# Photons to Represent a Wave

I fear that I have a fundamental misconception about the "wave particle duality" of light, but in a related question, the answerer said, in some sense, that a light wave propagates until it hits something, at which point in time it (can) act(s) like a photon. Which is fine to me, but there are a finite number of photons in a wave front, so what happens if you "run out" of photons in a wave front? Certainly the wave needs to interact with everything it touches, so if you have a wave that only effectively has one photon, and it "hits" two electrons, how does it interact with both? Say you have two electrons both a distance $R$ from a photon emitter, emitting circular waves. Or something like that.

• It’s easier to understand if you just forget about wave functions and focus on photons. Depending on how powerful the source, billions of photons will radiate outward spherically. The inverse square law tells you there will be less and less photons the further out you get. Nov 25, 2020 at 15:33

In case you "run out of photons", you must switch to probabilistic description of quantum mechanics.

Let's consider an extreme case:

You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to all directions and thus its wave function interacts with all the detectors at the same radius at the same time, no matter how distant they are.

The key point is, to detect a photon, its wave function must collapse to a single point in space and time, where it has to be detected. QM says the wave function will collapse in the entire universe simultaneously and the photon is always detected at a single place - only in one of your detectors.

The probability of detecting a single photon on single distant detector will be very low and such experiment has to repeated many times (with many single photons fired) ... the result will be that the rate of detection at each detector (at same radius) will be the same, but at single time, only single detector detects a photon.

• Oh yes, I know that! Thank you! My curiosity was with regards to a classic E&M wave, which interacts via photons. Such a wave must consist of some number of photons, so I was wondering what happens if you have more things to interact with than you do photons.
– user24082
Aug 24, 2014 at 20:31
• Anthony, I think that Tadeas has answered your question. If a single photon is emitted, then it can only be detected once. (Note that this is only the case if the photon is actually "detected" or absorbed. If a photon passes through a piece of glass, for example, being delayed slightly as it does so, then it can still be detected later. Is this the sort of interaction you are thinking about?) Your remark "Certainly the wave needs to interact with everything it touches" is not correct. Aug 25, 2014 at 21:17
• No no, I tried to clarify. My confusion, I suppose, is in what the difference between a classic E&M wave is, and it's quantum mechanical description as a collection of photons. As annav said below, a classic wave is a collection of some number of photons, I was wondering, in effect, how a finite number of photons emulate this seemingly continuous wave behavior.
– user24082
Aug 27, 2014 at 7:01

The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $$h\nu$$, where $$\nu$$ is the frequency of the classical wave. Have a look at this blog description of how this happens mathematically. The interference pattern of individual photons at a time is the same as the classical interference pattern because of this $$h\nu$$.

One photon does not a wave make in space. One photon can be described by a probability wave, which means the probability of being at an $$(x,y,z,t)$$, which manifests in the single photon at a time double slit experiments. It is an ensemble of photons that make up a classical wave. I like to think of it as analogous to a "stadium wave". One person does not a wave make.

• @annav But how many photons does it take to make a wave? I'm still confused, if there is some finite number of photons that can make up a wave, then if you have that same number, plus one, of electrons, wouldn't you "run out" of photons to interact? Again, I'm just considering a circular wave, hitting a ring of electrons.
– user24082
Aug 24, 2014 at 20:26
• Oh, I meant atoms! Not electrons! (I was hoping to think of an example where the photons would be absorbed) Is something considered a wave only if you have enough photons to interact on the scale you're considering?
– user24082
Aug 24, 2014 at 20:35
• @Anthony You don't have your wave to affect every atom in an absorber body to observe the absorption. The threshold is just that you have to have total energy of the wave to be $E\gg h\nu$, that is, to have the number of photons in it $N=E/h\nu$ very high. And that is called classical wave. Photons can be considered waves even if they don't make the classical wave. Aug 25, 2014 at 0:30
• @Anthony A classial wave composed by agreat number of photons can all be abosorbed impinging on a medium . There would be no reflection and the medium would look totaly blac. Evidently there must have been as many photon-atom interactions as there were photons. As matter consists of order 10^23 molecules per mole this is not impossible to happen, though most of the time there will be some reflection. The photon molecule interactions can be of various types and would be the content of another question. Aug 25, 2014 at 5:30
• @firtree "Photons can be considered waves even if they don't make the classical wave" Important to stress they are Probability waves, not energy-intensity-in-space waves that the classical EM wave is (Poynting vector). The individual probabilities of the photons synergistically create the classical wave front with its phases in space. Aug 25, 2014 at 5:39

"Running out" of photons simply means that your wavefront is absorbed or scattered in a different direction or something like that. Either way, the original wave is "consumed", so you loose intensity or photons, depending on which picture you like better.

