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Position Operator in Momentum Space?

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data
    So, I'm doing this problem from Townsend's QM book
    6.2

    Show that [tex] <p|\hat{x}|\psi> = i\hbar
    \frac{\partial}{\partial p}<p|\psi>
    [/tex]

    2. Relevant equations
    [tex] |\psi(p)> = \int_\infty^{-\infty} dp |p><p|\psi> [/tex]

    3. The attempt at a solution
    So,
    [tex] <p|\hat{x}|\psi> [/tex]
    [tex] = <p|\hat{x} \int_\infty^{-\infty} dp' |p'><p'|\psi> [/tex]
    [tex] = \hat{x} \int_\infty^{-\infty} dp' <p|p'> <p'|\psi> [/tex]
    [tex] = \hat{x} \int_\infty^{-\infty} dp' \delta (p-p') <p'|\psi> [/tex]
    [tex] = \hat{x} <p|\psi> [/tex]

    and here I've assumed, correctly I've found, that [tex] \hat{x} = i\hbar \frac{\partial}{\partial p}[/tex]
    However, knowing my professor if I just write that and say "tada." I'm not going to get a good grade on this problem set.

    How would I prove that the position operator takes that form in momentum space? I've found a proof that uses the more traditional notation with integrals and the like but I've read that it is pretty nice when done in Dirac notation, I'd appreciate any help/hint.

    Thank you.
     
  2. jcsd
  3. Jan 21, 2017 #2

    blue_leaf77

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    I wonder how you can claim to have found the answer correctly when there is a mistake in assuming that ##\langle p|\hat x | p' \rangle = \hat x \langle p|p' \rangle##.
     
  4. Jan 21, 2017 #3
    Hrm, thought I had seen something similar done in my book. Welp, then I guess I'm completely lost. I suppose it was done only after the momentum operator was written in position space. I cannot pull it out like that then until it has been rewritten?
     
  5. Jan 21, 2017 #4

    blue_leaf77

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    May be your memory deceived you. Certainly ##\delta(p-p')## is not equal to ##\partial_p \delta(p-p')##.
    It's easier if you use closure relation in position space rather than that in momentum space. Use it in ##\langle p |\hat x| \psi\rangle##.
     
  6. Jan 22, 2017 #5
    Ok, I originially was trying this but wasn't sure what to do after,

    [tex]
    \langle p |\hat x| \psi\rangle [/tex]

    [tex] = \int_\infty^{-\infty} \langle p |\hat x| x \rangle \langle x | \psi \rangle dx [/tex]
    [tex] = \int_\infty^{-\infty} x \langle p | x \rangle \langle x | \psi \rangle dx [/tex]
    I think that,
    [tex] \langle p|x \rangle = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{-ipx}{\hbar}} [/tex]
    and,
    [tex] \langle x|\psi \rangle = \int_\infty^{-\infty} dp \frac{1}{\sqrt{2 \pi \hbar }} e^{\frac{ipx}{\hbar}} \langle p | \psi \rangle [/tex]

    If I plug those in though I get a nightmare! Is this the right path?
     
  7. Jan 22, 2017 #6

    blue_leaf77

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    You don't need the last one actually, just the expression for ##\langle p|x\rangle##. Upon plugging in this to ##
    \int_\infty^{-\infty} x \langle p | x \rangle \langle x | \psi \rangle dx
    ##
    You will get terms like ##x \exp{(ipx/\hbar)}##. How can you get this expression from ## \exp{(ipx/\hbar)
    }## alone?
     
  8. Jan 22, 2017 #7
    Hrm, I'm still stuck. I have no idea what to do with something like this!?

    [tex] \frac{1}{\sqrt{2 \pi \hbar}} \int_\infty^{-\infty} x e^{\frac{-ipx}{\hbar}} \langle x | \psi \rangle dx [/tex]

    I don't know if it's because I'm just starting with all these integrals with Dirac notation or what, but I'm just so confused with stuff like this right now.

    integration by parts? it's the [tex] \langle x | \psi \rangle [/tex] that has me wierded out.
     
  9. Jan 22, 2017 #8

    blue_leaf77

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    ##\frac{\partial}{\partial p} \exp(-ipx/\hbar) = ??##
     
  10. Jan 22, 2017 #9
    That is [tex] -ix/\hbar exp(-ipx/\hbar) [/tex] I'm still not sure where that is going though. I see that that is part of my integrand.
     
  11. Jan 22, 2017 #10

    blue_leaf77

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    Yes, that means you can replace that with the expression involving the derivative in your integral.
     
  12. Jan 22, 2017 #11
    Wait, so does that just basically say... that [tex] x = i \hbar \frac{\partial}{\partial p} [/tex]
     
  13. Jan 22, 2017 #12

    blue_leaf77

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    Definitely no. What I meant is, since $$
    \frac{\partial}{\partial p} \exp(-ipx/\hbar) = \frac{-ix}{\hbar} \exp(-ipx/\hbar)$$
    you can replace the entire expression ##x \exp(-ipx/\hbar)## with ##i\hbar \frac{\partial}{\partial p} \exp(-ipx/\hbar) ##.
     
  14. Jan 22, 2017 #13
    Ok let me see that. I thought for a second I could just solve that expression for x. But then I wasn't sure how on earth you could just say that x was the position operator haha.
     
  15. Jan 22, 2017 #14
    I'm still stuck with an integral I just don't know what to do with.

    [tex] \frac{i\hbar}{\sqrt{2\pi\hbar}} \int_\infty^{-\infty} \frac{\partial}{\partial p} e^{\frac{-ipx}{\hbar}} \langle x|\psi \rangle dx [/tex]
     
  16. Jan 22, 2017 #15

    blue_leaf77

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    The integration variable is ##x## while that partial derivative operates on ##p##-dependent function. What can you do with the derivative? (If you don't know what to do you might want to review your calculus lecture).
     
  17. Jan 22, 2017 #16
    wait this looks like [tex] ih \frac{\partial}{\partial p} \psi(p) [/tex]
     
  18. Jan 22, 2017 #17

    blue_leaf77

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    Yes, that should be in the final step. But I guess you also need to know where to move the derivative. Check "Leibniz rule".
     
  19. Jan 22, 2017 #18
    oh... and that is it? yah?
     
  20. Jan 22, 2017 #19
    I mean, it's not the variable being integrated so it gets pulled out and turns into a regular derivative, I thought.
     
  21. Jan 22, 2017 #20

    blue_leaf77

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    Yes it can be pulled out of the integral but it's better to let it stay being a partial derivative. Later for more than one dimension you will see that the position and momentum operators cannot be reduced to ordinary derivative.
     
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