Relation between quantum tomography and optical Fresnel transform

TL;DR

This paper establishes a direct link between quantum tomography and optical Fresnel transforms, showing how Fresnel transformations relate to Radon transforms of the Wigner function in quantum optics.

Contribution

It demonstrates that Fresnel transformations in quantum optics correspond to Radon transforms of the Wigner operator, connecting optical Fresnel transforms with quantum state tomography.

Findings

Fresnel transformation maps position density to quantum tomogram.

Quantum Fresnel transform of states yields their wave functions.

The probability distribution for Fresnel quadrature phase equals the quantum tomogram.

Abstract

Corresponding to optical Fresnel transformation characteristic of ray transfer matrix elements (A;B;C;D); AD-BC = 1, there exists Fresnel operator F(A;B;C;D) in quantum optics, we show that under the Fresnel transformation the pure position density |x><x| becomes the tomographic density |x>_rs,rs_<x|, which is just the Radon transform of the Wigner operator, i.e., F|x><x|F^(+) = |x>_rs,rs_<x|= \int dx'dp'delta[x-(Dx'-Bp')]*Wigner operator where s, r are the complex-value expression of (A;B;C;D). So the probability distribution for the Fresnel quadrature phase is the tomography (Radon transform of Wigner function), and the tomogram of a state |phi> is just the wave function of its Fresnel transformed state F|phi>, i.e. rs_<x||phi>= <x|F^(+)|phi>. Similarly, we find F|p><p|F^(+) = |p>_rs,rs_<p|= \int dx'dp'delta[x-(Ap'-Cx')]*Wigner operator.