# 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 $[v_1,x_1]=i\hbar(v_1,x_1)\neq 0$.

• Welcome to the site! I have transcribed your images for you; in future please use explicit text entry for anything other than diagrams and similar pictures. (For the specific case of wikipedia, note that copy-pasting the math images will produce working LaTeX.) – Emilio Pisanty Sep 14 '17 at 14:59
• Please take note of the comment at the top of your linked page. The material on this page is seriously misleading with regard to the alpha matrix. It commutes with x because it operates only upon the sponsor coordinates of the wave function. – Lewis Miller Sep 14 '17 at 15:26
• That should have read spinor components. DAMN Autocorrect. – Lewis Miller Sep 14 '17 at 15:28
• @Lewis Miller Thanks,I still could not get the key point. – John Sep 14 '17 at 15:35
• 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...... – John Sep 15 '17 at 7:29