# What is the length of a photon?

Some questions that look kind of similar have been asked before, and I find the answers quite confusing. I intend to ask this question in a way that clearly shows what I'm asking.

Imagine the following thought-experiment. (I don't know whether it's practical to actually do the experiment or not.)

Set up the usual double-slit experiment with a photon source that sends one photon at a time. But replace the double slits with a single pinhole. Over a period of time you should observe a diffraction pattern that looks something like this:

Now in front of the pinhole, put a metal disk with slots cut in it, that can spin.

When a slot is entirely in front of the pinhole, it should have very little effect on photons that go through the pinhole. Maybe some minimal diffraction.

When the solid disk is entirely in front of the pinhole, hardly any photons should get through. A few might take the long route around the solid metal disk and through a slot, or around the outside edge of the disk.

When the edge of a slot is close to the pinhole, it might create an observable diffraction pattern different from the open one. You can observe those effects by rotating the disk to various angles and waiting for the diffraction pattern to form.

When the disk spins, then sometimes photons can get through the pinhole and other times they can't. Most photons are independent of all of the others. (If our photon-emitter releases them at random times, very occasionally two might go through the pinhole at similar times.)

The faster the disk spins, the shorter the times that photons can get through. We can vary two things -- the rotation speed of the disk, and the "wavelength" of the photons.

My assumption from a naive reading, is that photons are point particles that travel at lightspeed. So no matter how fast the disk spins, it should have no effect on the diffraction pattern.

But classically, radiation came from an accelerated charge. A single charge with an oscillatory motion would produce a wave, and if it oscillated a thousand times then the wave it produced might continue for a thousand wavelengths.

What happens in reality? Does our single-photon source produce point-particle photons which are entirely unaffected by the spinning disk except when an edge is close to the pinhole? Or does it produce photons that have a length, that can be interrupted by the spinning disk? If they are each a thousand wavelengths long, then at some wavelength and at some rotation speed they will all be affected. If they are one wavelength long, similarly at some speed they will all be affected.

Maybe the reality simply does not match up to point-particle photons that have special properties which make them each behave statistically like a wave. And it doesn't match up to literal waves either. If these concepts are useful teaching tools that don't really fit, it would be interesting to get a clear idea what to replace them with.

• In my opinion photons are linear dependent and wavelength represents a photon oscillating at a certain frequency through one full cycle. See “Single Edge Certainty” billalsept.com Commented Apr 11, 2021 at 15:36
• @BillAlsept I read your link. I notice you said "Contrary to scientific consensus, the fringe patterns observed in various slit, or edge experiments can be constructed from individual photons." I don't know whether your ideas are contrary to scientific consensus or not, but if they are it would explain why some people disagree with you. Commented Apr 13, 2021 at 10:44
• I challenge the anonymous downvoters to disprove my answer, which makes an unambiguous prediction of the experimental result. I also want to express again my aversion of anonymous, unmotivated downvoting. Such votes have a random effect on the quality of this site. Commented Apr 13, 2021 at 11:32
• @my2cts I agree, allowing downvoting without discourse runs contrary to the scientific process and is counterproductive for the community. Commented Apr 13, 2022 at 2:47
• Just an observation that one poster says a photon is infinitely long, another says it is not measured in terms of length and others provide examples of how it is measured.
– Paul
Commented Dec 22, 2022 at 20:39

## 6 Answers

A photon is not a point particle in the classical sense of the word (and neither is an electron, or any other fundamental 'particle'). Rather, it is a convenient word for describing some aspects of the electromagnetic field. It refers to the presence of energy in some particular 'mode' of the field. A mode can be thought of as an extended entity; it has a well-defined frequency and in free space it has a well-defined wavelength. It corresponds roughly to a plane wave.

In the type of diffraction /interference experiment in the question, the pattern observed on the screen is exactly as classical wave theory would predict, except that it appears as a set of dots rather than as completely continuous. However since the question concerns the pattern (i.e. the density distribution of the dots) we can use classical wave theory to answer the question.

