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Is the differential in the momentum operator commutative?

  1. Feb 25, 2016 #1
    As it says; I was looking over some provided solutions to a problem set I was given and noticed that, in finding the expectation value for the momentum operator of a given wavefunction, the following (constants/irrelevant stuff taken out) happened in the integrand:

    [tex] \int_{-\infty}^{\infty}\gamma(x)(\hat{p}(\gamma(x))dx = \int_{-\infty}^{\infty}\gamma(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\bigg(\gamma(x)\bigg)dx= \frac{\hbar}{i}\int_{-\infty}^{\infty}\frac{\partial}{\partial(x)}\bigg(\gamma^2(x)\bigg)dx[/tex]

    Are we allowed to commute functions of the same dependency on variables into the momentum operator if the momentum operator's differential is w.r.t. that same variable or something? How does this work?
     
  2. jcsd
  3. Feb 25, 2016 #2

    blue_leaf77

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    Are you sure you don't miss a factor of two somewhere in the equation?
     
  4. Feb 25, 2016 #3
    I just noticed that before you mentioned, and I can't figure out where it's coming from. There's now a factor of two in the denominator of a constant now outside the integral; I'm guessing some other operation was done that I didn't notice?

    (To be clear, the factor of two wasn't there, and now it is after that function is moved into the differential operator(?). It wasn't there prior, so it had to have come from some operation.)

    What you're left with, before the change I noted, is the integrand in this form:

    [tex] \gamma(x) \frac{d}{dx}\bigg(\gamma(x)\bigg)dx [/tex]
     
  5. Feb 25, 2016 #4

    blue_leaf77

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    It's just the chain rule, consider
    $$
    \frac{\partial}{\partial x} \gamma^2(x) = \frac{\partial \gamma^2(x)}{\partial \gamma(x)} \frac{\partial \gamma(x)}{\partial x}.
    $$
     
  6. Feb 25, 2016 #5

    dextercioby

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    Yes, a moderator can put the two in your OP. Yes, the derivative operator used in quantum mechanics has the same commutation properties as the derivative operator from multidimensional real calculus, namely the Leibniz property.
     
  7. Feb 25, 2016 #6
    Ah, thanks to both of you!
     
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