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There's a particle with mass $M$ moving in a static three-dimensional potential $V(\vec{r})$. The time-independent Schrödinger equation is $$-\frac{\hbar^2}{2M} \nabla^2 \psi + V(\vec{r})\psi = i\hbar\frac{\partial \psi}{\partial t}.$$ And the local probabilistic density $\rho$ and local probabilistic flux $\vec{j}$ are $$\rho = \psi^* \psi$$ $$\vec{\jmath} = \frac{\hbar}{2Mi} (\psi^* \nabla \psi - \psi \nabla \psi^{*}).$$

My problem is, how to show the following equation (the quantum mechanical interpretation of Newton's Second Law)? $$\frac{d}{dt} \int{M\,\vec{\jmath}\,d^3r} = -\int{(\nabla V)\rho\, d^3r}.$$ In fact, I've tried to solve this for days. The following part is my process until now: $$\mathrm{LHS} = \frac{\hbar}{2i} \int\frac{d}{dt}(\psi^* \nabla \psi - \psi \nabla \psi^{*}) \,d^3r,$$ where $$\frac{d}{dt}(\psi^* \nabla \psi - \psi \nabla \psi^{*}) = \frac{i}{\hbar} \left[ \hat{H}\psi^*(\nabla \psi) - \psi^* (\nabla\hat{H})\psi + \hat{H}\psi (\nabla \psi^*)-\psi (\nabla \hat{H})\psi^* \right].$$ Then I have trouble. I have no idea how to get to next stage.

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The Hamiltonian is proportional to the time derivative of the wave function, $\hat{H}=i\hbar\partial \psi/\partial t$.

Suppose you want to find the time derivative of the expectation value of operator $\hat{O}$. $$d\langle\hat{O}\rangle/dt=\frac{\partial}{\partial t}\int \ \psi^*\hat{O}\psi dr^3=\int \frac{\partial \psi^*}{\partial t}\hat{O}\psi +\psi^*\frac{\partial \hat{O}}{\partial t}\psi+\psi^*\hat{O}\frac{\partial \psi}{\partial t} dr^3, \\ d\langle\hat{O}\rangle/dt=\frac{1}{i\hbar}\int i \hbar \psi^*\hat{O}\psi +\psi^*\hat{O}ih\frac{\partial \psi}{\partial t} dr^3+\left\langle\frac{\partial \hat{O}}{\partial t}\right\rangle$$ To simplify, lets assume the operator has no explicit time dependence. Then the third term is zero. Now we substitute the Hamiltonian. $$ \frac{d\langle\hat{O}\rangle}{dt}=\frac{1}{ih}\int (-\hat{H}\psi)^*\hat{O}\psi+\psi^*\hat {O}\hat{H}\psi \ dr^3 $$ The Hamiltonian is Hermitian so $$\frac{d\langle\hat{O}\rangle}{dt}=\frac{1}{ih}\int -\psi^*\hat{H}\hat{O}\psi +\psi^*\hat{O}\hat{H} \ dr^3= \left\langle[\hat{O},\hat{H}]\right\rangle$$ QED.

Now apply it to the momentum operator, $\hat{O}=\hat{p}$, and $\hat{H}=\hat{p}^2/2m+\hat{V}$. $$ \hat{p}\hat{H}=\hat{p}\hat{p}^2/2m+\hat{p}\hat {V}\\ \hat{H}\hat{p}=\hat{p}^2\hat{p}/2m+\hat {V}\hat{p} $$ Powers of the momentum operator commute with each other,so the negative commutator of the Hamiltonian with the momentum operator is the commutator of the momentum operator with the potential operator.

$$\frac{d \langle\hat{p}\rangle}{dt}=\frac{1}{ih}\int -i\hbar\psi^*\frac{\partial V}{\partial x}\psi -i \hbar \psi^*V \frac{\partial \psi}{\partial x} +i\hbar \psi^* V \frac{\partial \psi}{\partial x} dr^3 = \left\langle-\frac{\partial V}{\partial x}\right\rangle$$

Alternatives I've seen will require vector calculus identifies.

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