# Does a single-photon spread while propagating?

Assume a single-photon propagates along z in a glass fiber that terminates, creating an interface glass-vacuum. After a certain length in vacuum we have a photon detector array that gives the (x,y) position of detection and the photon count (as far as I know such a device exists, e.g. from Hamamatsu). All this experiment is in a very dark environment and of course in vacuum.

In the fiber we can be confident that the photon is confined in the circle -3<x<3um, -3<y<3um, where this size is realistic for a 6um core diameter. After the fiber we do not know what happens, but the detector tells that the atom of the detector material absorbing the photon can be in a much bigger circle in the (x,y) plane. We call this spread in localization a diffraction. We repeat the experiment many times and are able to quantify that the single-photon is detected in a much bigger circle with radius r=R.

In principle, after this result, one can have two possible interpretations of what happens in vacuum:

1. The single-photon is very localized around a propagation direction, maybe not more than the pixel size, it just changes its direction of propagation when exiting the fiber core. The new direction has some randomness.
2. The single-photon is spreading during propagation so much that when impinging the camera it covers the entire region of radius R. Then why the atom A at r=0 rather than the atom B at r=R is absorbing it, is purely random.

Question 1: is there an experiment that can confirm one or exclude one or both of these pictures?

Question 2: is there an experiment that clarifies how extended in space in vacuum is a single-photon? Has the size of an electron, of an atom, of 10^n atoms?

Note: I do not want answers based on theories, quantum or non-quantum. I want a published experiment if any.

• A photon is just a small amount of energy. It's not a packet of energy that can be subdivided further. We can't observe a quarter photon here and three quarters over there. How many photons an absorbing surface or volume will absorb on average can be calculated with classical intensities (Maxwell). You don't even need a published experiment. Every time you see something with your own eyes this is what happens. The only intuitive problem here is the idea that there is something "below" the level of photons. There isn't. Commented Aug 3 at 10:26
• @FlatterMann That the energy of a single-photon (energy quantum) is the lowest possible for a specific optical frequency is a well known fact. But this does not have any obvious implication on the spatial extension of this physical thing ortogonally to the propagation direction. Has the size of an electron, of an atom, of 10^n atoms? What the experimental physics says about this question?
– Ang
Commented Aug 3 at 11:32
• Energy is the ability of a system to perform work on another system. It's a system property. When did we ever claim or teach that system properties have size and position? What you are calling size here is the size of the detector (measurement system). That has absolutely nothing to do with the energy that it detects. I can detect the same amount of energy with one atom or with a dinner plate. Nothing changes about the energy. Commented Aug 3 at 11:35
• @FlatterMann If you define a photon as a quantum of energy of the electromagnetic field, talking about position, does indeed not make sense. And when talking about a quantum particle of such sharply defined energy, it indeed would be delocalized across the entire space unless confined by some potential. However, when op talks about photons from my understanding he seems to just mean a quantum particle, that has a wave function with some small spread in momentum, energy and space.
– Zaph
Commented Aug 3 at 13:50
• @FlatterMann I just wanted to show some examples of what you called ”poor language choices from the early 1900s.” And that students are still today fooled with this wave-particle duality arguments. Well to me seemed more than poor language choices, rather seem wrong statements (except what W. Lamb states). Anyway, those statements are problematic as we have discussed here, in this very helpful discussion.
– Ang
Commented Aug 18 at 16:27

Both these situations are actually the same situation, just different ways of interpreting it. You don't really need a "published experiment" here because what happens is well known as a consequence of basic quantum theory.

Consider a single-mode fiber. Here, by definition, inside the fiber there is only one mode of propagation, and thus the photon can only be moving in one direction at any time: the direction along the length of the fiber.

Once outside the fiber, the wavefunction of the photon spreads out. Note: this does not mean the photon chooses a direction randomly. There is no randomness here, yet. The wavefunction (in the position basis) expands deterministically. The system enters a state of superposition, where the photon is in a superposition of being in different places with some probability amplitude.

Once the light hits the detector, there is no randomness either, yet. The system merely enters the state where all the sensors in the array are in a superposition of being activated or not activated.

