In the derivation of flux quantization for a superconducting ring, we say that $$\oint_C \nabla\theta . dl = \theta_{2} - \theta_{1}$$ Then we equate this value to $2\pi n$. The reason cited in Kittel for this is: "The probability amplitude $\psi$ is measurable in the classical approximation, so that $\psi$ must be single valued."

My question here is that why should the phase be integral multiple of $2\pi$? I know that when in the parameter space we subject a quantum mechanical system to a cyclic adiabatic change, we get an extra geometric phase with the wavefunction if the Hamiltonian has more than 1 parameter associated with it.

Can someone explain me that in this case what are the parameters on which Hamiltonian depends on and convince me that there will be no geometric phase arising here.


The phase you are dealing with here is not thought of as coming from some adiabatic changes in a Hamiltonian. To understand what where it comes from gauge invariance of the Lagrangian $$ \mathcal{L} = i \Psi^{\dagger}\partial_0 \Psi - \frac{1}{2m}\nabla \Psi^{\dagger}\cdot \nabla \Psi - \frac{g}{2}(n-\Psi^{\dagger}\Psi)^2. $$ This is invariant under a global $U(1)$ symmetry i.e $ \Psi \longmapsto \Psi e^{i\theta} $ where $\theta \in \mathbb{R} $. This describes superfluids and can be used to understand superconductors. Now there is a procedure to make this invariant under a local $U(1)$ symmetry i.e $ \Psi \longmapsto \Psi e^{i\theta(x)} $. This is where the phase in your question comes from.

What about the quantization condition? It turns out that the current density of the superconductor is $ J \propto (\nabla \theta - qA) $ where $A $ is the vector potential. No if I want to find the current I must integrate this. In the absence of a magnetic field we would arrive at your integral. Now Kittel does not explain this well but the probability amplitude is related to your current and that is what you measure in the lab. Now since the current is completely classical, this enforces the single valuedness of our wave function. Also since the angle appears inside a complex exponential the condition needed is the one you cited.

In summary, this has nothing to do with adiabatic changes in the parameters of the Hamiltonian it has to do with the properties of the Lagrangian and the concept of spontaneous symmetry breaking.

  • $\begingroup$ I am unable to understand the origin of phase difference because of the two concepts: 1) gauge invariance of Lagrangian and 2) adiabatic changes in parameters of Hamiltonian. I think even the Berry phase can be associated to the gauge invariance of the Lagrangian. $\endgroup$
    – Draco_1125
    Sep 8 '17 at 4:34
  • $\begingroup$ @ssp_user5275 Now I am confused because the question in the comment is different from the question you asked. The simple answer to the original question is that the phase is related to a classical variable namely the current and there is not a Hamiltonian that you are adiabatically adjusting. The phase comes from the fact that the Lagrangian has a symmetry that the ground state does not. Berry phases come from the fact that the manifold on which the parameters vary is not flat so there is a non trivial holonomy. $\endgroup$
    – Amara
    Sep 11 '17 at 20:12

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