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

+ 0 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
asked Dec 24, 2017 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user Cosmas Zachos
But Goldstones are pseudoscalars, do they break T and P?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
2SC is defined as u and d quarks pair within two colors say red and green.

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
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?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user Cosmas Zachos
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.

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
But in CFL, there is odd or even parity pairing, I think in odd parity pairing, it breaks P symmetry.

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart

1 Answer

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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).

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user Thomas
answered Dec 25, 2017 by tmchaefer (310 points) [ no revision ]
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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)

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user Thomas
I think the odd parity breaks P because, $P \psi^\dagger(t,x) \Delta(x) C \psi^*(t,x)P^{-1}= -\psi^\dagger(t,-x) \Delta(x) C \psi^*(t,-x),$ but it preserves T. But it does not make sense to talk about the C symmetry due to the fermi surface (?).

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user wonderich
@Thomas, Do you have a Ref for theta term breaks P? I thought theta term can preserve T for certain $\theta =n \pi$ values? Why does not theta term preserves P at $\theta =n \pi$ values?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
@Thomas, which best Ref for breaking P for theta term?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
The theta term involves $\vec{E}\cdot\vec{B}$, which is P-odd. Indeed, the the case $\theta=n\pi$ is special, because the phases of the original and the parity reflected states are equal, $e^{i\pi}=e^{-i\pi}$, so there is no explicit P violation. In this case, P can (and does) break spontaneously. This is also known to happen at zero baryon density.

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user Thomas
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see thi: physics.stackexchange.com/questions/376191

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user annie marie heart
@Thomas, I thought the odd parity pairing breaks P? Does odd parity pairing condensate break T or C?

This post imported from StackExchange Physics at 2020-10-30 22:44 (UTC), posted by SE-user wonderich

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