The rings in the pattern occur at angles from the pinhole as given by diffraction theory. For a monochromatic source (without the chopping effect proposed in the question) the first minimum is at an angle given by $$\theta \simeq 1.22 \lambda / a$$ where $$\lambda$$ is the wavelength and $$a$$ the radius of the pinhole.

If we now 'chop' the transmission through the pinhole, as proposed in the question, then the light emerging through the chopper is no longer monochromatic. It now has a range of frequencies whose spread $$\Delta \nu$$ is of the order of $$\Delta \nu \simeq 1 / \Delta t$$ where $$\Delta t$$ is the time for which the pinhole is open. Consequently the light emerging from the chopped pinhole has a range of wavelengths. The range is given by $$\Delta \lambda \simeq \Delta \nu \left| \frac{d\lambda}{d\nu} \right| = \frac{\lambda^2}{c} \Delta \nu = \frac{\lambda^2}{c \Delta t}$$ The effect of this range of wavelengths is to blur the interference pattern. The angle to the first minimum is proportional to the wavelength, so it gets blurred by $$\Delta \theta = 1.22 \Delta \lambda / a .$$ It is useful to compare this blurring to the angle itself: $$\frac{\Delta \theta}{\theta} = \frac{\Delta \lambda}{\lambda} = \frac{\lambda}{c \Delta t}.$$ Thus we find that when the pinhole is chopped so rapidly that it only allows through one wavelength at a time, then the diffraction pattern is completely blurred away.

To answer the overall question about "how long is a photon", the answer is that a truly monochromatic photon is infinitely long. That is, it is a way of referring to a state of excitation of a mode of infinite length and perfectly precise frequency. For more physically realistic cases the field excitation of not infinitely long; in this case one can imagine a pulse of light of some finite duration. More generally it is the coherence length that is the important quantity.

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Apr 13, 2021 at 18:00

It's called the coherence length. It depends on how the photon was created. A photon from a short-lived excited state has a shorter coherence length than a photon (of the same wavelength) from a long lived state.

You can measure it: the easiest way to visualise this is through an interference/diffraction experiment a bit different from the one you describe. You take a standard Michelson interferometer with a non-laser source and set it up to show a fringe pattern. Now increase the length of one of the arms. You will still see fringes, according to whether the arriving waves are in phase or out of phase. But as you increase the length further the pattern fades and eventually disappears, at the point when the wave travelling the longer path arrives at the detector so late after the wave travelling the shorter path that the photon waves do not overlap. The extra path length when the pattern disappears is the length of the photon.

• Thank you! I can imagine this easily with traditional wave theory. If you generate a single wavelet in water -- up and down once -- and the wave goes through a slit, you can get only three interference peaks. The center. There's a place where the up and the down cancel. There is one on each side where the up is not canceled, and beyond that a place where the down is not canceled. That's all. Given light waves that each only cancel with itself, we could get something similar. If each individual wave was only say 5 wavelets long, then.... Commented Apr 13, 2021 at 11:42

Consider the following:

There is no such thing as a photon 1000 wavelengths long, but there is such thing as an atom emitting 1000 photons, these are called quanta and are fundamental to photon/quantum theory, each one is unique and destined on its own path to be eventually absorbed by another atom/atoms/molecules. An atom takes time to emit a photon so a single atom can't overlap photons, but a laser with many excited atoms can have one "seed" photon create a cascade of many photons, these photons are separated spatially according to how the atoms are arranged.

Consider your wheel to have a pinhole instead of a slot and lets consider your wheel stopped but the pinholes slightly misaligned, obviously there will be attenuation but a few will make it thru both holes, the result is just another circular diffraction pattern as in your image but offset from your original one. If we average this over many revolutions of the wheel you are left with Gaussian (blob) distribution. You have effectively destroyed the single slit "interference" pattern! Your wheel is very similar to what has been attempted for the DSE, you are trying to determine path information ... which is just another way of saying that you're messing with the photons and altering their originally intended path.

Finally, if your wheel was really really fast you are NOT going to be able to change the photons or break them or alter their wavelength!, they will will just behave quantum mechanically or probabilistically .... some might get thru at speed V and even less at 2V but certainly none will get thru at speed c ....which is impossible for your wheel to go anyway.

