By Noether’s theorem, the generators of the Lorentz group are the zero components of the currents, i.e., the Lorentz charges:

$$S^{\alpha\beta} = S^{0,\alpha\beta}
= i\bar {\Psi}\gamma^{0}\eta^{\alpha \beta}\Psi
= \Psi^{\dagger}\eta^{\alpha \beta}\Psi $$

These charges generate the Lorentz transformations on the spinors by the canonical Poisson brackets:

$$\left \{ \Psi, \Psi^{\dagger} \right \}_{P.B.} = -i \mathbb{I}$$

(With all other Poisson combinations vanishing). The Poisson brackets can be obtained from the time derivative term in the Dirac Lagrangian:

$$i \Psi^{\dagger}\partial_0\Psi $$

Which implies that $i \Psi^{\dagger}$ is the canonical momentum of $\Psi$, thus satisfies the canonical Poisson brackets.

The action of the Lorentz charges correctly generates the Lorentz transformation:

$$\delta \Psi = \left \{ \frac{1}{2} \omega_{\alpha\beta }S^{\alpha\beta}, \Psi^{\dagger} \right \}_{P.B.} = \frac{1}{2}\omega^{\alpha \beta}\eta_{\alpha \beta}\Psi$$

This post imported from StackExchange Physics at 2014-12-14 11:30 (UTC), posted by SE-user David Bar Moshe