follows $${\bf b}(\vec{p},t,r)= (2 \pi\hbar)^{-3/2}\sqrt{ mc^2
\over E_{\vec{p}}} b_r(\vec{p},t), \hspace{1.0cm} {\rm with\ }
r=1,2,$$ and $${\bf b}(\vec{p},t,r)= (2 \pi\hbar)^{-3/2}\sqrt{ mc^2 \over
E_{\vec{p}}} d^*_{r-2}(-\vec{p},t), \hspace{1.0cm} {\rm with\ }
r=3,4;$$ and ${\bf b}^*$ is determined by the complex conjugate of these equations.

These variables suggest to redefine the solutions to the Dirac equation in the following form $$w_{\alpha}(\vec{p},r)={u_{\alpha}}^r(\vec{p}), \ \ \  {\rm for } \
\ r=1,2,$$ $$w_{\alpha}(\vec{p},r)= v_{\alpha}^{r-2}(-\vec{p}), \
\ \ {\rm for } \ \ r=3,4. \label{orelat}$$ From (\[relations\]) one gets the relations $${w}^{\dag}(\vec{p},r) w(\vec{p},r')=
{E_{\vec{p}} \over mc^2} \delta_{rr'} ,$$ $$\bar{w}(\varepsilon_r \vec{p},r)
w(\varepsilon_r' \vec{p},r')=\varepsilon_r \delta_{rr'},$$ $$\sum_{r=1}^4 w_{\alpha}(\vec{p},r) w_{\alpha'}^{\dag}(\vec{p},r)=
{E_{\vec{p}}\over mc^2 }\delta_{\alpha \alpha'}. \label{ident}$$

In terms of these variables Eqs. (\[fieldone\]) and (\[fieldtwo\]) read $$\psi_{\alpha}(\vec{x},t) = \sum_{r=1}^4 \int
{d^3p \over (2 \pi\hbar)^{3/2}} \sqrt{mc^2 \over E_{\vec{p}}} {\bf
b}(\vec{p},t,r)w_{\alpha}(\vec{p},r) \exp \big( i
\vec{p}\cdot\vec{x}/\hbar \big), \label{fpone}$$ $${\pi_{\psi}}_{\alpha} (\vec{x},t) = i \hbar
{\psi}_{\alpha}^{\dag}(\vec{x},t) = \sum_{r=1}^4 \int {d^3p \over
(2 \pi\hbar)^{3/2}} \ i \hbar \sqrt{mc^2 \over E_{\vec{p}}} {\bf
b}^{*}(\vec{p},t,r) {w}_{\alpha}^{\dag}(\vec{p},r)
\exp\big(-i\vec{p}\cdot\vec{x}/\hbar \big), \label{fptwo}$$ where ${\bf b}(\vec{p},t,r) = {\bf b}(\vec{p},r)\exp \big\{ -i
\varepsilon_r E_{\vec{p}}\ t /\hbar\big\}$ with $\varepsilon_r=1$ for $r=1,2$ and $\varepsilon_r=-1$ for $r=3,4$.

Substituting Eqs. (\[fpone\]) and (\[fptwo\]) into (\[comm\]) we find the Poisson brackets for the variables ${\bf b}$ y ${\bf
b}^{*}$ $$\{ {\bf b}(\vec{p},r), 