Now, observation. According to the collapse postulate, when you actually measure which photodetector was activated, you will measure one out of the many detectors being activated, and all other detectors not being activated. Exactly which detector you measure as being activated is random.

However, let's say that instead of photodetectors, you have an array of direction sensors, which tell you the angle of incidence of the photon once they hit the detector, but don't tell you anything about the spatial position of the photon. In this case, you would measure the photon randomly taking one direction out of many possible directions, once it exited the fiber.

You could repeat this experiment with either the spatial position detector or angle of incidence detector placed at any point distant from the fiber exit, and the results would be the same.

Ultimately, this is basically the same kind of experiment as the double-slit experiment. Depending on where you measure the photon and what you measure, different variables appear 'random'. But this is just because of how measurement works.

• You focus on the case where the photons wave function propagates unitarily. When thinking of the systems wave function, that might be true in general. How ever when trying to model the behavior with a single photons wave function there is the possibility of it getting perturbed by some unknown factors while propagating. Partial-tracing back to the single photons wave function would then destroy unitarity and one might think, that that might be what OP meant with his description of case 1. In that case it would be wrong to say that both are only different interpretations of the same thing.
– Zaph
Commented Aug 4 at 7:22
• OP said it propagates in vacuum. What unknown factors could perturb it? Commented Aug 4 at 16:02
• Well he asked for explicit experimental setups, so taking the possibility of non ideal effects due to limited experimental capabilities into account just seemed reasonable to me. What do I know what kind of interactions might realistically happen, when the wave leaves the fiber.
– Zaph
Commented Aug 4 at 18:23
• @Zaph The wave function does not belong to either the photon (which is a small amount of energy) or the system. It's a description of the ensemble of the system. Since ensembles are purely abstract/hypothetical constructs of the human mind, there can be no wave functions in nature. You can find foundational papers that derive wave functions from Kolmogorov's axioms for independent sampling. Unitarity and linearity of quantum mechanics follow directly from these axioms, which in mathematical terms are essentially just a special class of partitions of unity. Commented Aug 18 at 16:02

No such experiment has been done and I've read a lot of papers and experimented with many fibres/cameras/lasers. It is not a practical experiment ... theory says we can't measure a photon without destroying it ...so its not possible.

At the exit of the fiber there is a strong interaction of the EM field of space, the fiber end and the photon, i.e. all the electron motion in these materials and in the entire apparatus are random and contribute to the EM field... this is the essence of diffraction.

The other issue is no matter how small your fiber the photon will enter based on probability .... again the EM field rules all .... the smaller the hole the lower the probability. You can google weak measurement and/or weak interaction for photons ... this is a good method to get an idea of photon behaviour but it is indeed statistical.

Disclaimer I know you explicitly said, you don’t want theory based answers. But I find the question interesting and would like to share my thoughts on it anyways. Feel free to just ignore it, if you are not interested in this answer.

On my understanding of the Question The problem is, as you might well know, that the moment you measure the a particle it will collapse into in eigenstate of the corresponding observable. At that point all the information about the previous state of the particle is gone except, that it had some component of the eigenstate you measured. That means not only can’t you measure the path the evolution of a particles wave function took, you can’t even determine the wave function of the particle. The only thing you can do is try to prepare a lot of particles in very similar states, then measure each of them and try to reconstruct the probability density of that state, form you measurements.

So from that background I do understand your question in the following way: Is there a way to determine, whether the deviations of your repeated experiments do come from very similar quantum states, that just spread during unitary time evolution, or whether your setup is producing very different quantum states, that develop into very different states even before measuring. So to speak check if your measurement deviations are from quantum-uncertainty-effects or just artifacts of your setup.

My first idea of a procedure I think one thing one could do, is use some optical grid to create an interference pattern behind your setup. If you do create almost the same state in every measurement repeating the measurement often should yield the intensity profile expected behind the grid for a coherent light experiment. If you however create different states after leaving the fiber, I would expect you to just see some unstructured spread, as the interference patterns of you probability densities would differ every time you measure and therefore not converge to a pattern.

My second idea of a procedure One could try creating an entangled pair of photons and send each of them threw one of your described setups. If your setup only includes unitary time evolution (including the expected spreading of your wave-function after leaving the fiber), you would then expect perfect correlation of the directions your particles take due to momentum conservation. If however your state gets perturbed when leaving the fiber, such correlations should be destroyed.

