An operator-based proof of spectral decomposition of the Fourier transform in terms of the Hermite functions.
In my last post from quite a while ago, I derived a couple of simple but useful similarity relations between the position and momentum operators through the Fourier transform. These are: \(\begin{align} \hat{D}_X &= \iota \mathcal{F}^{-1} \hat{X} \mathcal{F} \label{eq.rel1} \\ \hat{D}_X &= -\iota \mathcal{F} \hat{X} \mathcal{F}^{-1} \label{eq.rel2} \end{align}\) Here, I use them to show how to determine the eigenvalues and a corresponding eigenbasis for the Fourier transform itself. As I mentioned in my last post, this is a re-telling of a well-known proof that I feel illuminates the power of using the operator formulation as a nice alternative to doing tedious integrals by hand.
In the proof that follows, I need another very useful relation between the position and derivative operators, one that does not explicitly contain the Fourier transform. For simplicity, I continue to operate on well-behaved, smooth and square-integrable functions belonging to $\mathcal{H} = \left\{f:\mathbb{R} \longrightarrow \mathbb{C} \left| \int_\mathbb{R} dx \left|f(x)\right|^2 < \infty\right.\right\}$, as in the previous post. This relation basically says that the derivatives of a function $f \in \mathcal{H}$ can be used to ‘generate’ functional values at different distances, or translations, from complete information of the function at a point $x_0$. By complete information, I mean the functional value and all its derivatives. In group theory -speak, the derivative is the generator of translations: \(\begin{align} \underbrace{e^{t\hat{D}_X}}_\text{translation}f(x_0) = f(x_0 + t) = \sum\limits_{n=0}^\infty \frac{t^n}{n!}\hat{D}_X^{(n)}f(x_0) \label{eq.groupgenerator} \end{align}\)
…which is also well known as the Taylor series expansion of $f(x)$ about $x_0$.
So as with the Wikipedia proof, I start with the generating function of the Hermite functions: \(\begin{align} e^{-\frac{x^2}{2} + 2xt - t^2} = e^{-\frac{1}{2} \left(x-2t\right)^2 + t^2} = \sum\limits_{n=0}^\infty \left[e^{-\frac{x^2}{2}} H_n(x)\right]\frac{t^n}{n} \label{eq.generatingfn} \end{align}\)
Here $H_n(x)$ are the Hermite polynomials, which by themselves don’t belong to $\mathcal{H}$ because, well, they’re polynomials and therefore not bounded on $(-\infty, \infty)$. So Eq. \eqref{eq.generatingfn} can be rewritten using Eq. \eqref{eq.groupgenerator} in linear operator form: \(\begin{align} \sum\limits_{n=0}^\infty \left[e^{-\frac{x^2}{2}} H_n(x)\right]\frac{t^n}{n} &= e^{-2t\hat{D}_X} e^{-\frac{x^2}{2}} e^{t^2} \end{align}\)
Taking the Fourier transform on both sides gives:
\(\begin{align}
\sum\limits_{n=0}^\infty \mathcal{F} \left[e^{-\frac{x^2}{2}}H_n(x)\right] \frac{t^n}{n!}
&= \mathcal{F}e^{-2t\hat{D}_X} e^{-\frac{x^2}{2}} e^{t^2} \notag \\
&= \left(\mathcal{F} e^{-2t\hat{D}_X} \mathcal{F}^{-1}\right) \left(\mathcal{F} e^{-\frac{x^2}{2}} \right) e^{t^2} \label{eq.breakup}
\end{align}\)
…where we have used the associative property of the linear operators and the fact that $\mathcal{F}^{-1} \mathcal{F} = \mathbb{I}$, the identity operator.
