Optical photon is an electromagnetic wave produced e.g. during deexcitation of an atom, carrying energy, momentum and angular momentum difference.

So how is this electromagnetic energy distributed in space (rho~|E|^2+|B|^2) - what is the shape and size of a single photon? What is the position distortion of such wavepacket?

Looking for literature, I have found started by Geoffrey Hunter, here is one of the articles: "Einstein’s Photon Concept Quantified by the Bohr Model of the Photon" https://arxiv.org/abs/quant-ph/0506231

Most importantly, he claims that such single optical photon has a shape similar to an elongated ellipsoid of length being wavelength λ, and diameter λ/π (?), providing reasonably looking arguments:

  1. Its length of λ is confirmed by:

– the generation of laser pulses that are just a few periods long;

– for the radiation from an atom to be monochromatic (as observed), the emission must take place within one period [10];

– the sub-picosecond response time of the photoelectric effect [11];

  1. The diameter of λ/π is confirmed by:

– he attenuation of direct (undiffracted) transmission of circularly polarized light through slits narrower than λ/π: our own measurements of the effective diameter of microwaves [8,p.166] confirmed this within the experimental error of 0.5%;

– the resolving power of a microscope (with monochromatic light) being “a little less than a third of the wavelength”; λ/π is 5% less than λ/3, [12];

Is it the proper answer?

Are there other reasonable answers, preferably experimental arguments?

Update: similar conclusions from the different author: https://arxiv.org/pdf/1604.03869

the length of a photon is half of the wavelength, and the radius is proportional to the square root of the wavelength

Update: 2021 "The size and shape of single-photon" http://dx.doi.org/10.4236/oalib.1107179

Attosecond chronoscopy brings hope to verify e.g. photon models experimentally - gathered: https://scholar.google.pl/scholar?cites=15193546925951882986&as_sdt=2005&sciodt=0,5&hl=en E.g. 2020 "Probing molecular environment through photoemission delays" https://www.nature.com/articles/s41567-020-0887-8

Attosecond chronoscopy has revealed small but measurable delays in photoionization, characterized by the ejection of an electron on absorption of a single photon. Ionization-delay measurements in atomic targets provide a wealth of information about the timing of the photoelectric effect, resonances, electron correlations and transport.

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    $\begingroup$ I keep finding threads with points of contact to my question physics.stackexchange.com/questions/612110/… Perhaps we could find one user willing to put all of them in harmony and less sparse form via Q/self A contribute. $\endgroup$
    – Alchimista
    Apr 30, 2021 at 11:48
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    $\begingroup$ 'photon is electromagnetic wave' This is not correct. The relation between a photon and an electromagnetic wave is similar to that between an electron and wave function. $\endgroup$
    – my2cts
    Apr 30, 2021 at 19:06
  • $\begingroup$ On a related note, just defining the position of a travelling photon is problematic; see physics.stackexchange.com/a/463454/123208 $\endgroup$
    – PM 2Ring
    Apr 30, 2021 at 19:09
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    $\begingroup$ The question asks about the size and shape of a photon and then says that experimental arguments are preferred. That seems to be asking "what is a photon like when it's not being measured, but please answer using only measurements." Are you really asking what a photon is like when it's not being measured? There is only one way to answer a question like that, which is to choose a well-tested theory (which in this case could be QED) and describe that theory's math. We can (and often do) use physical-ish words to describe the math, but it's still just describing the math. $\endgroup$ May 2, 2021 at 13:37
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    $\begingroup$ I completely agree that physics revolves around experiment, and we should be careful not to burden a theory with extra ideas that don't improve its agreement with experiment. That's exactly where my comment is coming from. Measurements don't tell us what properties a photon has; to do that, we would need to describe the theory's math. Measurements only tell us what happens when we measure things, and that depends on what we choose to measure (quantum physics' most important lesson). Either question is a good question. I'm just trying to help make sure the intent is clear. $\endgroup$ May 2, 2021 at 17:59

4 Answers 4


One must understand what theories are in physics. They are mathematical models , i.e. depend on solutions of differential equations usually, that are used to map existing data and predict new data. BUT the theories region of validity is different for different models.

Maxwell equations united electricity and magnetism and gave a road map for unification for theories, but it could not explain the photoelectric effect, the spectra of atoms or black body radiation.. These forced quantum mechanics to be proposed and accepted.

The Bohr model, which is the basis of your link, was the beginning of quantum mechanics, a phenomenological model, superseded by the solutions of Schrodinger's equation that explained the same data in a mathematically more rigorous manner.

Schrodinger's equation had to be modified because of special relativity, and quantum field theory gave a more rigorous prediction of the fine structures seen in the data.

At present the main stream model of physics accepts that the under lying nature is quantum mechanical, from which all other theories/models emerge. This is called the standard model of particle physics. This has axiomatically the table of elementary particles that is used for fitting all existing data and predicting future measurements in particle physics. You will see that the photon is a point particle, it has no extension in space only spin +/- 1, mass zero and energy correlated with the frequency of the classical Maxwell wave a large number of such photons make, E=hnu,h the Heisenberg constant.

