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I A few questions about colour in QCD...

  1. Jan 3, 2017 #1

    CAF123

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    I was watching some new lectures on QCD from Colorado and I have a few questions from what I heard -

    --The ##\lambda^a_{ij}## are generators of SU(3) in the fundamental representation so are 3x3 matrices. That is because the ij indices are colour indices and they act on a 3x1 vector in colour space (the colour wavefunction of the quarks).There are 8 generators (labelled by the index 'a') so the lambda are vectors in what the prof in the Colorado video calls 'gluon space'. This gluon space is spanned by eight independent non trivially transforming colour octet states in a eight dimensional real hilbert space so each of these states can be mapped to a unit vector in the real hilbert space, ie each one is a 8x1 unit basis vector.

    My understanding from the video was that the representation of the lambda in colour space are the 3x3 Gell-mann matrices acting on the colour component of the quark fields embedded in the fundamental representation. What is the representation of these lambda as vectors in the gluon space and what do they act on?

    --In the equation ##A^{\mu} = A^{\mu}_a \lambda^a/2## are we saying that the gluons are exactly the generators of SU(3) in the fundamental representation which give rise to non trivial colour transformations in colour space which when act on quark colour wavefunctions mix around the colours? This equation also tells us the gluons live in the lie algebra of SU(3) since the gluon ##A^{\mu}## can be expanded in the basis of generators ##T^a = \lambda^a/2##. The lie algebra is 8 dimensional but why do we say they transform under the adjoint representation of SU(3)? I guess it makes contact with the above in that we can write down each possible gluon as a basis vector in a eight dim space but what is it that they are transformed by?

    Thanks!
     
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  3. Jan 4, 2017 #2

    Ben Niehoff

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    The gluons live in the "adjoint representation" of the Lie algebra. This is the representation of the Lie algebra which acts on itself by commutation. Using the same a, b, indices for gluon space, the ##8 \times 8## representation matrices ##\rho(\lambda_a)_{bc}## are given by

    $$\rho(\lambda_a)_{bc} \lambda_c \equiv [\lambda_a, \lambda_b]$$
    up to minus signs and numerical factors which I'm ignoring.

    The a index in ##\lambda_a## is telling you which matrix, out of a collection of 8. One is ignoring the fundamental indices, which you could call i, j, and write it ##(\lambda_a)^i{}_j##. The quarks carry these fundamental indices. This is rather similar to the case of spinors vs. Dirac gamma matrices.
     
  4. Jan 5, 2017 #3

    CAF123

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    Hi @Ben Niehoff
    So, is what you have written here equivalent to the representation mapping ##\rho: su(3) \rightarrow su(3)## with ##\lambda^a \mapsto [\lambda^a, \lambda^b]##? If so, since representations are usually first discussed in the context of a homomorphic mapping between a group and a set of linear operators of dimension n acting on a n dim vector space, what is the equivalent representation mapping at the level of the lie group?

    I always remembered seeing the adjoint action being something of the form ##g x g^{-1}## with x and g members of the group so is the adjoint representation mapping at the level of the lie group ## \rho: SU(3) \rightarrow GL(V),## with ##x \mapsto g x g^{-1}## and x and g in SU(3)?

    If I transform ##A_{\mu}^a \lambda^a \rightarrow A_{\mu}^a \rho(\lambda^b) \lambda^a = A_{\mu}^a [\lambda^a, \lambda^b] = A_{\mu}^a i f^{abc}\lambda^c = i f^{abc} A_{\mu}^a \lambda^c##. This is to say the lambda's in the adjoint representation transform by multiplication with structure constants. Do I have my index placements right in this equation and is this correct?

    Thanks!
     
  5. Jan 5, 2017 #4

    Ben Niehoff

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    ##\rho## is the homomorphism from ##\mathfrak{su}(3)## into ##\mathfrak{gl}(8)## which maps ##\lambda_a \mapsto [\lambda_a , \cdot]##.

    You're thinking of the group adjoint ##\mathrm{Ad}_g##. ##\rho## is the algebra adjoint ##\mathrm{ad}_\lambda## for ##\lambda \in \mathfrak{su}(3)##. To see the relation between them, take ##g = 1 + \epsilon \lambda_a## and ##x = 1 + \epsilon \lambda_b## for small ##\epsilon##. Then work out ##g x g^{-1}## to linear order in ##\epsilon##.

    Your indices are inconsistent, so I don't know what you mean.
     
