In the Dirac equation, if the $\alpha$ is the mean velocity, why does it commute with $x,y,z,t$ if the velocity is related to the momentum? In the Wikipedia talk page for the Dirac equation I found the following passage:

The Dirac equation can be proved with the help of the correspondence principle. The energy and momentum of a particle can be expressed by the equation
  $$
E^{2}=p_{1}^{2}c^{2}+p_{2}^{2}c^{2}+p_{3}^{2}c^{2}+m^{2}c^{4}
$$
  This equation can be divided into {\displaystyle E} E on both sides. We obtain
  $$
E={\frac {v_{1}}{c}}p_{1}c+{\frac {v_{2}}{c}}p_{2}c+{\frac {v_{3}}{c}}p_{3}c+{\frac {v_{0}}{c}}m\,c^{2}
\tag 1
$$
  where $v_{0}={\sqrt {c^{2}-v^{2}}}$, and $v^{2}= v_{1}^{2} + v_{2}^{2} + v_{3}^{2}$;
Really ${\frac {p_{1}c}{E}}={\frac {mv_{1}c(c/v_{0})}{mc^{2}(c/v_{0})}}={\frac {v_{1}}{c}}$ and so on, and $ {\frac {mc^{2}}{E}}={\frac {mc^{2}}{mc^{2}(c/v_{0})}}={\frac {v_{0}}{c}}$;
The Dirac equation has the form
  $$
i\hbar {\frac {\partial \psi }{\partial t}}=(\alpha _{1}{\hat {p}}_{1}c+\alpha _{2}{\hat {p}}_{2}c+\alpha _{3}{\hat {p}}_{3}c+\alpha _{0}\,m\,c^{2})\psi 
\tag 2
$$
  where $\alpha_{j}$ is matrix $(j=0,1,2,3,)$. We obtain from $(1)$ and $(2)$: $v_{j}/c\rightarrow \alpha _{j}$.
In fact, in quantum mechanics it shows that the relativistic velocity operator $v_{\nu }=dx_{\nu }/dt;(\nu =1,2,3)$ has the form $v_{\nu }=c\alpha _{\nu }$, ie is a matrix operator (see textbook LA Borisoglebsky "Quantum mechanics", Minsk, publishing house "University", 1988, p.340-342).
Really
  $$
{\frac {dx_{\nu }}{dt}}={\frac {\partial x_{\nu }}{\partial t}}+[H,x_{\nu }]
$$
  where
  $$
H=\alpha _{\nu }p_{\nu }c+\alpha _{0}m\,c^{2}
$$
  Since the operator $x_{\nu }$ does not depend on time, it will be $dx_{\nu }/dt=[H,x_{\nu }]$. We get
  $$
{\frac {dx_{\nu }}{dt}}=[(\alpha _{\mu }p_{\mu }c+\alpha _{0}m\,c^{2}),\,x_{\nu }]
$$
  The matrix $\alpha _{\mu }$ commutes with $x_{\nu}$, so that the matrix $\alpha _{\mu }$ can be factored out. Finally we have
  $$
dx_{\nu }/dt=c\alpha _{\mu }[p_{\mu },x_{\nu }]=c\alpha _{\mu }\delta _{\mu \nu }=c\alpha _{\nu }
$$

Why does the $\alpha_\mu$ commute with $x_\nu$? If $\alpha_\mu$ means the velocity,then the Poisson bracket $(v_1,x_1)=\frac{\partial{v_1}}{\partial{p_1}}=\frac{c^2}{E}\neq 0$ and the commutator $[v_1,x_1]=i\hbar(v_1,x_1)\neq 0$.
 A: You say the velocity is related to the momentum. What you probably have in mind is the classical relation
$$
\mathbf{P}=m\mathbf{V}
$$
This relation does not hold true in quantum mechanics. Velocity and momentum are not proportional operators. Therefore there is not contradiction between
$$
[\mathbf{V}_j,\mathbf{X}_k]=0 \quad \quad \textrm{and} \quad \quad [\mathbf{P}_j,\mathbf{X}_k]=-ih\delta_{jk}
$$
@Adam pointed this out implicitly.
A: It’s important to understand what the “Dirac wave function" represents. An electron of momentum $p$ with spin +1/2 in the $z$-direction is represented by the quantum state

$$|p,\uparrow\rangle\quad.$$
A general state of momentum $p$ will be represented as a superposition
$$c_1 |p,\uparrow\rangle + c_2 |p,\downarrow\rangle$$
which can be packaged nicely into a spinor
$$\chi(p) = \begin{pmatrix}c_1\\c_2\end{pmatrix}\quad.$$
Under Lorentz transformations, the momentum will change, but the spinor will also rotate. This transformation is not very nice. The solution is to multiply this spinor by a boost matrix:
$$\psi^\alpha(p) = (\sqrt{p\cdot \sigma})^{\alpha}{}_{\alpha'}\,\chi^{\alpha'}(p)\quad.$$
The new spinor $\psi(p)$ transforms nicely under Lorentz transformations $\Lambda^\alpha{}_{\alpha'}$, and inverting the boost:
$$(\sqrt{\Lambda p\cdot \sigma}\;{}^{-1})^{\alpha}{}_{\beta}\,\Lambda^{\beta}{}_{\beta'}(\sqrt{p\cdot \sigma})^{\beta'}{}_{\alpha'}=R{}\,^{\alpha}{}_{\alpha'}$$
gives precisely the needed rotation of the spinor. This construction is not unique. We can also construct a spinor in the conjugate representation of the Lorentz group:
$$\widetilde\psi^\alpha(p) = (\sqrt{p\cdot \bar\sigma}\,)^{\alpha}{}_{\alpha'}\,\chi^{\alpha'}(p)\quad$$
so that now $\widetilde \psi$ transforms under $\Lambda\!{}^*{}^\alpha{}_{\alpha'}$. These spinors satisfy
$$p\cdot \bar\sigma \,\psi(p) = m \,\widetilde\psi(p)\quad.$$
We can Fourier transform to position space, where the equation will read
$$\bar\sigma^\mu \partial_\mu \psi(x)= m\widetilde \psi(x)\quad.$$
These wavefunctions are more conveniently packaged into a single wave function
$$\Psi(x) = \begin{pmatrix}\psi(x)\\\widetilde{\psi}(x)\end{pmatrix}$$
that now satisfies the Dirac equation
$$\gamma^\mu \partial_\mu \Psi(x) = m\, \Psi(x)\quad.$$
But now remember that $\Psi(x)$ is just a convenient way of packaging the coefficients $\chi(p)$ and is not a quantum state at all. Thus, the matrices $\gamma^\mu$ commute with $X$ because they are not quantum operators.
