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The simplest derivation of position operator for momentum space

  1. Oct 14, 2014 #1
    Might be simple but I couldn't see. We can easily derive momentum operator for position space by differentiating the plane wave solution. Analogously I want to derive the position operator for momentum space, however I am getting additional minus sign.

    By replacing $$k=\frac{p}{\hbar}$$ and $$w=\frac{E}{\hbar}$$ into the plane wave solution, we get
    $$\Psi=e^{ipx/\hbar-iEt/\hbar}$$
    Then taking the derivative with respect to momentum,
    $$\frac{\partial\Psi}{\partial p}=\frac{ix}{\hbar}\Psi$$
    Then I get,
    $$\hat{x}=-i\hbar \frac{\partial}{\partial p}$$

    It has additional minus sign. Where is my mistake and/or how do I derive the position operator for momentum space in the simplest way?
     
  2. jcsd
  3. Oct 17, 2014 #2

    stevendaryl

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    Well, you need to think what is the relationship between a function in position-space and in momentum space. This is actually clearest if you use Dirac bra-ket notation, but if you haven't learned that, we can do without.

    If you have a wave function [itex]\psi(x)[/itex] in position space, the corresponding function in momentum space is [itex]\tilde{\psi}(p)[/itex]. The two are related as follows:

    [itex]\psi(x) = \frac{1}{\sqrt{2 \pi}} \int e^{i p x} \tilde{\psi}(p) dp[/itex]
    [itex]\tilde{\psi}(p) = \frac{1}{\sqrt{2 \pi}}\int e^{-i p x} \psi(x) dx[/itex]

    (I'm leaving out the h-bars for simplicity; I hope you can figure out where they should go. I think every [itex]p[/itex] should be [itex]p/\hbar[/itex])

    So notice that the inverse transform has a minus sign in the exponent. That makes it so that [itex]\hat{x} = +i \frac{d}{dp}[/itex] while [itex]\hat{p} = -i \frac{d}{dx}[/itex]

    [edit: changed the conventions to make forward and reverse transforms more symmetric]
     
    Last edited: Oct 17, 2014
  4. Oct 17, 2014 #3

    stevendaryl

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    You can directly verify using these equations that:
    [itex]-i \frac{d}{dx} \psi(x) = \frac{1}{2 \pi} \int p e^{i p x} \tilde{\psi}(p) dp[/itex]
    [itex]+i \frac{d}{dp} \tilde{\psi}(p) = \frac{1}{2 \pi} \int x e^{- i p x} \psi(x) dx[/itex]
     
  5. Oct 17, 2014 #4

    dextercioby

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    The quantum mechanical convention is to put a sqrt(2pi) for each part of a Fourier transform, the conventionally direct one from coordinate space to momentum one and the reverse one which is from momentum space to coordinate space.
     
  6. Oct 17, 2014 #5

    stevendaryl

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    I seem to remember that. I changed it.
     
  7. Oct 17, 2014 #6

    atyy

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    No one else uses this convention except physicists. That's because physicists love non-rigourous lazy ways of thinking, and are always trying to make their lives easy. :D By having the sqrt(2pi), it's the same forwards and back, and one doesn't have to remember where to put the 2pi.
     
  8. Oct 20, 2014 #7
    Thank you for the answer. So, in order to derive position opeator for momentum space, I have to start with the wavefunction in momentum space. Not in position space. That was my mistake as far as I understand.
     
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