In the photon by photon double slit experiment like done here, It cannot tell us whether or not there are some photons that are not detected by camera but they actually hit the wall in some kind of hidden form ?
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$\begingroup$ Well, if the photons are not detected by the screen, we can't really say whether they hit or not, can we? To do that we'd have to detect them. $\endgroup$– AsherMar 27, 2016 at 19:56
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$\begingroup$ Yes, some energy is always lost, but it wouldn't be hidden. We can always do calorimetry on the total energy going in and out (and we try to do just that in experiments that actually produce important physics, which the double slit does not). It wouldn't make a major difference to the result, though. $\endgroup$– CuriousOneMar 28, 2016 at 4:02
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1 Answer
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Calibration of the detector with a well characterized source permits the quantum efficiency of the detector (camera) to be determined. Some losses are geometric: the active area of each pixel may not extend to the edge of the neighboring pixels. Some of the most sensitive single photon detectors available today have a quantum efficiency of 80%; these are silicon avalanche systems, and are capable of counting individual photons at better than 100 MHz.
So some are lost. Can we tell which ones? Yes, if the camera is detecting a member of an entangled pair we can check if there is a click from the camera for every click detected by a separate detector. These are called heralded photons, because the herald announces their presence; depending on the geometric relationship they can even be heralded in advance of their arrival at the camera.
So the answer to your first question is: yes, some will be missed; and the second question: yes, we can even tell (sometimes) which ones were missed.
With improved detection schemes the results can be improved.
answered Mar 27, 2016 at 22:47
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$\begingroup$ is there any way to account for the total number of expected photons from a source? In other words if 1000 photons on average are expected to impact but only 50% of that actually do what does that tell us? $\endgroup$ Mar 27, 2016 at 23:20
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$\begingroup$ Spontaneous down conversion is a very inefficient and unpredictable technique, and it's currently the best we have.. so no, we don't know how many there are. $\endgroup$ Mar 27, 2016 at 23:36