Editing Hermite polynomials (section)

===Hermite functions as eigenfunctions of the Fourier transform===
The Hermite functions {{math|''ψ''<sub>''n''</sub>(''x'')}} are a set of [[eigenfunction]]s of the [[continuous Fourier transform]] {{mathcal|F}}. To see this, take the physicist's version of the generating function and multiply by {{math|''e''<sup>−{{sfrac|1|2}}''x''<sup>2</sup></sup>}}. This gives
<math display="block">e^{-\frac12 x^2 + 2xt - t^2} = \sum_{n=0}^\infty e^{-\frac12 x^2} H_n(x) \frac{t^n}{n!}.</math>

The Fourier transform of the left side is given by
<math display="block">\begin{align}
 \mathcal{F} \left\{ e^{ -\frac12 x^2 + 2xt - t^2 } \right\}(k)
  &= \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{-ixk}e^{-\frac12 x^2 + 2xt - t^2}\, dx \\
  &= e^{-\frac12 k^2 - 2kit + t^2 } \\
  &= \sum_{n=0}^\infty e^{ -\frac12 k^2 } H_n(k) \frac{(-it)^n}{n!}.
\end{align}</math>

The Fourier transform of the right side is given by
<math display="block">\mathcal{F} \left\{ \sum_{n=0}^\infty e^{-\frac12 x^2} H_n(x) \frac {t^n}{n!} \right\} = \sum_{n=0}^\infty \mathcal{F} \left \{ e^{-\frac12 x^2} H_n(x) \right\} \frac{t^n}{n!}.</math>

Equating like powers of {{mvar|t}} in the transformed versions of the left and right sides finally yields
<math display="block">\mathcal{F} \left\{ e^{-\frac12 x^2} H_n(x) \right\} = (-i)^n e^{-\frac12 k^2} H_n(k).</math>

The Hermite functions {{math|''ψ<sub>n</sub>''(''x'')}} are thus an orthonormal basis of {{math|''L''<sup>2</sup>('''R''')}}, which ''diagonalizes the Fourier transform operator''.<ref>In this case, we used the unitary version of the Fourier transform, so the [[eigenvalue]]s are {{math|(−''i'')<sup>''n''</sup>}}. The ensuing resolution of the identity then serves to define powers, including fractional ones, of the Fourier transform, to wit a [[Fractional Fourier transform]] generalization, in effect a [[Mehler kernel]].</ref> In short, we have:<math display="block">\frac{1}{\sqrt{2\pi}} \int e^{-ikx} \psi_n(x) dx = (-i)^n \psi_n(k), \quad \frac{1}{\sqrt{2\pi}} \int e^{+ikx} \psi_n(k) dk = i^n \psi_n(x)</math>