The term in the first parenthesis on the RHS of Eq. \eqref{eq.breakup} may be analyzed as:
\(\begin{align}
\mathcal{F} e^{-2t\hat{D}_X} \mathcal{F}^{-1} &= \sum\limits_{n=0}^\infty \frac{(-2t)^n}{n!} \mathcal{F} \hat{D}_X^{(n)} \mathcal{F}^{-1} \notag \\
&= \sum\limits_{n=0}^\infty \frac{(-2t)^n}{n!} \underbrace{
\left(\mathcal{F} \hat{D}_X \mathcal{F}^{-1}\right)
\left(\mathcal{F} \hat{D}_X \mathcal{F}^{-1}\right)
\ldots
\left(\mathcal{F} \hat{D}_X \mathcal{F}^{-1}\right)
}_{n\text{ times}} \notag \\
&= \sum\limits_{n=0}^\infty \frac{(-2t)^n}{n!} \left(\iota \hat{X}\right)^n \tag{from Eq. \eqref{eq.rel1}} \\
&= e^{-\iota 2t \hat{X}} \notag
\end{align}\)
while the term in the second parenthesis of Eq. \eqref{eq.breakup} is simply the well-known Fourier transform of the bell curve with a standard deviation of $1$:
\(\begin{align}
e^{-\frac{x^2}{2}} \underset{\mathcal{F}^{-1}}{\stackrel{\mathcal{F}}{\rightleftharpoons}} e^{-\frac{k^2}{2}}
\end{align}\)
with the result that Eq. \eqref{eq.breakup} becomes:
\(\begin{align}
\sum\limits_{n=0}^\infty \mathcal{F} \left[e^{-\frac{x^2}{2}}H_n(x)\right] \frac{t^n}{n!}
&= \left(e^{-\iota 2t \hat{X}} e^{-\frac{k^2}{2}}\right) e^{t^2} \notag \\
&= e^{-\iota 2t k -\frac{k^2}{2}} e^{t^2} \notag \\
&= e^{-\frac{k^2}{2}}e^{-2(\iota t) k + (\iota t)^2} \notag \\
&= e^{-\frac{k^2}{2}} \sum\limits_{n=0}^\infty H_n(k) \frac{(\iota t)^n}{n!} \label{eq.final}
\end{align}\)
Comparing coefficients of $t^n$ on either side of Eq. \eqref{eq.final}, we get:
\(\begin{align}
\boxed{
\mathcal{F}\left[e^{-\frac{x^2}{2}}H_n(x)\right] = \iota^n \left[e^{-\frac{k^2}{2}}H_n(k)\right]
} \label{eq.result}
\end{align}\)
i.e., the Hermite functions $e^{-\frac{x^2}{2}}H_n(x)$ form an eigenbasis of the Fourier transform operator $\mathcal{F}$ with corresponding eigenvalue $\iota^n = e^{\iota n \frac{\pi}{2}}$.
Clearly, the eigenvalues $\left\{e^{\iota n \frac{\pi}{2}}\left|n \in \mathbb{N}\right.\right\} = \left\{\iota, -1, -\iota, 1\right\}$ are heavily degenerate, and therefore the Hermite functions do not form a unique eigenbasis of $\mathcal{F}$.
In fact, an infinity of other orthogonal eigenbases can be generated by (i) taking any or all of the four eigen-subspaces spanned by the sets $S_p = \left\{e^{-\frac{x^2}{2}}H_{4n+p}(x)\left|n \in \mathbb{N}\right.\right\}$ for $p = 0, 1, 2, 3$ and (ii) performing a high-dimensional ‘rotation’ or a unitary transformation on the basis functions therein.
So one of many ways to write the Fourier transform in terms of orthogonal eigenfunctions is:
\(\begin{align}
\mathcal{F}(k, x )f(x) \equiv \underbrace{
\sum\limits_{n=0}^\infty
e^{\iota n\frac{\pi}{2}}
\int_\mathbb{R} dx~e^{-\frac{x^2 + k^2}{2}}H_n(k) H_n(x)
}_\mathcal{F} f(x) \label{eq.springboard}
\end{align}\)
It turns out that the expression in Eq. \eqref{eq.springboard} (known as the spectral decomposition of $\mathcal{F}$) is a nice entry into the generalization into the fractional Fourier transform.
In particular: consider the linear operator (assuming it exists), whose eigenbasis is the set of Hermite functions just like with $\mathcal{F}$, but the corresponding eigenvalues are powers of, not $e^{\iota \frac{\pi}{2}}$, but the more general unit phasor $e^{\iota \alpha}$ with $\alpha \in \left[0, 2\pi\right]$.
The corresponding operator \(\mathcal{F}_\text{fr}\) on the left hand side of Eq. \eqref{eq.springboard} would depend on the fractional angular parameter $\alpha$.
This fractional-order Fourier transform (FrFT) may be defined by:
\(\begin{align}
\mathcal{F}_\text{fr}(k, x\left|\alpha\right.)f(x) \equiv
\sum\limits_{n=0}^\infty
e^{\iota n\alpha}
\int_\mathbb{R} dx~e^{-\frac{x^2+k^2}{2}}H_n(k) H_n(x)
f(x) \label{eq.frft}
\end{align}\)
As I mentioned in my introductory post, there is a nice, easy-to-read 1937 PNAS article by Condon that I really like, which approaches the FrFT from a purely group-theoretic approach without referring to the spectral decomposition at all.
It nicely derives and parcels all the group-like and spectrum-like properties of this important integral transform in a systematic way.
For my part, I eventually dive into the adaptation of the FrFT formalism in optics and coherent diffraction, in my next post.