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Gluon octet visualisation?

  1. Aug 12, 2012 #1
    Hi, it's my first post :)

    I've been working on a learning app about subatomic particles for some time, and now we'd need to represent gluons.

    I want to do this correctly by representing the eight possible gluons in the octet, but don't really know how to read the name of a gluon.

    I know that, for example, (red-antiblue + blue-antired)/sqrt2 has a 50/50 probability of being either red-antiblue or blue-antired. That is if I understood correctly.

    But for gluons which have a minus sign instead of a plus, is it still 50/50?

    And for those with a minus i in front of? Like -i(red-antiblue - blue-antired)/sqrt2, how do you calculate the probabilities of it being red-antiblue or blue-antired?

    And finally (red-antired + green-antigreen - 2blue-antiblue)/sqrt6, what's the color charge probability for it?

    Thanks for your help!
  2. jcsd
  3. Aug 13, 2012 #2
    Also, is it correct to say this?

  4. Aug 14, 2012 #3
    I think for a layman explanation this will do.

    But on the fundamental level it's wrong - it's way too simplified.

    Red-antired gluons don't exist (because color neutral states do not interact), similarly g-gbar and b-bbar gluons do not exist. So you are left with six possible color pairs, such as red-antiblue. But there should be EIGHT, not six, different gluons.
    That's why we have to use mind-bending constructs like "-i(red-antiblue - blue-antired)/sqrt2" gluon - only they give us linearly independent eight gluons which can't ever be mixed to result in a color neutral state.

    Secondly, color coupling constant is ~1, which means that loopy Feynman diagrams contribute strongly to the interaction (unlike electromagnetism, where they don't). Basically it means that emitted gluon is quite likely to emit and absorb a few virtual gluons and/or quark pairs before it reaches the other quark. Which makes the picture much messier.
  5. Aug 14, 2012 #4
    Yes, gluons do interact between themselves, but I thought it'd be easier not to include this in the explanation, but only mention it.

    So, for the mind-bending constructs, like you said, are for what? Except for normalization or linear independance... does that mean they have the same "color charge probability" than those without the "-i" sign?

    You said there aren't red-antired gluons, but the equations make me think there is a probability that it exists... maybe it just doesn't exist within a normal proton/neutron? Maybe it is used in mesons? I don'T know, but there is a (red-antired - blue-antiblue)/sqrt2 gluon.

    And finally, what for the last one of the eight, the one with three "probabilities" as I understand it, divided by sqrt6?
  6. Aug 15, 2012 #5
    I'd recommend something similar to nikkkom's approach, but even simpler:

    Here's a really simplified (but slightly wrong) explanation: each gluon has a color and an anti-color. So there are red-antigreen gluons, blue anti-blue gluons, blue-antired gluons, and so on. Then e.g. a red quark can emit a red-antigreen gluon and turn green. The red anti-green gluon might then hit a green quark and turn it red. Or it might hit an anti-red antiquark and turn it anti-green.

    Of course the reason this is slightly wrong is that if you count up the possibilities here there are 9 possible gluons. In reality there are only 8 gluons, and that is because the "linear combination" (red anti-red + blue anti-blue + green anti-green) doesn't actually exist. However even explaining the meaning of this statement requires a bunch of linear algebra, and I would recommend just ignoring it. Unless you are talking to people who are going to deduce that the existence of an extra color singlet gluon would cause the existence of an extra unobserved long-range force between baryons, I think you should just ignore this technicality.

    If you want to know, in gluons like -i(red anti-green - green anti-red)/sqrt(2) factors of i and minus signs don't affect the probabilities, so that gluon has a 50% chance to be red anti-green and a 50% chance to be green anti-red. And in (red-antired + green-antigreen - 2blue-antiblue)/sqrt(6), there's a 1 in 6 chance to be red anti-red, a 1 in 6 chance to be green anti-green, and a 4 in 6 chance to be blue anti-blue. The probabilities are proportional to the square of the coefficient of each possibility (1 squared vs 1 squared vs 2 squared).
  7. Aug 15, 2012 #6
    Thanks man! That was actually really helpful!
  8. Aug 16, 2012 #7
    If such gluon could exist, then color neutral particles like neutrons (any hadron in fact) would be interacting via strong force, which contradicts observations.
  9. Aug 16, 2012 #8
    Hey, look, someone who understands this stuff :D

    I have a somewhat off-topic question. I read in Wiki and other sources that current understanding of QCD says that glueballs - hadrons build from gluons, not quarks - are possible.

    Well, I don't understand how it is possible: looks like any combination of gluons alone will have an overall non-neutral color charge, and therefore such particle can not be isolated.