• Unitary time evolution simply means that the number of ensemble members in our ensemble doesn't change. There is no deeper physical meaning to it than that. It doesn't tell us anything about the individual measurement. The measurement itself is the transfer of a quantum of energy from the quantum system to the measurement system. There is no such thing as "collapse". It is energy flowing from one system to another. There is no great mystery behind it than that, either. What confuses people about quantum mechanics is the "abstract state" notion when it's actually all about energy flow. Commented Aug 3 at 19:48
• @FlatterMann When describing Quantum mechanics theoretically, we do use wave functions developing unitarily (,which means more then just perservation of the number of different states in ones ensemble). Now its true that we can’t measure them directly, but assuming they exist (and collapse) is still what gives us the best predictions of the real measurements. I see no use in claiming they are not a real thing.
– Zaph
Commented Aug 4 at 7:35
• I have never measured a state. I have measured quanta of energy, momentum, angular momentum and charge. One has to differentiate between the abstract bookkeeping of the purely theoretical ensemble state and actual physics, which are these individual energy, momentum, angular momentum and charge exchanges. There is no need to insert something like "collapse" between the two. Quanta are real, quantum mechanical state is the law of large numbers limit of actual frequentist histograms. Would you call a histogram (a table on paper) part of nature? A wavefunction is even less real than that table. Commented Aug 5 at 0:57

Please resist the urge to think of photons as particles. Photons are quantized excitations of spatial modes of the electromagnetic field.

When thinking about these kinds of questions always think about the classical EM field explanation first. In the classical EM field we have a wave guided along by the fiber. When it exits the fiber the mode couples out of the fiber and into free space so it rapidly diffracts. If you put a detector or a screen some distance from the fiber tip you will see an approximately Gaussian spot on the detector/screen. Now if you turn the power propagating through the fiber lower and lower and you make sure your detector is a pixel array of single photon detectors, you will start to see the pixels only light up one at a time. But if you wait very long and plot the distribution of photons counted you will see the distribution matches the classical mode profile.

The fact that we continuously tune from the classical EM wave picture to the "photon" picture makes me start to think the photon is spreading out, it is not "localized but travelling in a random direction". But we can come up with an experiment that lends even more credence to this point of view.

Back to classical EM picture. We have the fiber and we have light propagating down it and exiting. Now instead of placing a screen some distance from the fiber we place instead two new fiber inputs.

                           \  \  \   \ >---- New Fiber 1 Input
Original Fiber Output ----< |  |  |  |
/  /  /   / >---- New Fiber 2 Input


Now take the outputs of the two fibers and combine them on a beamsplitter. and look at one or both of the outputs. Also arrange the ability to precisely tune (to sub-wavelength precision) the distance between one of the fiber outputs and the beamsplitter.

                             Detector 1
|
|
|
o

New Fiber 1 Output ----<         /     o--- Detector 2

v
|
|
New Fiber 2 Output


As you tune the distance between one of the fiber outputs and the splitter you will see interference fringes on the two detectors. Recall this is a classical experiment.

Now turn down the light so you are back in the single photon regime. If, in the classical experiment, you were on a dark fringe on detector one and a bright fringe on detector two, you will never see a photon detected on detector one and you will always see one on detector two.

This says there is spatial coherence between the output of the first fiber at the locations of the inputs of the latter two fibers. In my mind this ruins the theory that the photon is localized but travelling in a specific but random direction. At LEAST, given this analog of the Young's double slit experiment, you have to allow that the "localized" bullets of light (I really dislike viewing it this way) travel along multiple paths at once. But at this point just admit it's a quantized electromagnetic field.

Ok, finally after writing this answer, I embarrassingly realize that it would be much simpler to just put a double slit at the output of the fiber and observe the direct Young's double slit interference pattern. First classically with bright light, but then quantum mechanically as the light level is turned down. This question is asking exactly the question which is answered by Young's double slit experiment. The answer is that the photon "interferes" with itself. So we cannot just think of it as a particle travelling in a random direction. There is necessarily more to the theory than that. For me it is best explained by thinking of photons as being quantized excitations of spatial modes of the electromagnetic field. With emphasis on the field nature of the EM field.

