17
$\begingroup$

Consider a typical optical focusing system: A small light source, then a collimating lens, then a focussing lens, and then a detector (e.g. CCD).

Assume that source intensity is so low that only one photon enter collimating lens per second. Today's modern technology is able to produce single-photon light sources. Assume that the dark room (in which the experiment is done) is 100% dark, i.e. detector detects only the photons coming from the source.

A photon that originated from the source is detected by the CCD. Is the detected photon the same as the one which originated from source?

Between source and detector, there are two lenses. When the photon enters the first collimating lens, it will interact with the electrons inside the molecules of the material from which the lens is made, but it will not interact with nuclei of different elements present in the molecules. Is the photon that came out of this collimating lens the same as the photon which entered the collimating lens?

What is the reason that the photon falling on the edge of the focussing lens gets deviated by some angle (i.e. gets focussed) and falls on the detector?

$\endgroup$
  • 20
    $\begingroup$ Can you write your name on that photon? If you can't, the entire question makes no sense. $\endgroup$ – CuriousOne May 5 '16 at 7:22
  • 2
    $\begingroup$ @AnubhavGoel: It means that something that can not be individualized can not be an individual? $\endgroup$ – CuriousOne May 5 '16 at 7:34
  • 6
    $\begingroup$ Even when you have just one photon you don't know that it's the same photon. I can make two urns for you and hand you a white ball to put into the first urn. Then I ask you to pull a ball out of the second. It's white... is it the same ball? $\endgroup$ – CuriousOne May 5 '16 at 7:41
  • 10
    $\begingroup$ @AnubhavGoel: It's not about that, at all. It is simply about there being no logical way to get from "One photon here, one photon there." to "It's the same photon.". The phrase "the same" requires means to distinguish individual objects, which, for photons, don't exist. $\endgroup$ – CuriousOne May 5 '16 at 7:57
  • 3
    $\begingroup$ @AnubhavGoel: Then that's what the OP should have asked. $\endgroup$ – CuriousOne May 5 '16 at 8:37
21
$\begingroup$

It is a matter of definition of "same".

Classically one can define "same" condition of particles by labels stuck on them. Light classically is a wave, and same needs a new definition. We apply the everyday definition by identifying the light beam with the source. The light leaving the sun is the same light arriving on earth. The light reflected from the moon is the same light. The light from a candle falling on a mirror is the same light. One can only label light from its source, imo.

Quantum mechanically light is an emergent phenomenon from a confluence of photons, and photons are elementary particles. Quantum mechanical calculations have been very successful in describing elementary particle experiments and are used extensively with success in cosmological models. The simplest basic calculation is a scatter of a particle on another particle , and the photon electron scatter can be represented as:

photon electron

This diagram is used to calculate the probability of the interaction happening, which is the mathematical modeling at the quantum level. It is the dominant term in calculating the cross section for a photon hitting an electron at energies below particle creation.

Now the question becomes "is it the same photon entering and leaving the scatter diagram". The position is not different then when considering light beams, which also cannot be labeled. The source is a photon on an electron, the output is a photon and an electron. One can define the photon's existence from the source and call it "the same".

When one goes to the apparatus you describe, the electron line becomes off shell, the interaction with the fields of the lattices it goes through, but the logic is the same. The scatter may be elastic, or inelastic and there is a probability for each case. Through the lens the probability is high to scatter elastically in the direction defined by the macroscopic optical ray.

Thus, if I were doing the experiment, and got a photon hit on the CCD , and had a source of photons, I would identify it as the same from the source. Of course not all CCD hits come from the light source, as there are cosmics and surrounding radiation, but that will be the noise level of the experiment.

$\endgroup$
  • 1
    $\begingroup$ But what is that stuff which makes this quantum mechanical probability so high that it compels that single photon to keep macroscopic "sameness"? Does not it seem extremely improbable to come out of the very dense jungle of molecules of the lens in a very "special" direction at the end ? $\endgroup$ – atom May 5 '16 at 10:27
  • 2
    $\begingroup$ If you go to the trouble to read the Motl blog I referred in the beginning of my answer, you will see that the macroscopic wave is built up by the complex conjugate part of the wavefunctions , which depend on the A potential of the macroscopic wave. Materials that are transparent have a specific crystal structure that allow for the continuation of the coherence in the confluence of the macroscopic beam, i.e. the complex conjugate parts of the individual photon wavefunctions retain phases . If it does not allow it, it is not transparent, what happens then will be scattering back and absorption $\endgroup$ – anna v May 5 '16 at 10:37
  • 3
    $\begingroup$ @atom That's a common misunderstanding. Quantum physics is anything but random - non-deterministic and random are two entirely separate concepts. The (very complex) wave function is absolutely entirely physically real, and completely deterministic and non-random - that's what tells you the possible "paths" the "photon" can take (the possible observations you can make of the system), and how likely each one is. It's not like the photon stops every second and thinks "well, should I turn 180° now?" It's more like a flow of a river system - all the branches contribute to the final result. $\endgroup$ – Luaan May 5 '16 at 10:59
  • 3
    $\begingroup$ @atom you are quite wrong . Classical mechanics emerges from quantum mechanics, the same way that a building emerges from bricks and cement. The underlying frame of nature is quantum mechanics. This is the present peer reviewed standard position of physics. Motl is a physicist and was a strong contributor in this site until he got fed up for some reason. He had the highest value in reputation here at the time. A physicist is one who knows physics, and believe me, Motl knows physics. $\endgroup$ – anna v May 5 '16 at 11:52
  • 2
    $\begingroup$ Perhaps the real take-home here is that the "photons" of QED diagrams (and, hence, of the Standard Model) are not really the same things as the "photons" we first meet in atomic-scale quantum mechanics. For example, QED photons don't have frequency as an intrinsic property, but QM photons do. The QM photon is an ensemble phenomenon made up of many interfering histories involving possible QED photons. $\endgroup$ – hmakholm left over Monica May 5 '16 at 23:20
11
$\begingroup$

