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

Question

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

Comments

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

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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

Answer 1

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

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

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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.

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

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Thanks Thomas, +1, it looks you are knowledgeable in some way.

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

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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?

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

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And the quark masses are dynamically generated, which have nothing to do with the Goldstone counting, am I correct?

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

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Yes, one octet eaten by Higgs mechanism, and quark gaps generated as in BCS.

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

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But my second octave Goldstone bosons all survive to produce massless modes. You should update your answer with more clarity, otherwise I will write an answer myself. ;)

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

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No, eaten by Higgs mecahnism.

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

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Here is what I mean: I thought my first octave Goldstone bosons are eaten by gluons (thus gluons are Higgsed), and my second octave Goldstone bosons all survive to produce massless modes. Do you agree?

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

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Yes, one octet eaten by Higgs mechanism, one octet produces Goldstone bosons. Evene simpler: Only global symmetries produce GBs. We have $SU(3)_L\times SU(3)_R\times U(1)\to SU(3)_F$, corresponding to [8] +[1] Gbs.

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

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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

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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

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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

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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

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@annieheart I added a postcsript.

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

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Thanks, but you havent explained the physical meanings of all the Goldstone modes (8+1=9)?

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

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@Thomas, nice discussions, the [8] +[1] Goldstone bosons here are spin-0 scalars, aren't they? They have nothing to do with the un-Higgised spin-1 gluon-photon mixture, yes? So what are [8] +[1] Goldstone bosons, physically, in terms of spin-0 quark-pair condensate, yes or no?

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

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@annieheart More edits.

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

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Thanks, but how about the [1] Goldstone mode? Is that spin-0 particle? what is it?

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

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The usual superfluid GB mode from breaking U(1)_B

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

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Can I think of it as some kinds of phonons in QCD? I mean how do I write down [1] Goldstone mode from using QCD Lagrangian?

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

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Also do you see that all the quark pair condensate need to be Lorentz scalar instead of Lorentz pseudo-scalar ? or even non-relativistic scalar only?

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

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Goldstone mode is derivatively coupled to baryon current, $\bar{q}\gamma_\mu q\sim f\partial_\mu\phi$.

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

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ok, which best short Ref are you looking at for this?

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

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@Thomas, I am puzzled by one thing. If we have a two quark condensation $<qq>\neq 0$, isnt that implying that we need its Hermitian conjugate $<\bar{q}\bar{q}>\neq 0$ as well? But the interpretation of $<\bar{q}\bar{q}>\neq 0$ could be that the two quark "hole" pair condense. Isn"t that the same thing as two anti-quark condensation?

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

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@Thomas Namely, $<\bar{q}\bar{q}>\neq 0$ $\Leftrightarrow$ two quark "hole" pair condense $\Leftrightarrow$ two anti-quark condensation? So why do we dont encounter two anti-quark condensation in your answer?

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

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@annieheart Quarks are Dirac fermions, so the the $qq$ condensate is a $4\times 4$ matrix. These components can be interpreted as quark-quark, hole-hole, anti-quark-anti-quark , anti-hole-anti-hole pairing. Near the Fermi surface we get quark-quark and hole-hole, but anti-quark or anti-hole pairing.

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

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@ Thomas, Are you sure what you are writing " Near the Fermi surface we get quark-quark and hole-hole, but anti-quark or anti-hole pairing" ???

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

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Should be .. but not ..

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

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@Thomas, you may shed light on this too -- I guess you know the full answer: physics.stackexchange.com/questions/376164

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

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For some reason, I still do not understand where can I find this Ref on your "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 annie marie heart

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Can you please illuminate why is that the case and give a Ref?

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

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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

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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