What happens if one adds magnetic flux quanta into a Byers-Yang type system (http://en.wikipedia.org/wiki/Byers-Yang_theorem)?
The wavefunction does not change according to the gauge transformation described in the article.
But what happens to the hamiltonian of such system?
If I am not wrong, then substituting the A = A' + \nabla \chi vector potential into a Pauli equation will produce some new terms, which I think will change the energy...
The energy does change. The spectrum stays the same but if you turn on the magnetic field it creates electric field and excites the particle to the first excited state. The ground state energy is the same with the additional flux quantum. However, the old ground state becomes the first excited state of the system with an additional flux.
Shouldn't the electric field appear only on a time variation of the vector potential?
I mean - the moment of introduction of the magnetic flux quantum - would make an electric field appear due to the change occurring.
But as far as I understand - that electric field should vanish once the MFlux quantum is "stable" and present...
What makes me surprised in this is - should it be possible to have an ambiguity between energy states for one wavefunction?
To me it seems wrong.
But then either such ambiguity is indeed allowed, or there is no change in energy, and I can't explain the additional terms in the Hamiltonian.
Electric field appears only while you are varying the flux and performs work. This is the work required to lift your system from the ground state to the first excited state. When the flux is already there there is no electric field and the system is in the eigenstate of the Hamiltonian which is the first excited state.
This is one wave function but written for different Hamiltonians (or if you wish for different forms of the same Hamiltonian). The Hamiltonian without flux and the Hamiltonian with one flux quantum are related by a gauge transformation. The same transformation relates the eigenfunctions corresponding to the same energy. In particular, you have to identify \psi=1 and \psi =e^{i\phi}.
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