Starting with $⟨x⟩=\int {\Psi }^{\ast }(x,t)x\Psi (x,t)dx$ and the time-dependent Schrödinger equation, show that $\frac{d⟨x⟩}{dt}=\int {\Psi }^{\ast }\frac{i}{\hslash }(\hat{H}x-x\hat{H})\Psi dx$ Given that $\hat{H}=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+V(x)$ show that $\hat{H}x-x\hat{H}=-2\frac{{\hslash }^2}{2m}\frac{d}{dx}=-\frac{{\hslash }^2}{m}\frac{i}{\hslash }{\hat{P}}_x=-\frac{i\hslash }{m}{\hat{P}}_x$ Finally, substitute this result into the equation for $d⟨x⟩/dt$ to show that $m\frac{d⟨x⟩}{dt}=⟨{\hat{P}}_x⟩$ Interpret this result.

Step by Step Solution: Step 1. Apply the time derivative operator to the given expression for $⟨x⟩$: Step 2. $\frac{d}{dt}⟨x⟩=\frac{d}{dt}\int {\Psi }^{\ast }(x,t)x\Psi (x,t)dx$ Step 3. Apply the product rule of differentiation: Step 4. $\frac{d}{dt}⟨x⟩=\int \frac{\partial {\Psi }^{\ast }}{\partial t}x\Psi dx+\int {\Psi }^{\ast }x\frac{\partial \Psi }{\partial t}dx$ Step 5. Substitute the time-dependent Schrödinger equation $i\hslash \partial \Psi /\partial t=\hat{H}\Psi$ and $-i\hslash \partial {\Psi }^{\ast }/\partial t=\widehat{H^{\ast }}{\Psi }^{\ast }$ into the above expression: Step 6. $\frac{d}{dt}⟨x⟩=\frac{i}{\hslash }\int {\Psi }^{\ast }x\hat{H}\Psi dx-\frac{i}{\hslash }\int {\Psi }^{\ast }\widehat{H^{\ast }}x\Psi dx$ Step 7. Rearrange the operators $\hat{H}$ and $x$ in the first term and substitute the expression for $\hat{H}$: Step 8. $\frac{d}{dt}⟨x⟩=\frac{i}{\hslash }\int {\Psi }^{\ast }(\hat{H}x-x\hat{H})\Psi dx$ Step 9. Substitute the expression for $\hat{H}x-x\hat{H}$, given in the question: Step 10. $\frac{d}{dt}⟨x⟩=\frac{i}{\hslash }\int {\Psi }^{\ast }(-\frac{i\hslash }{m}{\hat{P}}_x)\Psi dx=\frac{1}{m}\int {\Psi }^{\ast }{\hat{P}}_x\Psi dx$ Step 11. Thus, we have $m\frac{d}{dt}⟨x⟩=⟨{\hat{P}}_x⟩$ Final Answer: m \frac{d}{d t} \langle x \rangle = \langle \hat{P}_{x} \rangle