Gluon Scattering - Colored Feynman Rules for Yang Mills Theory

1. Jun 1, 2012

maverick280857

Hi,

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

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

Suppose we adopt the following normalization

$$\text{tr}(T^a T^b) = \frac{1}{2}\delta^{a b}$$

Then, we have

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

so that

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

Also,

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

which implies that

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

However, the author explicitly states that

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

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

Last edited: Jun 1, 2012
2. Jun 1, 2012

Bill_K

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.

3. Jun 4, 2012

maverick280857

Thanks Bill.

How is

$$\frac{\langle 1 2\rangle^4}{\langle 1 2\rangle \langle 2 3\rangle \langle 3 4\rangle \langle 4 1\rangle} = \frac{p_1 \cdot p_2}{p_2 \cdot p_3}$$?

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..