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What is the smallest number of photons needed to make a "light wave"? In other words, how many (coherent?) photons start to exhibit classical behavior?

For example, how many photons are needed to get linear polarization? (Single photon has circular polarization.)

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jul 21 '16 at 18:07
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Even though there is a single photon in a volume of your choice the light is still a wave.

An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just imagine one photon is being split up by the beam splitter and combined on the detector to give the interference pattern.

After several hours of the exposure people have recovered the classical interference pattern is generated (as if one photon has interfered with itself).

Hence the interference pattern (the classical proof that light is a wave) is just our perception, It remain wave all the time whether it is one photon or one million.

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    $\begingroup$ I was about to give you a +1 for your first sentence, which is true, but unfortunately, the rest is unreflected conventional wisdom at the high school level and it is not even borderline correct. Dimming the lights is not a physics experiment that can tell us anything about quantum electrodynamics. The electromagnetic field at these energies and amplitudes doesn't interact with itself, which means that the brightness is not a physical parameter that changes the system. The only QED interaction here is in the detector and that's the one thing that is not being analyzed by this scenario. $\endgroup$ – CuriousOne Jul 20 '16 at 18:25
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    $\begingroup$ I understand where it comes from and it's not your fault, but, like I said, the textbooks are teaching the wrong thing here. An interferometer can test the wave nature of light, but that's a classical experiment. It says nothing about quantization. The spotty pattern on the plate tests the interaction of the chemicals on the plate with the em field, but it does so completely independent of the pattern. You could project a spotty image of George Washington or Pokemon on a photographic plate and arrive at the same conclusions. The interferometer is a completely superfluous element. $\endgroup$ – CuriousOne Jul 20 '16 at 18:44
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    $\begingroup$ Photons don't take paths. You can't write your name on one, then throw it into the experiment and then find the one with your name on at different parts of it at different times. You have to learn to live without the corporal existence of the photon. It is not backed up by a single experiment. Worse, still, photon statistics behaves like Bose-Einstein, which is for indistinguishable integer spin quanta, which is different from Fermi-Dirac for indistinguishable fractional spin quanta, which is different from the statistics for distinguishable macroscopic objects. $\endgroup$ – CuriousOne Jul 20 '16 at 18:47
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    $\begingroup$ Light always behaves like a quantum field. The only question is whether you are actually testing features of that quantum field or not. In the photoelectric effect you are testing them because you are focused on light-matter interaction which involve electrons. In the dim interferometer you are not testing them because you are focused on the free theory, which is as boring as it comes for QED. What you are not doing, is to test QED if no matter is involved, unless you have a gamma-gamma collider. The way I can tell that you don't is because nobody has built such a facility, yet. $\endgroup$ – CuriousOne Jul 20 '16 at 19:51
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    $\begingroup$ You don't need to read about QED for this purpose. If you want to know what happens when you are hanging on to the particle picture too hard, though, you can read Feynman's "QED: The strange theory of light and matter", in which he gives a painless introduction to path integrals. There you will see what a particle interpretation of quantum mechanics has to look like to reproduce the correct predictions of quantum mechanics. I call it the Santa Clause interpretation of QM because the path integral kernel has to visit all points in all possible ways in one night. :-) $\endgroup$ – CuriousOne Jul 20 '16 at 21:31
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Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion.

Qualitatively

spin angular momentum

Left and right handed circular polarization, and their associate angular momenta.

The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum field theory to understand it. In a summary, the wave function of a photon is controlled by a quantized maxwell equation, and the complex wavefunction has the information and phases necessary to build up the classical electric and magnetic field of the classical electromagnetic wave.

The number of photons for a given frequency of light can be estimated by dividing the classical power by the energy of each individual photon. An order of magnitude estimate of when the classical behavior appears can be seen in this double slit experiment

ds1

single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

Single photons below 200 in number look practically random on the photo One sees that already with 1000 photons the interference of the classical type is evident.

This also demonstrates the probabilistic nature of the photon's spatial behavior, as the classical interference pattern measures the probability of finding a photon at (x,y). At the same time, the macroscopic, point nature of a single photon , a dot on the ccd , is evident.

