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In Schrieffer's book "Theory of Superconductivity", there is said when he deals with multiple connected superconductors (and discuss London equations), that if one takes the line integral of the potential vector on the boundary of a hole where there is a magnetic flux inside we have : $$\int{\vec dl\cdot \vec A} = \Phi$$ Then, if we go inside the superconductor ($\vec B=0$ thus $\vec A = \nabla \chi$ where $\chi$ is a scalar function) and again take the line integral of $\vec A$, we deduce that $\chi$ must be multi-valued : $$\int{\vec dl\cdot \vec A} = \Delta \chi = \Phi$$ But it is stated that now, if the magnetic flux is known in each hole, $\vec A$ is determined and the physics is saved. But for me, it is not clear that if we know how much $\chi$ raises with every $2 \pi$ rotation around a hole, we don't know how it varies...

Could someone help me on this ?

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Indeed, the only thing you can say is that the phase varies by the magnetic flux quantity after one loop. That's all.

Since the phase must be periodic, you also know that the flux inside a superconducting loop is quantised in integer number of the flux quantum $\Phi_{0}=\pi \hbar/e$.

You don't need to know more, since only a phase-difference matters. You can make the phase evolving the way you want along the loop, the only constraint is that after one turn you have $\Delta \chi = \chi\left(2\pi\right) - \chi\left(0\right) = 2\pi n \Phi / \Phi_{0}$, with $n$ some integer. Physical effects are associated with $\Delta \chi$, never with $\chi$ only.

In quantum mechanics, it is the same as saying that the wave function $\left\vert\Psi\right\rangle$ is unphysical, but only $\left\vert\left\langle x \vert\Psi \right\rangle\right\vert^{2}$ can be measured (for instance, you can replace $\left\langle x\right|$ by whatever bra you prefer).

In classical physics, this is the same thing as for a voltage drop: you don't need to know the voltage at any point, you just need to know that at the two end of a wire, you have a non-zero voltage drop.

So yes, superconductivity is a quantum mechanics problem at the macroscopic scale.

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It is incorrect to say that $A=\nabla \chi$ everywhere inside the superconducting ring as it is not a simply connected region. You could take two different $\chi_i$'s in two simply connected regions such the hole is not in any one of them.

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