Prove that $\frac{d \langle p \rangle}{dt} =\langle \frac{ - \partial V}{ \partial x} \rangle$ Knowing that the Schrödinger Equation is:
$$i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi$$
The change of momentum in function of time is:
$$ \frac{d\langle p \rangle}{dt} = -i\hbar \int_{-\infty}^{\infty} \frac{\partial}{\partial t} \Psi^* \frac{ \partial \Psi}{ \partial x} dx$$
I got the expression for $\frac{\partial}{\partial t} \Psi^* \frac{ \partial \Psi}{ \partial x}$ :
$$\frac{\partial}{\partial t} \Psi^* \frac{ \partial \Psi}{ \partial x} = \frac{i\hbar}{2m}(\Psi^*\frac{\partial^3 \Psi}{\partial^2x^3} - \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x}) - \frac{i}{\hbar} |\Psi|^2 \frac{\partial V}{\partial x} $$
The next step is to integrate this expression. Concretely, I am struggling to integrate this part:
$$-\frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x}$$ 
I thought about integrating by parts as follows:
$$-\int \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} = -\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \Psi\int \frac {\partial^3 \Psi^*}{\partial x^3} dx ...$$
But I did not get the equality.
May you help me out?
 A: We have to let the time derivative act on each of $\Psi^*$ and $\Psi$ individually by the product rule. Then we can switch the order of partial derivatives, do a partial integration, insert the right hand side of the Schrödinger equation, and rearrange terms:
$$\begin{align}
\frac{d}{dt} \langle p \rangle
&= \frac{d}{dt} \int_{-\infty}^{\infty} \Psi^* \left( -i\hbar \frac{\partial}{\partial x} \right) \Psi \, dx \\
&= \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial t} \left( -i\hbar \frac{\partial}{\partial x} \right) \Psi \, dx
+ \int_{-\infty}^{\infty} \Psi^* \left( -i\hbar \frac{\partial}{\partial x} \right) \frac{\partial \Psi}{\partial t} \, dx \\
&= \int_{-\infty}^{\infty} \left( -i\hbar \frac{\partial \Psi^*}{\partial t} \right) \frac{\partial \Psi}{\partial x} \, dx
- \int_{-\infty}^{\infty} \Psi^* \frac{\partial}{\partial x} \left( i\hbar \frac{\partial \Psi}{\partial t} \right) \, dx \\
&= \int_{-\infty}^{\infty} \left( -i\hbar \frac{\partial \Psi^*}{\partial t} \right) \frac{\partial \Psi}{\partial x} \, dx
+ \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x} \left( i\hbar \frac{\partial \Psi}{\partial t} \right) \, dx \\
&= \int_{-\infty}^{\infty} \left( -\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V\Psi^* \right) \frac{\partial \Psi}{\partial x} \, dx
+ \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x} \left( -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi \right) \, dx \\
&= -\frac{\hbar^2}{2m}\int_{-\infty}^{\infty} \left( \frac{\partial^2\Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2} \right) \, dx
+ \int_{-\infty}^{\infty} \left( \Psi^* V \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^*}{\partial x} V \Psi \right) \, dx \\
\end{align}$$
By doing a partial integration of the second term in the first integral we see that the first integral actually vanishes:
$$
\int_{-\infty}^{\infty} \left( \frac{\partial^2\Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^*}{\partial x} \frac{\partial^2\Psi}{\partial x^2} \right) \, dx 
= \int_{-\infty}^{\infty} \left( \frac{\partial^2\Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} - \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial\Psi}{\partial x} \right) \, dx 
= 0
$$
Now we do a partial integration of the second term in the second integral:
$$\begin{align}
\int_{-\infty}^{\infty} \left( \Psi^* V \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^*}{\partial x} V \Psi \right) \, dx
= \int_{-\infty}^{\infty} \left( \Psi^* V \frac{\partial \Psi}{\partial x} 
- \Psi^* \frac{\partial}{\partial x} \left( V \Psi \right) \right) \, dx \\
= \int_{-\infty}^{\infty} \left( \Psi^* V \frac{\partial \Psi}{\partial x} 
- \Psi^* \frac{\partial V}{\partial x} \Psi - \Psi^* V \frac{\partial \Psi}{\partial x} \right) \, dx 
= - \int_{-\infty}^{\infty} \Psi^* \frac{\partial V}{\partial x} \Psi \, dx 
= \left\langle \frac{\partial V}{\partial x} \right\rangle
\end{align}$$
Thus,
$$
\frac{d}{dt} \langle p \rangle = -\left\langle \frac{\partial V}{\partial x} \right\rangle
$$
A: Continuing from the following expression in your procedure,
$$ \Psi^*\frac{\partial^3 \Psi}{\partial x^3} - \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} $$
We apply the integration in $(-\infty, \infty)$ on the above expression.
$$
\int_{-\infty}^{\infty} \Psi^*\frac{\partial^3 \Psi}{\partial x^3} - \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \, dx
$$
Taking the second term in the integral and integrating by parts
\begin{align}
-\int_{-\infty}^{\infty} \frac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} \, dx
= -\left[\frac{\partial \Psi}{\partial x} \frac{\partial \Psi^*}{\partial x}\right]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} \frac{\partial^2 \Psi}{\partial x^2} \frac{\partial \Psi^*}{\partial x} \, dx
\end{align}
Note that we consider that far away at infinity nothing interesting happens (as this not a periodic domain case) so we consider both the wave function and its derivatives vanish at the $\infty$ and $-\infty$ (for more on this see https://physics.stackexchange.com/questions/30228/must-the-derivative-of-the-wave-function-at-infinity-be-zero). 
So the first term in the above expression vanishes! Taking the second term and doing the integration by parts again we have,
$$
\int_{-\infty}^{\infty} \frac{\partial^2 \Psi}{\partial x^2} \frac{\partial \Psi^*}{\partial x} \, dx 
= \left[\frac{\partial^2 \Psi}{\partial x^2} \Psi^*\right]_{-\infty}^{\infty}
- \int_{-\infty}^{\infty} \frac{\partial^3 \Psi}{\partial x^3} \Psi^* \, dx = - \int_{-\infty}^{\infty} \Psi^* \frac{\partial^3 \Psi}{\partial x^3} \, dx
$$
Just interchanged the product inside integral of the second term above at the last equality. See that we again have taken the wave function to be vanishing at infinities. 
Now, we can observe that the obtained result is just the negative of the first term in the expression we started with. So, we get a zero overall at the end and remaining with the term $- \left\langle \frac{\partial V}{\partial x} \right\rangle$.
For generalization, see https://en.wikipedia.org/wiki/Ehrenfest_theorem
