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?


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:

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

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  • $\begingroup$ "...Which implies that $i\Psi^{+}$ is the canonical momentum of $\Psi$, thus satisfies the canonical Poisson brackets...", - did you make this conclusion by the connection $$ L_{d} = \pi \partial_{0}\Psi - H_{d}, $$ where $H$ is given from Dirac equation, $$ i\partial_{0}\Psi = H\Psi , \quad H = (\gamma \cdot \hat {\mathbf p}) + m, \quad H_{d} = \bar {\Psi} H \Psi ? $$ $\endgroup$ – user8817 Oct 15 '13 at 16:43
  • $\begingroup$ And what physical sense has bracket $$ \delta \Psi = \left[\frac{1}{2}\omega_{\alpha \beta}S^{\alpha \beta}, \Psi^{+}\right]_{P.B.}? $$ $\endgroup$ – user8817 Oct 15 '13 at 18:05
  • $\begingroup$ Oh yes, I understand. It is generator of infinitesimal transformations through Poisson's brackets. $\endgroup$ – user8817 Oct 16 '13 at 8:42
  • $\begingroup$ Unfortunately, it generates false infinitesimal transformations for $i\Psi^{+}$. $\endgroup$ – user8817 Oct 18 '13 at 18:40
  • $\begingroup$ If the problem is a sign problem, please notice that the $\Psi$s are Grassmann variables and they acquire minus signs when they are commuted. $\endgroup$ – David Bar Moshe Oct 18 '13 at 19:37

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