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Why 8 gluons?

  1. Jan 12, 2005 #1

    I found a puzzling reference in my EP-course, that basically says :

    "Gluon cary a dual charge, one colour and one anticolour. This means one can have 3²=9 combinations."

    So far, so good. But then :

    "We have to substract one colourless combination. This leaves us with 8 gluons."

    Huh? Why one colourless combination? I can see three :


    Which would leave us with 6 gluons, which is obviouly wrong. Can anyone explain this to me?

    Thanks in advance,

  2. jcsd
  3. Jan 12, 2005 #2


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    'Colorless' means color singlet in this context, of which there is only one type:

    [tex] \Psi_c =\frac{1}{\sqrt{3}} (r\bar{r}+b\bar{b}+g\bar{g}) [/tex]
  4. Jan 12, 2005 #3


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    Technically, SU(3) has 8 generators, while U(3) has 9. Why SU() instead of U()?
  5. Jan 12, 2005 #4
    My prior knowledge of this matter is rather limited.
    I see what the colourless combination looks like. What about the others? What do they look like?
  6. Jan 12, 2005 #5
    I haven't got the slightest idea. Something to do with inversion symmetry? Parity? :confused:
  7. Jan 12, 2005 #6


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    Roughly, there would be nine, but there are algebraic relations since the matrices of SU(3) are unitary. This knocks out one degree of freedom. The situation is that if you know any eight, you can solve for the ninth, so there are really only eight independent generators.
  8. Jan 12, 2005 #7
    Well, this is quite vague. What is meant here is that indeed you only need eight generators in order to describe the entire symmetry group (local SU(3)-colour-symmetry). These generators really are the conserved quantity at hand, which is in this case colour. Basically this means that when particles with colour interact (like gluons and quarks and...) the total net-colour needs to be conserved (this works just like energy conservation).
    A coloursinglet is a state describing a particle of which the "colour-part" cannot change when you perform colour-tranformations on it (think of these colourtransformations as rotations and they are the actual inhabitants of the SU(3)-colour-world). A particle that is a singulet is observable. So quarks on themselves are NOT observable for this reason : they carry an overall colour (R,G or B). Now watch it: a red-antired quark may seem white but it is NO coloursingulet. And being really or truly white means being a coloursinglet.

    In grouptheory, you can prove that if a local symmetrygroup has dimension N, there will always be N²-1 generators that generate the transformations under which the physical quantities (like a lagrangian) need to be invariant...

    But this is all math. A more physical interpretation is this one : If the wavefunction of some gluon is completely white, then there is no colour and no interaction can go on. Basically this state is a singulet which means that it can never change (no change in charge, and so on...). Now for a gluonstate to be white there is ONE condition: the particle cannot have any preference for any colour. This means that the gluon must be red-antired, green-antigreen, AND blue-antiblue.

    Now incorporating some normalization-constants we have that : the
    white gluon is (red-antired + blue-antiblue + green-antigreen)/sqrt(3).

    For example there are two kinds of wavefunctions that are not actually white: (red-antired -green-antigreen)/sqrt(6)
    and (red-antired + green-antigreen -2*blue-antiblue)/sqrt(6).

    These two gluons can interact without changing
    the color of a quark, but they are not completely white.Indeed, colourcharges interact via the exchange of colour. So the singlet state cannot interact because it cannot change its colours. that's why it is a singlet !!!Now, a red-antired gluon is indeed white BUT NO SINGLET because the colour can be changed. Indeed the total colour is white but so is the total colour of a blue-antiblue gluon so it can change into that for example. You see the difference between being white and being TRULY white ???That is the main point

    An analogous things happens in the quantum information theory. Suppose you have a wavefunction that is a superposition of spin up and down. Suppose that the probability for measuring the spins along some axis is 1/2 then you really know nothing at all do you ???This same thing happens with the TRULY white wavefunction. For this reason all combinations that yield white must be included

    Check out my journal for more info...


    ps : veel succes met de QFT-studie Dimitri...geniet ervan... :wink:
    Last edited: Jan 12, 2005
  9. Jan 12, 2005 #8
    BTW this is a very good question Dimitri...

  10. Jan 12, 2005 #9

    Because SU(3) is a local symmetry-group and U(3) is global. the rule is that the number of generators of a local group of dimension N is N²-1. Besides the -1 comes from the fact that SU(N)-matrices are unitary and traceless

    Last edited: Jan 12, 2005
  11. Jan 12, 2005 #10


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    WHAT??????????? :surprised :surprised :surprised :surprised :surprised :surprised :surprised

    Jesus,Marlon,it's your area of expertise.QCD,remember??Group theory,remember??????????????????How can u say that????

    SU(N) matrices are unitary and have determinant "+1".Generators of the matrices which form the Cartan subalgebra (of the Lie algebra su(n) of the group SU(N)) are traceless,not the matrices form SU(N).

    That's a shocker,Marlon... :rolleyes: Didn't expect that...And u're saying all things about me... :wink: :rolleyes:


    PS.U've written that thing at almost 2 a.m.Yet fatigue it's not an excuse.Not in this case... :yuck:
  12. Jan 13, 2005 #11
    Hi DIMITRI, oe ist ?????????

    Hi Dimitri, I indeed made a mistake in stating that SU(N) matrices are traceless. Of course they are NOT...Sorry for that...

    Any unitary matrix U can be written as U =exp(iH) where H is a matrix with the property that [tex]H^{+} = H[/tex]

    Given the fact that det(U) = exp(iTr(H)) where Tr denotes the trace, we can conclude that all generators of SU(N) are hermitic and traceless N*N-matrices. Because of hermiticity you have N² matrices and the need to be traceless gives an extra condition so in toto you have N²-1 matrices that obey these two conditins at the same time. Thus N²-1 generators for SU(N)...

    That is the complete picture...

    ps : let niet op de onvolledige verklaring van dextercioby, de laatste dagen doet die niets anders als mij willen verbeteren... :rolleyes:

  13. Jan 13, 2005 #12
    Thanks alot Marlon. Mind you, this is only an introductory course. Next semester I'll have a decent QED course... Then I'll be around with more questions :tongue2:
  14. Jan 13, 2005 #13
    And we shall be ready to help you out at any given time on any given place... :cool:

  15. Jan 15, 2005 #14


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    Just a comment: it's not a question of local vs gauge symmetry whether the number of generator is N^2-1. It's just a question of which group is involved. SU(N) has N^2-1 generators, U(N) has N^2 generators. This is the math part.

    Now, which of this group is a local (gauge) symmetry group is a matter of experimental verification, not of mathematics. It turns out QCD is SU(3), but
    Nature could have chosen U(3) (in which case the world would be quite different!)

  16. Jan 15, 2005 #15
    I agree...

    Depends on how you look at it. I do see your point and i agree. However, just look at how the eightfoldway was born. Basically the SU(3) representation was born because via tensor-products of this fundamental representation we could construct the mathematical formalism (using baryons and mesons) that would correspond to the results of the eightfolds...

    Just a some way to look at it and this is what i intended.

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