Consider a Mach-Zehnder interferometer where in one arm we place a filter with a transmission coefficient f << 1. The system is set up such that at one detector site we have constructive interference and at the other side we have destructive interference. Demonstrate that the expected number of detected photons N needed before a deviation of k sigma's from the case f = 0 is obtained, is given by N = k^2/(2 f). Therefore, had we just measured the transmission coefficient by shining light through the filter, we could expected to detect N f = k^2/2 photons, so the interferometer doesn't seem to offer any advantage.
Consider how the presence of noise that leads to a spurious background photon detection rate, can change this conclusion.© BrainMass Inc. brainmass.com March 5, 2021, 1:51 am ad1c9bdddf
In a Mach-Zehnder interferometer, laser light moves through a beam-splitter, the two beams are then combined via another beam splitter. At each beam splitter one side side the reflected light will be phase shifted by pi, while the beam that moves through or is reflected at the other side will not be phase shifted. This means that at th detection site at one end one beam will be shifted by pi relative to that beam moving through, while the other beam will not undergo a phase shift regardless of whether it moves through or is reflected. This means that if there is constructive interference at one detector, there will be destructive interference at the other detector.
The amplitude for a photon to pass through the filter is A = sqrt(f), this yields the transmission probability per incident photon of |A|^2 = f. Any phase factors here can be ignored, because that would show up at the detector sites and has been taken into account so that we end up with constructive or destructive ...
The noise reduction using interferometry is examined. The photons interferometer advantages are analyzed.