Your question is based on the assumption that a photon is a fundamental object i.e. that photons are something we can point to and say here is photon 1, here is photon 2, and so on. The trouble is that quantum field theory particles are somewhat elusive objects. This is particularly so for particles like photons that are their own antiparticles because such particles can be freely created and destroyed. At least fermions like electrons are protected by conservation of lepton number.

In general energy propagating in a quantum field looks like a particle only when energy is being transferred into or out of the field i.e. when a photon is created or destroyed. Outside of these events it's hard to point to anything that looks like a photon.

So I don't think your question has an answer, because it isn't meaningful to talk about a single photon except when some interaction is occurring.

$\endgroup$
  • $\begingroup$ Interactions can been seen when it collides lens. $\endgroup$ – Anubhav Goel May 5 '16 at 8:18
  • 2
    $\begingroup$ The interaction with the lens is more of a collective effect rather than a single photon interaction $\endgroup$ – John Rennie May 5 '16 at 8:30
  • 1
    $\begingroup$ @JohnRennie: Consider a LASER beam. It is stream of billions of photons; all photons moving in almost straight line. There is no interaction of photons with anything else. Still we can MEANINGFULLY say photons are traveling in almost straight line. So isn't it meaningful to talk about a single photon ? $\endgroup$ – atom May 5 '16 at 9:44
  • 3
    $\begingroup$ @atom: no, see for example What is the relation between electromagnetic wave and photon?. There are other related questions on the site, but right now I can't find them. $\endgroup$ – John Rennie May 5 '16 at 10:09
10
$\begingroup$

Photons are boson, so it follows the Bose-Einstein statistics which is only true if the particles are truly indistinguishable. If you can distinguish between two photons, then it will follow the classical Boltzmann statistics which is not what happen in experiments. That means photons with same properties are the same.

Even in your situation with photon from source to destination, there are no ways to tell if the photon is the original photon. One possibility is that there are vacuum fluctuation creating a pair of photon and one of them hitting the detector. The "original photon" the other fly away. It is also possible that one of them annihilate with the "old photon", and the "new photon" is now pretending to be the "old photon".

How we distinguish photon is by their properties. Being said, you can actually imprint different properties to the photon to make them different from others, such as wavelength, polarization and angular momentum. Photon with different properties are indeed different so you can distinguish them and the statistics is different. They may interact with the matter such as lens in which it is being absorbed and re-emitted repeated. As long as the properties does not change, there are no way you can tell if it is different from the "original" photon because such question is invalid.

A special property is the entanglement between photon which does provide pretty good way to distinguish if the photon is "original". You can actually tell if the photon arrive at the detector are the same as the original one statistically (by testing the entanglement) because it requires all the processes in between coherence. So it can separate the case with the detector click of the photon from vacuum fluctuation.

The fundamental question is that if two objects are really the same by definition (in any way, really any, there are no physical way to tell them apart), are they the indistinguishable? And the answer is yes, you cannot distinguish them by the physical law.

$\endgroup$
  • $\begingroup$ Does your second paragraph hold even though there's no such thing as an anti-photon? $\endgroup$ – OrangeDog May 5 '16 at 20:37
1
$\begingroup$

Borrowing the concept of a "black box" from engineering, we have a photon "going in" into the box and a photon coming out of the box.
We take a "picture" of the photon going in (its amplitude and wavelength), and we take a "picture" of the photon coming out (its amplitude and wavelength), and we compare the "pictures." If the "pictures" are equal, then we can say it is the same photon. Keep in mind that inside the "black box" we could simply have a fiber-optic conduit, or we could have a detector-transmitter pair, or some other photo-electric device inside. Since we don't know what's inside the black box, all we can do is compare the characteristics of the input and output photons to determine if they are (or not) equal. If they are equal, then we would be justified to conclude that the output photon is the same photon that went in, but just delayed.

$\endgroup$
  • $\begingroup$ How would one consider this question from the point of QFT, Quantum Field Theory, which, as I understand it, includes the concept that all photons are simply fluctuations in a single universe-pervasive photon field? $\endgroup$ – John Fistere May 13 '16 at 21:32

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