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Understanding SU(5) gauge boson matrix?

  1. Jul 25, 2015 #1
    I have questions regarding the 24 gauge bosons of the SU(5) model. I keep seeing this matrix popping up in the documents I'm reading with no real explanation of why:

    SU(5)_gauge_boson_matrix.png

    First of all I'm wondering how this is constructed, which means I'm wondering what the [itex]V_{\mu}^{a}[/itex] look like (I already have the 24 generators [itex]T^{a}[/itex]).

    Actually what motivates me wanting to do this is that there are supposed to be 24 gauge bosons in this theory - the 12 of the standard model plus another 12. I can see the [itex]W^{\pm}[/itex],and also the [itex]W^{3}[/itex] and [itex]B[/itex] bosons are in there which mix to give the [itex]Z^{0}[/itex] and [itex]\gamma[/itex], so that accounts for 4 of the SM bosons. However, looking at the 3x3 in the top left of this matrix, I can see 9 [itex]g_{ij}[/itex] - but there are only supposed to be 8 gluons. So there's an extra g in there that I don't understand?

    Furthermore, I've so far counted 4+9=13 bosons. There are supposed to be 24 so I need another 11 from somewhere but I only see 3 [itex]X^{i}[/itex] bosons and 3 [itex]Y^{i}[/itex] bosons, which gives another 6 (which is not enough, adding up to 19 bosons in total). Or if you count their antiparticles as a further 6 this gives 25 bosons, which is one too many. What is going on here?

    Thanks.
     
  2. jcsd
  3. Jul 25, 2015 #2

    Orodruin

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    The gluon matrix needs to be traceless or you are doing U(5) instead of SU(5). This is where you get a degree of freedom to much. The ##V^a_\mu## are just the components of the gauge field. And yes, you can count ##Y_i## and ##\bar Y_i## as different degrees of freedom, just as you do for ##W^\pm##.
     
  4. Jul 25, 2015 #3
    So I need [itex](g_{11} - \frac{2B}{\sqrt{30}}) + (g_{22} - \frac{2B}{\sqrt{30}}) + (g_{33} - \frac{2B}{\sqrt{30}}) = 0[/itex].

    Which one of the [itex]g_{ij}[/itex] is not supposed to be there so that I end up with only 8 of them?
     
  5. Jul 25, 2015 #4

    Orodruin

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    No. You need ##g_{11} + g_{22} + g_{33} = 0##. This removes one degree of freedom from ##g_{ij}##. The ##B##'s in the upper 3x3 cancel against the ##B##'s in the lower 2x2 when taking the trace. You need the trace of the full 5x5 to be equal to zero or you are not in SU(5) (note that ##\det(e^{iV}) = e^{i{\rm tr}(V)}##!).
     
  6. Jul 25, 2015 #5
    Ahh ok I can see that the [itex]B[/itex] and the [itex]W^{-}[/itex] terms add up to zero when considering the whole 5x5 thing, but I don't understand why the bit in bold is true?
     
  7. Jul 25, 2015 #6

    Orodruin

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    Because if you let the gs be general, then the matrix is not traceless. You therefore need to apply the condition on the gs which reduces your available degrees of freedom by one.

    This is just basic linear algebra, if you have a 25 dimensional vector space and impose one linear constraint on it, you end up with a 24-dimensional one. It works the same way when you do SU(2) and SU(3), the reason you do not have 9 gluons is that the su(3) Lie algebra is traceless.
     
  8. Jul 25, 2015 #7
    Is it that by requiring that [itex]g_{11} + g_{22} + g_{33} = 0[/itex] means, say, [itex]g_{33}[/itex] has to be a combination of [itex]g_{11}[/itex] and [itex]g_{22}[/itex] or similar?
     
  9. Jul 25, 2015 #8

    fzero

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    Yes. In fact we can make the parameterization clearer if we just set ##g_{ab} = f_{abc} G^c##, where ##f_{abc}## are the SU(3) structure constants and ##G^a## are the gluons.

    Edit: For the sake of argument, then
    $$\begin{split}
    & g_{11} = G^3 +\frac{1}{\sqrt{3} }G^8 \\
    & g_{22} = -G^3 +\frac{1}{\sqrt{3}} G^8 \\
    & g_{33} = -\frac{2}{\sqrt{3}} G^8. \end{split}
    $$
     
    Last edited: Jul 25, 2015
  10. Jul 25, 2015 #9
    Excellent, many thanks.
     
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