In quantum mechanics, the velocity field which governs phase space, takes the form(adsbygoogle = window.adsbygoogle || []).push({});

\begin{equation}

\boldsymbol{\mathcal{w}}=\begin{pmatrix}\partial_tx\\\partial_tp\end{pmatrix}

=\frac{1}{W}\begin{pmatrix}J_x\\J_p\end{pmatrix}

=\begin{pmatrix}\frac{p}{m}\\-\frac{dV}{dx}-\sum\limits_{l=1}^\infty

\frac{(i\hbar/2)^{2l}}{(2l+1)!}\frac{\partial_p^{2l} W}{W}\frac{d^{2l+1}

V}{dx^{2l+1}}\end{pmatrix}\

\end{equation}

For reference see Eq.(17) of https://journals.aps.org/pra/abstract/10.1103/PhysRevA.95.022127

The total derivative of p which is now a function in x, p and t, takes the form

\begin{equation}

\label{eq:Total_Derivative}

\frac{dp}{dt}=\partial_tp+\frac{dx}{dt}\partial_xp+\frac{dp}{dt}\partial_pp

=\partial_tp+\frac{p}{m}\partial_xp+\frac{dp}{dt}\; .

\end{equation}

In the above equation

\begin{equation}

\frac{dx}{dt}=\frac{p}{m}\text{ and }\partial_pp=1\; .

\end{equation}

My question is whether Eq.(\ref{eq:Total_Derivative}) is correct or not, and can I therefore conclude that

\begin{equation}

\partial_tp=-\frac{p}{m}\partial_xp=\frac{J_p}{W}\; .

\end{equation}

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# A Total derivative of momentum in quantum mechanics

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