84
$\begingroup$

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

$\endgroup$
3
  • 2
    $\begingroup$ 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 ? $\endgroup$
    – Benj
    Aug 23, 2017 at 13:14
  • $\begingroup$ Photons have gravitational pull, right? Does that answer this question? $\endgroup$
    – user541686
    Aug 25, 2017 at 3:07
  • 1
    $\begingroup$ Well it is impossible to interact with something without reacting with it. Otherwiese it would violate heisenberg's uncertainty principle. $\endgroup$ May 14, 2020 at 1:48

3 Answers 3

107
$\begingroup$

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.

$\endgroup$
6
  • 28
    $\begingroup$ Side note: This work (among others) led to the 2012 Nobel prize in physics. $\endgroup$
    – Arthur
    Aug 23, 2017 at 10:18
  • 1
    $\begingroup$ As additional information, see physics.aps.org/articles/v11/38 $\endgroup$ Apr 23, 2018 at 16:23
  • $\begingroup$ Does the detection change the state of the photon? I presume it must. If so, how? If not, why not? $\endgroup$
    – rghome
    Oct 12, 2020 at 8:30
  • $\begingroup$ 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? $\endgroup$ Apr 19, 2021 at 5:47
  • $\begingroup$ @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. $\endgroup$ Apr 20, 2021 at 13:17
37
$\begingroup$

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.

$\endgroup$
10
  • 14
    $\begingroup$ Reflection is not technically an absorbtion and emission of a photon in a particular way? $\endgroup$ Aug 22, 2017 at 20:44
  • 7
    $\begingroup$ @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. $\endgroup$ Aug 23, 2017 at 3:54
  • 5
    $\begingroup$ @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. $\endgroup$ Aug 23, 2017 at 7:39
  • 10
    $\begingroup$ @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. $\endgroup$ Aug 23, 2017 at 7:45
  • 5
    $\begingroup$ @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). $\endgroup$ Aug 23, 2017 at 14:20
-3
$\begingroup$

A photon can (in theory) be measured as a slight impulse change on a solar sail, i.e. mirror.

$\endgroup$
3
  • 14
    $\begingroup$ I suspect the downvotes here are because conservation of momentum requires the re-emitted photon to be of different wavelength than the incident photon. $\endgroup$ Aug 23, 2017 at 19:18
  • $\begingroup$ The so called photon that is reflected off a solar sail will obviously have lower energy, energy conservation and Maxwell demand it. $\endgroup$
    – barry
    Apr 29, 2021 at 20:48
  • $\begingroup$ Non demolition and the integrity of the photon are incompatible, energy conservation is sacred! $\endgroup$
    – barry
    Apr 29, 2021 at 20:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.