Physics Asked by eveningdance on July 28, 2021
Good morning/evening.
In Peskin & Schroeder chapter 16 on gauge invariance, the gauge boson self interaction vertex rules are given. For three gauge bosons, the relevant interaction term in the Lagrangian is
$$mathcal{L}_{YM} supset g ,f^{ijk}A_{mu}{}^{(j)} A_{nu}{}^{(k)} partial^{mu} A^{nu}{}^{(i)} $$
I have rewritten this term using the total asymmetry of the structure constants:
$$ mathcal{L}_{YM} supset dfrac{g}{6} f^{ijk} left[ A_{mu}{}^{(j)} A_{nu}{}^{(k)} left(partial^{mu} A^{nu}{}^{(i)} – partial^{nu} A^{mu}{}^{(i)}right) + A_{mu}{}^{(i)} A_{nu}{}^{(k)} left(partial^{mu} A^{nu}{}^{(j)} – partial^{nu} A^{mu}{}^{(j)}right) + A_{mu}{}^{(k)} A_{nu}{}^{(i)} left(partial^{mu} A^{nu}{}^{(j)} – partial^{nu} A^{mu}{}^{(j)}right)right] $$
I know that the derivative of the field will make the momenta appear in the expression. The problem is that I do not understand which gauge field momentum appears from which derivative, and how to go from this expression to the answer, which is
$$ g f^{abc} left[ g^{mu nu} (k-p)^rho + g^{nu rho} (p-q)^mu + g^{rho mu} (q-k)^nu right] $$
where the momenta and indices are taken according to the attached diagram.
Thanks in advance for any clarification!
To see which momentum appears, its useful to fix some configuration as Peskin & Schroeder suggest. After writing out the Lagrangian, the term linear in g is begin{align} -igf^{abc}partial_mu A_nu^aA^{b,mu}A^{c,nu} end{align} This term appears (after some index renaming) four times and cancels the factor $frac{1}{4}$ from the Lagrangian. To get the vertex factor, pull down the indices with the metric. Then the vertex factor for the configuration $A_mu^a(k)$, $A_nu^b(p)$ and $A_rho^c(q)$ is begin{align} -gk^nu f^{abc}eta^{murho} end{align} as prefactor of $A_mu^a(k)A_nu^b(p)A_rho^c(q)$. Now you have to sum all possible permutations of this expression, which are 6 in this case. You can do this for example by switching $A$'s and renaming Minkowski and gauge indices to have the same index order as in the first considered term. As an example: begin{align} -g(k^nueta^{murho}f^{abc}A_mu^aA_nu^bA_rho^c+k^rhoeta^{munu}f^{acb}A_mu^aA_rho^cA_nu^b) end{align} It is important here to keep track of the gauge fields, since it is always the momentum of the first A in the expression that gives the momentum Switching now $A_nu^b$ in the above expression to be in front means that it will contribute with p. Now bringing all structure constants into some chosen order, e.g. $f^{abc}$, will give you the signs that appear in the resulting formula
Answered by Moritz on July 28, 2021
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