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When I have read about literatures, the non-BCS type superconductor always involve some kind of pairing which the energy gap $\Delta$ is not a constant but a function of momentum $k$: $\Delta=\Delta(k)$, while the relation is always directly given without mention how to derive. Hence, I would like if there is anyone here helping me to understand how, say p-wave and d-wave, and $p_x+ip_y$ superconductor has such kind of energy gap?

My understanding is that the electron-phonon interaction is not simply a constant, but what is the precise electron-phonon interaction term in these kind of superconductors, and how this thing really matters the energy gap is unknown to me.

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  • $\begingroup$ The gap equation is $\Delta(k_0)=-\sum_k \frac{\Delta(k)}{E_k}V_{kk_0}$, where $E_k=\sqrt{\epsilon_k^2+\Delta(k)^2}$. While the interaction $V$ is assumed to be a constant in BCS theory, it is in general a function of momentum, resulting in a momentum-dependent gap. You may be interested in the review paper by Carbotte: Rev. Mod. Phys. 62, 1027 (1990). $\endgroup$
    – leongz
    Jul 12, 2016 at 1:13

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As you pointed out, the phonon-mediated BCS-type superconductors exhibit a gap $\Delta_0$ which is isotropic in $k$-space, we call it an s-wave gap. As @leongz pointed out, it comes from the fact that the electron-phonon interaction used in the BCS model does not depend a momentum ; inserting it into the gap equation gives a s-wave gap.

The precise microscopic mechanism giving rise to superconductivity in unconventional superconductors is still a subject of intense debate. Although many people suspect that electron-phonon interaction plays a role in pairing, it is highly likely that it is not the only important ingredient. Models involving antiferromagnetic correlations, spin density waves or nematicity among others are studied.

These interactions can have a more complicated structure than the electron-phonon in $k$-space as well as a frequency dependence. This can give rise to a $k$-dependent gap ; depending on the symmetries of the gap, we will call it different names. For instance a $p$-wave gap would be one that picks up minus sign under the transformation $k \rightarrow - k$, while a $d$-wave would be invariant under this transformation but pick up a minus under a $\frac{\pi}{2}$ rotation.

EDIT regarding comments below.

See this nice picture of atomic orbitals (increasing values of $l$ from top to bottom, different figures on the same line correspond to different values of m) :

enter image description here

You clearly see that the l=0 orbital is isotropic. The l=1 orbitals have the property that if you transform them with inversion symmetry ($x \mapsto -x$ for the first one, for instance) they pick up a minus sign. You also see that the $l=2$ orbitals can pick up a minus sign if rotated by $90$ degrees along a certain axis, but remain unchanged by inversion symmetry. By analogy with atomic orbitals, we will classify the superconducting gaps according to their behavior with respect to certain transformations in $k$-space.

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  • $\begingroup$ That informs a lot. BTW, did this name just due to the symmetry operation? I mean, yes of course that different relative orbital angular momentum $L=0, 1, 2$ has different symmetry group, but only from the symmetry rather than actual angular momentum consideration to give the name seems less convincing (to me). $\endgroup$
    – RoderickLee
    Jul 12, 2016 at 14:24
  • $\begingroup$ You mean names like s,p,d etc ? This terminology comes from atomic physics, more specifically from the usual shape of peaks the spectroscopists get from the l=0, l=1, l=2,l=3 peaks. s=sharp, p=principal, d=diffuse, f=fundamental. $\endgroup$
    – Dimitri
    Jul 12, 2016 at 14:32
  • $\begingroup$ Not this... I know why it called spdf, but I just wondering why we can from the symmetry to directly say it's $L=0, 1, 2$-wave pairing, not from real angular momentum model. (maybe I still haven't express my question clear...) $\endgroup$
    – RoderickLee
    Jul 12, 2016 at 15:06
  • $\begingroup$ See the edit to my post, hope it awnsers your question. $\endgroup$
    – Dimitri
    Jul 12, 2016 at 16:04
  • $\begingroup$ Thx for the explanation. $\endgroup$
    – RoderickLee
    Jul 13, 2016 at 15:01

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