For the case of a single photon source: One photon can only interact with one electron. However, there are more complex cases, where the electrons could be coupled (like in Cooper pairs), then of course both electrons would somehow "feel" the photon. Or you can think of higher order processes. For example the photon could couple to one electron and form a polariton, which then could interact with another electron.

• The first paragraph correctly answers the original question, but the second introduces a misconception. A lone electron can't absorb an photon (though it can scatter from one, as Compton showed), since that process always has a reference frame where energy is not conserved. Photons are emitted and absorbed by atoms (and other more exotic interacting systems). While many atomic transitions can be approximated as a spectator nucleus, some spectator electrons, and one participant electron, that's just an approximation — one which fails badly for Cooper pairs and similar small systems.
– rob
Aug 24, 2014 at 8:15
• @rob engineer never said that an elecron should absorb the photon, neither s/he said that the electron is free. Aug 24, 2014 at 11:31
• @firtree The current version says "One photon can only interact with one electron," which is false (even though it's a useful approximation).
– rob
Aug 24, 2014 at 14:09
• @rob It is much less false than if it would say "a photon can be absorbed by an electron", so your criticism is still overdone. In QM (not QFT) exactly this approximation is used. Aug 24, 2014 at 14:13
• @firtree Fair enough.
– rob
Aug 24, 2014 at 14:19

Tadeas Bilkas answer let me think about the sence of all and all time citing the quantum mechanics. I write his answer in terms of common mechanics and get the same result: You have an emitter of balls which radiates just one single ball but in a spherical area. You place a lot of baskets some meters apart (with same distance) from the emitter. Mathematics says that if the source throw the balls strong probabilistic in all (horizontal) directions there exist a mathematical wave function by which help you know the probability that one of the baskets will be hit. You can call this the collaps of wave function but you don't have. If you don't call it collaps you are able to agree, that a single photon is all time a single part, and not only in the moment of observation. Something what is not observable you can call a wave function or you can call it a single part. Then ever you observe it you find a particle.

The quantum mechanics came into the play in your question about a single photon and two electrons. To count this case you realy need quantum mechanics. This is because the photon such as the electrons are located in space and interact like wave distributions around their centres.

And this is not a contradiction to the above said. A spherical probabilistic source has hidden parameters which we describe with a mathematical wave function. But to say that the wave collapses then the particle hits one of the possible points insists, that the wave in the over points instantaneously reduces to zero. That means that information will be leaded with velocity higher c.

To observe electron (or photon) you interact with them by the help of other photons (or electrons) and then you have to use quantum mechanics.

• I would almost agree. Here you do not have interference effects so the result can be explained classically (even with single particles). I just like to use the underlying math we use to interpret QM - but this "reduction of wave function in all space at once" is just math. There can (is) other math which describes it other ways. What we care about are the outputs of experiments - that is QM. Aug 25, 2017 at 9:55

there are a finite number of photons in a wave front, so what happens if you "run out" of photons in a wave front?

You are trying to make sense out of this. But it is not intended to make sense.

Modern physics is designed to be compatible with experiment. It does not need to also fit a simple mental model.

We cannot measure everything we want to. We cannot measure everything about subatomic particles all at once. So when we measure one thing, the things that we can't measure at the same time can be thought to fit a probability distribution. QM describes the measurements and the probability distributions both, resulting in a probability distribution for later measurements. This fits experimental results. But it's hard to think in terms of experimental probability distributions, and if you think about deterministic cases then you are thinking about things we cannot measure experimentally.

Electromagnetic radiation can be measured only when it is absorbed by atoms that get changed in detectable ways. The atoms are quantized. So all our measurements will be quantized. We can theorize that whenever a detection event happens, one photon has been absorbed. We can theorize about atomic events that each emit one photon. We can even theorize that it's the same photon that (we theorize) was emitted by one atom that gets absorbed by a detector. This is all handwaving assumption. The math does not demand those assumptions to generate the correct probability distributions.

I am going to suggest a thought experiment. I will suggest a plausible outcome. Maybe an expert will announce that my outcome is wrong.

Imagine you generate a single photon. You have a detector at a distance that gives it a 1% chance to detect that photon. We agree this is possible.

What if instead we put 100 detectors at 10 times the distance. Each of them has a 0.01% chance to detect the photon. Their chances to detect the photon are independent, so there is one chance in $$10^8$$ that any two of them will detect it. (There's just about a 99% chance that none of them will detect it. There's about a 0.99% chance that one will detect it. The chance that two or more will detect it is about 0.01%.)

There's a chance that 100 different detectors all detect the same photon? Yes. A very small chance. All the detections are independent.

But doesn't physics prove that when one photon is created by one source, that photon can only be absorbed by one detector? No, I don't think that has been experimentally proven. But maybe someone will prove me wrong.