    Where I am mistaken?
  10. Aug 16, 2012 #9
    Properly understanding this requires group theory, but I've had a go at explaining it below:

    When we say something "has neutral color charge," what we actual mean is "doesn't change under any color gauge transformations." What are color gauge transformations? Well, one example of a color gauge transformation would be the transformation: red becomes green, antired becomes antigreen, green becomes minus red, antigreen becomes minus antired. This is essentially switching the "red" and "green" color labels, so that whatever you called a red quark before you now call a green quark, or whatever you called a blue-antigreen gluon before you now call a blue-antired gluon (there's also that funny minus sign). You can switch any two color labels in this way as a color gauge transformation, or you could even do a cyclic relabeling like: red->green, green->blue, blue->red, and same for the anticolors. Actually, there are infinitely many more exotic gauge transformations, like ones that turn red into (red + green)/sqrt(2) and green into (red - green)/sqrt(2), or transformations that behave differently at different points in space and time.

    So for example, a red quark is not color neutral because e.g. if you do the above gauge transformations it turns into a green quark. So this matches up with our intuition that quarks "carry color charge."

    Now, you can verify that none of the gluons are invariant under *all* color gauge transformations. For example, (red-antired + blue-antiblue - 2*green-antigreen) is invariant under switching red and blue, but changes if you switch red and green. So the gluons aren't "color neutral" and this is why gluons can emit and absorb other gluons. You can also verify that the nonexistent gluon (red-antired + blue-antiblue + green-antigreen) actually *would* be invariant under all color transformations, and this is basically why it doesn't exist.

    Before getting to glueballs, let's talk about mesons and baryons. Mesons consist of a quark and an anti-quark. The quark has a color and the antiquark has an anticolor. The color state of a meson is denoted (red-antired + blue-antiblue + green-antigreen). The meaning of this is basically "if you pull apart a meson, 1/3 of the time you will find a red quark and an antired antiquark, 1/3 of the time you will find a blue quark and an antiblue antiquark, and 1/3 of the time you will find a green quark and an antigreen antiquark." We can't actually pull apart mesons to look at the quarks inside, but this is the general idea. Note that this is the same color state as the nonexistent gluon--it's color neutral. It has to be, because as you know all particles that can exist in isolation must be color neutral.

    Next, baryons. Baryons, e.g. the proton, consist of three quarks, and each quark has a color. Baryons have the color state (rgb - rbg + gbr - grb + brg - bgr), where I've abbreviated r = red, etc. The meaning of this is that if you "pulled apart" the proton, 1/6 of the time the first quark would be red, the second green, and the third blue, etc. This state is also unchanged if you do any color transformation. For example, let's try the one that switches red and green. It takes r to g and g to -r. So

    (rgb - rbg + gbr - grb + brg - bgr) goes to (g(-r)b - gb(-r) + (-r)bg - (-r)gb + bg(-r) - b(-r)g)

    Pulling out the minus signs to the front of each term this is

    (-grb + gbr - rbg + rgb - bgr + brg) and you can verify that this is just a rearrangement of the original state, so it didn't actually change. This is actually the only color-invariant state of three quarks, so it must be the color state of all baryons.

    So, what about glueballs? Let me pretend for simplicity that there are only two colors, r and g. Then there are three gluons, denoted ##(r \bar{g} + g \bar{r}), i(r \bar{g} - g \bar{r}), (r \bar{r} - g \bar{g})##. Here the overbars denote anticolors. Let's call these three gluons types 1, 2, and 3.

    Then a color-invariant state of two gluons might be denoted (11 + 22 + 33). The meaning here is that if you "pulled apart" the glueball, 1/3 of the time you'd get two type 1 gluons, 1/3 of the time two type 2 gluons, and 1/3 of the time two type 3 gluons.

    Is the 11 + 22 + 33 state color invariant? With only two colors in play, the only simple color transformation is the one that takes r to g and g to -r. You can verify that this transformation turns a type 1 gluon into a type 2 and a type 2 into a type 1 (up to signs and factors of i which I'm not going to work out). It leaves the type 3 gluon unchanged (though this gluon still isn't color neutral because there are other more exotic color transformations under which it changes). So our simple transformation takes (11 + 22 + 33) into (22 + 11 + 33) which is of course the same state. The state is indeed color invariant (though really we'd have to prove this for all the more exotic possible color transformations).

    So that's a long but still shallow introduction to the meaning of "color neutral" and how a glueball can be color neutral. To get the full story, learn group theory :)
  11. Aug 16, 2012 #10
    And in any case, there will be one red, one green, and one blue quark, right? Just like electron can be in various states, but it always has charge -1e, never 0 or +1e. Electron's charge never change. It's not a probabilistic measurable.

    This part is confusing.

    Unlike baryon case above, where we always (in all six cases) get one red, one green, and one blue quark - they are just reordered, it looks like gluons in your glueball *don't* have same colors in these three cases.

