Why the velocity operator commutes with position (Dirac equation)

[zhouhao]

$\hat{v}_i=c\hat{\alpha}_i$ commute with $\hat{x}_i$,
$E^2={p_1}^2c^2+{p_2}^2c^2+{p_3}^2c^2+m^2c^4$
But in classical picture,the poisson braket $(v_i,x_i)=\frac{\partial{(\frac{c^2p_i}{E})}}{\partial{p_i}}=\frac{c^2}{E}+p_i\frac{\partial{(\frac{c^2}{E})}}{\partial{p_i}}=\frac{c^2}{E}-\frac{p_1c^2}{E^2}\frac{\partial{E}}{\partial{p_1}}=\frac{c^2}{E}-\frac{{p_1}^2c^4}{E^3}\neq 0$
In quantum mechanic,commutor [$\hat{v}_i,\hat{x}_i$]=$i\hbar(v_i,x_i)=\frac{c^2}{{\hat{H}}}-\frac{{\hat{p}_1}^2c^4}{\hat{H}^3}$,I do not know what the result is and how to deal with the formula,but it not equal zero...

 

[zhouhao]

In classic mechanic,if we regard $v_i$ as a field $v_i(t,x_1,x_2,x_3,x_4)$,
the poisson braket $(v_i,x_i)=\frac{\partial{v_i}}{\partial{p_i}}=\frac{\partial{v_i}}{\partial{x_i}}\frac{\partial{x_i}}{\partial{p_i}}+\frac{\partial{v_i}}{\partial{t}}\frac{\partial{t}}{\partial{p_i}}=0$,

so the velocity and position commute.

I do not know how to explain this...