How Is Non s-Wave Superconductivity Gap Parameter Chosen?

[nbo10]

In BCS you make the assumption that the effective electron-electron interaction is constant within a small shell around the fermi surface and zero otherwise. From this you get a constant spherical gap.

In non s-wave SC there is a specific form for the gap ie, $$\Delta_0 = [ \cos (k_x a) - \cos (k_y a)]$$. I know that you can calculate the gap parameter using group theory and the underlying symmetry of the crystal lattice. But is there a way to choose an arbitary form for $$k \cdot k^\prime$$, self consistently solve the gap equation and arrive at a d-wave gap or any other non s-wave gap?

How do you choose the effective electron-electron interaction? Is it based on the atomic orbitals that are believed to be responsible for SC? I've done a few searches and all I find is a assumption for the form of the gap. Thanks

 

[ZapperZ]

In BCS you make the assumption that the effective electron-electron interaction is constant within a small shell around the fermi surface and zero otherwise. From this you get a constant spherical gap.

In non s-wave SC there is a specific form for the gap ie, $$\Delta_0 = [ \cos (k_x a) - \cos (k_y a)]$$. I know that you can calculate the gap parameter using group theory and the underlying symmetry of the crystal lattice. But is there a way to choose an arbitary form for $$k \cdot k^\prime$$, self consistently solve the gap equation and arrive at a d-wave gap or any other non s-wave gap?

How do you choose the effective electron-electron interaction? Is it based on the atomic orbitals that are believed to be responsible for SC? I've done a few searches and all I find is a assumption for the form of the gap. Thanks

This isn't an easy question to answer because this is still something being worked on. Something like the t-J method using mean-field approximation can drop the d-wave symmetry onto your lap (this assertion is still controversial). In many instances, the symmetry is inserted by hand because that is the product of experimental observation. There's a persuasive reason why this is having that d-wave symmetry - the valence shell of the Cu in the CuO plane where superconductivity is thought to reside. The transition metals have d-orbitals valence shell.

Note that for the Ruthenates, you have a p-wave symmetry for the pair.

Zz.