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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?

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    $\begingroup$ 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. $\endgroup$
    – sammy gerbil
    Commented Apr 1, 2017 at 23:58
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    $\begingroup$ @ 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? $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 0:02
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    $\begingroup$ 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. $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 0:14
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    $\begingroup$ maybe then you should edit your question and add details to highlight your conceptual difficulties... $\endgroup$
    – ZeroTheHero
    Commented Apr 2, 2017 at 1:01

1 Answer 1

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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, $(qq)(\bar{q}\bar{q})$.

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  • $\begingroup$ Thanks but can you count carefully? See that ((u,c,d)×(r,g,b)x(anti colors)) -- do they have more than 9 quarks? $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 0:49
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    $\begingroup$ This is very dense matter, made of quarks. It is reasonable to ask if quarks have gaps, because quarks near the Fermi surface are long-lived quasi-particles. But anti-quarks are far off-shell, and it is not clear what we might mean by anti-quark gaps. $\endgroup$
    – Thomas
    Commented Apr 2, 2017 at 1:28
  • $\begingroup$ Thanks Thomas, +1, it looks you are knowledgeable in some way. $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 1:35
  • $\begingroup$ But $SU(3)_{color} \times SU(3)_L \times SU(3)_R \times U(1)_B$ symmetry has $8*3+1=25$ generators, is broken down to the global diagonal $SU(3)_{color+L+R}$ which has 8 generators. Thus the maximum Goldstone bosons should be $25-8=17$. Am I understand correctly that 17 Goldstone bosons equal to your 8 (gives mass to 8 gluons)+8 (Octave Goldstone)+1 (massless Goldstone of U(1))? So $$17=8+8+1,$$correct? $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 1:52
  • $\begingroup$ And the quark masses are dynamically generated, which have nothing to do with the Goldstone counting, am I correct? $\endgroup$
    – ann marie cœur
    Commented Apr 2, 2017 at 1:54

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