Instance of Ehrenfest's Theorem

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$$ \frac{\text d \langle {p} \rangle}{ \text{d} t} =\left\langle - \frac{ \partial V }{\partial x} \right\rangle .$$

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$$\frac{\text d \langle {p} \rangle}{ \text{d} t} $$ $$= \frac{\text d}{\text d t} \int\limits_{-\infty}^{\infty} \Psi^* \left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) \Psi ~\text d x$$ $$= \frac{\hbar}{i}\int\frac{\partial }{\partial t} \left(\Psi^*\frac{\partial\Psi}{\partial x}\right)~\text d x $$ $$= \frac{\hbar}{i}\int \frac{\partial \Psi^*}{\partial t}\frac{\partial \Psi}{\partial x} +\Psi^*\frac{\partial }{\partial x}\frac{\partial \Psi }{ \partial t} ~\text {d} x$$ $$= \frac{\hbar}{i}\int\left( -\frac{i\hbar}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}+\frac{i}{\hbar} V\Psi^*\right)\frac{\partial \Psi}{\partial x}+\Psi^* \frac{\partial}{\partial x} \left(\frac{i\hbar}{2m} \frac{\partial^2 \Psi }{\partial x^2} -\frac{i}{\hbar}V\Psi \right)~\text d x$$ $$=\int\left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{\partial x^2}\right) \frac{\partial \Psi}{\partial x}+\Psi^*\frac{\partial}{\partial x}\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}-V\Psi\right)\text d x$$

$$=\left. \left(V\Psi^*-\frac{\hbar^2}{2m}\frac{\partial^2 \Psi^*}{ \partial x^2} \right) \Psi \right|_{-\infty}^{\infty}-\int\left(\frac{\partial}{\partial x} (V\Psi^*)-\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}\right)\Psi \text d x+ \qquad\qquad\left.\Psi^*\left(\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} - V\Psi\right)\right|_{-\infty}^\infty-\int\frac{\partial\Psi^*}{\partial x} \left( \frac{\hbar^2}{2m}\frac{\partial ^2 \Psi}{\partial x^2} - V \Psi \right)\text d x$$

$$=0 + \int\left(\frac{\hbar^2}{2m}\frac{\partial^3 \Psi^*}{ \partial x^3}-\frac{\partial}{\partial x} (V\Psi^*)\right) \Psi \text d x+0+\int\frac{\partial\Psi^*}{\partial x}\left(V\Psi - \frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2}\right) \text d x$$ $$=\int\frac{\hbar^2}{2m} \left(\frac{\partial^3\Psi^*}{\partial x^3}\Psi -\frac{\partial\Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2}\right)+\frac{\partial\Psi^*}{\partial x}(V\Psi )-\frac{\partial}{\partial x}(V\Psi^*)\Psi\text d x $$

$$\vdots$$ $$=\int \frac{\hbar^2}{2m}\left( \Psi^* \frac{\partial^3 \Psi}{\partial x^3} -\frac {\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \right)+\left( V \Psi^* \frac{\partial \Psi}{\partial x}-\Psi^* \frac{\partial}{\partial x} (V \Psi) \right)~\text d x$$ $$\vdots $$ $$= \int \left( V \Psi^* \frac{\partial \Psi}{\partial x}-\Psi^*V \frac{\partial \Psi}{\partial x} - \Psi^* \frac{\partial V}{\partial x} \Psi \right)~\text d x$$ $$= \int\limits_{-\infty}^{\infty} -\Psi^* \frac{\partial V}{\partial x} \Psi ~\text {d} x$$ $$=\left\langle - \frac{ \partial V }{\partial x} \right\rangle $$

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Edit:

Problem 1.7 (Introduction to Quantum Mechanics, 2edJ -David G. Griffiths)

Calculate $ \frac{\text d \langle {p} \rangle}{ \text{d} t}$

The solutions manual for the text provides this incomplete method, I haven't worked out all the details

Answer 1

You're effectively applying the Ehrenfest theorem (see the section "General example"), but you're not making use of the fact that the momentum operator commutes with the kinetic energy (which is essentially just the square of the momentum operator). The two terms involving the kinetic energy are complex conjugates of each other, and thus, since they're real, they cancel. The same for the two terms involving the potential energy that involve the derivative of the wave function and not of the potential.

Answer 2

Just two expand on joriki's answer. Your statement is not true for general Hamiltonian but only for Hamiltonian of the form (kinetic energy which does not depend on position and potential energy which does not depend on time) $$H = T(p) + V(x).$$

Then using the fact that $p = -i \partial_x$ and $\dot p = i [H,p]$, you can show easily that $$ \frac{dp}{dt} = i \underbrace{[T(p), p]}_{=0} + i [V(x), p] = - V'(x). $$ So you equation is even valid before taking the expectation value.