The Fresnel diffraction integral is a fractional Fourier transform
Light propagation can be viewed as a process of continual fractional Fourier transformation. As a beam advances, its amplitude distribution evolves through fractional transforms of increasing order.
Let $\mathbf{u} = (u,v)$ represent the coordinates transverse to the $z$-axis, which is the principal direction of propagation. Let $k = 2\pi/\lambda$ be the wavenumber along that axis.
We can show that the complex amplitude of the field is $$U(\mathbf{u},z) = e^{j(k\Delta z-\alpha)}\cos\alpha\exp\!\left(\frac{jk}{2\Delta z}\sin^2\alpha\|\mathbf{u}\|^2\right)\mathcal{F}_{-\alpha}\{\tilde{U}_0\}(\mathbf{x},z),$$ where $\alpha$ is the fractional order of the transform ($0 < \alpha < \pi$).
The $e^{jk\Delta z}$ represents the constant-phase contribution that is often neglected, the $e^{j\alpha}$ term is the Gouy phase shift, and the remaining exponential represents a quadratic approximation to a diverging spherical wavefront with radius $\mathscr{R}_{\alpha} = \Delta z/\sin^2\alpha$.
Hence propagation over a distance $\Delta z$ involves a fractional Fourier transform. When $\alpha = \pi$, the fractional Fourier transform is equivalent to the usual Fourier transform.
The connection to Talbot images is really interesting, and I intend to update this post and the associated notes in the future with further comments.
Aaron J. Lemmer
PhD Candidate, Mechanical and Aerospace Engineering
My research interests include adaptive optics and super-resolution microscopy.