# Vertex factor for $\frac{g}{4} (A_{\nu}A^{\nu})^2$ in QED

Suppose we had an interaction term $\frac{g}{4} (A_{\nu}A^{\nu})^2$ in QED. What would the vertex rule be? I believe it would have four vector indices such as $g^{\mu\rho}g^{\sigma\nu}$ and it would be symmetric in the four indices; but I do not know how to derive this exactly.

• Funny that this question showed up here precisely during the time when students were working on my take-home final exam that included this question. – Matt Reece Dec 16 '17 at 2:49

You can write the interaction as $$\frac{g}{4} A^\rho A^\nu A^\sigma A^\mu g_{\rho \nu} g_{\sigma \mu}$$

Suppose you have a 4-point function to compute with the following external polarization vector

$$\epsilon^{\mu_1}(p_1),\,\epsilon^{\mu_2}(p_2)\,,\epsilon^{\mu_3}(p_3)\,,\epsilon^{\mu_4}(p_4)$$

The rule is then

$$i \frac{g}{4}\epsilon^{\mu_1}(p_1)\,\epsilon^{\mu_2}(p_2)\,\epsilon^{\mu_3}(p_3)\,\epsilon^{\mu_4}(p_4)g_{\rho \nu} g_{\sigma \mu}\left(\delta^{\mu_1}_\rho\delta^{\mu_2}_\nu\delta^{\mu_3}_\sigma\delta^{\mu_4}_\mu + \textit{perm} \right)$$

where perm means all possible permutation of $\mu_1,\mu_2,\mu_3,\mu_4$.

EDIT

This is how I compute the permutation.

In the following I try to simplify the notation and I'll set

$$\delta^1_\mu=\delta^{\mu_1}_\mu\\ \delta^{1\,2}_{\sigma\,\mu} = \delta^1_\sigma\delta^2_\mu + \delta^1_\mu\delta^2_\sigma\\ g_{12}=g_{\mu_1\,\mu_2}$$

I fix one leg and do the permutation on the other legs, and then I just permute the result by permuting the fixed leg. For example,

I choose to fix $\mu_1$. The result is

$$g_{\rho\nu}g_{\sigma\mu} \delta^1_\rho\left[\delta^2_\nu \delta^{3\,4}_{\sigma\,\mu}+\delta^2_\sigma \delta^{3\,4}_{\nu\,\mu}+\delta^2_\mu \delta^{3\,4}_{\sigma\,\nu}\right]$$

then I have to do the permutation of $2,3,4$ which is equivalent to do the permutation of $\nu,\sigma,\mu$.

The result is then

\begin{align} &g_{\rho\nu}g_{\sigma\mu} \left[\delta^{1\,2}_{\rho\nu}\delta^{3\,4}_{\sigma\,\mu}+\delta^{1\,2}_{\rho\sigma}\delta^{3\,4}_{\nu\,\mu}+\delta^{1\,2}_{\rho\mu}\delta^{3\,4}_{\sigma\,\nu}+\delta^{1\,2}_{\nu\sigma}\delta^{3\,4}_{\rho\,\mu}+\delta^{1\,2}_{\mu\nu}\delta^{3\,4}_{\sigma\,\rho}+\delta^{1\,2}_{\sigma\mu}\delta^{3\,4}_{\rho\,\nu}\right]\\ &=8g_{12}g_{34} + g_{\rho\nu}g_{\sigma\mu} \left[\delta^{1\,2}_{\rho\sigma}\delta^{3\,4}_{\nu\,\mu}+\delta^{1\,2}_{\rho\mu}\delta^{3\,4}_{\sigma\,\nu}+\delta^{1\,2}_{\nu\sigma}\delta^{3\,4}_{\rho\,\mu}+\delta^{1\,2}_{\mu\nu}\delta^{3\,4}_{\sigma\,\rho}\right] \end{align}

I'll just compute one term inside the brakets (actually, the other ones are equals)

$$g_{\rho\nu}g_{\sigma\mu} \delta^{1\,2}_{\rho\sigma}\delta^{3\,4}_{\nu\,\mu} = 2(g_{13}g_{24}+g_{14}g_{23})$$

Then, I end up with

$$\frac{ig}{4}8(g_{12}g_{34} + g_{14}g_{32}+g_{13}g_{42})$$

• Okay, given 24 permutations and a factor of $\frac14$ out front, would that yield a vertex rule of $6ig (g^{\rho\nu}g^{\sigma\mu} + g^{\rho\sigma} g^{\nu\mu} + g^{\rho\mu}g^{\sigma\nu}$? Or am I over counting? – user168146 Dec 14 '17 at 14:26
• I meant $2ig(g^{\rho\nu} g^{\sigma\mu} + g^{\rho\sigma} g^{\nu\mu} + g^{\rho\mu} g^{\sigma\mu})$. – user168146 Dec 14 '17 at 14:49
• @AgriculturalEnergy I edited my answer. By matching my convention with yours, I get the same factor of 2. – apt45 Dec 14 '17 at 15:19