field theory - Spin tensor and Lorentz group operator in bispinor case

For infinisesimal bispinor transformations we have

$$\delta \Psi = \frac{1}{2}\omega^{\mu \nu}\eta_{\mu \nu}\Psi , \quad \delta \bar {\Psi} = -\frac{1}{2}\omega^{\mu \nu}\bar {\Psi}\eta_{\mu \nu}, \quad \eta_{\mu \nu} = -\frac{1}{4}(\gamma_{\mu}\gamma_{\nu} - \gamma_{\nu}\gamma_{\mu}). \qquad (.1)$$ Then, by compairing $(.1)$ with transformation by the generators of the Lorentz group, $$\delta \Psi = \frac{i}{2}\omega^{\mu \nu}J_{\mu \nu}\Psi ,$$ we can make the conclusion that in bispinor representation $$J_{\mu \nu} = -i\eta_{\mu \nu}. \qquad (.2)$$ By the other way, from Noether theorem we can get spin tensor, $$S^{\mu, \alpha \beta} = \frac{\partial L}{\partial (\partial_{\mu}\Psi)}Y^{\alpha \beta} + \bar {Y}^{\alpha \beta}\frac{\partial L}{\partial (\partial_{\mu}\bar {\Psi})}.$$ Then, by having $(.1)$ and Lagrangian $$L = \bar {\Psi}(i \gamma^{\mu}\partial_{\mu} - m)\Psi ,$$ it's easy to show that $$S^{\mu, \alpha \beta} = i\bar {\Psi}\gamma^{\mu}\eta^{\alpha \beta}\Psi .$$ It's clearly that I can get $(.2)$ by $$S^{\alpha \beta} = \int S^{\mu, \alpha \beta}dx_{\mu},$$ but for me it's not obvious how to compute it. Can you help me?

Answer

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:

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