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I want to calculate the commutation of $[\hat{x},\vec{\alpha}\;\vec{p}]$. This boils down to $$[\hat{x},\vec{\alpha}\;\vec{p}] = i\hbar\hat{\alpha_x}+\left[\hat{x},\hat{\alpha_x}\right]\hat{p_x} +\left[\hat{x},\hat{\alpha_y}\right]\hat{p_y} +\left[\hat{x},\hat{\alpha_z}\right]\hat{p_z}$$

My problem is calculating $[\hat{x},\hat{\alpha_x}]$, $[\hat{x},\hat{\alpha_y}]$ and $[\hat{x},\hat{\alpha_z}]$. Are they just zero? If so why?

Also is there a way I can represent $\hat{x}$ explicitly so that I can multiply it with $\hat{\alpha}$?

$$\hat{\pmb{\alpha}} = \begin{bmatrix} 0 & \pmb{\sigma} \\ \pmb{\sigma} & 0 \\ \end{bmatrix} $$

Where $\pmb{\sigma}$ the Pauli matrices.

Edit(Answer): Since the matrices are arrays of numbers they do commute with $\hat{x}$ hence $[\hat{x},\vec{\alpha}\;\vec{p}] = i\hbar\hat{\alpha_x}$

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    $\begingroup$ Dirac matrices, Pauli matrices, etc... are arrays of numbers, not operators, so they all commute with Hilbert space operators. $\endgroup$ – Cosmas Zachos May 23 at 14:22
  • $\begingroup$ Oh, so that would actually make them zero. Yeah that makes sense. Just a follow up question: If we consider $\frac{\partial{<\hat{x}>}}{\partial{t}} = -\frac{ic}{\hbar}<[\bar{x},\vec{\alpha}\;\vec{p}]>$ what physical interpretation does $\alpha$ take? $\endgroup$ – fielder May 23 at 14:39
  • $\begingroup$ possible duplicates: physics.stackexchange.com/q/476458/84967, physics.stackexchange.com/q/104241/84967 $\endgroup$ – AccidentalFourierTransform May 23 at 15:00
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    $\begingroup$ Velocity? Is this part of your homework? $\endgroup$ – Cosmas Zachos May 23 at 15:08
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Special thanks to Cosmas Zachos!

Since the matrices are arrays of numbers they do commute with $\hat{x}$. Furthermore we have $$\frac{\partial{<\hat{x}>}}{\partial{t}} = -\frac{ic}{\hbar}<\left[\hat{x},\vec{\alpha}\;\vec{p}\right]>$$

Now $\left[ \hat{x},\vec{\alpha}\;\vec{p}\right] = \vec{\alpha}[\hat{x},\vec{p}] + [\hat{x},\vec{\alpha}]\vec{p} = i\hbar\hat{\alpha_x}$ hence $$\frac{\partial{<\hat{x}>}}{\partial{t}} = -\frac{ic}{\hbar}<i\hbar\hat{\alpha_x}> = c\hat{\alpha_x}$$ This means we can classically interpret $\pmb{\vec{\alpha}}$ as the velocity.

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