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I Momentum operator on positon/momentum representation

  1. Apr 23, 2016 #1

    dyn

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    Hi. I have come across the following step in a derivation of the harmonic oscillator groundstate wavefunction using ladder operators

    ∫ <x | p | p><p | o > dp = ∫ p<x | p><p | o > dp = -iħ d/dx ∫ <x | p><p | 0>dp

    I am confused about how the -iħ d/dx arises. I thought the p produced when the p operator acts on |p> is an eigenvalue not an operator ? And it also seems as though the equation is in the momentum representation not the position one and -iħ d/dx is the momentum operator in the position representation ?
    Thanks
     
  2. jcsd
  3. Apr 23, 2016 #2

    blue_leaf77

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    ##p \langle x|p \rangle = \hbar k e^{ikx} = -i\hbar \partial_x e^{ikx} = -i\hbar \partial_x \langle x|p \rangle##.
    It's in position representation because you project the state onto a position basis ##\langle x|##.
     
  4. Apr 24, 2016 #3

    dyn

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    Thanks for your reply but i'm still confused. I understand how to get to the p <x |p > but what is the p ? An eigenvalue or an operator ?
     
  5. Apr 24, 2016 #4

    blue_leaf77

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    Yes, an eigenvalue and hence not an operator.
     
  6. Apr 24, 2016 #5

    dyn

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    Thanks again. So how does the eigenvalue p become the momentum operator on position space ?
     
  7. Apr 24, 2016 #6

    blue_leaf77

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    That's what I wrote in post #2.
     
  8. Apr 25, 2016 #7

    dyn

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    Thanks. I thought ħk is the momentum eigenvalue for a free particle and eikx is the momentum eigenfunction for a free particle ? Why do these still apply to a harmonic oscillator ?
     
  9. Apr 25, 2016 #8

    blue_leaf77

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    Yes.
    As the poster, you should be the one who knows in which occasion the author of the resource from which you found that equation did this derivation. I can only guess that probably he only wants to show that acting a momentum operator to a ground state of harmonic oscillator (basically can also be to any state) and then projecting the resulting state onto position basis translates into a calculus operation, namely the first derivative of the ground state wavefunction in position basis.
     
    Last edited: Apr 26, 2016
  10. Apr 26, 2016 #9

    dyn

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    I found the equation in some notes deriving the ground state wavefunction of the harmonic oscillator using ladder operators. I can see how the maths works but I don't understand why free particle momentum eigenvalues and eigenfunctions can be applied to a harmonic oscillator.
     
  11. Apr 26, 2016 #10

    blue_leaf77

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    The equation you found is just a mathematical manipulation to show that
    $$
    \langle x |\hat{p}| 0\rangle = -i\hbar \frac{\partial}{\partial x} \langle x | 0\rangle
    $$
    and in doing so it turns out that you end up using the wavefunction for a free particle. Anyway, a wavefunction written in position or momentum basis can never get away completely from the free particle wavefunction. If you remember, in shifting from momentum to position space or vice versa, you have to Fourier transform the wavefunction and the complex exponential function of the form ##\exp(ikx)## is inherent in the Fourier transform formulas.
     
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