Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I often read about s-wave and p-wave superconductors. In particular a $p_x + i p_y$ superconductor - often mentioned in combination with topological superconductors.

I understand that the overall Cooper pair wavefunction may have orbital momentum = 0 (s-wave) or orbital momentum = 1 (p-wave) where the first one is spherically symmetric.

Now what does the splitting in a real ($p_x$) and imaginary ($p_y$) part mean? Why is it written in this form and why is that important (e.g. for zero Majorana modes) ?

share|improve this question

1 Answer 1

Symmetry of the superconducting gap

First of all, a bit of theory. Superconductivity appears due to the Cooper paring of two electrons, making non-trivial correlations between them in space. The correlation is widely known as the gap parameter $\Delta_{\alpha\beta}\left(\mathbf{k}\right)\propto\left\langle c_{\alpha}\left(\mathbf{k}\right)c_{\beta}\left(-\mathbf{k}\right)\right\rangle $ (the proportionality is merely a convention that will not matter for us) with $\alpha$ and $\beta$ the spin indices, $\mathbf{k}$ some wave vector, and $c$ the fermionic destruction operator. $\Delta$ corresponds to the order parameter associated to the general recipe of second order phase transition proposed by Landau. Physically, $\Delta$ is the energy gap at the Fermi energy created by the Fermi surface instability responsible for superconductivity.

Since it is a correlation function between two fermions, $\Delta$ has to verify the Pauli exclusion principle, which imposes that $\Delta_{\alpha\beta}\left(\mathbf{k}\right)=-\Delta_{\beta\alpha}\left(-\mathbf{k}\right)$. You can derive this property from the anti-commutation relation of the fermion operator and the definition of $\Delta_{\alpha\beta}\left(\mathbf{k}\right)$ if you wish. When there is no spin-orbit coupling, both the spin and the momentum are good quantum numbers (you need an infinite system for the second, but this is of no importance here), and one can separate $\Delta_{\alpha\beta}\left(\mathbf{k}\right)=\chi_{\alpha\beta}\Delta\left(\mathbf{k}\right)$ with $\chi_{\alpha \beta}$ a spinor matrix and $\Delta\left(\mathbf{k}\right)$ a function. Then, two possibilities

  • $\chi_{\alpha\beta}=-\chi_{\beta\alpha}\Leftrightarrow\Delta\left(\mathbf{k}\right)=\Delta\left(-\mathbf{k}\right)$ this situation is referred as the spin-singlet pairing

  • $\chi_{\alpha\beta}=\chi_{\beta\alpha}\Leftrightarrow\Delta\left(\mathbf{k}\right)=-\Delta\left(-\mathbf{k}\right)$ this situation is referred as the spin-triplet pairing.

Singlet includes $s$-wave, $d$-wave, ... terms, triplet includes the famous $p$-wave superconductivity (among others, like $f$-wave, ...).

Since the normal situation (say, the historical BCS one) was for singlet pairing, and because only the second Pauli $\sigma_{2}$ matrix is antisymmetric, one conventionally writes the order parameter as $$ \Delta_{\alpha\beta}\left(\mathbf{k}\right)=\left[\Delta_{0}\left(\mathbf{k}\right)+\mathbf{d}\left(\mathbf{k}\right)\boldsymbol{\cdot\sigma}\right]\left(\mathbf{i}\sigma_{2}\right)_{\alpha\beta} $$ where $\Delta_{0}\left(\mathbf{k}\right)=\Delta_{0}\left(-\mathbf{k}\right)$ encodes the singlet component of $\Delta_{\alpha\beta}\left(\mathbf{k}\right)$ and $\mathbf{d}\left(\mathbf{k}\right)=-\mathbf{d}\left(-\mathbf{k}\right)$ is a vector encoding the triplet state.

Now the main important point: what is the exact $\mathbf{k}$-dependency of $\Delta_{0}$ or $\mathbf{d}$ ? This is a highly non-trivial question, to some extend still unanswered. There is a common consensus supposing that the symmetry of the lattice plays a central role for this question. I highly encourage you to open the book by Mineev and Samokhin (1998), Introduction to unconventional superconductivity, Gordon and Breach Science Publishers, to have a better idea about that point.

