Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Superconductivity: difference between s-wave and d-wave

  1. Dec 16, 2009 #1
    I'm very new to superconductivity and I've tried searching for the difference between s-wave and d-wave superconductivity but to no avail. I find it's often mentioned but never explained. I assume it's fairly basic, but if anyone has an explanation or some references I can check, it would be much appreciated.
     
  2. jcsd
  3. Dec 16, 2009 #2

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    It would help if you explain a little bit more on what you have read, and what exactly is the part that you don't understand.

    Let's start first of all with the most obvious difference - the symmetry. The standard "s-wave" is spherically symmetric (there are other s-wave symmetry that aren't, but the s-wave order parameter as used in the standard BCS model is spherically symmetric). The "d-wave" being mentioned with respect to the cuprate superconductors is the [itex]d_{x^2 - y^}[/itex] symmetry. It has 4 lobes with alternating phase.

    So right off the bat, you can see to major differences: (i) the "geometry" of the order parameter and (ii) the fact that the d-wave gap has alternating phase sign, while the s-wave gap has only a single phase sign.

    Zz.
     
  4. Dec 16, 2009 #3
    So how exactly does this symmetry affect the conductivity? Does this mean that in s-wave symmetry the Cooper pairs travel on the s orbitals?
     
  5. Dec 16, 2009 #4

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    It means that it may have a preferred direction of electrical transport. But this is a naive answer because charge transport depends on a whole lot of other factors beyond the order parameter.

    Note that the symmetry mentioned is in reciprocal space!

    Zz.
     
  6. Dec 16, 2009 #5

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    Just to emphasise what ZapperZ said, d- and s- refers to the symmetry of the superconducting wavefunction (well, the order parameter), it has nothing to do with atomic orbitals if that is what you are referring to. However, it is worth pointing out that although we are in k-space there IS obviously a connection to real space as well, in e.g. YBCO the + and - lobes are in the ab-plane of the crystal.

    The d-wave symmetry affects the transport in many different ways, although the effect in a homogeneous bulk conductor are actually quite subtle. However, if make Josephson junctions or SQUIDs out of a d-wave superconductor and orient the electrodes in such as way that you have transport from e.g. a node to a lobe or from a +lobe to a -lobe (known as a pi-junction since this gives an intrinsic pi-shift of the phase) the effects become much more prominent.
     
  7. Dec 16, 2009 #6
    I have a question concerning this. First, when talking about s-wave versus d-wave superconductivity, is this describing the symmetry of the order parameter. Also, some people insist that the "order parameter" is about the symmetry of the gap, while other people refer to it as the symmetry of the cooper pair wave function. Which is it? Or are both stances ok because the gap and wavefucntion symmetries are always related (I'm not saying this is true, you tell me.) Second, are these s-wave and d-wave states identical to those derived from the quantum mechanical treatment of angular momentum. Specifically, does angular momentum and the superconducting Hamiltonian commute so that you can form eigenstates for the SC wavefn from the spherical harmonics.
     
  8. Dec 16, 2009 #7

    f95toli

    User Avatar
    Science Advisor
    Gold Member

    Yes, the gap and the order parameter are related. If you measure the superconducting gap along different axis of the crystal you will see that it follows the symmetry. A good example would be a Josephson junction in a node-lobe arrangement, this has to first order zero critical current because there is no gap on the node side (although it can still carry a "higher order" current, mediated via the Andreev states that form at the interface, but that is a different story).

    Remember that all d-wave superconductors are high temperature superconductors, we don't understand the mechanism that gives rise to HTS so we don't know what the Hamiltonian looks like. At the moment it is probably best to think about "d" simply as a name of the symmetry, that just happens to have the same shape as l=2. Also, it is possible that there are admixtures of other symmetries in there are as well, meaning the total symmetry might be d+is or d+s, i.e. with small s-wave components. This was a hot topic a few years ago when some experiment claimed to see an s-component (although most people now belive it to be pure d)

    That said, there are LTS p-wave superconductors where we do know the Hamiltonian, but I don't remember what the BCS Hamiltonian looks like for them.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook