Stone's One Parameter Unitary Group Theorem and the Fourier transform Stone's theorem on one parameter unitary groups asserts a one-to-one correspondence between strongly continuous one parameter groups of unitary operators $\mathcal{H}\to\mathcal{H}$ on a Hilbert space $\mathcal{H}$ and self-adjoint operators $\mathcal{H}\to\mathcal{H}$ i.e. for each unitary group 
$$\left\{U:\mathbb{R}\times\mathcal{H}\to\mathcal{H};\begin{array}{ll}U(t)\,U^\dagger(t)=\mathrm{id}&\forall\,t\in\mathbb{R}\\ U(t)\,U(s)=U(s+t)&\forall\,s,\,t\in\mathbb{R}\\\lim\limits_{t\to t_0}U(t)\,X = U(t_0)\,X&\forall t_0\in \mathbb{R};\;X\in\mathcal{H}\end{array}\right\}$$
there is precisely one self adjoint $P:\mathcal{H}\to\mathcal{H}$ such that $U(t) = e^{i\,P\,t}$.
My question is simple: with $\mathcal{H}$ the separable Hilbert space of complex $\mathbf{L}^2$ functions on $\mathbb{R}$, is there any such one parameter group which includes the Fourier transform? More informally: can we deform the identity operator into the Fourier transform through a one parameter family of unitary operators? I suspect the answer is no, but cannot see a reason for it.
 A: The Fourier transform $\mathcal{F}$ has a spectral decomposition
$$
            \mathcal{F} = P_{1}+iP_{i}+(-1)P_{-1}+(-i)P_{-i}
$$
where $P_{\lambda}$ are mutually disjoint orthogonal projections onto the eigenspaces of $\mathcal{F}$ corresponding to $\lambda\in \sigma(\mathcal{F})=\{1,i,-1,-i\}$. Hence, you can define
$$
             U(t) = P_{1}+e^{it\pi/2}P_{i}+e^{it\pi}P_{-1}+e^{3it\pi/2}P_{-i}.
$$
This is a unitary group with $U(0)=I$ and $U(1)=\mathcal{F}$. The hermite functions form a complete orthonormal basis of eigenfunctions of $\mathcal{F}$ with eigenvalues $\{ 1,i,-1,-i\}$.
http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions
A: Another answer to this can be constructed, as the Fractional Fourier Transform, from the Linear Canonical Transformation, a family of integral transforms (which includes Fourier, Laplace and Fresnel transforms as special cases). Given suitable restrictions on the set of functions the integral tranform acts on, the family is isomorphic to the group $\mathrm{SL}(2,\,\mathbb{R})\cong \mathrm{Sp}(2,\,\mathbb{R})$.
Under a member of this family of integral transforms, we have $(x:\mathbb{C}\to\mathbb{C}) \mapsto (X_\gamma:\mathbb{C}\to\mathbb{C})$ where:
$$X_\gamma(u) = \left\{\begin{array}{lcl}\sqrt{-i} \cdot e^{i\,  \pi \,\frac{d}{b}\, u^{2}} \int_{-\infty}^\infty e^{-\frac{2}{b}\, \pi\,i\, u\,t}\,e^{i \, \pi\, \frac{a}{b}\, t^2}\, x(t) \; \mathrm{d}t&&b\neq 0\\\sqrt{d} \cdot e^{i \, \pi \,c\,d\,u^{2}} x(d\,u)&&b=0\end{array}\right.;\quad \gamma=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)\in \mathrm{SL}(2,\,\mathbb{R})$$
and the integral transforms from the family compose by matrix-multiplying their $\gamma$ parameters; that is, if we denote the transform $x(t)\mapsto X_\gamma(t)$ by $X_\gamma(t) = \mathscr{L}(\gamma,\,x)(t)$ then:
$$\mathscr{L}(\zeta,\,\mathscr{L}(\gamma,\,x))(t)=\mathscr{L}(\zeta\,\gamma,\,x)(t);\quad\zeta,\,\gamma \in \mathrm{SL}(2,\,\mathbb{R})$$
The Fourier transform is given by $\gamma = \left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$. So, clearly we can specialize the above ideas to  the fractional Fourier transform where $\gamma = \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$; the Fourier transform results with $\theta=\pi/2$:
$$X_{\gamma(\theta)}(u)= \left\{\begin{array}{lcl}\sqrt{-i} \cdot e^{i\,  \pi \,\cot\theta\, u^{2}} \int_{-\infty}^\infty e^{2\,\csc\theta\, \pi\,i\, u\,t}\,e^{-i \, \pi\, \cot\theta \,t^2}\, x(t) \; \mathrm{d}t&&\sin\theta\neq 0\\i\, x(-u)&&\cos\theta=-1\\x(u)&&\cos\theta=+1\end{array}\right.$$
which is a unitary transform when $\sin\theta\neq0$ when it acts on a tempered distribution, because it is the composition of four unitary transformations in the following order:


*

*Tempered distribution $x(t)$ is mapped to $e^{-i \, \pi\, \cot\theta \,t^2}\,x(t)$, where we consider $e^{-i \, \pi\, \cot\theta \,t^2}$ to be a function in the non-distributional sense;

*Fourier transformation

*$x(t)\mapsto x(2\,\csc\theta\, \pi\,t)$

*$x(t)\mapsto \sqrt{-i} \cdot e^{i\,  \pi \,\cot\theta\, t^{2}}\,x(t)$, where, again, we consider $e^{i\,  \pi \,\cot\theta\, t^{2}}$ to be a function in the non-distributional sense.


Unitarity when $\sin\theta=0$ is clear.
