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.
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