# How to prove dp/dt = -dV/dx? Quantum mechanics [closed]

I got this problem from a book called Introduction to quantum mechanics, griffin 2nd edition. and I did not get why the solution says

first term integrates to zero, integration by parts twice?!

Please see the solution below! Thanks everyone for helping!

## closed as off-topic by Emilio Pisanty, Waffle's Crazy Peanut, ManishearthSep 8 '13 at 8:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Emilio Pisanty, Waffle's Crazy Peanut
If this question can be reworded to fit the rules in the help center, please edit the question.

• This question appears to be off-topic because it is about interpreting a book, and not about physics. – Manishearth Sep 8 '13 at 8:15

$$\int\left(\frac{\partial}{\partial x}f(x)\right)\ g(x)\ \text dx=\int\ f(x)\left(-\frac{\partial}{\partial x}g(x)\right)\ \text dx,$$
$$\int\left(\frac{\partial^2}{\partial x^2}\Psi^*\right)\frac{\partial}{\partial x}\Psi\ \text dx =\int\left(\frac{\partial}{\partial x}\Psi^*\right)\left(-\frac{\partial^2}{\partial x^2}\Psi\right)\ \text dx=\int\Psi^*\frac{\partial^3}{\partial x^3}\Psi\ \text dx.$$
The fact that you can shift the sqare of the derivative $\frac{\partial^2}{\partial x^2}$ around like this is already suggested by $\Delta=\sum_{n=1}^3\frac{\partial^2}{\partial x_n^2}$ being an observable in several contexts. If it's an observable, then it's hermitean and $\langle \Delta\Phi|\Phi\rangle=\langle \Phi|\Delta\Phi\rangle$.