• Nice answer (+1) Could you elaborate on just the statement "when it exits the fiber the mode couples out of the fiber and into free space so it rapidly diffracts", please? This seems contrary to my intuition about light. E.g. A beam of laser light will travel pretty much in a straight path without diffraction in free space, right? Light rays in from sunlight, torches, etc. also do not seem to diffract as they travel in free space... Commented Aug 8 at 5:02
• ‘Photons are quantized excitations of spatial modes of the electromagnetic field.’ How does that differ from any other quantum particle? Commented Aug 8 at 6:35
• @James Look up Gaussian beams. it is impossible to have a "column" of light propagating through free space with a fixed radius. For "typical" beams, the closest thing we have is a Gaussian beam. The beam has a point in space along its axis where it is smallest (the waist) and it spread out from there. If the waist is large compared to its wavelength the spreading is modest, but if the waist is small, like at the output of a single-mode fiber, the spreading is aggressive. One can quarrel over whether to call this effect "diffraction". The point is it is a result of the wave equation (1/2) Commented Aug 8 at 12:06
• @James Well what I just explained about a Gaussian beam says that a beam of light does spread away from its axis so that its amplitude is reduced as it propagates. I'm sure you can have a relatively directed water wave but both water and light waves "spread" as they propagate because of the wave equations they satisfy. So they can't carry their energy in an infinite "beam". Commented Aug 8 at 13:14
• @James please ask a separate question. Your question is about classical propagation of general waves and electromagnetic waves specifically. I'm not sure what you mean by long range propagation, but water wave packets can propagate for hundreds of miles. Commented Aug 8 at 15:14

All received answers contain some crucial aspects. I try to compile a single answer containing all these aspects.

When we change the apparatus, the probability of the photons being absorbed in a certain point in space changes. So we'll have different macroscopic/collective patterns depending on the apparatus characteristics. It appears that the way how energy flows through the apparatus depends on the apparatus.

Question 1: is there an experiment that can confirm one or exclude one or both of these pictures?

Answer: No, what happens between the source (the fiber-vacuum interface) and the detector is not possible to be disclosed directly. If one count is recorded by the instrument we can state that the energy quantum was transferred to the detector. We can state that the transfer of energy happened in one specific pixel in the detector array. We could move the detector at differnt distances z from the end of the fiber with the intention to trace a trajectory of the photon. But this cannot be done because we do not have any possibility to guarantee that the next photon will be an exact copy of the previous one.

Question 2: is there an experiment that clarifies how extended in space in vacuum is a single-photon? Has the size of an electron, of an atom, of 10^n atoms?

Answer: No, there isn’t. It seems obvious to represent the photon as something that has the size comparable to the electron cloud of the atom absorbing it. But it is really arbitrary how we want to picture that. It can be pictured as a ball with the size of a pixel but also as something that extends over the entire detection area and then the absorbing process happens with a certain probability distribution in a specific point $$(x_0, y_0)$$. Depending on what kind of apparatus we use to localize where the absorption process happens, we get different results, therefore the extension of a quantum is just an arbitrary property or, in other words, not a property of the quantum.

• The "photon" has the same extent as a classical EM wave passing through the same apparatus. Commented Aug 5 at 20:43
• @Jagerber48 if you edit your answer and add great level of details or link an article that proves your statement I will be happy to delete my answer and accept yours instead.
– Ang
Commented Aug 6 at 15:57
• I've expanded my answer. Think about doing a Young's double slit experiment with the output of the fiber. Commented Aug 8 at 4:41
• @Jagerber48 A photon is an amount of energy, momentum and angular momentum. These are system properties. System properties don't have "extent". They didn't have that in classical mechanics, either. A classical EM wave is an approximation of many photons, IF they form a suitable classical state. If they don't, then there is no classical EM wave that can describe the physics of that particular experiment correctly. Commented Aug 8 at 12:28
• @Flatterman Classically, and quantum mechanically, the momentum and angular momentum of the electromagnetic field are spatially varying. Commented Aug 8 at 12:33