Calculating the time derivative of <p>

[SplinterCell]

Homework Statement

Problem 1.7 in Griffiths "Quantum Mechanics" asks to prove
$$\frac{d\left \langle p \right \rangle}{dt}=\left \langle -\frac{\partial V}{\partial x} \right \rangle$$

Homework Equations

Schrödinger equation

The Attempt at a Solution

I was able to arrive at the correct expression for the integrand:
$$\frac{\partial }{\partial t} \left(\Psi^\ast \frac{\partial \Psi}{\partial x} \right )=\frac{i \hbar}{2m}\left[\Psi^\ast \frac{\partial^3 \Psi}{\partial x^3} -\frac{\partial^2 \Psi^\ast }{\partial x^2}\frac{\partial \Psi}{\partial x}\right ]-\frac{i}{\hbar}\left |\Psi \right |^2 \frac{\partial V}{\partial x}$$
The only thing left here is to prove that the first term integrates to zero. The solution manual suggests integrating by parts twice. However I fail to get zero. Here's what I tried:
$$\int \left(\Psi^\ast \frac{\partial^3 \Psi}{\partial x^3} -\frac{\partial^2 \Psi^\ast }{\partial x^2}\frac{\partial \Psi}{\partial x} \right ) dx = \Psi^\ast \frac{\partial^2 \Psi}{\partial x^2} - \int \left (\frac{\partial \Psi^\ast}{\partial x} \frac{\partial^2 \Psi}{\partial x^2}\right ) dx - \int \left (\frac{\partial \Psi}{\partial x} \frac{\partial^2 \Psi^\ast}{\partial x^2}\right ) dx =\\= \Psi^\ast \frac{\partial^2 \Psi }{\partial x^2} - \frac{\partial \Psi}{\partial x}\frac{\partial \Psi^\ast}{\partial x}$$
Which isn't necessarily equal to zero. Any suggestions?

 

[Orodruin]

Your boundary terms should be zero. You are just taking the primitive function, not actually integrating.

 

[PeroK]

$$ \Psi^\ast \frac{\partial^2 \Psi }{\partial x^2} - \frac{\partial \Psi}{\partial x}\frac{\partial \Psi^\ast}{\partial x}$$
Which isn't necessarily equal to zero. Any suggestions?

It's not sufficient for $\Psi$ to be square integrable, but if it's well enough behaved then those boundary terms will be zero. In other words, assume that $\Psi$ and all its derivatives tend to zero at $\pm \infty$.

 

[SplinterCell]

It's not sufficient for $\Psi$ to be square integrable, but if it's well enough behaved then those boundary terms will be zero. In other words, assume that $\Psi$ and all its derivatives tend to zero at $\pm \infty$.

Your boundary terms should be zero. You are just taking the primitive function, not actually integrating.

Thank you both. I forgot that the expectation value is actually an integral from -infinity to +infinity and not some indefinite integral.