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Theory questions on superconductivity

  1. May 12, 2013 #1
    Hi just completed a chapter on Superconductivity in a solid-state physics book. I have a few remaining questions. I don't expect they can be easily answered, but I appreciate if someone happens to know. Most of the questions regard how the equations have been motivated. I guess that's kind of nitpicky but I like to understand the history of it.

    (1) In the Cooper pair interaction, why is the el-el interaction ignored for all el- states with energies much greater than the typical phonon (where el- energy is measured relative to the Fermi level). I understand two el- states whose energy difference is greater than a typical phonon won't interact directly. But that doesn't make a statement about their absolute energies. Would I be right to say this is because we don't expect the bound Cooper state to contain these states in the solution and so it's mathematically simpler to just zero the potential for the interaction out at higher energies?

    (2) Is there a better way to motivate the BCS variational form? This form is Prod[ uk + vk c_(-k)* c_k ]|0> in standard notation. Since a Cooper pair is actually linear superposition of c_(-k)*c_k states, this doesn't quite match as a "filling of Cooper pairs." Can the BCS form be derived from a Slater determinant of Cooper pairs? Perhaps all of them in the same eigenstate (since they colloquially behave like bosons).

    (3) The most characteristic property of a superconductor is the zero resistance. The explanation of this is that "Changing the momentum of a single pair with respect to the common momentum requires a cost in energy equal to the binding energy. As soon as the number of pairs with random momenta increases, the energy penalty to sustain the situation becomes prohibitively large: any scattering process, that occasionally breaks or restores Cooper pairs, tends to restore the situation in which the pairs have common momentum." Is this supposed to be obvious? I just don't see why it becomes increasingly hard to break Cooper pairs as you continue to break them. The first one to break cost the binding energy. Why not also the last one?

    (4) In Ginzburg-Lindau, is psi supposed to represent the wave-function of a Cooper pair (as in all Cooper pairs condense into the same state)? If not, where did they come up with the idea of a |(p-eA)psi|^2 term in the free energy?

    (5) In the Josephson effect, why do begin with the assumption that the order parameter is constant in the superconductor and decays in the insulator? Isn't it the other way around when we consider the Meissner effect (i.e. boundary condition psi = 0 at surface and rises to bulk value in superconductor).

    Thanks,
    Sam
     
  2. jcsd
  3. May 12, 2013 #2

    mfb

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    I think it is the Meissner effect. Zero resistance follows from the Meissner effect, but the opposite direction is not true.

    I think that is related to entropy: if you have more and more unoccupied bound states, any random process is more and more likely to fill them.

    I have no idea about the other questions.
     
  4. May 12, 2013 #3
    I suppose you're right.

    I see it claimed that the resistance is exactly zero (or close enough that it would exceed the lifetime of the universe to see current decay). It might be difficult to get that out of an entropic argument about occupancies, which might be made in other contexts as well. Somehow disturbing one pair requires affecting the energies of all the pairs. The net energy change is then too large to be effected by a phonon transfer etc. Or something like that.
     
  5. May 13, 2013 #4

    DrDu

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    That's quite a list of good questions and I can only comment on some of them:
    1. I don't quite understand the first question. One reason we don't consider the scattering of electrons outside a shell of width of the Debye frequency is that this is very unlikely at the low temperatures where superconductivity is observed.
    2. You can in deed work with an antisymmetrized product of Cooper pairs and it is easy to obtain it from the ansatz of BCS. See e.g. Schrieffers book "Superconductivity".
    3. AFAIK, the zero resistivity cannot be reduced to the discussion of the Meissner effect as the latter only concerns the reaction to magnetic fields. The point with breaking up of Cooper pairs is the following: Let's assume you have a condensate carrying momentum, then the states paired have momenta k+s and -k+s. Now scattering such a pair into states with momenta k and -k means generating an excited state unless you create so many of them that they form a new condensate carrying no current. But this would require to break up almost all of the original superconducting state, which is extremely improbable and costs a lot of energy.
     
  6. May 13, 2013 #5

    marcusl

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    The physical picture resembles the classical E&M picture of evanescent waves decaying exponentially from a waveguide (optical fiber, say) into the surrounding medium. The point is made especially clearly in a Josephson junction made by forming a constriction in a superconducting wire. The critical current in the link is observed to be higher than that in a long superconductor of same material having the link's cross-section. This is because spillage of the bulk wave functions into the short link from each side helps the link support a high current.

    The Meissner effect is different because the external magnetic field effectively exerts a pressure on the SC, therefore penetrating inwards a short distance.
     
  7. May 13, 2013 #6

    mfb

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    The entropy argument gives some Cooper pairs, which then lead to zero resistance up to some current limit. This current limit decreases with increasing temperature...
     
  8. May 14, 2013 #7

    DrDu

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    Where did you find these boundary conditions?
     
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