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I reorganized the question to clarify exactly what it is that I'm asking.

Suppose an experiment is performed where a particle detector records 50 particles per second, on average.

Absent any other considerations, it seems easy enough to come up with any number of theories to explain these results. For example, either of the following two theories would seem to be acceptible:

  1. assume the emitter produces 100 particles per second, implying a detector efficiency of 50%
  2. assume the emitter produces 200 particles per second, implying a detector efficiency of 25%

The only constraint in designing a consistent model is that the particle production rate times the detector efficiency must equal the number of particles detected. It seems that I can assume any emission rate that is greater than or equal to my actual measured detection rate.

In other words, we seem to be free to multiply the presumed emission rate by any positive factor, as long as we reduce the efficiency of the detector by the same factor.

The context of the question is this: detector efficiency is crucial to the argument in every Bell Test experiment I'm familiar with. But as far as I can tell, assumptions about detector efficiency are determined in practice so as not to violate the tenets of Quantum Theory. If that's the case, then the argument becomes circular and Bell tests only provide evidence that QT axioms are consistent with experiment, but doesn't decide between QT and other possible theories.

This line of thought led me to wonder whether the proportionality between macroscopic and sub-atomic energy and mass constants (for example, the Compton wavelength) aren't similarly under-determined.

To be clear, I'm asking a question about detector theory, and am not concerned about accuracy or calibration.

Thanks in advance, and let me know if my question can be improved, or if it's ambiguous in any way.

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    $\begingroup$ Your assumption that we don't know anything about the emitter is almost always wrong. $\endgroup$ – probably_someone Jan 16 '18 at 19:19
  • $\begingroup$ I'm not assuming we don't know anything, I'm asking if there is an argument that forces us to accept published detector efficiency numbers for logical reasons alone, or must the argument rely on some of the specific postulates/axioms of Quantum Theory. $\endgroup$ – philwalk Jan 16 '18 at 19:23
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    $\begingroup$ There are independent ways to get the luminosity of an emitter. For example, just tracking the power consumption and heat output of the emitter (while knowing the energy of the output particles) should give you a pretty good idea of its average luminosity. Once you have the average luminosity, you can constrain the detector efficiency. The problem is not underdetermined. $\endgroup$ – probably_someone Jan 16 '18 at 19:26
  • $\begingroup$ Heat measured in joules is based on kilograms, not atomic mass. I believe we use QT to translate between the two. $\endgroup$ – philwalk Jan 16 '18 at 19:31
  • $\begingroup$ You can get atomic masses using only classical electromagnetism, by using a mass spectrometer. The only thing you need for that is the fact that charge is quantized, a fact that can easily be proven using the (entirely classical) Millikan oil drop experiment. $\endgroup$ – probably_someone Jan 16 '18 at 19:33
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  1. assume the emitter produces 100 particles per second, implying a detector efficiency of 50%

    1. assume the emitter produces 200 particles per second, implying a detector efficiency of 25%

The only constraint in designing a consistent model is that the particle production rate times the detector efficiency must equal the number of particles detected. It seems that I can assume any emission rate that is greater than or equal to my actual measured detection rate.

Cart before the horse? The calibration of the detector, i.e. the efficiency, is done by measurements independent of the experiment under study.

In finding the efficiency of a detector to be used in an experiment, one does not assume an emission rate randomly. One uses a source with a known lifetime and, yes, the quantum mechanical models fitting the particular matter of the detector.

After the efficiency versus energy of a detector is known, general experiments can be devised to use this detector, i.e. independent of the particular experiments,sources and calculations used for calibration. The basic assumption is that a photon of a particular energy, talking of photon detectors, would register in the detector independent of the source. One would use the efficiency to get at the real numbers from a new source.

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  • $\begingroup$ You say that quantum mechanical models are used to fit the particular matter of the detector. Is that the only way derive the result, or can it be determined without recourse to quantum theory? $\endgroup$ – philwalk Jan 16 '18 at 21:28
  • $\begingroup$ By "the cart", do you mean my speculative simple-minded detector efficiency calculation, with the "horse" being the calibration process? But if calibration depends on quantum mechanics, then perhaps quantum theory could have inherited different historical values for the various classical constants, leading to a no less consistent theory, but with a different (but proportionally consistent) calibration process. $\endgroup$ – philwalk Jan 16 '18 at 21:45
  • $\begingroup$ decay curves are measured for nuclei and characterize them, without knowing what gave rise to the curve , the undelying theory does not enter, only the probability of decay is seen, fitted with mathematics. This can be fitted to classical physics plots, en.wikipedia.org/wiki/Exponential_decay $\endgroup$ – anna v Jan 17 '18 at 5:10
  • $\begingroup$ the mathematics of quantum mechanics affects the details of the detector, so as to get a good fit with data, the decay curve in my above comment.*The assumption/postulate is that a detector, once calibrated, is independent of the source it is measuring*. The use of a model for the calibrating source behavior with radiation is in order to fit an observed curve and are higher order corrections. The model happens to be quantum mechanical. If a classical physics model one would fit, it would make no difference as the nuclei of the detector do not change . $\endgroup$ – anna v Jan 17 '18 at 5:18
  • $\begingroup$ See en.wikipedia.org/wiki/Natural_units , for possible manipulation of units $\endgroup$ – anna v Jan 17 '18 at 5:34
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Yes, detectors need to be calibrated properly. Of course. It would certainly be circular if a detector was calibrated against itself or some other detector whose efficiency is equally uncalibrated. There are well-developed ways to properly calibrate particle detectors within specified bounds, especially for anything photon-related. The precise ways through which this is achieved are numerous. I have included relevant articles/sources below.

