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.

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    $\begingroup$ 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 physics.stackexchange.com/questions/133780 of interest for you, as well as arxiv.org/abs/cond-mat/0503400 $\endgroup$
    – FraSchelle
    Commented Feb 13, 2015 at 6:04
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    $\begingroup$ 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 $U\left(1\right)/\mathbb{Z}_{2}$. $\endgroup$
    – FraSchelle
    Commented Feb 13, 2015 at 6:08
  • $\begingroup$ @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 $\tilde {\psi}_{n}$? $\endgroup$
    – Name YYY
    Commented Feb 13, 2015 at 13:04
  • $\begingroup$ 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. $\endgroup$
    – FraSchelle
    Commented Feb 13, 2015 at 14:29
  • $\begingroup$ 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. (...) $\endgroup$
    – FraSchelle
    Commented Feb 13, 2015 at 14:35


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