The form of the fourier transform I love the most (because it is very symmetric) is:(adsbygoogle = window.adsbygoogle || []).push({});

[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?

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# Question about momentum space fourier transform

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