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Quark color 2-dimensional?

  1. Sep 27, 2009 #1
    I'm trying to understand quark color. Everyone seems to say that there are three quark colors, say Red, Blue, Green, that add up to uncolored. So why not just pick a basis, say Red and Blue, so that the color Green is equivalent to -1 Red plus -1 Blue?
     
  2. jcsd
  3. Sep 27, 2009 #2
    Quarks are fermions, so if you have three in the same state except for their color degree of freedom, the colors must all be different (orthogonal in color space). Therefore you wouldn't be able to write one color in terms of two others. There are such particles that have the same flavor quarks (uuu or ddd etc.) with zero orbital angular momentum ground states, so that the above requirement applies.
     
  4. Sep 28, 2009 #3
    Thanks, that makes a lot of sense.

    To what extent does a +1 Red, +1 Blue, +1 Green particle differ in color charge from a 0 Red, 0 Blue, 0 Green particle? Based on a three dimensional color space, I would say these are different, yet they both seem to be dumped into the same "colorless" category without distinction.
     
  5. Sep 28, 2009 #4

    arivero

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    Consider the antiparticle too.
     
  6. Sep 28, 2009 #5
    Okay. A 0 Red, 0 Blue, 0 Green particle can be its own antiparticle, but a +1 Red, +1 Blue, +1 Green particle cannot be its own antiparticle. Is this what you mean?

    Is there a name for the +1 Red, +1 Blue, +1 Green color charge combination, like +1 Colorless or +1 Colorfull? That way, free particles would have a quantum number expressing their total color charge in Colorfull units.
     
  7. Sep 28, 2009 #6
    I'm wondering now if this is just the baryon number?
     
  8. Sep 29, 2009 #7
    The problem here is that combining a red charge, a blue charge, and a green charge does not necessarily make and uncolored object. There are 6 linearly independent states with that composition; and, out of these, there is only one that can be identified as colorless - the state which is fully antisymmetric under the exchange of colors:

    [tex]|singlet\rangle = \frac{1}{\sqrt{6}}\left(|rgb\rangle+|brg\rangle+|gbr\rangle-|bgr\rangle-|rbg\rangle-|grb\rangle\right)[/tex].

    Of the other five states, one belongs to the 10 dimensional representation of SU(3), which is fully symmetric under exchange of colors and two each of the other four belong to two copies of the 8 dimensional adjoint representation, which has mixed symmetry.

    The reason we can get away with acting as if the sum of the charges is simply 0 is that any free particle must be uncolored; and, since the quarks making up a proton or neutron (or other baryon) are fermions, their total state must be antisymmetric under exchange, while the combined spin/flavor wavefunctions always turn out to be fully symmetric.
     
  9. Sep 29, 2009 #8
    Thanks for the additional explanation. For some reason, I'm having trouble understanding the color charge concept.
     
  10. Sep 29, 2009 #9

    arivero

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    Right. But well, rgb is "colorless", it is not an "object" :biggrin:. Lets define a quantum "object" as an entity fulfilling the spin-statistics rule, ie fully symmetrc if it is a boson, fully antisymetric if it is a fermion.
     
  11. Sep 29, 2009 #10

    arivero

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    Indeed.

    For the same token, you could have asked if "+1 Green" is the same charge that "-1 Red -1 Blue". If it were, you could organize mesons and baryons in supermultiplets. In fact there was some work in this sense, near 1968.
     
  12. Sep 29, 2009 #11
    You're picking at subatomic nits. Clearly my point was that an object made up of a particle carrying a red charge, one carrying a green charge, and one carrying a blue charge is only uncolored if the wave function is fully antisymmetric under the exchanges of colors. And that there are five linearly independent states with the same color composition which don't satisfy this. These statements are independent of spin-statistics. They apply just as well to squarks (if they exist) as to quarks. So, there really was no ambiguity to my statement.
     
  13. Sep 30, 2009 #12

    arivero

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    Well, you said that for a boson you must need a fully symmetric wavefunction, for a fermion a fully antisymmetric. I am sorry I was confused because of it, I apologize.

    This is related to my question on antiparticles, ie in which sense -R-B is +G. Point is, you must to use the decomposition of representations
    [tex] \bar 3 \otimes \bar 3 = \bar 6 \oplus 3 [/tex]
    and choose 6 or 3 depending if you wish symmetry or antisymmetry. So while both cases are combinations, in one case you get again the fundamental representation. I do not remember if it is the symmetric or the antisymmetric case.

    For three quarks it is more complicated because [tex] \bar 3 \otimes 3 = 8 \oplus 1 [/tex] so

    [tex]3 \otimes 3 \otimes 3 = ( 6 \oplus \bar 3) \otimes 3 = ( 6 \otimes 3) \oplus ( \bar 3 \otimes 3) = ( 6 \otimes 3) \oplus 8 \oplus 1 = 10 \oplus 8 \oplus 8 \oplus 1 [/tex]

    Well there is the singlet combination, yes, which is going to be the sum you told above. And according http://en.wikipedia.org/wiki/Quark_model
    "The decuplet is symmetric ... the singlet antisymmetric and the two octets have mixed symmetry."
    as you said.

    So OK, you need to get antisymmetry to get the singlet. It is amusing that some particles in the colour decuplets and octets are not to be considered white.

    Edit: I see I am basically repeating your argument. Still I think that the case of two quarks to do an antiquark (or reciprocally) is interesting and it deserved some lines, so I have fitted it here.
     
    Last edited: Sep 30, 2009
  14. Sep 30, 2009 #13
    Actually, what you're discussing now is the result of the flavor wavefunctions. When considering only baryons constructed from u, d, and s quarks, there's an approximate SU(3) flavor symmetry in addition to the standard gauge symmetries. It's the flavor decomposition that leads to the decuplet, octets, and singlet structure that shows up in the baryon spectrum. (Mathematically, the decomposition is the same as what I discussed above; but, it's coming from a different property of the quarks.) All free baryon states are color singlets; but, they have varying symmetries under SU(3) flavor.
     
  15. Oct 1, 2009 #14

    Vanadium 50

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    There is an implicit assumption here that is not correct - that color charges add like numbers: e.g. a red-antiblue gluon carries twice the charge of a red quark. In fact, color charges are not numbers, they are matrices: that's what is meant by SU(3) symmetry. The actual "charge" then depends on the process, and by doing the calculation you might well find that the red-antiblue gluon has 17/9ths the charge of a red quark (or something like that).

    The SU(3) symmetry explains what arivero et al. are discussing. A colorless object is a "singlet" under SU(3) - all colorless objects act the same in QCD. The three quarks form a triplet, often called red, green and blue, although we could just have easily called them provelone, limburger and muenster. The next larger representation is an octet, so if I have a quark (in a triplet), and an antiquark (also in a triplet), I have 3 x 3 = 9 combinations, or one octet plus one singlet. That's why two combinations that have as much anti-red as red (et al.) still have non-zero color charge. If charge added like numbers (e.g. the symmetry is U(1) and not SU(3)), these would be colorless. Because color charge is a matrix, this is not the case.
     
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