  6. Jan 5, 2017 #5

    CAF123

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    Thanks!
    I see - so in comparing the two representation maps (##\text{Ad}_g## and ##\text{ad}_{\lambda}##) is ##\mathfrak{gl}(8)## the same as the lie algebra of the target space ##GL(8)##? ( e.g ##\mathfrak{su}(3)## is the lie algebra of the lie group ##SU(3)## and likewise ##\mathfrak{gl}(8)## is the lie algebra of ##GL(8)##, where ##GL(8)## is the lie group of 8x8 linear operators). If that is true, I have also seen what we call ##\rho## written as the map between ##\mathfrak{su}(3)## and ##\text{End}(8)##, where ##\text{End}## is the endomorphisms of ##V = \mathfrak{su}(3)##. I guess End is equivalent to the lie algebra of GL(8) but if so, why would one introduce it and what does it mean? Also, could you say what you mean by the map ##\lambda^a \mapsto [\lambda^a, \cdot]##? I haven't come across the notation with the dot in the commutator before.

    I got ##gxg^{-1} = x ## to first order in ##\epsilon##. The lie group adjoint is ##\text{Ad}_x: SU(3) \rightarrow GL(8)## with ##x \mapsto gxg^{-1}## but what does ##gxg^{-1} = x## to first order mean?


    Ah sorry, since ##A_{\mu} = A_{\mu}^a \lambda^a## I was trying to transform the ##\lambda_a## using the lie algebra adjoint discussed above. The ##\lambda_a## transform under the adjoint representation by 8x8 matrices ##\rho(\lambda_a)_{bc}## but I suppose once I understand the ##[\lambda_a, \cdot]## notation I might be able to fix this.
     
    Last edited: Jan 5, 2017
  7. Jan 23, 2017 #6

    CAF123

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    Hi @Ben Niehoff
    Could you explain why something like ##\rho( \lambda_a)_{cd}(\lambda_b)_d## is not a valid indicial rewriting of the matrix/vector product ##\rho(\lambda_a)(\lambda_b) = [\lambda_a, \lambda_b]##?

    Thanks!
     
  8. Jan 23, 2017 #7

    Ben Niehoff

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    Perhaps you can explain what you think ##(\lambda_b)_d## means.
     
  9. Jan 23, 2017 #8

    CAF123

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    I'd say it meant the ##d##th component of ##\lambda_b##? And b labels what basis vector of the lie algebra we are talking about.
     
  10. Jan 24, 2017 #9

    Ben Niehoff

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    The ##\lambda_a## are the Gell-Mann matrices, or at least they were in your initial post.
     
  11. Jan 24, 2017 #10

    CAF123

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    I see. Since the adjoint rep of SU(3) is eight dimensional, 3x3 hermitean traceless matrices (i.e the Gell mann matrices) or 8x1 real vector can transform in adjoint rep. If we consider the latter then what would be the correct transformation law? if I did something like $$\rho(\lambda_a)_{cd} (\lambda_b)_d = [\lambda_a, \lambda_b]_c = i f_{abd} (\lambda_d)_c$$ this doesn't allow me to identify what ##\rho## is.
     
  12. Jan 26, 2017 #11

    CAF123

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    Any ideas on why ^^ is not the right way to write the indices out but ##\rho(\lambda_a)_{bc}\lambda_c= [\lambda_a, \lambda_b]## is? perhaps the way I wrote it above is not consistent with ##\rho## being a lie algebra homomorphism.
     
  13. Feb 1, 2017 #12

    CAF123

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    Ok, I think I'm piecing things together - could you write out explicitly the factors of minus signs and numerical factors (I presume factors of i) that you ignored? I think the equation ##\rho(\lambda_a)_{bc} = -if_{abc}## is correct (seen it in the literature) and from the equation ##\rho(\lambda_a)_{bc} \lambda_c = [\lambda_a, \lambda_b]## that you wrote, I can rewrite the rhs as ##if_{abc}\lambda_c## and thus conclude ##\rho(\lambda_a)_{bc} = if_{abc}## which is incorrect by a minus. So factors are important so I can make comparison.

    Thanks!
     
  14. Feb 1, 2017 #13

    Ben Niehoff

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    Factors of i are entirely conventional. I use anti-hermitian generators so that I don't have to write factors of i everywhere, and my real Lie algebras look real.

    The sign of ##\rho## may well be the opposite of what I wrote. It's not really important. Just pick a convention and stick with it.
     
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