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It is said that light behaves like a wave until it is measured, then it behaves like a particle. Photons (the particles) then have to be defined by the measuring device. It is my understanding that all single photon detectors rely (at best) on light moving a single electron from one atom orbital to another (or even eject the electron).

Therefore it's impossible to detect arbitrarily low quantities of light because that would require detectors with arbitrarily small difference in electron orbitals and that thermal energy would create orbital jumps resulting in noise preventing a meaningful measure.

Are there techniques that allow to somehow work around this problem ?

Manu de Hanoi
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  • I recommend reading up on the photoelectric effect. (Especially the "emission mechanism" section of that page) – Andrew Nov 13 '20 at 16:57
  • Please see this answer of mine and links therein that has single photon double slit experimental pictures https://physics.stackexchange.com/questions/592672/destructive-interference/592683#592683 – anna v Nov 13 '20 at 17:52
  • @Andrew please be more specific, I think my question already shows a fair understanding of the photoelectric effect – Manu de Hanoi Nov 14 '20 at 14:13
  • @annav I'm aware of the double slit experiment but can't see how it's related to detector sensitivity – Manu de Hanoi Nov 14 '20 at 14:15
  • "Therefore it's impossible to detect arbitrarily low quantities of light" -- What do you mean by "quantities" here? One of the key lessons of the photoelectric effect is that you can detect a single photon (arbitrarily low intensity light) so long as it has the right frequency so that an individual photon can overcome the work function of the material. – Andrew Nov 14 '20 at 14:16
  • @ManudeHanoi did you look at the experiment one photon at a time? the detector is the photosphor screen has a footpring of a single photon, that can be photographed, fohttps://physics.stackexchange.com/questions/590491/type-of-camera-that-is-used-in-a-double-slit-experiment/590508#590508r example. – anna v Nov 14 '20 at 16:00
  • @Andrew I tried to explain in my question that when the work function of the material is small enough, thermal energy can activate it and create noise. – Manu de Hanoi Nov 15 '20 at 15:36
  • @ManudeHanoi There is a table of work functions for common materials on wikipedia, which are all in the range of a few eV. At room temperature, $kT$ is about 1/40 eV. So thermal fluctuations ionizing a material is pretty rare. Are you interested in a system outside of "typical" materials and room temperature? – Andrew Nov 15 '20 at 15:41
  • @Andrew I read that photon detectors are all having "dark noise" i.e. false positives. I believe the modern ones also need refrigeration. I therefore concluded that the noise was from heat. The work function of photon detectors is smaller than typical because high voltage is applied between the plates of the photo mutiplier. – Manu de Hanoi Nov 15 '20 at 15:50
  • I guess I just don't understand the sense of the question. Any detector is going to have noise. That doesn't mean you can't make a measurement. My only point is that single photons are routinely detected, at least for some range of frequencies. – Andrew Nov 15 '20 at 15:56

2 Answers2

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There is some confusion of what "single photon" really means in the context of single-photon detection.

  • When applied to photon counters, single photon means means the precision rather than a quantity of photons. Assuming Poissonian statistics the relative error in counting photons decreases as $1/\sqrt{N}$, where $N$ is the number of photons. That is the number of photons can be counted with precision up to a single photon, provided we have much more than one photon to detect.
  • Alternatively, one may talk about detecting very weak fields, potentially containing very few or even one photon. In this case the device may be sufficiently sensitive to produce a signal, if only one photon is present. But, as should be clear from the previous point, we may err by at least one photon in this case.
  • Finally, it is important to note that in both cases above we discussed only the energy of photons. Photons however may also differ by their polarization and direction (or other quantum numbers, if in a cavity). Some problems, such as studying Compton scattering, require detecting single photon of a particular direction, which requires compromising on the precision in photon counting or sensitivity.
my2cts
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Roger V.
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  • it does not answer the question but helps me seeing clearer , thanks – Manu de Hanoi Nov 14 '20 at 14:04
  • @ManudeHanoi the first part of your question is about the precision of photon counters (which is what I tried to answer), whereas the second is about frequency resolution, which has little to do with the first. Perhaps, you could reformulate the question or ask a new one, more precise. – Roger V. Nov 14 '20 at 15:23
  • There is a vocabulary/semantics/logic problem here. How could you detect an EM wave of less than a photon considering that we can't detect less than a photon because there is not such technology ( my question is clearly about such technology), and because, as I understand, there is no theoretical support for the detection of less than a photon. Because QM say that the smallest quantity of light is a photon. It seems to me that QM is entirely dependant on a technical point. There I wrote it. Please dont get my question blocked – Manu de Hanoi Nov 15 '20 at 15:46
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It is important to understand that the wave energy is quantised. The wave equation merely tells you where that quantum of energy might be found. When it is found, that event is localised and we say that we have detected a photon.

This quantisation of light waves as particle-like photons was proposed by Einstein to explain the photoelectric effect, for which he received his Nobel prize.

Very low levels of light take the form of sporadic quanta, each travelling as its own little wave. This causes what is known as "shot noise" in the detector, which fires sporadically as each such wave comes along and its quantum of energy gets absorbed.

Guy Inchbald
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  • My point is that quantization of wave energy is only proven experimentally using single photon detectors and that single photon detectors are limited in sensitivity by the quantization of electron orbitals (see the photoelectric effect). it's like having a single equation with 2 unknowns – Manu de Hanoi Nov 14 '20 at 14:08