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Question about momentum space fourier transform

  1. Sep 14, 2011 #1
    The form of the fourier transform I love the most (because it is very symmetric) is:

    [tex]f(x) = \int_{-\infty}^\infty g(\xi)e^{2\pi i x \xi}\,d\xi[/tex]
    [tex]g(\xi) = \int_{-\infty}^\infty f(x)e^{-2\pi i x \xi}\,dx[/tex]

    If we take [tex]\xi = p[/tex] then we get:

    [tex]f(x) = \int_{-\infty}^\infty g(p)e^{2\pi i x p}dp = \frac{1}{2\pi}\int_{-\infty}^\infty g(p) e^{ixp}\,dp[/tex] (rescaling p in the second equality)
    [tex]g(p) = \int_{-\infty}^\infty f(x)e^{-2\pi i x p}dx=\frac{1}{2\pi}\int_{-\infty}^\infty f(x) e^{-ixp}\,dx[/tex] (rescaling x in the second equality)

    However, in numerous references I see that:

    [tex]f(x) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^\infty g(p)e^\frac{ixp}{\hbar}\,dp[/tex]
    [tex]g(p) = \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^\infty f(x)e^\frac{-ixp}{\hbar}\,dx[/tex]

    Why does the reduced Planck's constant come into it at all, and why do we have a square root?
     
  2. jcsd
  3. Sep 14, 2011 #2

    Ken G

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    Gold Member

    You can't have xp in an exponential, the argument of the exponent must be unitless. So it's really xk, not xp, where k is the "wave number." It is related to p by p=hbar*k, so that's how the hbar gets in there. The square root is just an arbitrary way to normalize the Fourier transform such that it has the same form as its inverse, and apply the transform and then its inverse needs to get you back to the original function.
     
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