• @JThomas There are many old theories and new theories about photons, there are many theories to address many different aspects of photons, there are many many more opinions than theories on what these theories mean! The photon quanta is fundamental ... I don't know of any theories otherwise ... but I have seen a few opinions. Also some of the old theories are better than some of the newer ones! An atom can stay in an exited state for a quite a while (ms) but once the probabilities in the EM field align the emission time is very very short, wavelength over c is a good approximation. Commented Apr 13, 2021 at 3:30
• @JThomas according to QM orbits are not circular, they are probability distributions. Commented Apr 13, 2021 at 3:32
• There is most certainly such a thing as a photon 1,000, 10,000 or a million wavelengths long by the most-standard understanding (which is to equate the coherence length with the photon length). An infinitely long photon is called a momentum eigenstate (plane wave) photon. The "long photon" limit (monochromatic, single-frequency (thus single momentum)) photons are a staple of quantum scattering calculations: see eg. R. Loudon, The Quantum Theory of Light or journals.aps.org/pra/abstract/10.1103/PhysRevA.76.062709 or
– Dast
Commented Apr 13, 2021 at 11:36
• @Dast How long is a photon is most typically described as in arxiv.org/pdf/0803.2596.pdf. Also what is being discussed here is a 1000 photons, not a special long mode of a single photon. Commented Apr 13, 2021 at 15:43
• @PhysicsDave. Thanks for sharing the paper. The idea of "length" they introduce at the beginning (the one implied by fig.2 or fig.3) is exactly the one I was talking about. The timescale they emphasise throughout the paper (decay time of source atom) is the inverse of the photon's bandwidth (frequency spread). This timescale they refer to mutliplied by the speed of the photon (c in vacuum) is the coherence length I referred to. I think the OP is asking for a length scale to associate with a single photon (What is the Length of a photon) so a long mode photon seems a relevant thing to mention.
– Dast
Commented Apr 13, 2021 at 17:13

Time and energy in quantum mechanics are conjugate parameters, linked by The Uncertainty Principle where: (delta t) x (delta E) is greater than or equal to Planck's Constant/4 Pi. My understanding is that...if you had a one photon 'at-a-time' source, the time location/coordinate of that one photon and its property of energy would be linked and bound by The Uncertainty Principle above. You can never be certain where that one photon is, or the exact energy that it has. So, I tend to think of a photon as yes a point particle, but with each photon somewhere 'Uncertain' within a 'packet' of say delta t in 'length', and with an amplitude energy of say delta E in the 3 spatial dimensions. One photon would illuminate one spot only on the screen. As the number of photons sent increases, The Uncertainty Principle tells us that they hit different places (coordinates) with different energies (properties). What we see then is a wave pattern of light and dark fringes. This actually shows a Probability Wave for where the photons are most likely to end up, or not. This is the meaning of 'wave-particle duality', for photons, or indeed for particles such as electrons in an electron diffraction tube where this duality may be easily demonstrated. The wave pattern and the intensity/energy of the fringes, diminishing from the centre outwards, mathematically is described by Schroedinger's second-order differential equation. Hope this is helpful.

• If a photon is a point particle that travels at lightspeed with some uncertainty about its location or time, I would expect most of them to go through a slot in the disk with essentially no interaction, or else fail to go through the solid disk. So the spinning disk should have little effect on the diffraction pattern no matter how fast it spins. But if the photon is more like a traveling line segment, that takes time to leave its source and just as much time to pass a point, then the disk may affect it. Do you know which is true? Commented Apr 11, 2021 at 17:18
• My guess would be that the spinning disk will make the fringes dimmer. Say the slits are a tenth of 360 degrees, then I would expect the intensity of the fringes to be a tenth of those seen without a disc in place. Commented Apr 11, 2021 at 17:28
• Thank you. If I understand, if the slots take half the area of the disk that photons could reach, then only half of them will get through so the fringes will be dimmer. Half as many photons would result in half the detections, though that might not be half the perceived brightness. Commented Apr 11, 2021 at 17:32
• Radiative Lifetimes can be measured in a gas at low pressure in a tube. Excite the electrons with an alternating field from a coil wrapped around the tube. Modulate the alternating field. Measure the phase lag between the input and photon output signals. This gives the Radiative Lifetime. Commented Apr 13, 2021 at 8:53
• Thank you! When I look up "radiative lifetime" I mostly get things about semiconductors. When an electron is put into a higher orbital and not stimulated, it takes a random time before it radiates. I don't see how this relates to my question, but there's a lot of interesting information there which relates to a lot of things. Commented Apr 13, 2021 at 11:03

