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

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2015-12-05 at 11.47.41 pm.png

    2. Relevant equations
    Screen Shot 2015-12-05 at 11.48.08 pm.png

    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.
    Screen Shot 2015-12-05 at 11.51.11 pm.png
     
  2. jcsd
  3. Dec 5, 2015 #2
    Oh I think I got it. It seems like we have to assume the derivatives of a wave function are all 0 at infinity.
     
  4. Dec 5, 2015 #3

    nrqed

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    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!
     
  5. Dec 5, 2015 #4
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