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  Quark pair superconductor: Even parity is favorred than odd parity

+ 1 like - 0 dislike
33 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
asked Dec 25, 2017 in Theoretical Physics by annie marie heart (450 points) [ no revision ]

1 Answer

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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 tmchaefer (280 points) [ no revision ]
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

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