# Infinite-fermion condensed $\langle \psi^n \rangle$ superconductor as $U(1)_f$ QED

I find it puzzling to consider the relation between (1) and (2).

(1) The superconductor obtained from an infinite number of fermion condensation $\langle \psi^n \rangle$ with a $n \to \infty$ with a fermion $Z_n$ parity symmetry. The $n$ fermion condense superconductor has an order parameter $$<ψ^n> = <ψ ψ ψ .. ψ> \neq 0.$$ This means the vacuum (or ground state) of this universe (as a superconductor) can have $ψ^n$ $n$-fermions poping up from vacuum freely.

Basically it is induced by Higgs-Yukawa coupling term to as $ψ^n \Phi + U(\Phi)$ where $U(\Phi)$ gives a Higgs potential. There is a global symmetry $$ψ \to e^{i \frac{2 \pi j}{n}} ψ$$ where $j \in \mathbb{Z}_n$. So there are many $n$ fermions create and annihilate from vacuum due to this term $$<ψ^n> \Phi + <(ψ^\dagger)^n> \Phi^\dagger.$$

(2) QED (Quantum electrodynamics) with a fermion number conservation global symmetry $U(1)_f$. QED has this global symmetry $$ψ \to e^{i \theta} ψ$$ where the gauge field $A$ can also transform. The $U(1)_f$ global symmetry is part of the $U(1)$ gauge symmetry of QED.

Now, the fact that when $n=2$ we have a usual $Z_2$ Bardeen Cooper Schrieffer superconductor, that we make

$U(1)_f$ breaks down to $Z_{2f}$,

which sends the $U(1)$ symmetry ($ψ$ to $e^{i \theta} ψ$) to $Z_{2 f}$ symmetry ($ψ \to - ψ$) is OK.

However, when we have $n \to$ infinity, it looks to me that we can almost restore

$Z_{n f}$ to $U(1)_f$ for large $n$.

Question: Isn’t that $<ψ^n> = <ψ ψ ψ .. ψ>$ condensation relation to a $U(1)$ QED theory puzzling?

I mean of course, we know $U(1)_f$ is not exactly the same as $Z_{nf}$. But they are similar in some way when $n \to \infty$.

How would you understand their relations and comment these? [I suspect that there could be some topological differences between the twos.]

Note Add: Here the $<ψ^n> = <ψ \cdot dψ \cdot d^2ψ ..\cdot d^{\#} ψ>$ may involve higher derivative terms, in order to have the Grassman valued fields $ψ$ nonzero in $<ψ^n>$. For $n=2$, we only need the usual Cooper pairing coupling.

Note Add 2: Here the fermionic condensate $<ψ^n>$ may be also adjusted to bosonic condensate $<\varphi^n>$. For $n=2$, we only need the usual two boson $\varphi$ pair condensed. We don't need higher derivative terms for boson because $\varphi$ is not Grassman valued.

• I'm not sure I understand your gauge transform. To me, $j$ can be in N, not necessarily in Zn. If I'm correct your questions and concerns fade out, as far as I can see. – FraSchelle Sep 13 '17 at 23:42