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  • $\begingroup$ The question when the classical double slit image appears is just a matter of taste, there is no physics in that, whatsoever. It's like asking if and estimate for a classical velocity with a 50% error bar is still a classical velocity, or not. $\endgroup$ – CuriousOne Jul 20 '16 at 19:57
  • $\begingroup$ Is the "event" of two photons passing the slits at the same time equivalent to two subsequent events of a single photon passing through the slits? In other words, when a "beam of light" (many photons) pass the slits, is it "the same" as when one photon at a time passes the slits? $\endgroup$ – Sparkler Jul 20 '16 at 20:39
  • $\begingroup$ @Sparkler . Photon photon interaction are very much repressed so , yes it is equivalent to a simple superposition of the wave functions . The square will give the probability distribution. en.wikipedia.org/wiki/Two-photon_physics. Note that the "the same time" for two photons is also very improbable. It is their great number that allow the build up of the classical wave in time and space, numbers given in another answer. $\endgroup$ – anna v Jul 21 '16 at 5:04
  • $\begingroup$ @Sparkler Yes, because the photons do not interact in any way (okay, they interact gravitationally, but gravity is so weak we can ignore it here). What is classically described as an interference in a wave is well described by Feynman's path integrals - basically, light behaves as if it took into account all the possible paths it could take, with no regard to the path it actually took (because no such path exists). Double-slit experiment can't distinguish between the two, it still assumes the photon is a particle. A better example would be a refraction grating/mirror. $\endgroup$ – Luaan Jul 21 '16 at 8:52
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    $\begingroup$ @Luaan higher order photon photon interactions are still much stronger than gravity-photon. For high energy photons they are even designing a gamma gamma collider. $\endgroup$ – anna v Jul 21 '16 at 12:20
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Let us look at coherent states $$ |\alpha\rangle~=~e^{-|\alpha|^2/2}e^{\alpha a^\dagger}|0\rangle $$ $$ e^{-|\alpha|^2/2}\sum_{n=0}^\infty \frac{(\alpha)^n (a^\dagger)^n}{n!}|0\rangle $$ If you have a classical system it means overlap between states is small. We then look at over lap $\langle\alpha'|\alpha\rangle$ $$ \langle\alpha'|\alpha\rangle~=~e^{-(|\alpha|^2~+~|\alpha'|^2)/2}\sum_{m,n=0}^\infty \langle 0|\frac{(\alpha)^m a^m}{m!}\frac{(\alpha'^*)^n (a^\dagger)^n}{n!}|0\rangle $$ $$ =~e^{-(|\alpha|^2~+~|\alpha'|^2)/2}\sum_{m,n=0}^\infty\frac{(\alpha)^m}{m!}\frac{(\alpha'^*)^n }{n!}\langle m|n\rangle~=~e^{-(|\alpha|^2~+~|\alpha'|^2)/2}\sum_{n=0}^\infty\frac{(\alpha)^n(\alpha')^n}{n!^2}. $$ The key factor to look at is $e^{-(|\alpha|^2~+~|\alpha'|^2)/2}$ and what when does this go to zero. Below is a diagram of the meaning of $\alpha$.

enter image description here

For the value of large momentum this factor $e^{-(|\alpha|^2~+~|\alpha'|^2)/2}$ is small.

There is no hard boundary between the quantum and classical worlds. However, let us consider light with a wavelength $\lambda~=~400nm$ which is near the middle of the optical range. The energy of a photon is $E~=~7.8\times 10^{-20}$j. Let us consider a $1000$ watt light source, which is comparable to direct sunlight. This light is about $1.3\times 10^{22}$ photons per second. Then in one second this amounts to that many photon and the momentum $p~=~E/c$ is then $3\times 10^{-6}kgm/s$. The momentum of each photon is about $2.5\times 10^{-28}$kgm/s. Now consider expanding the momentum in this diagram by that expansion and consider that this factor $e^{-(|\alpha|^2~+~|\alpha'|^2)/2}~\sim~e^{-10^{-28}}$ which is pretty small! This pretty clearly puts it is a classical domain.

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  • $\begingroup$ I added polarization as an example. I'm not sure the answer addresses polarization..? $\endgroup$ – Sparkler Jul 20 '16 at 18:53
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Light never completely behaves as a particle. Light never completely behaves as a wave. As pointed out by hsinghal, the Michelson interferometer showed that, even at the "single photon" level, we still see wave behaviors. These behaviors are well modeled by quantum mechanics, which treats light as neither a pure wave nor a pure particle.

