- #1

zhouhao

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##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...