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

  1. Sep 29, 2010 #1
    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?
     
  2. jcsd
  3. Sep 29, 2010 #2

    ZapperZ

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    Try doing it, and see if what you think you understand matches reality!

    Zz.
     
  4. Sep 29, 2010 #3
    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.
     
  5. Sep 29, 2010 #4

    ZapperZ

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    We shouldn't be surprised with quantum tunneling and quantum entanglement either, because the basic ideas are also very simple. Yet, we still do!

    Zz.
     
  6. Sep 30, 2010 #5

    DrDu

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    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.
     
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