As you "add photons" the "light is a wave" approximation gives you better and better results. However, as for "how many photons are needed to make a light wave," the answer depends on just how good you want that light wave model to approximate the behavior you see. This answer is entirely dependent on the quality of your sensory apparatus. Once the quantum behaviors of the photons cease to be measurable by your particular apparatus, it is reasonable to declare it is moving as a light wave because you have no way to distinguish the results you see from those predicted by a light wave.

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    $\begingroup$ Light doesn't ever behave like a particle or a wave. It behaves like a quantum field. People need to stop talking about it the way their great-grandfathers talked about it for a dozen years before Dirac wrote up with the correct explanation in the early 1930s! We have been over this wave-particle duality nonsense almost as long as we have been over the aether. $\endgroup$ – CuriousOne Jul 20 '16 at 18:50
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    $\begingroup$ @CuriousOne As far as I can tell, this answer agrees with your point of view completely. Nowhere does it say light behaves like a wave, and nowhere does it say light behaves like a particle. It does say "quantum mechanics ... treats light as neither...". Sounds like you and the poster are in alignment on this. $\endgroup$ – garyp Jul 20 '16 at 19:09
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    $\begingroup$ @CortAmmon: The point is that one has to talk about the quantum field as a completely new phenomenon. It's not enough to say it's "not quite this and not quite that". No, it's very much one thing, and that one thing is 0% this old thing and 0% that old thing. Dirac pointed this out very early, yet there is a lot of pussyfooting around in the teaching that keeps trying to pretend that we somehow can treat this new thing like a mixture of two old things. We can't. That's not what the experiments are telling us. $\endgroup$ – CuriousOne Jul 20 '16 at 19:59
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    $\begingroup$ @CuriousOne One only has to talk that way in a few circumstances. The OP just happens to be in one of those case. In 99.999% of my life, I can treat light as either a wave or a particle, and get results equally accurate as those coming from a quantum based theory because my sources of error dwarf the inaccuracy of my model. The trick is understanding at what point you have to cut the ties on those simpler models and use a more advanced model like QM. Even then, we can't really say QM is "right," from an ontological perspective -- it's merely consistent with what we see in the data. $\endgroup$ – Cort Ammon Jul 20 '16 at 20:01
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    $\begingroup$ @CuriousOne What if I challenged you that there are no quanta either? After all, quantum mechanics does not define the behavior of the universe, it merely models it to a degree we find sufficient. $\endgroup$ – Cort Ammon Jul 20 '16 at 20:06
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I'm afraid that linear polarization is not as interesting an example as you may have hoped.

First, the answer: in quantum optics, whether or not a quantum state exhibits linear polarization is independent of the photon count for that state. A single-photon state can be linearly polarized.

Now, the explanation: in quantum electrodynamics (QED) it is convenient (especially if you want to perform any actual calculations!) to quantize the field in terms of circularly polarized quanta we usually call photons. However, so long as you are only interested in the electromagnetic field ("quantum optics") it is equally valid [see footnote] - and in this case a better choice - to quantize in terms of linearly polarized quanta. (When I was studying quantum optics, we usually called these photons too, though I'm not sure whether that is considered technically correct or not.)

Specifically, in a thought experiment about a one-dimensional cavity with an ideal linearly polarizing filter part way along, the most natural quanta divide into three groups: those with the linear polarization that the filter passes through, those on the left of the filter with the linear polarization that the filter reflects, and those on the right of the filter with the linear polarization that the filter reflects.

There's nothing mysterious about this, because translating between a state described in terms of linearly polarized quanta and a state described in terms of circularly polarized quanta is trivial. If I remember correctly, in open space (or a simple cavity) a state containing exactly one linearly polarized quanta is just an equal superposition of the state containing a single photon with right circular polarization and the state containing a single photon with left circular polarization. (The orientation of the linear polarization is determined by the phase between the two component states.)

In an actual experiment, of course, you would not expect to see classical polarization at very low (single photon) intensities, because we don't have any ideal polarizing filters to experiment on. However, the photon count necessary to make the experiment work would depend not on the nature of light but on the exact mechanics of the polarizing filter itself.


PS: I am not familiar enough with QED to be absolutely certain, but as far as I know it is still true that you could in principle work with linearly polarized quanta, it just isn't a useful choice if you want to perform any actual calculations.

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