Commutator of momentum and velocity operator in relativistic quantum mechanics I am currently reading up on Dirac's 'The Principles of Quantum Mechanics'. 
It is stated there, that for the motion of a free particle with Hamiltonian \begin{equation}H=c(\boldsymbol{\alpha},\boldsymbol{p}+\rho_3 mc^2)\end{equation}
we see at once that the momentum commutes with $H$ and is thus a constant of the motion. Further, the $x_1$-component of the velocity is \begin{equation} \dot{x}_1=[x_1,H]=c\alpha_1. \end{equation} and therefore we see a different relation between velocity and momentum in relativistic quantum mechanics, that they don't commute.
I was able to calculate the commutators $[p_1,H]$ and $[x_1,H]$. But even though I thought intuitively it should be clear, that $[\alpha_1,p_1]\neq0$, when I try to calculate it explicitly, I stumble. The $\alpha_1$-matrix written out should be \begin{equation} \alpha_1=c \left(
        \begin{array}{cccc}
            0 & 0 & 0 & 1  \\
            0 & 0 & 1 & 0  \\
            0 & 1 & 0 & 0  \\
            1 & 0 & 0 & 0  \\
        \end{array} \right)
\end{equation} with eigenvalues $\pm c$.
But now when I calculate 
\begin{equation}
\alpha_1 p_1 u^1=\left(
        \begin{array}{cccc}
            0 & 0 & 0 & 1  \\
            0 & 0 & 1 & 0  \\
            0 & 1 & 0 & 0  \\
            1 & 0 & 0 & 0  \\
        \end{array} \right) \left(
  \begin{array}{c}
  p_1\\
  0\\
  p_1 a\\
  p_1 b
 \end{array}
 \right)= \left(
  \begin{array}{c}
  p_1 b\\
  p_1 a\\
  0\\
  p_1
 \end{array}
 \right)
\end{equation}
i get the same as
\begin{equation}
p_1\alpha_1  u^1=p_1\left(
        \begin{array}{cccc}
            0 & 0 & 0 & 1  \\
            0 & 0 & 1 & 0  \\
            0 & 1 & 0 & 0  \\
            1 & 0 & 0 & 0  \\
        \end{array} \right)\left(
  \begin{array}{c}
  1\\
  0\\
  a\\
  b
 \end{array}
 \right)=p_1\left(
  \begin{array}{c}
  b\\
  a\\
    0\\
  1
 \end{array}
 \right)
\end{equation}
I feel like I'm overseeing something really basic here. 
 A: 
But even though I thought intuitively it should be clear, that $[\alpha_1,p_1]\neq 0$ [...]

Why do you think this?  $p_1$ is a differential operator, and $\alpha_1$ is a constant matrix.

Velocity in the sense of $\dot x = i[H,x]$ is not something which can be measured at any moment in time - it requires two successive position measurements.  One can find a simultaneous eigenbasis of $\alpha_1$ and $\mathbf p$ by simply writing the Dirac spinor components as spatial plane waves
$$\psi = \pmatrix{a(t)\\b(t)\\c(t)\\d(t)} e^{i\mathbf p \cdot \mathbf x}$$
and then choosing an eigenbasis of $\alpha_1$, so $a(t) = \pm d(t)$, and  $b(t) = \pm c(t)$.  Such a basis might be
$$ c_1(t)e^{i\mathbf p \cdot \mathbf x} \pmatrix{1\\0\\0\\1}, \ c_2(t)e^{i\mathbf p \cdot \mathbf x} \pmatrix{1\\0\\0\\-1}, \ c_3(t)e^{i\mathbf p \cdot \mathbf x} \pmatrix{0\\1\\1\\0}, \ c_4(t)e^{i\mathbf p \cdot \mathbf x} \pmatrix{0\\1\\-1\\0}$$
One cannot, however, impose the additional demand of harmonic time depenence, because $\alpha_1$ does not commute with the Hamiltonian. Operationally, at the next instant $\alpha_1$ will be different, which means that this basis will no longer be an eigenbasis of it.
