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?


1 Answer 1


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:

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

  • $\begingroup$ "...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, 2013 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, 2013 at 18:05
  • $\begingroup$ Oh yes, I understand. It is generator of infinitesimal transformations through Poisson's brackets. $\endgroup$
    – user8817
    Oct 16, 2013 at 8:42
  • $\begingroup$ Unfortunately, it generates false infinitesimal transformations for $i\Psi^{+}$. $\endgroup$
    – user8817
    Oct 18, 2013 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. $\endgroup$ Oct 18, 2013 at 19:37

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