Superconductivity: difference between s-wave and d-wave

In summary, the difference between s-wave and d-wave superconductivity lies in the symmetry of the order parameter. The "s-wave" is spherically symmetric while the "d-wave" has 4 lobes with alternating phase. This affects the electrical transport in a superconductor and can be seen in Josephson junctions. The gap and the order parameter are related, and the d-wave state is not identical to the quantum mechanical treatment of angular momentum. The exact mechanism behind d-wave superconductivity is still unknown.
  • #1
tramar
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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.
 
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  • #2
tramar said:
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.

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.
 
  • #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?
 
  • #4
tramar said:
So how exactly does this symmetry affect the conductivity?

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.

Does this mean that in s-wave symmetry the Cooper pairs travel on the s orbitals?

Note that the symmetry mentioned is in reciprocal space!

Zz.
 
  • #5
tramar said:
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?

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.
 
  • #6
ZapperZ said:
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.

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 wavefunction 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.
 
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  • #7
BANG! said:
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 wavefunction symmetries are always related (I'm not saying this is true, you tell me.)

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).

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.

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 believe 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.
 

Related to Superconductivity: difference between s-wave and d-wave

1. What is the difference between s-wave and d-wave superconductivity?

S-wave and d-wave superconductivity refer to the different types of superconducting pairing symmetry in a material. In s-wave superconductors, the electrons are paired with opposite spin, while in d-wave superconductors, the electrons are paired with alternating spin. This results in different behaviors and properties of the superconducting state.

2. How do s-wave and d-wave superconductors differ in terms of critical temperature?

The critical temperature, also known as Tc, is the temperature at which a material becomes a superconductor. S-wave superconductors tend to have a higher Tc compared to d-wave superconductors. This is because the s-wave pairing symmetry allows for stronger electron pairing and thus a higher transition temperature.

3. Can you explain the Meissner effect in s-wave and d-wave superconductors?

The Meissner effect is the expulsion of magnetic fields from the interior of a superconductor. In s-wave superconductors, the Meissner effect is strong and the magnetic field is completely excluded from the material. However, in d-wave superconductors, the Meissner effect is weaker and some magnetic field can penetrate the material.

4. How do s-wave and d-wave superconductors differ in terms of their energy gap?

The energy gap is the amount of energy required to break apart a pair of electrons in the superconducting state. In s-wave superconductors, the energy gap is constant throughout the material. In d-wave superconductors, the energy gap varies in different directions, leading to an anisotropic behavior.

5. Are there any practical applications for d-wave superconductors?

While s-wave superconductors have been more widely used in practical applications, d-wave superconductors have potential uses in high-temperature superconducting devices such as quantum computers and magnetic resonance imaging machines. They also have potential for use in high-speed data transmission and energy storage systems.

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