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Gluon Scattering - Colored Feynman Rules for Yang Mills Theory

  1. Jun 1, 2012 #1

    I'm reading Appendix 1 of Section N2 (Gluon Scattering) in "Quantum Field Theory in a Nutshell" by Anthony Zee. The generators for SU(N) have the usual algebra

    [tex][T^a, T^b] = i \epsilon^{a b c}T^c[/tex]

    Suppose we adopt the following normalization

    [tex]\text{tr}(T^a T^b) = \frac{1}{2}\delta^{a b}[/tex]

    Then, we have

    [tex]\text{tr}([T^a, T^b]T^c) = \text{tr}(i f^{a b e} T^{e} T^{c}) = i f^{a b e}\text{tr}(T^e T^c) = \frac{i f^{a b e} \delta^{e c}}{2} = \frac{i f ^{a b c}}{2}[/tex]

    so that

    [tex]f^{a b c} = - 2 i \text{tr}([T^a, T^b]T^c)[/tex]


    [tex]\text{tr}([T^a, T^b][T^c, T^d]) = \text{tr}(i f^{a b \alpha} T^{\alpha} i f ^{c d \beta} T^{\beta}) = -f^{a b \alpha} f^{c d \beta}\text{tr}(T^\alpha T^\beta) = -\frac{1}{2}f^{a b e} f^{c d e}[/tex]

    which implies that

    [tex]f^{a b e} f^{c d e} = -2 \text{tr}([T^a, T^b][T^c, T^d])[/tex]

    However, the author explicitly states that

    [tex]f^{a b e} f^{c d e} = -4 \text{tr}([T^a, T^b][T^c, T^d])[/tex]

    I don't get this additional factor of 2. What am I missing? Is there an error somewhere?

    Thanks in advance!
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 1, 2012 #2


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    Science Advisor

    Right on, the factor should be 2, not 4. For example check it for a = c = 1, b = d = 2. You get

    f123f123 = 2 tr(T3T3)

    The LHS is 1 and the trace of (T3)2 is indeed 1/2, so your form is correct.
  4. Jun 4, 2012 #3
    Thanks Bill.

    How is

    [tex]\frac{\langle 1 2\rangle^4}{\langle 1 2\rangle \langle 2 3\rangle \langle 3 4\rangle \langle 4 1\rangle} = \frac{p_1 \cdot p_2}{p_2 \cdot p_3}[/tex]?

    For a given set of momenta, doesn't one have to find out what <1|2>, <3|4> etc are explicitly? How does this simple relationship arise? I think I missed it somehow. I can't get the LHS to equal the RHS. Any ideas?

    Normally, papers/books seem to derive it up to the LHS..
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