I read in a book "Physics - Resnik and Halliday" the explanation of Type-I Superconductors {cold ones} that:

The Electrons that make up current at super-cool temperatures move in coordinated pairs. One of the electrons in a pair may electrically distort the molecular structure of the superconducting material as it moves through, creating nearby short-lived regions of positive charge.the other electron in the the pair may be attracted to the positive spot. According to the theory the coordination would prevent them from colliding with the molecules of the material and thus would eliminate electrical resistance

Is this the only explanation or can somebody give me a more intuitive explanation that also takes into the problem of defect scattering as in the case of resistance and also explains the Type-II superconductors {hot ones}

P.S. What are "coordinated pairs"?

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    $\begingroup$ en.wikipedia.org/wiki/Cooper_pair $\endgroup$ May 21, 2012 at 10:12
  • $\begingroup$ @Hans-PeterE.Kristiansen i know there's an article on Wikipedia i want an explanation that is understandable for a high-school student $\endgroup$ May 21, 2012 at 10:15
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    $\begingroup$ Sorry - It really is not a simple topic. There is not manny shortcuts. Study hard for 10 years, and maybe you will understand some of it. No one in the world understand superconductivity 100%. $\endgroup$ May 21, 2012 at 10:29

2 Answers 2


When you scatter an electron you change it's energy. So if it wasn't possible to change the energy of an electron you couldn't scatter it. This is basically what happens in superconductors.

In a metal at room temperature the electrons have a continuous range of energies. This means if I want to change the energy of an electron by 0.001eV, or even 0.000000001eV there's no problem doing this. This is a big difference from an isolated atom, where the electrons occupy discrete separated energy levels. If you try to change the energy of the electron in a hydrogen atom by 0.001eV you can't do it. You have to supply enough energy to make it jump across the energy gap to the next energy level.

In superconductors the correlation between the electrons effectively turns them into bosons and they all fall into the lowest energy state. However the correlation also opens a gap between the energy of this lowest state and the energy of the next state up. This is why defects in the solid can't scatter electrons, and why they conduct with no resistance.

For an electron to scatter off a defect (or anything else) in the conductor you have to supply enough energy to jump across the gap between the lowest energy level and the next one up. However the energy available isn't great enough for this, and this means the defects can't scatter the electrons and that's why they superconduct.

The trouble is you're now going to ask for an intuitive description of why the electron correlations open a gap in the energy spectrum, and I can't think of any way to give you such a description. Sorry :-(

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    $\begingroup$ Hi, nitpicking again (sort of a compliment, i.e. I follow your answers). One has to distinguish between elastic and inelastic scattering. The latter one results in a change of energy, related to dissipation related to resistance. But in the fundamental way the BCS ground state is formulated in terms of (elastic) scattering. $\endgroup$ Apr 10, 2015 at 12:39

That description is very misleading and is something that seems to be taught everywhere. It doesn't seem to make sense, as with the electrons being so close together you would expect a large repulsion. For an attractive force to exist, the lattice would have to screen the charge very quickly.

A much better picture emphasizes the electron phonon interaction. With the lattice being deformed slightly by one electron which then provides a slight attractive force that can be felt by another electron. In this image, the electrons don't have to be physically close to each other - they can be within a few hundred nanometers of each other. The only requirement is that they are close to each other in k space.

PS - I'll add more to this later.

  • $\begingroup$ wait a sec i havent written the description word to word ill edit it........ $\endgroup$ May 21, 2012 at 10:07
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    $\begingroup$ Professor Ronald Griessen illustrates this nicely in the lecture materials available on his website nat.vu.nl/~griessen. I've put the illustration here for convenience: i.imgur.com/PcNUbei.png $\endgroup$
    – user129412
    Apr 28, 2017 at 13:36

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