Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 0 like - 0 dislike
2049 views

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

+ 2 like - 0 dislike
  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 ]
Most voted comments show all comments
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
Most recent comments show all comments
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

Your answer

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):
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$\varnothing$ysicsOverflow
Then 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).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...