    It does not sound right to me: it's like saying that electron 1/2 of the time can have charge -1e, but other 1/2 it can be +1e.
  12. Aug 17, 2012 #11
    Yes, in this language there will always be a red, a blue, and a green quark in a baryon.

    Indeed. This is much like a meson, where if you pull apart the quark and antiquark you have a 1/3 chance of getting red-antired, 1/3 of getting green anti-green, and 1/3 of getting blue-antiblue.

    It may not be intuitive, but it's the way of things. I'll just say that your analogy is not right--a superposition of +1 electric charge and -1 electric charge is not electrically neutral in the group theoretic way I was talking about above for color. Your intuition is right that such a thing would not be electrically neutral, but your intuition is wrong in thinking that the glueball state I wrote above has the same problem.

    Similar things go on with quark flavors too. For example the neutral pion is a 50/50 superposition of an up-antiup pair and a down-antidown pair. If you pulled it apart, half the time you'd end up with an up quark and antiquark, and half the time with a down quark and anti-quark.
  13. Aug 17, 2012 #12


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    I don't think the linear algebra is too complicated.

    We start with nine 3*3 matrices [itex]\tilde{T}^a[/itex], [itex]a=0 \ldots 8[/itex]. These matrices are now used as basis vectors; one could e.g. introduce the matrices such that they have zeros everywhere except at one position.

    Using these matrices as basis vectors we can introduce nine "gluons" [itex]A^a[/itex] as linear combinations of these matrices.

    [tex]A = \sum_{a=0}^8 A^a\,\tilde{T}^a[/tex]

    Now we transform to a new basis

    [tex]\tilde{T}^a \to T^a[/tex]

    such that new matrices [itex]T^a[/itex], [itex]a=1 \ldots 8[/itex] are trace-free, except for a=0, i.e.

    [tex]\text{tr}\,T^a =0;\quad a=1\ldots 8 [/tex]

    Expanding the vector in terms of this new basis reads

    [tex]A = A^0\,T^0 + \sum_{a=1}^8 A^a\,T^a[/tex]

    The new matrix for a=0 is simply the 3*3 identity matrix.

    Now we want to have trace-free gluon fields, that means we set [itex]A^0[/itex] = 0 and we finally arrive at

    [tex]A|_{\text{tr}=0} = \sum_{a=1}^8 A^a\,T^a[/tex]

    The strange linear combinbations are introduced via this basis transformation, i.e. in terms of the new matrices. The condition that the gluon fields are trace-free eliminates one "color-neutral" gluon.

    A well-known representation for the matrices can be found here.

  14. Aug 17, 2012 #13
    Ok, now I can see how at least one kind of color neutral glueball is possible: a "mesonic" one which consists of two gluons with opposite color charges. A "gluonium" if you will :D
  15. Aug 17, 2012 #14


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    No, we havn't talked about this yet.

    What were are trying to understand was the adjoint 8-dim. representation of SU(3) in which one single gluon lives. For glueballs we have to understand how to couple e.g. two adjont reps. i.e. tw gluons (or more) in order to get one singulet (the singulet is something like the identity matrix which we have eliminated in the adjoint rep.). This is like Clebsch-Gordon and angular momentum or spin coupling, but more complicated b/c of SU(3) instead of SU(2)
  16. Aug 17, 2012 #15
    Please elaborate. Are you really saying that glueball can exist in a state where its overall color charge is probabilistic?

    This is quite hard to believe.

    Charge ##(r \bar{g} + g \bar{r})## is not the came charge as ##i(r \bar{g} - g \bar{r})##. How isolated particle can have its charge float like this?

    If yes, then why electron is not allowed to do this?

    up+anti-up combination has all its quantum numbers cancelled, so I see no problem with it mixing with down+anti-down.
  17. Aug 18, 2012 #16
    The glueball is just color neutral, period. The glueball state I wrote down is invariant under color transformations. In group theoretic terms it is a "color singlet," which is the real meaning of "color neutral."
  18. Aug 18, 2012 #17
    Yes, you told me that. I am curious how you arrived to that conclusion. I don't understand this part:

    "Then a color-invariant state of two gluons might be denoted (11 + 22 + 33). The meaning here is that if you "pulled apart" the glueball, 1/3 of the time you'd get two type 1 gluons, 1/3 of the time two type 2 gluons, and 1/3 of the time two type 3 gluons."

    Why "it might be denoted (11 + 22 + 33)"? This supposition does not look valid to me. Why gluons type 1 are allowed to mix with gluons type 2? They have different (color) charges. Again, it's as if you'd say "electron can mix with a positron". My intuition says "No, it can't".
  19. Aug 18, 2012 #18


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    It is not mixing of particles with antiparticles. So if you try to find an analogy in the lepton sector, it is similar to mixing of positronium (electron+positron) with (true) "muonium" (muon+antimuon) - if all leptons would be massless, this would be allowed.
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