It seems that the quark pair superconductor can be odd or even parity pairing respect to the parity $P$.

Say that the even parity has the form: $$ \langle\psi C \gamma^5 \psi\rangle $$

the odd parity has the form: $$ \langle\psi C \psi\rangle $$ There is no difference at perturbative computation. $C$ is charge conjugate matrix.

But the literature seems to suggest that instanton effect favors the even parity not the odd parity. I look into the literature but the original paper seems not to assert that claim. Refs cited here

Do you have either a simple and intuitive or a rigorous analytic explanation of the claim?


1 Answer 1


This requires an actual calculation (and getting the signs right), see for example https://arxiv.org/abs/hep-ph/9810509 .

There are some simple heuristics. For example, there is a successful quark-scalar-diquark model of the nucleon. Lattice QCD practioners know that the nucleon couples strongly to $$ \eta_S = \psi (\psi C\gamma_5 C) $$ but not to $$ \eta_{PS} = \gamma_5\psi ( \psi C\psi) $$

  • $\begingroup$ Thanks, you final form is kind of abstrat, not sure what it implies... $\endgroup$ Dec 25, 2017 at 4:24
  • $\begingroup$ +1, but do you agree that the gap function in $k$ space has $\Delta(k)=-\Delta(k)$ for even parity, and $\Delta(k)=+\Delta(k)$ for odd parity? This is kind of counter intuition but I wanted to make sure your odd even parity means the same thing $\endgroup$ Dec 25, 2017 at 4:26
  • $\begingroup$ No, there is no difference in the symmetry of the gap function. This is just a relative phase. The two order parameters are $\psi_L\psi_L\pm \psi_R\psi_R$. $\endgroup$
    – Thomas
    Dec 25, 2017 at 4:54
  • $\begingroup$ but when you convert to the k space, like in the BdG equation, you should see the potential difference of $\Delta(k)$. See any condensed matter BdG equation. $\endgroup$ Dec 25, 2017 at 5:02
  • $\begingroup$ No. The $J^\pi=0^\pm$ order parameters have the same gap function. The simplest way to get an odd gap function is to consider $p$ wave pairing, $J^\pi=1^\pm$. $\endgroup$
    – Thomas
    Dec 25, 2017 at 6:03

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