al onestone Posted May 4, 2012 Share Posted May 4, 2012 In the ZWM study of early 1991, Mandel (U of Rochester) achieves "induced coherence withouyt induced emmission" between two optically pumped downconversion crystals (figure provided). The explaination of the experiment goes like this: The pump beam goes to the 50/50 beamsplitter (BS) and then pumps one of two downconversion crystals (DC1 and DC2) which transform the photon into two photons (signal and idler) with each having roughly half the frequency/energy. The idler from DC1 is forced to transmit through the DC2 crystal so that it retains coherence between the output of the two crystals. The two signal outputs from DC1 and DC2 are combined at a detector Ds . There are two types of interference here, first is the interference in the coincidence counter (CC). When one of the optical paths is varied there is interference as a function of the combined phases (φs + φi). The coincidence interference results from indistinguishability between the combined pathways from DC1 to Ds and Di -or- from DC2 to Ds and Di . We cannot tell which way the pump photon went after the beam splitter. The second interference occurs at detector Ds and is a simple function of the phase. The interference that occurs at this detector is a lower order. It is simply the indistinguishability between the optical paths from the pump photon to detector Ds . If the detector has a count you cannot tell if it came from DC1 or DC2 depending on which path it took after the beam splitter. The coincidence interference(4th order) requires that all four optical paths to the two detectors be identical (or at least differing by no more than a coherence length). The interference at Ds (2nd order) requires the two optical paths from BS to DC1 to Ds and from BS to DC2 to Ds be identical. By placing the neutral density filter (NDF) between DC1 and DC2 in the idler path i1 , the interference at detector Ds vanishes. Why? Because the paths to Ds from the pump are now distinguishable. How? If the pump photon goes to DC1 and downconverts then the signal will go to Ds but the idler will be attenuated by the filter and you will not get a coincidence count at Di . If the pump photon goes to DC2 and downconverts then the signal will go to Ds and the idler will go to Di and there will be a coincidence count. So the presence or non-presence of a coincidence count at Di is the measurement which will determine whether the pump photon went to DC1 or DC2 to downconvert. So by inserting the NDF into the idler path, Mandel has eliminated the interference at detector Ds . And it is worth noting that the interference effect at Ds can only be negated by a distinguishability measurement which occurs at a second detector Di which is spatially separated from Ds , and that these measurements don’t actually need to be made to negate the interference. Mandel notes that these measurements at Di need only be in principle possible to negate the interference pattern. He got the same results without even making the distinguishing measurements. Absolutely Brilliant! My question is, what is happening at detector Di when there is no NDF placed in the idler path? Is there interference (2nd order) at this detector also? Mandel refferred to this detector as "superfluous" to the idler detector, so he gave no analysis of its counting rate. If there is interference at Di then one should be able to negate this effect by placing a NDF in the signal path s1 which could be done very close to the detector Ds. In this case we would be affecting what measurements are made at Di by performing a change in the preparation at Ds? Relevant papers of the study: X. Y. Zou, L. J. Wang, and L. Mandel, Phys. Rev. Lett. 67, 318 (1991). L. J. Wang, X. Y. Zou, and L. Mandel, Phys. Rev. A 44, 4614 (1991). Link to comment Share on other sites More sharing options...
al onestone Posted May 25, 2012 Author Share Posted May 25, 2012 This does not work. The second paper clarifies the issue. The counting rate at Di does not display interference because it is not a function of phase. For whatever reason, after the math only Ds displays interference. Link to comment Share on other sites More sharing options...
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