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  Goldstone bosons, quark and gluon masses counting in color-flavor locking QCD

+ 1 like - 0 dislike
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Consider QCD, with three flavors of massless quarks, we like to focus on the possible Cooper paired phases.

For 3 quarks $(u,c,d)$ and 3 colors $(r,g,b)$, the Cooper pairs cannot be flavor singlets, and both color and flavor symmetries are broken. The attractive channel favored by 1-gluon exchange is known as “color-flavor locking.” A condensate involving left-handed quarks alone locks $SU(3)_L$ flavor rotations to $SU(3)_{color}$, in the sense that the condensate is not symmetric under either alone, but is symmetric under the simultaneous $SU(3)_{L+color}$ rotations. A condensate involving right-handed quarks alone locks $SU(3)_R$ flavor rotations to $SU(3)_{color}$. Because color is vectorial, the result is to breaking chiral symmetry. Thus, in quark matter with three massless quarks, the $SU(3)_{color} \times SU(3)_L \times SU(3)_R \times U(1)_B$ (the last one is baryon) symmetry is broken down to the global diagonal $SU(3)_{color+L+R}$ group.

question:

1) How many quarks among nine ($(u,c,d) \times (r,g,b)$) have a dynamical energy gap? What are they?

2) How many among the eight gluons get a mass? What are they?

3) How many massless Nambu-Goldstone bosons there are? What are they? How to describe them?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
asked Apr 2, 2017 in Theoretical Physics by annie marie heart (500 points) [ no revision ]
This looks rather like a "homework" question, although very advanced. This is not a homework help site - we do not do your assignments for you. Please show your attempt to answer these questions, and ask about a conceptual difficulty.

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user sammy gerbil
@ sammy gerbil, your comment is very abusive and wronged. It is absolutely not a homework question --- just an extended question from reading the Wikipedia. And it is 100% not a homework question that a professor ever assigns to you . Otherwise can you answer? Other than saying this agressive words out of nowhere?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
The mid paragraph and the formulation of this question are my attempt to answer the question. The Goldstone theorem tells us that the Goldstone boson lives on the coset space of |original group/unbroken group|, but for this example, it is subtler because Goldstone boson can be eaten by gauge fields. So this counting is more subtle. I have my own counting, but I do not like to bias the readers. Also I do not know it is correct.

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
maybe then you should edit your question and add details to highlight your conceptual difficulties...

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user ZeroTheHero

1 Answer

+ 4 like - 0 dislike

These questions are answered in the original literature:

1) All quarks are gapped. The nine quarks arrange themselves into an octet with gap $\Delta$ and a singlet with gap $2\Delta$.

2) All gluons are gapped.

3) There is an octet of Goldstone bosons related to chiral symmetry breaking, and a singlet associated with $U(1)$ breaking.

Postscript:

i) When pair condensates form there is a gap in the excitation spectrum of single quarks (this is just regular BCS). However, the gapped excitations may be linear combinations of the microscopic quark fields. In the present case the nine types of quark fields ($N_c\times N_f=9$), form an octet and a singlet of an unbroken $SU(3)$ color-flavor symmetry.

ii) Pair condensation and the formation of a gap take place near the Fermi surface. There is no Fermi surface for anti-quarks (if $\mu$ is positive and large), and therefore no pairing and no gaps.

iii) There is both a $U(1)$ GB (associated with the broken $U(1)_B$) and a masssless $U(1)$ gauge boson (associated with the $U(1)_{Q}$ gauge symmetry that is not Higgsed).

iv) The [8] GB correspond to spontaneous breaking of chiral symmetry. In ordinary QCD these would be quark-anti-quark states, but at high density anti-quarks decouple. A detailed analysis shows that the GBs are predominantly 2-particle-2-hole states, $(qq)(\bar{q}\bar{q})$.

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user Thomas
answered Apr 2, 2017 by tmchaefer (280 points) [ no revision ]
Most voted comments show all comments
Thanks but can you count carefully? See that ((u,c,d)×(r,g,b)x(anti colors)) -- do they have more than 9 quarks?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
Concern 17=8+8+1, “All quarks are gapped. The nine quarks arrange themselves into an octet with gap Δ and a singlet with gap 2Δ.” The superconducting gap are paired of quarks instead of individual quarks. So what do you mean by nine quarks are gapped? Also quark-quark Cooper pair <qq> also implies anti-quark-anti-quark <q ̄q ̄> can form a Cooper pair, isnt’t it? So what exactly are the 9 different gaps you are talking about from (physical indices)?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
Once we are in CFL, it may not be useful to talk about the individual quarks, no? But instead quark pairs?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
am still confused by what you said. Is the 9 Goldstone modes containing 8 mesons? But here we are in CFL that has no Chiral Symmetry Breaking, so why do we have mesons? I thought we do NOT have 8 “massless” mesons as Goldstone modes? So what exactly are Goldstone modes here, within superconducting gap 2Δ?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
Is the remained 1 mode from U(1) a Goldstone mode of a gluon-photon mixture? Why or why not?

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user annie marie heart
Most recent comments show all comments
This was noticed in arxiv.org/abs/hep-ph/9908227 and is described in standard review, for example Sect V.C in arxiv.org/abs/0709.4635

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user Thomas
The basic point is that in ordinary QCD the object that condenses are $q\bar{q}$ pairs, and GB are chiral rotations of the condensate. In high density QCD the basic condensates are pairs of particles and holes, $(qq)$ and $(\bar{q}\bar{q})$. Now I notice that a $(qq)$ in the $\bar{3}$ of color and flavor transforms just like a single $\bar{q}$ and the $(\bar{q}\bar{q})$ transforms like a single $q$. This means I can construct a GB field that transforms like the usual one, but the microscopic content is $(qq)(\bar{q}\bar{q})$.

This post imported from StackExchange Physics at 2020-10-28 19:05 (UTC), posted by SE-user Thomas

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