amplitude, it's a complex object by itself. (...) – FraSchelle Feb 13 '15 at 14:35
  • (...) In Weinberg's idea (however I could try to think like him :-), I guess it means you could do $\tilde{\psi}\rightarrow\tilde{\psi}e^{i\theta}$ and that $\theta$ is not constrained. In principle he is right, since you can do as many gauge transformations as you want, moreover it allows to write (21.6.5), with the introduction of $L_{s}\left[A_{\mu}-\partial_{\mu}\phi\right]$, which is all you need to exhibit the Higgs-physics for superconductors: Meissner, fluxoid and Josephson as he does later on in this chapter, see e.g. (21.6.22) and following. – FraSchelle Feb 13 '15 at 14:49
  • (...) As far as I can see, the important point in all this discussion is that the Goldstone mode $\phi$ becomes a dynamical variable like the vector-potential, and the only way it can enters dynamics is as $A_{\mu}-\partial_{\mu}\phi$. I'd say a bit crudely that the replacement $A_{\mu}\rightarrow A_{\mu}-\partial_{\mu}\phi$ is automatic for condensed-matter physicists once you accept the Ginzburg-Landau functional. Weinberg gives the complete argument for high-energy physicists, in order to squeeze the philosophical question: is Ginzburg-landau a valid functional ? – FraSchelle Feb 13 '15 at 14:57

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