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

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Homework Help Overview

The discussion revolves around the derivation of the momentum-to-the-power-of-n operator in quantum mechanics, focusing on the mathematical treatment of wave functions and their derivatives at infinity.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the substitution of wave functions in momentum space and the implications of integrating by parts. Questions arise regarding the behavior of wave functions and their derivatives at infinity, particularly whether they can be assumed to be zero.

Discussion Status

Some participants suggest that the assumption of derivatives of wave functions being zero at infinity is necessary for the derivation. Others express skepticism and seek examples that contradict this assumption, indicating a mix of interpretations and ongoing exploration of the topic.

Contextual Notes

There is a reference to a proof for the case of the momentum operator < p_x >, and a link is provided to an external discussion regarding the behavior of wave functions at infinity.

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Homework Statement


Screen Shot 2015-12-05 at 11.47.41 pm.png


Homework Equations


Screen Shot 2015-12-05 at 11.48.08 pm.png


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.
Screen Shot 2015-12-05 at 11.51.11 pm.png
 
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Oh I think I got it. It seems like we have to assume the derivatives of a wave function are all 0 at infinity.
 
Happiness said:
Oh I think I got it. It seems like we have to assume the derivatives of a wave function are all 0 at infinity.
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!
 

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