However, Bell-type experiments that test the polarization-correlation of coincident photons are not as simply affected by this type of absolute-scale problem (as are, say, thermometers). In these experiments, it is the property of one measurement (of a photon) relative to another. Also, if there were significant differences between the two (or more) detectors involved, we would expect the correlations to go down, which is why Bell-type experiments test their results against theoretical upper bounds - classical versus quantum. The goal of Bell-type experiments is to produce results that could not be explained using any sort of "classical" reasoning - a phrase that needs to be sharply defined, but which certainly includes detector issues.

Here is a recent publication and related popular article on the topic.

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  • $\begingroup$ thanks for the response, and for the links, I will read the article. But see my additional paragraph above, intended to clarify the question. I'm asking about the seemingly unavoidable under-determination of any emitter-detector theory by the emperical evidence. $\endgroup$ – philwalk Jan 14 '18 at 16:22
  • $\begingroup$ From the popular article: "In 2013, taking advantage of new types of photodetectors with intrinsic quantum efficiencies over 90%, two experiments closed the detection loophole". However, it doesn't address the question of what assumptions were made in arriving at the 90% number. I'm still downloading the files referred to in the publication, so I don't know whether they address the question or not. $\endgroup$ – philwalk Jan 14 '18 at 20:54
  • $\begingroup$ @philwalk One problem you are going to run into is that the calibration of scientific instruments is a boot-strapping process. You use a set of reasonable well calibrated instruments and a clever process to get another instrument better calibrated. But to understand the basis of the system you have to track the process back through dozens of iterations and hundreds of years. $\endgroup$ – dmckee --- ex-moderator kitten Jan 16 '18 at 19:56
  • $\begingroup$ the reason detector efficiency is relevant to Bell Tests AFAICT is because if the detector efficiency were exactly half (or less) than is currently assumed, all the interesting effects (e.g., non-locality) would no longer be implied by experimental results, and Bells inequality would not be violated by experiment. $\endgroup$ – philwalk Jan 16 '18 at 21:53
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The following is not a direct answer to the asked question about efficiency, but about the context (Bell test) which you mention in your question. I think this context can only be related to the precise question asked through a misunderstanding, which I hope to clarify here.

I think saying “detector efficiency is crucial to the argument in every Bell Test” is not true, even if detector efficiency indeed plays a crucial role in the design of a Bell experiment, and allows to predict its expected success or failure: detector efficiency plays no role at all in analysing the results, because, by definition quantum mechanics itself is what is tested in a Bell test.

More precisely, in “old” Bell tests (say, 1982–2015), where “the fair sampling assumption” was used, the only assumption made on the efficiency of the detectors is constant during the experiment, or at least, is not varied in an adversarial way implying coordination of the various photodetectors of the same side. In the recent (2015) loophole-free Bell tests, even the latter assumption is not made on the detector efficiency, since the detectors essentially have an answer almost each time the source sends “something” (quantum mechanics tells us that “something”$=$“a pair of entangled particles”, but the analysis of the Bell test only cares that a detector clicks at each side when the (metaphorical) “on”-button is pressed, and it does not care about what it actually is.)

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  • $\begingroup$ when you say "each time the source sends something" you're assuming a specific detector efficiency (otherwise, you wouldn't know how many particles the source had sent). However, you probably know that, since you disclaimed that you were responding about detector efficiency. I admit I don't understand the assumptions in the 2015 loophole-free tests, but I'm working on it. $\endgroup$ – philwalk Jan 18 '18 at 15:01
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    $\begingroup$ 1. For the 2015 test, I guess it is better for you to 1st avoid the Delft experiment (with NV centres in diamond), which is in a very specific set-up. 2. A Bell test is operated in a pulsed regime, where a pair of particle is emitted at well defined moments, say every 10 ms. This is very different from 100/s on average. In each time-slot, Alice and Bob check whether they have a click on not, and which detector have clicked. 3. Not specifying what “something” is, I do’nt assume anything (it could be vacuum). An assumption’d make it an assumption of the efficiency of the source. $\endgroup$ – Frédéric Grosshans Jan 18 '18 at 16:34
  • $\begingroup$ But I do’nt need such assumption, because if the source and/or the detectors would be of bad efficiency, they would simply fail to exhibit the correlations needed to pass the Bell tests. $\endgroup$ – Frédéric Grosshans Jan 18 '18 at 16:35
  • $\begingroup$ Thanks for the clarifications and the additional details. The next comment is only intended to clarify the idea that inspired the question, although I'm starting to understand why there is confidence in the published detector efficiency values. $\endgroup$ – philwalk Jan 19 '18 at 17:25
  • $\begingroup$ With respect to your last statement, I can understand that bad detector efficiency might cause it to fail to exhibit the correlations needed to pass the test, but my concern about the source is that it might be MORE efficient than expected, and that it produces more particles than theory predicts. This would lead to overestimation of the efficiency of the detector, to underestimation of the total number of pairs, and finally to overestimation of the correlated proportion. $\endgroup$ – philwalk Jan 19 '18 at 17:26

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