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I am thinking about a detector that would beep if light passes through it. Is it possible?

Qmechanic
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Arik
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    Can we assume you want the light beam to have no interaction at all with the "sensor gate", and this "gate" to be of a certain size ? By "no interaction", I mean that the beam passes through the sensor without touching anything. And by "a certain size" I mean something that could be used at home or in any daily appliance without needing expensive machinery. Am I right ? – Benj Aug 23 '17 at 13:14
  • Photons have gravitational pull, right? Does that answer this question? – user541686 Aug 25 '17 at 03:07
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    Well it is impossible to interact with something without reacting with it. Otherwiese it would violate heisenberg's uncertainty principle. – MiltonTheMeme May 14 '20 at 01:48

3 Answers3

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It is indeed possible, as demonstrated by the group of Serge Haroche in 1999 using so-called quantum non-demolition Ramsey interferometry. The idea was to observe the presence or absence of a photon in a cavity by observing its interaction with atoms.

This beautiful experiment relies heavily on the behaviour of quantum superposition of atomic states. A simplified explanation is that the presence of a photon in the cavity results in an additional relative phase shift in one term of the superposition of atomic states, and this additional phase shift can be detected. Since all the measurements are done on atomic rather than photonic states, one can infer (and thus detect) the presence of the photon without actually absorbing it.

ZeroTheHero
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    Side note: This work (among others) led to the 2012 Nobel prize in physics. – Arthur Aug 23 '17 at 10:18
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    As additional information, see https://physics.aps.org/articles/v11/38 – ZeroTheHero Apr 23 '18 at 16:23
  • Does the detection change the state of the photon? I presume it must. If so, how? If not, why not? – rghome Oct 12 '20 at 08:30
  • This is news to me. does this mean that you could insert one of these detectors into a two-slit interference pattern device and get "which-slit" information without destroying the interference pattern? – niels nielsen Apr 19 '21 at 05:47
  • @nielsnielsen I am not an expert in this technique, but I don't think you can use it towards what you have in mind. The photon is highly frequency constrained by a cavity and the transit time of the rubidium atoms in the cavity adjusted be exactly 1/2 Rabi cycle. There is a good not-so-technical discussion of the experiment in "The quantum challenge" by G. Greenstein and A. G. Zajonc, where the setup better described. Beyond the technical difficulties, I don't see how you could do it and if one could, smart people Haroche et al would have done it already. – ZeroTheHero Apr 20 '21 at 13:17
  • @ZeroTheHero, thanks for the reference- NN – niels nielsen Apr 20 '21 at 15:28
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Yes, according to a paper by the Rempe group [Science 342, 1349 (2013)], photons can be detected after reflection by an optical resonator that contains a prepared atom in a superposition of two states. The reflection of the photon then results in a certain projection of the state that can be probed to detect the incident photon indirectly.

Paul
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    Reflection is not technically an absorbtion and emission of a photon in a particular way? – Todd Wilcox Aug 22 '17 at 20:44
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    @ToddWilcox To the extent that that is true (nonzero, but also with nontrivial subtleties), so is any interaction with a dielectric medium such as e.g. air (and you can draw a very solid equivalence between the two). If the physics in this answer counts as 'absorption' to you, then the only situation in which photons don't get 'absorbed' is when travelling through vacuum. That's a valid viewpoint but it guts most of the physical content and it's in pretty direct contradiction with our usual intuitive understanding of the meaning of the word. – Emilio Pisanty Aug 23 '17 at 03:54
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    @EmilioPisanty For me, the "usual intuitive understanding" is in pretty direct contradiction with what actually occurs. For example, absorption/reemission is the reason why "light travels slower" in media different from absolute vacuum.This is not difficult, only often "forgotten". I think this is why the question is so interesting and I will try to take a look at the reference from ZeroTheHero. – Knut Gjerden Aug 23 '17 at 07:39
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    @KnutGjerden As I said, it's a matter of viewpoint so neither view is wrong, but it's important to note that the virtual transitions involved in refraction and reflection do not change the photon-number content of the field (i.e. $\hat N$ commutes with the interaction hamiltonian) whereas the processes normally classed under absorption do change the photon number. So you can't really fault the people who reserve the term for the latter to the exclusion of the former. – Emilio Pisanty Aug 23 '17 at 07:45
  • @EmilioPisanty Seems like it would be nice for the asker to clarify what they consider absorption and not. – Todd Wilcox Aug 23 '17 at 13:43
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    @Todd As I said, if you count virtual transitions as absorption and re-emission, the only way for a photon to avoid this 'absorption' is to propagate in a complete vacuum without any interaction at all. That means that the answer would be trivially no, and the question would be deeply uninteresting. So I think it's pretty safe to accept answers with virtual transitions so long as the photon number doesn't change (other than the projective measurement as required by QM). – Emilio Pisanty Aug 23 '17 at 14:20
  • @EmilioPisanty I see your point. Well formulated. – Knut Gjerden Aug 24 '17 at 07:44
  • @ToddWilcox Taking a particle view it could elastically scatter, or be absorbed and another one emitted. However, if the emitted one is identical to the absorbed one, there's no way to tell the difference. – OrangeDog Aug 24 '17 at 13:38
  • @EmilioPisanty: This caveat also applies to the other answer, about quantum non-demolition Ramsey interferometry, right? We only ever observe photons when they interact (and since they are not charged, they interact directly - i.e. absorption and emission, and not by exchanging something). I think this observation is not completely trivial, if you are interested foundational questions. It tells us we never observe photons, we only ever see photon interactions. We cannot tell what happens between two interactions. – jdm Aug 24 '17 at 18:58
  • @jdm Yes, the conversation about real vs virtual transitions gets is equally applicable to Zero's answer. – Emilio Pisanty Aug 25 '17 at 10:35
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A photon can (in theory) be measured as a slight impulse change on a solar sail, i.e. mirror.

user2394284
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    I suspect the downvotes here are because conservation of momentum requires the re-emitted photon to be of different wavelength than the incident photon. – Carl Witthoft Aug 23 '17 at 19:18
  • The so called photon that is reflected off a solar sail will obviously have lower energy, energy conservation and Maxwell demand it. – barry Apr 29 '21 at 20:48
  • Non demolition and the integrity of the photon are incompatible, energy conservation is sacred! – barry Apr 29 '21 at 20:56