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

+ 1 like - 0 dislike
51 views

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?

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user annie marie heart

+ 2 like - 0 dislike

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 $$\eta_{PS} = \gamma_5\psi ( \psi C\psi)$$

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user Thomas
answered Dec 25, 2017 by (310 points)
Thanks, you final form is kind of abstrat, not sure what it implies...

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user annie marie heart
+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

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user annie marie heart
No, there is no difference in the symmetry of the gap function. This is just a relative phase. The two order parameters are $\psi_L\psi_L\pm \psi_R\psi_R$.

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user Thomas
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.

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user annie marie heart
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, $J^\pi=1^\pm$.

This post imported from StackExchange Physics at 2020-10-30 22:45 (UTC), posted by SE-user Thomas

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.