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Calculations with Dirac deltas and position/momentum operators

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the matrix elements of [itex] \hat{x} [/itex] in momentum space. That is, evaluate [itex] \langle p | \hat{x} | \psi(t) \rangle [/itex] in terms of the momentum space wave equation [itex] \langle p | \psi(t) \rangle [/itex].

    2. Relevant equations

    [tex] \langle x | p \rangle = \frac{1}{\sqrt{2 \pi \hbar}} e^{ipx/ \hbar} [/tex]
    [tex] I = \int | x \rangle \langle x | dx [/tex] [tex] I = \int | p \rangle \langle p | dp [/tex]

    3. The attempt at a solution

    So when we insert the identity, we obtain our integral: [tex] \int_{-\infty}^{+\infty} dx \langle p | \hat{x} | x \rangle \langle x | \psi(t) \rangle[/tex]

    The thing is, I know we have to insert the identity again to obtain:

    [tex] \int_{-\infty}^{+\infty} dx dp' \langle p | \hat{x} | x \rangle \langle x | p' \rangle \langle p' | \psi(t) \rangle [/tex]

    (Assume it's a double integral.)

    This was necessary so that we could get our ket psi in terms of the momentum basis. We then act [itex]\hat{x}[/itex] on [itex] |x \rangle [/itex] and get an x in our integral. Then we insert the Fourier transforms and obtain:

    [tex] \int_{-\infty}^{+\infty} dx dp' x e^{ipx/ \hbar} e^{-ip'x/ \hbar} \psi(p', t) = \int_{-\infty}^{+\infty} dx dp' x e^{i(p-p')x/ \hbar} \psi(p', t) [/tex]

    I'm a little lost here. I see that the exponential will become a Dirac delta, which will bring p into the psi (making it the form it should be). I can also see where the differential form of x in the p basis comes out of the exponential. But I don't see how the integral with respect to x will disappear.

    **EDIT** I've solved it. No need to respond.
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2

    gabbagabbahey

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    You need to be careful when you say stuff like "then we insert the Fourier transform" and "the exponential will become a Dirac delta". Inaccurate claims like these can result in los marks on your assignments.

    It would be better to say something like "inserting our expression for [itex]\langle x | p \rangle[/itex], we see that the integral is in the form of a Fourier transform", and "integrating the exponential over all [itex]x[/itex], results in a Dirac delta distribution" .
     
  4. Sep 16, 2012 #3
    Thanks for the input. This is actually the revelation I had last night when I was solving this problem. I realized I was misunderstanding how the Dirac delta comes about.

    Also, thanks for the clarification on the Fourier transform. Is it incorrect to say that it *is* the Fourier transform since we're missing the weighting function and the integral? It still is unclear to me what the relationship between [itex]\langle x | p \rangle[/itex] and the Fourier transform actually is (and for some reason I've been conditioned to simply refer to it *as* the Fourier transform), save the exponential.
     
  5. Sep 16, 2012 #4

    gabbagabbahey

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    [itex]\langle x | p \rangle[/itex] is just a complex exponential (multiplied by some constant factor). It can be expressed as the Fourier transform of some function (a Dirac delta distribution, multiplied by some constant factor) with an appropriate definition of Fourier transform (something like [itex]\mathcal{F}\left[f(x')\right]\equiv \frac{1}{ \sqrt{ 2 \pi } } \int_{- \infty }^{ \infty } f(x')e^{-i\frac{p}{\hbar}x'}dx'[/itex] would work with [itex]f(x') = \frac{1}{\sqrt{ \hbar }} \delta(x'-x)[/itex]), but this is not what is usually meant when we talk about something being a Fourier transform.

    What is usually meant is that by inserting an appropriate form of the identity operator, a state expressed in the position basis can be represented in the momentum basis by taking its Fourier transform since

    [tex]\psi(p) \equiv \langle p | \psi \rangle =\int_{- \infty }^{ \infty } \langle p | x\rangle \langle x | \psi \rangle dx = \frac{1}{ \sqrt{ 2 \pi \hbar } } \int_{- \infty }^{ \infty } e^{-i \frac{ px }{ \hbar }} \psi(x) dx[/tex]

    which is in the form of a Fourier transform, give or take some constant factors.

    When we say something is a Fourier transform, we usually mean that it is in the form of a Fourier transform integral like the above.
     
  6. Sep 16, 2012 #5
    Thanks a lot for the clear response. It certainly makes sense -- once you illustrated the connection between x and p -- why it is only correct to refer to the integral expression as the transform.
     
  7. Sep 17, 2012 #6

    gabbagabbahey

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    The definition of Fourier transform is an integral of that form, so that is why whenever you see an integral of that form, you can call it a Fourier transform and most physicists will know what you are talking about.

    Of course, any suitably behaved function can be represented as the Fourier transform of some other function (its inverse transform), but it doesn't make much sense to refer to every suitably behaved function as a Fourier transform, since it doesn't tell you much.
     
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