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I think the following is true, please correct me if I'm wrong. For coherent states, such as that come out of a laser, the intensity in a single mode is a Poisson distribution with mean given by the classical intensity. To give a stupid example, let's say the intensity is 100 photons / second, and the exposure time is 1 second. Then the intensity will be sampled from a Poisson distribution with mean = 100. The variance will then of course also be 100.

When we do off-axis holography, we do spatial heterodyne detection and we can measure the field. I believe the noise model is now complex gaussian, with mean given by the complex field. What is the variance of this measurement? how does it depend on the exposure time? I use exposure time and integration time interchangeably here.

Any references would also be helpful!

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  • $\begingroup$ In all light sources (including lasers) there is photon shot noise which is the square root of intensity .... so if intensity is 100 then noise is +/- 10 where 10 is one sigma deviation in the std distribution. So for 1 second you measure 100 and the next second it might be 90 or 110. $\endgroup$ Commented May 13 at 1:36
  • $\begingroup$ The spatial heterodyning affects the spatiotemporal degrees of freedom. It should not affect the particle-number degrees of freedom. So the photon statistics should still be given as Poisson statistics. $\endgroup$ Commented May 13 at 3:27
  • $\begingroup$ @flippiefanus While the intensity of reference + object beam has poisson statistics, you use post processing to extract the amplitude of the object beam. This amplitude has different statistics. $\endgroup$ Commented May 13 at 8:10
  • $\begingroup$ Can you model the post processing in terms of a mathematical process? $\endgroup$ Commented May 14 at 2:56
  • $\begingroup$ @flippiefanus Yes, so in digital off-axis holography you can model the reference beam as a plane wave that hits the detector at an angle. This causes a stripey interference pattern. You then take the fourier transform (fft) of the image. Next, you cut out the pattern at the first order. This first order contains a perfect copy of the amplitude in fourier space. You finally shift this copy back to the origin and perform an inverse fourier transform. But considering this path might be more work. I think you can think of it as just two quadrature measurements, which is also what chatgpt said $\endgroup$ Commented May 21 at 7:36

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