Derivation of the momentum-to-the-power-of-n operator

1. Dec 5, 2015

Happiness

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
First substitute $\Phi(p,t)$ in terms of $\Psi(r,t)$ and similarly for $\Phi^*(p,t)$, and substitute $p_x^n$ in terms of the differentiation operator
$< p_x^n>\,=(2\pi\hbar)^{-3}\int\int e^{ip.r'/\hbar}\Psi^*(r',t)\,dr'\int(i\hbar)^n(\frac{\partial^n}{\partial x^n}e^{-ip.r/\hbar})\Psi(r,t)\,dr\,dp$

Then integrate by parts with respect to $x$. The integrated part vanishes since $\Psi(r,t)$ equals to 0 at infinity. We have
$=(2\pi\hbar)^{-3}\int\int e^{ip.r'/\hbar}\Psi^*(r',t)\,dr'\int(i\hbar)^{n-1}(\frac{\partial^{n-1}}{\partial x^{n-1}}e^{-ip.r/\hbar})\Big[-i\hbar\frac{\partial}{\partial x}\Psi(r,t)\Big]\,dr\,dp$

Then I'm stuck because it seems like we need to integrate by parts with respect to $x$ again for another $(n-1)$ times in order to get
$=(2\pi\hbar)^{-3}\int\int e^{ip.r'/\hbar}\Psi^*(r',t)\,dr'\int e^{-ip.r/\hbar}\Big[(-i\hbar)^n\frac{\partial^n}{\partial x^n}\Psi(r,t)\Big]\,dr\,dp$.

But now the integrated part no longer vanishes since $\frac{\partial}{\partial x}\Psi(r,t)$ and higher derivatives may not be 0 at infinity. So we can't get the above that we need.

The proof for the case of $< p_x>$ is attached below for reference.

2. Dec 5, 2015

Happiness

Oh I think I got it. It seems like we have to assume the derivatives of a wave function are all 0 at infinity.

3. Dec 5, 2015

nrqed

Right. And it is natural. I, for one, cannot imagine a continuous function who would go to zero at infinity but whose derivative would not. If someone has such an example, I would love to see it!

4. Dec 5, 2015