# Quark pair superconductor: Even parity is favorred than odd parity

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?

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 $$\eta_{PS} = \gamma_5\psi ( \psi C\psi)$$
• +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 Dec 25, 2017 at 4:26
• No, there is no difference in the symmetry of the gap function. This is just a relative phase. The two order parameters are $\psi_L\psi_L\pm \psi_R\psi_R$. Dec 25, 2017 at 4:54
• 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. Dec 25, 2017 at 5:02
• 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, $J^\pi=1^\pm$. Dec 25, 2017 at 6:03