The $p_{x}+\mathbf{i}p_{y}$ superconductivity

For what bothers you, the $p_{x}+\mathbf{i}p_{y}$ superconductivity is the superconducting theory based on the following "choice" $\Delta_{0}=0$, $\mathbf{d}=\left(k_{x}+\mathbf{i}k_{y},\mathbf{i}\left(k_{x}+\mathbf{i}k_{y}\right),0\right)$ such that one has $$ \Delta_{\alpha\beta}\left(\mathbf{k}\right)\propto\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)\left(k_{x}+\mathbf{i}k_{y}\right)\equiv\left(k_{x}+\mathbf{i}k_{y}\right)\left|\uparrow\uparrow\right\rangle $$ which is essentially a phase term (when $k_{x}=k\cos\theta$ and $k_{y}=k\sin\theta$) on top of a spin-polarized electron pair. This phase accumulates around a vortex, and has non-trivial properties then.

Note that the notation $\left|\uparrow\uparrow\right\rangle $ refers to the spins of the electrons forming the Cooper pair. A singlet state would have something like $\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle $, and for $s$-wave $\Delta_0$ is $\mathbf{k}$ independent, whereas $\mathbf{d}=0$.

  • Note that the $p$-wave also refers to the angular momentum $\ell=1$ as you mentioned in your question. Then, in complete analogy with conventional composition of angular momentum (here it's for two electrons only), the magnetic moment can be $m=0,\;\pm1$. The natural spherical harmonic for these states are then $Y_{\ell,m}$ with $Y_{1,\pm1}\propto k_{x}\pm\mathbf{i}k_{y}$ and $Y_{1,0}\propto k_{z}$, so it should be rather natural to find the above mentioned "choice" for $\mathbf{d}\left(\mathbf{k}\right)$. I nevertheless say a "choice" since this is not a real choice: the symmetry of the gap should be imposed by the material you consider, even if it is not yet satisfactorily understood.
  • Note also that only the state $m=+1$ appears in the $p_{x}+\mathbf{i}p_{y}$ superconductivity. You might wonder about the other magnetic momentum contribution... Well, they are discarded, being less favourable (having a lower transition temperature for instance) under specific conditions that you have to know / specify for a given material. Here you may argue about the Zeeman effect for instance, which polarises the Cooper pair. [NB: I'm not sure about the validity of this last remark.]

Relation between $p_{x}+\mathbf{i}p_{y}$ superconductivity and emergent unpaired Majorana modes

Now, quickly, I'll try to answer your second question: why is this state important for emergent unpaired Majorana fermions in the vortices excitations ? To understand that, one has to remember that the emergent unpaired Majorana modes in superconductors are non-degenerate particle-hole protected states at zero-energy (in the middle of the gap if you prefer). Particle-hole symmetry comes along with superconductivity, so we already validate one point of our check list. To make non-degenerate mode, one has to fight against the Kramers degeneracy. That's the reason why we need spin-triplet state. If you would have a singlet state Cooper pair stuck in the vortex, it would have been degenerate, and you would have been unable to separate the Majorana modes, see also Basic questions in Majorana fermions for more details about the difference between Majorana modes and unpaired Majorana modes in condensed matter.

A more elaborate treatment about the topological aspect of $p$-wave superconductivity can be found in the book by Volovik, G. E. (2003), Universe in a Helium Droplet, Oxford University Press, available freely from the author's website http://ltl.tkk.fi/wiki/Grigori_Volovik.

  • Note that Volovik mainly discuss superfluids, for which $p$-wave has been observed in $^{3}$He. The $p_{x}+\mathbf{i}p_{y}$ superfluidity is also called the $A_{1}$-phase [Volovik, section 7.4.8]. There is no known $p$-wave superconductor to date.
  • Note also that the two above mentionned books (Samokhin and Mineev, Volovik) are not strictly speaking introductory materials for the topic of superconductivity. More basics are in Gennes, Tinkham or Schrieffer books (they are all named blabla... superconductivity blabla...).
share|improve this answer
A small mistake: the matrix index in the expression of $\Delta(k)$ by using $\Delta_0$ and $\mathbf{d}$ should be taken after the matrix multiplication. –  Tengen Apr 27 '13 at 9:17
@Tengen Thanks a lot. I corrected. –  FraSchelle Apr 27 '13 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.