dispersive propagation & Fractional FT

--- tags: dispersive propagation & Fractional FT ---  $$ \def\moy#1{{\langle #1 \rangle}} \def\abs#1{{\left| #1 \right|}} \def\dd{\mathrm{d}} \def\ed{\mathrm{e}} \def\norm#1{{\parallel #1 \parallel}} \def\ket#1{{| #1 \rangle}} \def\bra#1{{\langle #1|}} \def\braket#1#2{{\langle #1 | #2 \rangle}} \def\un{\small1\normalsize\kern-.33em1} $$ # dispersive propagation & Fractional FT ### The fractional Fourier transform Following Victor Namias in J. Inst. Maths Applies (1980) 25, 241-265, the fractional Fourier transform $\mathcal{F}_\alpha(x)$ of $f(x)$ is given by (Eq. 3.6): $$ \mathcal{F}_\alpha f(x) = \frac{\exp\left[i\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)\right]}{\sqrt{2\pi\sin\alpha}} \exp\left[-\frac{i}{2}\cot\alpha x^2\right] \int\exp\left(-\frac{i}{2}\cot\alpha y^2 +\frac{ixy}{\sin\alpha}\right) f(y) dy $$ which can also be written as $$ \mathcal{F}_\alpha f(x) = \frac{\exp\left[i\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)\right]}{\sqrt{2\pi\sin\alpha}} \exp\left[\frac{i}{2}\tan\alpha x^2\right] \int\exp\left[-\frac{i}{2}\cot\alpha\left(y-\frac{x}{\cos\alpha}\right)^2\right] f(y)dy $$ The definiition of the inverse FFT is obtained by complex conjugaison of this equation with $\alpha$ real. ### The propagation in a dispersive fiber Following Keisuke Goda and coll. in PHYSICAL REVIEW A 80, 043821 (2009) (Eq. 14), the field amplitude $u(z,T)$ after a propagation distance $z$, at time $t = T + \beta_1 z$ (where $\beta_1$ is the inverse of the group velocity) is given by: $$ u(z,T) = \exp\left(-i\frac{T^2}{2\beta_2 z}\right) \int \exp\left[i\frac{\beta_2 z}{2}\left(\omega - \omega_s -\frac{T}{\beta_2 z}\right)^2\right] \tilde{u}(0,\omega-\omega_s)d\omega $$ where $\tilde{u}(0,x)$ is the Fourier transform of $u(0,x)$. With the change of variable $y = \omega-\omega_s$ we obtain $$ u(z,T) = \exp\left(-i\frac{T^2}{2\beta_2 z}\right) \int \exp\left[i\frac{\beta_2 z}{2}\left(y -\frac{T}{\beta_2 z}\right)^2\right] \tilde{u}(0,y)dy. $$ Defining $$ \beta_2z \equiv \cot\alpha, \quad T \equiv \frac{x}{\sin\alpha}, $$ we obtain $$ u(z,T) = \exp\left(-i\frac{x^2}{\sin(2\alpha)}\right) \int \exp\left[i\frac{\cot \alpha}{2}\left(y -\frac{x}{\cos\alpha}\right)^2\right] \tilde{u}(0,y)dy $$ comparing with the definition of the fractional Fourier transform, we have, $$ u(z,T) = \sqrt{2\pi\sin\alpha}\exp\left[i\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)\right]\exp\left[\frac{i}{2}\cot(\alpha) x^2\right] \mathcal{F}_{\alpha}^{-1} \left[\tilde{u}(0,y)\right]. $$ We conclude that the proppagation over a distance $z$ in a dispersive fiber does not correspond exactly to an inverse fractional Fourier transform. There is an additional quadratic phase $\exp\left[\frac{i}{2}\cot(\alpha) x^2\right]$. The same problem occurs in the case of Fresnel diffraction (see "Fresnel diffraction and the fractional-orderFourier transform" by Pierre Pellat-Finet in OPTICS LETTERS / Vol. 19, No. 18 / September 15, 1994). In this reference Pellat-Finet claims that observing the diffraction pattern on a spherical surface instead on a plane solves the problem of the additional quadratic phase. Can we use a similar trick ?