# Derivation of the velocity for the expectation value of position in quantum mechanics

I am currently reading Griffiths book on quantum mechanics and I don't understand the derivation for the time derivative of the expectation value of the position.

The part am I stuck is after an integration by parts I have to calculate $$x(\psi^*\dfrac{d\psi}{dx} - \psi\dfrac{d\psi^*}{dx})\big|^{\infty}_{-\infty}$$.

I know that for the wave function to be normalized, it needs to go to zero as x approaches infinity. With that in mind, I calculate $$\infty*0-(-\infty*0)$$ which is supposed to equal $$0$$. What am I doing wrong?

• maybe consider if $x\psi(x)$ or $x\psi’(x)$ need to go to $0$? – ZeroTheHero Jan 5 at 16:15
• @ZeroTheHero The only thing I can think of is that in the book it says that $\psi$ goes to zero faster than $\dfrac{1}{\sqrt{x}}$. Does this mean that $\psi$ goes to zero faster than $x$ goes to infinity, therefore $x\psi$ goes to zero? – alexk745 Jan 5 at 16:49