# Distinguishing Insulator, Metal, Superconductor by a flux insertion argument

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@EverettYou I have the following argument to distinguish Insulator, Metal and Superconductor.

For simplicity let's consider electrons on a circle
and thread one quantum of flux (e$\Phi_0 = 2\pi$) through it (or make a "large" gauge transformation on the system). I will ignore the spin of the electrons. Then do the following arguments make sense ?

a) In an insulator the electrons are localized and they do not see the flux at all. So nothing happens in an insulator.

b) In metals electrons are nearly free with the dispersion relation $E(k_n) = k_n^2/(2m)$ .
As the flux is inserted adiabatically, each momentum level shifts and at the end of the process the spectrum is mapped to itself according to $k_n \to k_{n+1}$. The end results is a particle-hole excitation at the fermi surface.

c)In a superconductor the electrons are delocalized and most importantly each $k_n$ is paired with $-k_n$. The pairing will be lost if the electrons behave independently, because $k_n+ \alpha (\Phi)$ and $- k_n + \alpha(\Phi)$ have different energy if the dispersion relation is $E(k) = k^2/(2m)$, where $\alpha(\Phi) = \Phi/\Phi_0$ , $0 < \Phi < \Phi_0$ . However, if the dispersion relation is also changed according to something like $E(k, \alpha) = (k-\alpha)^2/(2m)$, then there will always be pairing about $k = \alpha$. This is equivalent to saying that instead of changing their states independently, the system of electrons starts to move as a whole, as we insert the flux.

Of course for the superconductor I already knew the final result (that if a flux is inserted inside the hole of a superconducting ring, a supercurrent starts to flow along the ring).
I worked out the argument backward and I have the feeling that I might have missed some subtle (or not so subtle) point.

It will be helpful for me to know whether the above arguments are correct.

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