Question on possibility of uniquely defining the FRFT via certain properties

Question

I was working around with the fractional Fourier transform (FRFT) when the mathematics undergrad found out, by brute-force computations, that the derivative of the FRFT with respect to the parameter can be written as a closed form formula- explicitly, $d/da(F^a(f)(y))$$ = i(1/8d^2/dy^2-π^2y^2/2$$+π/4)F^a(f)(y)$.

Then I remembered that the first definition of the FRFT, using "only" a certain set (Hermite–Gaussian functions) of orthonormal eigenfunctions of the FT, is not actually "unique", especially when attempting to discretise the transform.

So I'm asking the following question instead- do unitarity, index (assumed to be real) continuity, index additivity, reduction to the identity (resp. Fourier transform) operator when the index is 0 (resp. 1), and the formula above uniquely define the FRFT, as operators on Schwartz functions?

NB: the version of the FRFT I'm using is the one used by the current revision of the Wikipedia page for the FRFT.

Answer 1

Answering my own question; turns out that the index additivity and "reduction to FT" conditions are not necessary at all. Ignoring said conditions-

Suppose $T_a$ exists and satisfies all of the other properties. Then, the kernel of $T_a$ can be represented as a tempered distribution $K_a(x,y)$, by the Schwartz kernel theorem. Let $H$ denote the Hermite-Gaussian function, normalised to "ordinary frequency". Then, $K_a(x,y) = Σ_n,_m H_n(x)H_m(y)Q_n,_m(a)$ for some function $Q_n,_m$ of $a$. Note that $Q$ is strictly non-zero; should it be zero for some ${n_0, m_0, a_0}$, no $T_{a_0}[f](y)$ can be equal to $H_{m_0}(y)$(as no Hermite function can be a linear series, finite or infinite, of Hermite functions of other orders), which makes the whole transform non-unitary.

Let $Q_n,_m(a)=e^{iM_n,_m(a)}$, $M$ not necessarily being real. Applying the condition that $d/daK_a[f](y) $$= i(1/8d^2/dy^2-π^2y^2/2+π/4)K_a[f](y)$ to the Hermite expansion of both sides and appreciating that $(1/8d^2/dy^2-π^2y^2/2+π/4)$$H_m(y)$ $=-πm/2$$H_m(y)$, $d/daM_n,_m(a)$$=-πm/2$ and $M_n,_m(a)=-πma/2+S_n,_m$, $S$ being constant of $a$.

Now throw in the "reduction to the identity" condition- $δ(x-y)=Σ_m H_m(x)H_m(y)=Σ_n,_m H_n(x)H_m(y)e^{iS_n,_m}$. Since Hermite expansion coefficients are unique, and $S_n,_m$ is constant of $a$, $e^{iS_n,_m}$ must be $δ_{nm}$ for all $n$, $m$.

Therefore, $K_a(x, y) = Σ_m H_m(x)H_m(y)(-i)^{am}$ for all $a$, which is exactly the definition of the FRFT. Q.E.D.