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Momentum and fourier series

  1. Oct 23, 2009 #1
    Both (i d/dx) and (-i d/dx ) are Hermitian. For some reason (-i d/dx ) is chosen to be the momentum operator, and the consequences are that [x,p]=ih (and not -ih), and that [tex]e^{ipx} [/tex] is an eigenvalue of momentum p (and not -p).

    Is there any fundamental reason why [x,p] can't be -ih, and [tex]e^{ipx} [/tex] can't have an eigenvalue -p, so that the momentum operator can be (i d/dx) ?

    When dealing with Fourier series, [tex]f(x)=\int d^3p \mbox{ } f(p) e^{-ipx} [/tex] would be incorrect, right? It would have to be [tex]f(x)=\int d^3p \mbox{ } f(p) e^{ipx} [/tex] if you choose (-i d/dx )?

    Do most math books use [tex]f(x)=\int d^3p \mbox{ } f(p) e^{-ipx} [/tex] or [tex]f(x)=\int d^3p \mbox{ } f(p) e^{ipx} [/tex] for their definition of a Fourier series? Which convention do you use for a Fourier series?
     
  2. jcsd
  3. Oct 24, 2009 #2

    Ben Niehoff

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    There is no fundamental reason. In fact, all quantum mechanics (indeed, all mathematics) remains valid under the substitution, everywhere, of i -> -i. The reason is that there is some ambiguity in defining i in the first place. There are two square roots of -1, after all.

    The convention I use for Fourier integrals is

    [tex]f(x) = \int \frac{d^3p}{(2\pi)^3} \; f(p) e^{ipx}[/tex]
     
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