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Fourier transform. Impulse representation.

  1. Mar 9, 2014 #1
    ##\varphi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dx\psi(x)e^{-\frac{ipx}{\hbar}}##. This ##\hbar## looks strange here for me. Does it holds identity
    ##\int^{\infty}_{-\infty}|\varphi(p)|^2dp=\int^{\infty}_{-\infty}|\psi(x)|^2dx=1##???
    I'm don't think so because this ##\hbar##. So state in impulse space is not normalized.
     
  2. jcsd
  3. Mar 9, 2014 #2

    jtbell

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    Staff: Mentor

    Yes, it does. The ##\hbar## in the transform is required in order to make this true. You may be thinking of the standard Fourier transform:
    $$A(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\psi(x)e^{-ikx}dx}$$
    Changing variables from k to ##p = \hbar k## introduces the extra ##\hbar## in the constant factor.
     
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