Gordon identity manipulation?

[Higgsy]

The Gordon identity allows us to solve using

$$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}u_{s}(\textbf{p}) = \bar{u}_{s'}(\textbf{p}')[(p'+p)^{\mu} -2iS^{\mu\nu} (p'-p)_{\nu}]u_{s}(\textbf{p}) $$

But how would we solve for

$$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) $$

Would a rederivation of the Gordon identity be required?

 

[eljose79]

Hello,

Thank you for your question. The Gordon identity is a useful tool in solving equations involving spinors and gamma matrices. In order to solve for the expression $ 2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) $, we can use a similar approach to the one used in the original Gordon identity.

First, we can write the expression as:

$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) = 2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}\gamma^{0}u_{s}(\textbf{p}) $

Next, we can use the identity $ \gamma^{0}u_{s}(\textbf{p}) = v_{s}(\textbf{p}) $ to rewrite the expression as:

$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) = 2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}\gamma^{0}v_{s}(\textbf{p}) $

Now, we can use the Gordon identity to solve for this expression. The Gordon identity states that:

$$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}\gamma^{0}v_{s}(\textbf{p}) = \bar{u}_{s'}(\textbf{p}')[(p'+p)^{\mu} -2iS^{\mu\nu} (p'-p)_{\nu}]v_{s}(\textbf{p}) $$

Therefore, using this identity, we can solve for the expression $ 2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) $ as:

$2m \bar{u}_{s'}(\textbf{p}')\gamma^{\mu}v_{s}(\textbf{p}) = \bar{u}_{s'}(\textbf{p}')[(p'+p)^{\mu} -2iS^{\mu\nu} (p'-p)