To Prove:

$$\bar{U}(P_2) \gamma^\mu U(P_1)= \frac{1}{2M}\bar{U}(P_2)[(P_1+P_2)^\mu+\iota\sigma^{\mu\nu}(P_2-P_1)_\nu]U(P_1)$$

We use $2g^{\mu\nu}=[\gamma^\mu,\gamma^\nu]_+$ and $\sigma^{\mu\nu}=\frac{\iota}{2}[\gamma^\mu,\gamma^\nu]_-$ to get: $$\bar{U}(P_2) \gamma^\mu U(P_1)= \frac{1}{2M}\bar{U}(P_2)[(P_1+P_2)^\mu-\frac{1}{2}\gamma^\mu\gamma^\nu(P_2-P_1)_\nu+\frac{1}{2}\gamma^\nu\gamma^\mu(P_2-P_1)_\nu]U(P_1)$$
$$\bar{U}(P_2) \gamma^\mu U(P_1)= \frac{1}{2M}\bar{U}(P_2)[(P_1+P_2)^\mu-\frac{1}{2}\gamma^\mu\gamma^\nu(P_2-P_1)_\nu+\frac{1}{2}(2g^{\mu\nu}-\gamma^\mu\gamma^\nu)(P_2-P_1)_\nu]U(P_1)$$ contraction of indices gives: $$\bar{U}(P_2) \gamma^\mu U(P_1)= \frac{1}{2M}\bar{U}(P_2)[(P_1+P_2)^\mu-\frac{1}{2}\gamma^\mu\gamma^\nu(P_2-P_1)_\nu+(P_2-P_1)^\mu-\frac{\gamma^\mu\gamma^\nu}{2}(P_2-P_1)_\nu]U(P_1)$$ rearranging: $$\bar{U}(P_2) \gamma^\mu U(P_1)= \frac{1}{2M}\bar{U}(P_2)[2P_2^\mu-\gamma^\mu\gamma^\nu(P_2-P_1)_\nu]U(P_1) $$ now using Dirac's equation $(\iota\gamma^\mu\partial_\mu-M)\Psi=0, \bar{\Psi}(\iota\gamma^\mu\partial_\mu+M)=0, gives $
$$\bar{U}(P_2)\gamma^\mu U(P_1)=\frac{1}{2M}\bar{U}(P_2)[2P_2^\mu+2\gamma^\mu M]U(P_1)$$
$$\bar{U}(P_2)\gamma^\mu U(P_1)=\bar{U}(P_2)\frac{P_2^\mu}{M}U(P_1)+\bar{U}(P_2)\gamma^\mu U(P_1)$$
Please tell me how to proceed further

This post imported from StackExchange Physics at 2014-12-06 09:31 (UTC), posted by SE-user Akshansh Singh