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Why can momentum be expressed as an operator

  1. Jan 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Im having a hard time figuring out how in quantum mechanics things such as momentum can be expressed as an operator.

    I know the simple algebra to get the relation. Starting with the 1D solution to wave equation[itex]\Psi=e^{i\omega x}[/itex] then differentiating that with respect to x and replacing the resulting k with the de Broglie relation you get [itex]\frac{\partial \Psi}{\partial x}=i\frac{p}{\hbar}\Psi [/itex].

    From here I can sort of see that p may =[itex]-i\hbar \frac{\partial }{\partial x} [/itex] but I just dont get how a value such as momentum can be a operator.

    Is it that to get the momentum from the wave function you need to perform this operator on the wavefunction, and since every term in the Schrodinger equation has the wavefunction in it, they simply represent momentum as the operator? Even if this is true, there has to be more to it than being a simple variable substitution right?
  2. jcsd
  3. Jan 22, 2014 #2


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    hi bfusco! :smile:
    technically, momentum is the eigenvector of the operator, not the operator itself

    however, the operator needs a name, so we call that "momentum" too! :rolleyes:

    if this was biology … where they pay such detailed attention to nomenclature … we'd probably call it "momentumase" :wink:
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