# 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!

• 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 $\langle \Delta\Phi|\Phi\rangle=\langle \Phi|\Delta\Phi\rangle$.