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I Why the velocity operator commutes with position (Dirac equation)

  1. Sep 14, 2017 at 10:26 AM #1
    ##\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........
     
  2. jcsd
  3. Sep 15, 2017 at 2:32 AM #2
    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......
     
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