Transformation of Dirac spinors

[Gene Naden]

So I am working through Lessons in Particle Physics by Luis Anchordoqui and Francis Halzen, the link is https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf

I am in the discussion of the Dirac equation, on page 21, trying to go from equation 1.5.49 to 1.5.51. And I get stuck.

Equation 1.5.49 is $S^{-1}(\Lambda)\gamma^{\mu}S(\Lambda)\Lambda_{\mu}{^\nu}=\gamma^{\nu}$
where Lambda is an infinitesimal Lorentz transformation and S is the corresponding transformation of the wave function, also infinitesimal, given by $S=1-\frac{i}{2}\omega_{\mu \nu}\Sigma^{\mu \nu}$.

I am not sure I understand $\omega$ . I think it is the parameters of the transformation and that the equation is supposed to be true for all $\omega$.

Equation 1.5.51 is $[\Sigma^{\mu\nu},\gamma^\rho]=-i(g^{\mu\rho} \gamma^\nu-g^{\nu\beta}\gamma^\mu)$. This is the one I am having trouble reproducing.

I see the metric tensor g appears in the result. Perhaps this is from the relation $\gamma^\mu\gamma^\nu+\gamma^\nu \gamma^\mu=2g^{\mu\nu}$

 

[George Jones]

trying to go from equation 1.5.49 to 1.5.51.

Multiply both sides of (1.5.49) by $S\left(\Lambda\right)$ on the left. Use (1.3.16) to get the infinitesimal version of $\Lambda_{\mu}{}^\nu$. Since each $\omega$ is infinitesimal, take the product of any two terms that each involve an $\omega$ to be zero. Keep free and summed indices straight. I also used $2\omega_{\alpha \beta} = \omega_{\alpha \beta} + \omega_{\alpha \beta} = \omega_{\alpha \beta} - \omega_{\beta \alpha}$.

How far can you get?

 

[Gene Naden]

Your suggestions got me a lot closer! I got it except for a minus sign on the overall result.
(1.3.16) $\Lambda^{\mu}_{\nu} = \delta^{\mu}_{\nu} + \omega^{\mu}_{\nu}$
(1.5.49) $S^{-1}\gamma^\mu S \Lambda^{\nu}_{\mu} = \gamma^\nu$
(1.5.50) $S=1-\frac{i}{2} \omega_{\mu\nu}\Sigma^{\mu\nu}$
$\gamma^\sigma \omega^\nu_\sigma - \frac{i}{2} \gamma^\sigma \omega_{\alpha\beta} \Sigma^{\alpha\beta} \delta^\nu_\sigma = - \frac{i}{2}\omega_{\alpha\beta} \Sigma^{\alpha\beta} \gamma^\nu$
$\gamma^\sigma \omega^\nu_\sigma - \frac{i}{2} \gamma^\nu \omega_{\alpha\beta} \Sigma^{\alpha\beta} = - \frac{i}{2}\omega_{\alpha\beta} \Sigma^{\alpha\beta} \gamma^\nu$
$\gamma^\sigma \omega^\nu_\sigma = \frac{i}{2} \gamma^\nu \omega_{\alpha\beta} \Sigma^{\alpha\beta} - \frac{i}{2}\omega_{\alpha\beta} \Sigma^{\alpha\beta} \gamma^\nu$
$= \frac{i}{2} (\gamma^\nu \omega_{\alpha\beta} \Sigma^{\alpha\beta} - \omega_{\alpha\beta} \Sigma^{\alpha\beta}) \gamma^\nu$
$= \frac{i}{2} \omega_{\alpha\beta}(\gamma^\nu \Sigma^{\alpha\beta} - \Sigma^{\alpha\beta} \gamma^\nu)$
$\gamma^\sigma \omega^\nu_\sigma= \frac{i}{2} \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
But $\omega^\nu_\sigma= g ^{\rho\nu} \omega_{\rho\sigma}$
So $\gamma^\sigma g ^{\rho\nu} \omega_{\rho\sigma}=\frac{i}{2} \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$2\omega_{\alpha \beta} = \omega_{\alpha \beta} - \omega_{\beta \alpha}$
$\gamma^\sigma g ^{\rho\nu} \frac{1}{2}(\omega_{\rho\sigma}-\omega_{\sigma\rho})=\frac{i}{2} \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i\gamma^\sigma g ^{\rho\nu} (\omega_{\rho\sigma}-\omega_{\sigma\rho})= \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i (g ^{\rho\nu}\gamma^\sigma\omega_{\rho\sigma}-g ^{\rho\nu}\gamma^\sigma\omega_{\sigma\rho})= \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i (g ^{\rho\nu}\gamma^\sigma\omega_{\rho\sigma}-g ^{\sigma\nu}\gamma^\rho\omega_{\rho\sigma})= \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i (g ^{\rho\nu}\gamma^\sigma-g ^{\sigma\nu}\gamma^\rho)\omega_{\rho\sigma}= \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i (g ^{\alpha\nu}\gamma^\beta-g ^{\beta\nu}\gamma^\alpha)\omega_{\alpha\beta}= \omega_{\alpha\beta}[\gamma^\nu, \Sigma^{\alpha\beta}]$
$-i (g ^{\alpha\nu}\gamma^\beta-g ^{\beta\nu}\gamma^\alpha)= [\gamma^\nu, \Sigma^{\alpha\beta}]$
$[\Sigma^{\alpha\beta},\gamma^\nu]=-i(g ^{\beta\nu}\gamma^\alpha - g ^{\alpha\nu}\gamma^\beta)$
$[\Sigma^{\mu\nu},\gamma^\beta]=-i(g ^{\nu\beta}\gamma^\mu - g ^{\mu\beta}\gamma^\nu)$
This is equation (1.5.51) except for the sign.

 

[Gene Naden]

So I redid the math... it is a bit simpler and the sign matches the reference.