# Ward Identity for Pair Pair Annihilation

The following Feynman Diagrams for the process $$e^++e^-\to\gamma+\gamma$$ are:

Knowing this I wrote the amplitude matrix:

$$M = -e^2\epsilon^{*}_{\mu}(p_3)\epsilon^{*}_{\nu}(p_4)\bar{u}(p_2)\left[\frac{\gamma^\mu \left(\displaystyle{\not} p_1 -\displaystyle{\not} p_3 - m\right)\gamma^\nu}{(p_1 - p_3)^2 -m^2}+\frac{\gamma^\nu \left(\displaystyle{\not} p_1 -\displaystyle{\not} p_4 - m\right)\gamma^\mu}{(p_1 - p_3)^2 -m^2} \right]u(p_1).$$

I define $$M$$ as: $$M = M^{\mu \nu}\epsilon^{*}_{\mu}(p_3)\epsilon^{*}_{\nu}(p_4),$$ where $$M^{\mu \nu} = -e^2\bar{u}(p_2)\left[\frac{\gamma^\mu \left(\displaystyle{\not} p_1 -\displaystyle{\not} p_3 - m\right)\gamma^\nu}{(p_1 - p_3)^2 -m^2}+\frac{\gamma^\nu \left(\displaystyle{\not} p_1 -\displaystyle{\not} p_4 - m\right)\gamma^\mu}{(p_1 - p_3)^2 -m^2} \right]u(p_1).$$

Now, I want to prove Ward's Identity for this case so:

$$(p_3)_\mu M^{\mu \nu} = 0 \iff -e^2\bar{u}(p_2)\left[\frac{\displaystyle{\not} p_3\left(\displaystyle{\not} p_1 -\displaystyle{\not} p_3 - m\right)\gamma^\nu}{(p_1 - p_3)^2 -m^2}+\frac{\gamma^\nu \left(\displaystyle{\not} p_1 -\displaystyle{\not} p_4 - m\right)\displaystyle{\not} p_3}{(p_1 - p_3)^2 -m^2} \right]u(p_1).$$

How do I go from here?

In the screenshot above, we see an example for Compton scattering. I tried to use the same idea but I don't understand why from equation (3) to equation (4) the indices in the gamma matrices change.

Hope you can help me.

• Check the amplitude, $\mu$ and $\nu$ are misplaced. – Youran Jun 7 at 10:37
• Why do you say that $\mu$ and $\nu$ are misplaced? My equation has the same indices as equation (3) in the screenshot. – RFeynman Jun 7 at 21:20
• From $\epsilon_\mu(p_3)$ we see that $\mu$ is for the lower vertex. So the numerator of the first diagram should be $\gamma^\nu(p-p+m)\gamma^\mu$. Maybe it is clearer to add the diagrams for equations (3) (4) and (5). By the way, it should be $-m$ rather than $+m$ in the numerator and the denominator of the second diagram is $(p_1-p_4)^2$ not $(p_1-p_3)^2$. – Youran Jun 8 at 4:37