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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.

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    $\begingroup$ 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? $\endgroup$ – Cosmas Zachos Dec 24 '17 at 21:23
  • $\begingroup$ But Goldstones are pseudoscalars, do they break T and P? $\endgroup$ – annie heart Dec 24 '17 at 21:25
  • $\begingroup$ 2SC is defined as u and d quarks pair within two colors say red and green. $\endgroup$ – annie heart Dec 24 '17 at 21:26
  • $\begingroup$ 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? $\endgroup$ – Cosmas Zachos Dec 24 '17 at 21:28
  • $\begingroup$ 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. $\endgroup$ – annie heart Dec 24 '17 at 21:30
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  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).

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  • $\begingroup$ Thanks +1, I thought that if 2SC and CFL have odd parity pairing, they also break P and T? No? Why not odd parity? I cannot see why from the Wiki or literature? $\endgroup$ – annie heart Dec 25 '17 at 3:24
  • $\begingroup$ see thi: physics.stackexchange.com/questions/376191 $\endgroup$ – annie heart Dec 25 '17 at 3:26
  • $\begingroup$ @Thomas, I thought the odd parity pairing breaks P? Does odd parity pairing condensate break T or C? $\endgroup$ – wonderich Dec 25 '17 at 6:35
  • $\begingroup$ @Thomas, it looks to me the odd parity pairing breaks P but preserves T. How about the C symmetry? $\endgroup$ – annie heart Dec 25 '17 at 17:36
  • $\begingroup$ Yes, odd parity breaks P, but the correct ground state is not odd parity unless the theta angle is non- zero (which breaks P explicitly) $\endgroup$ – Thomas Dec 25 '17 at 19:01

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