I apologize for the cumbersome calculations.

Let's have $\Psi$, $i\Psi^{\dagger}$, which are canonical coordinate and impulse in space of solutions of Dirac equation. It can be showed that they have infinitesimal transformations $$ \delta \Psi = \frac{1}{2}\omega^{\mu \nu}\eta_{\mu \nu}\Psi , \quad i\delta \Psi^{\dagger} = -\frac{i\omega^{\mu \nu}}{2}\Psi^{+}\gamma^{0}\eta_{\mu \nu}\gamma^{0}, \quad \eta_{\mu \nu} = -\frac{1}{4}(\gamma_{\mu}\gamma_{\nu} - \gamma_{\nu}\gamma_{\mu}), \qquad (.1) $$ and $\omega^{\mu \nu}$ are local coordinates of the Lorentz group in bispinor representation.

I wanted to use Poisson's (i.e., introducing Poisson's bracket with all convenient consequences) formalism for Dirac equation. So after introducing Poisson's brackets $$ [A, B]_{P} = \frac{\partial A}{\partial \Psi}\frac{\partial B}{i\partial \Psi^{\dagger}} - \frac{\partial A}{i\partial \Psi^{\dagger}}\frac{\partial B}{\partial \Psi} $$ I also decided to introduce generating function for canonical transformations (I have problem with them; see the question). For function $F(\Psi , i\Psi^{+}{'})$ the canonical equations are following: $$ \Psi {'} = \frac{\partial F}{i\partial \Psi^{\dagger}{'}}, \quad i \Psi^{\dagger} = \frac{\partial F}{\partial \Psi}. $$ So for infinitesimal transformations $$ F = \Psi i\Psi^{\dagger}{'} + G(\Psi , i\Psi^{\dagger}{'}), $$ where $G$ is called generator, I got $$ \Psi{'} = \Psi + \frac{\partial G}{i\partial \Psi^{\dagger}{'}} \approx \Psi + \frac{\partial G}{i\partial \Psi^{\dagger}}, \quad i\Psi^{+} = i\Psi^{\dagger}{'} +\frac{\partial G}{\partial \Psi} \Rightarrow $$ $$ \delta \Psi = \frac{\partial G}{i\partial \Psi^{+}}, \quad i\delta \Psi^{\dagger} = -\frac{\partial G}{\partial \Psi} \Rightarrow $$ $$ \delta \Psi = [\Psi , G]_{P}, \quad i\delta \Psi^{\dagger} = [i\Psi^{\dagger}, G]_{P}. \qquad (.2) $$ As result I can get the generator for $(.1)$ by using $(.2)$.

For $\delta \Psi$, $$ \frac{\partial G}{i\partial \Psi^{\dagger}} = \frac{1}{2}\omega^{\mu \nu}\eta_{\mu \nu}\Psi, \qquad (.3) $$ for $i\delta \Psi^{\dagger}$, $$ \frac{\partial G}{\partial \Psi} = -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\gamma^{0}\eta_{\mu \nu}\gamma^{0}. \qquad (.4) $$ Then I expected that the density of the generators of the Lorentz group in bispinor case, which has the expression $$ i\omega\Psi^{\dagger}\eta_{\mu \nu}\Psi , $$ can be naturally interpreted as the generator of infinitesimal transformations. So by substituning it into $(.2)$ I got $$ \delta \Psi = \frac{1}{2}\omega^{\mu \nu}\eta_{\mu \nu}\Psi , \quad i\delta\Psi^{\dagger} \neq -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\gamma^{0}\eta_{\mu \nu}\gamma^{0} , $$ which contradicts with $(.3), (.4)$. Also, it seems that system $(.3), (.4)$ doesn't have solution.

Where is the mistake?

An addition.

My attempt to solve $(.3), (.4)$.

From $(.3)$ I can get $$ G = \frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\eta_{\mu \nu}\Psi + G_{1}(\Psi) $$ and by substituting it in $(.4)$ it follows that $$ -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\eta_{\mu \nu} - \partial_{\Psi}G_{1} = -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\gamma_{0}\eta_{\mu \nu}\gamma_{0}. $$ It is possible to transform the right part of this expression by using commutator $$ [\gamma_{0}, \eta_{\mu \nu}] = \frac{1}{2}\left( g_{0 \nu}\gamma_{\mu} - g_{0\mu}\gamma_{\nu}\right): $$ $$ -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\eta_{\mu \nu} - \partial_{\Psi}G_{1} = -\frac{i\omega^{\mu \nu}}{2}\Psi^{\dagger}\eta_{\mu \nu} - i\omega^{\mu 0}\gamma_{\mu}, $$ so $$ \partial_{\Psi}G_{1}(\Psi) = i\omega^{\mu 0}\Psi^{\dagger}\gamma_{\mu}, $$ which leads to the problems.

  • $\begingroup$ By $\Psi^+$ did you mean $\Psi^\dagger$? Or is it a completely separate quantity? $\endgroup$
    – David Z
    Commented Oct 19, 2013 at 17:14
  • $\begingroup$ @DavidZ . I meaned hermitian conjugate to $\Psi$ bispinor. So Dirac conjurated bispinor is $\bar \Psi = \Psi^{+}\gamma^{0}$. $\endgroup$
    – user8817
    Commented Oct 19, 2013 at 17:16
  • $\begingroup$ OK, gotcha. That would normally be indicated as \Psi^\dagger, $\Psi^\dagger$. Perhaps you'd like to edit accordingly. $\endgroup$
    – David Z
    Commented Oct 19, 2013 at 17:19


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