# chiral symmetry condensate and 2SC, CFL breaking C, P and T symmetry?

Because we know that chiral symmetry condensate causes the chiral symmetry breaking, and it produces Goldstone modes of pseudo-scalars, so I believe that chiral symmetry breaking also breaks the T symmetry.

question: Do we have similar statements for C, P and T symmetry? For other QCD phases? Say for chiral symmetry breaking phases, 2SC, CFL breaking of color superconductor?

Namely,

1. Does chiral symmetry breaking break C, P and T symmetry?

2. Does 2SC two quark color/flavor break C, P and T symmetry?

3. Does CFL three color/flavor locking break C, P and T symmetry?

By playing around, I know some partial answers, like chiral symmetry breaking breaks T, but the CFL preserves all C,P,T. But I like to hear from the experts, just to confirm the correct answer.

• The chiral symmetry condensate has the quantum numbers of a mass term, so it does not break T, C, or P. How did you get that misimpression? Did you define 2SC? – Cosmas Zachos Dec 24 '17 at 21:23
• But Goldstones are pseudoscalars, do they break T and P? – annie heart Dec 24 '17 at 21:25
• 2SC is defined as u and d quarks pair within two colors say red and green. – annie heart Dec 24 '17 at 21:26
• A pseudoscalar goldston does not break P or T. The strong interactions which break chiral symmetry do not break P or T or C. How are you getting these misimpressions? Can you show your calculation? – Cosmas Zachos Dec 24 '17 at 21:28
• If pseudo scalar condense in ChSB, then it will break P and T. But I think pseudo scalar does not condense, as you pointed out, perhaps do not condense in ChSB. – annie heart Dec 24 '17 at 21:30

1. Chiral symmetry breaking does not imply breaking of either $C,P$ or $T$. The order parameter is a scalar.
2. (and 3.) Both 2SC and CFL are phases at finite baryon density, so $C$ is broken explicitly. $P,T$ are unbroken (the order parameter is a scalar).