How to move from the aperture to the far screen (from identity to Fourier transform in a smooth way)? In diffraction experiments, the pattern on a screen is the fourier transform of the aperture when it is a Fraunhofer diffraction (when the screen is far from the aperture).
When the distance is null (screen just behind the aperture) the
shape of the pattern is the shape of the aperture. so we have two operators (Id and FT) transforming aperture into pattern.
I wonder if in between we could not find operators depending on the distance.
Is it given by the propagator of quantum mechanics?
 A: You could do this using the angular spectrum method.
Let the field at the aperture (in the $x'y'$ plane at $z=0$) be $E_0(x',y')$. We then wish to find the field $E_z(x,y)$ in the $xy$ plane at $z$.
First we decompose $E_0$ into a superposition of plane waves travelling at different angles. This can be done by the Fourier transform $\mathcal{F}[E_0]$ where the Fourier variables are $k_x$ and $k_y$, the components of the wave vector in the $x$ and $y$ directions respectively. $\mathcal{F}[E_0](k_x,k_y)$ tells us the amplitude of the plane wave in the superposition with a given $k_x$ and $k_y$ and is known as the angular spectrum.
Each of these plane waves propagates a distance $z$ and so is multiplied by a phase $e^{ik_zz}$, where $k_z$ is the component of the wave vector in the $z$ direction. Thus at $z$, the new angular spectrum (i.e. the amplitude of each of the plane waves in the superposition) is $e^{ik_zz}\mathcal{F}[E_0]$.
Finally, we want to add up all these plane waves to give us the field $E_z$ in the $xy$ plane at $z$. This is simply an inverse Fourier transform $E_z=\mathcal{F}^{-1}\left[e^{ik_zz}\mathcal{F}[E_0]\right]$ (with Fourier variables $x$ and $y$). Written as an operator then, we get $\mathcal{O}=\mathcal{F}^{-1}e^{ik_zz}\mathcal{F}$.
To show that $\mathcal{O}$ becomes (proportional to) a single fourier transform $\mathcal{F}$ in the far-field Fraunhofer regime is not particularly easy, but you could look e.g. here. On the other hand, it is easy to see that $\mathcal{O}$ becomes the identity $I$ for $z=0$.
