Why pesistent current in a normal metal ring is a surprise?

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wdlang
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i cannot understand why persistent current in a normal metal ring threaded by a magnetic field is a surprise.

the hamiltonian is

[tex]H=\frac{1}{2I}\left(-i \frac{\partial}{\partial \theta}-A\right)^2[/tex]

and the eigenstates are

[tex]\phi_m(\theta)=\frac{1}{\sqrt{2\pi}} e^{i m \theta}[/tex]

with eigenvalues

[tex]E_m=\frac{1}{2I}(m-A)^2[/tex].

It is ready to see that generally every eigenstate carries a current, a persistent one.

so why people think it is a surprise?
 
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ZapperZ said:
Try doing it, and see if what you think you understand matches reality!

Zz.

of course, in reality, the situation is more complicated, e.g., the potential is not uniform, there is decoherence.

but i think the basic idea is just too simple.
 
In a many particle state, the contributions of the states with different m nearly cancel and there is only a very tiny fraction of that effect that survives. It forms the basis of diamagnetism.