Often, as in some answers here, the terms photon and electromagnetic wave are used as synonyms. Yet they mean very different things. An em wave has wavelength, frequency and coherence length and time. A photon just has energy, momentum and spin. It is a dimensionless point particle as far as we know. The wave describes the probability of finding photons with given energy, momentum and spin at at given place and time. It is similar to an electron wave function in this respect.

In your thought experiment the transmission is not only position dependent, as in the traditional slit, but also time dependent. Classically this means that your original monochromatic wave is amplitude modulated and now consists of a band of frequencies. In photon terms you will observe a distribution of photon energies, each following their own probability distribution or diffraction pattern. However, in practice such effects are unobservable since mechanical chopping cannot be done at optically relevant frequencies. You will just observe the same interference pattern as without chopping, but chopped.

• I challenge the anonymous downvoters to disprove my answer, which makes an unambiguous prediction of the experimental result. I also want to express again my aversion of anonymous, unmotivated downvoting. Such votes have a random effect on the quality of this site. Commented Apr 13, 2021 at 11:21
• @my2cts I am not one of your dvotes. But my guess for the reason is (1) You are right, but a better answer would persuade me that it is right, by providing evidence (eg links, quotes) or a convincing argument. Not just a flat statement of "this is the case as far as we know". (2) Photons have several senses of "length": wavelength, coherence length and its possible they may also have finite radius. You only discuss the last of these. The OP's title question is asking for radius, but their proposed experiment picks out the coherence length, while wavelength is the one everyone has heard of.
– Dast
Commented Apr 13, 2021 at 17:23
• @Chiral Anomaly This does not suggest a hidden-variables interpretation anymore than calling an electron a point particle does so. This is mainstream physics. See en.wikipedia.org/wiki/Electron Commented Apr 25, 2021 at 19:36
• What Wikipedia describes as an upper bound on the size of an electron is a result about the scale at which electrons have been probed without finding any evidence of compositeness (the first part of my comment). Good texts are careful to explain this, but not everybody appreciates this nuance. Without clarification, a person might misinterpret the point-particle language to mean something like "at any given instant, the electron is localized at a specific point whether or not we're measuring it's location," which is the classic example of a hidden variables interpretation. Commented Apr 25, 2021 at 23:17
• @ChiralAnomaly The photon size is zero in the following sense: 'Moreover, the interactions of the particle can be represented as a superposition of interactions of individual states which are localized. ... is in this sense that physicists can discuss the intrinsic "size" of a particle: The size of its internal structure, not the size of its wavepacket. The "size" of an elementary particle, in this sense, is exactly zero.' en.wikipedia.org/wiki/Point_particle Commented Apr 25, 2021 at 23:45

We can determine photon duration given:

1. sunlight power flux of 1370 $$W / m^{2}$$
2. sunlight photon flux of $$10^{18} / m^{2}$$

Watts $$\left( J/s \right)$$ per photon of sunlight $$= 1.37 \cdot 10^{15}$$

Energy of green light photon (mid-point of sunlight spectrum, average energy) $$= 4.3 \cdot 10^{-19} J \left( h \cdot f \right)$$

Divide energy of photon by power of photon to determine time duration per photon $$\left( J / W \right) = 3.13 \cdot 10^{-4} s$$.

Multiply the time found above by $$c$$ to get length: 93,837.4 $$m$$ per photon.

This applies to all photons, not just visible light, since $$h$$ is in $$J s/cyc$$

Additionally, dividing $$h$$ by this time gives us the energy per cycle of field displacement