The fact that this is the mainstream theory, does not mean that other quantum mechanical models , for particular studies, cannot be successful in describing particular data, as the quantum field theory used for quantum optics. Or superconductivity, or...

The paper you quote though is from a conference "Quantum Theory: Reconsideration of Foundations, Vaxjo, Sweden, June 6-11, 2005" , which by its title is attempting to find theories beyond the mainstream one. That is why it has few citations also and has no peer review link. Such a theory, even if the mathematics is correct, can only be for a special region of the variables, because it cannot be incorporated in the standard model, which needs the particles in the table to be point particles.

So no, it is not the proper answer for a photon of mainstream, and I suspect it is not a proper answer for quantum optics, which in material have "photons" , because it talks about the Bohr model, not photons in material.

P.S. When I am in doubt I have found the MIT open courses useful., there is one for atomic and optical physics, from what I see. There are book recomendations

At the point where physics research is now, in mainstream physics , the photon is a point particle. To see how the interference pattern arise from point particles, see this experiment, it is all about quantum mechanical probabilities.

  • $\begingroup$ "The photon is point particle" would mean infinite energy density - so why it doesn't create a black hole? I think you are referring to Feynman diagrams - perturbative approximation, but the question is what is objectively: without approximation, accordingly to experiments like mentioned above. Real optical processes are not instant, but take e.g. attoseconds for photoemission ( science.sciencemag.org/content/328/5986/1658 ) - what happens during this time, how (nonperturbatively) EM field evolves? Similar photon sizes are here: arxiv.org/ftp/arxiv/papers/1604/1604.03869 $\endgroup$
    – Jarek Duda
    Apr 30, 2021 at 5:46
  • $\begingroup$ Mainstream accepts that because the QFT of the SM works to great precision , the axiomatic acceptance of all point particles in the table is validated., In QM there is no density for the point particles , of anything, because it is all Quantum Mechanical Probabilities. When measuring a point particle there is a probability of being at a specific (x,y,z,t). There are no singularities in QM. . How classical fields emerge see here motls.blogspot.com/2011/11/… $\endgroup$
    – anna v
    Apr 30, 2021 at 6:22
  • $\begingroup$ your last link does not work. The delay should be within the width of the probability distributions for the interaction. All energy levels have a width after all $\endgroup$
    – anna v
    Apr 30, 2021 at 6:30
  • $\begingroup$ Sure QFT, SM work to great precision, but referring to perfect point particles I understand you are referring to approximation as perturbative series in momentum space - asking for size, shouldn't you use nonperturbative in position space instead? I am not asking about density of particles, but about density of energy of EM field of single photon. During such observed 22as delay in photoemission, EM wave will travel ~6nm. Here is better link for this article with independent similar claim: arxiv.org/abs/1604.03869 $\endgroup$
    – Jarek Duda
    Apr 30, 2021 at 6:38
  • $\begingroup$ The last is within quantum optics, as far as I gather and has not been peer reviewed yet. The photons of quantum optics can have a size, it is a different QFT. I am talking of the standard model QFTfrom which, mainstream physics posits that the elementary particles are point particles, that the photon wave function is from a quantized maxwell equation cds.cern.ch/record/944002?ln=en and the creation and annihilation operators on the photon field create point particles according to the theory. To describe a single particle, whether photon or electron or... one needs the wavepacket form $\endgroup$
    – anna v
    Apr 30, 2021 at 7:55

So how is this energy distributed in space - what is the shape and size of single photon?

Energy is never properly localized because it depends on a difference of potentials, for exemple the gravity force of the earth depends on the mass of the earth and the mass of the object it pulls. Energy is bound to "systems" of objects interacting with each others through "fields".

Most importantly, he claims that such single optical photon has shape similar to elongated ellipsoid of length being wavelength λ, and diameter λ/π (?)

What makes the shape of a photon ? What does it mean to say that it has a diameter of λ/π and yet can be "attenuated" by going "through slits narrower than λ/π" ? So if it can squeeze through such slit does it mean the size is elastic ? How do you define the meaning of "size" in that instance ?

Light has properties of wavelength, amplitude and polarization, a photon is a concept that boils down to "it's the light that a single electron emits or absorbs when changing atomic orbitals" meaning a photon is only defined in regard to its interaction with an electron bound to an atom. So each time you try to talk about photons in mid flight, you're in trouble because the only way to detect a given photon is by absorbing it with an atom, and then this photon is no more

  • $\begingroup$ Isn't photon EM field configuration? If so, there should be rho ~ |E|^2 + |B|^2 energy density ( en.wikipedia.org/wiki/… ) - some distribution of its energy. What can we tell about shape/size of this energy distribution? $\endgroup$
    – Jarek Duda
    Apr 30, 2021 at 8:00
  • $\begingroup$ You can only detect energy when it does work. And the location of the work done is hard to pinpoint. We can however give a geometry to the electric field for light and that geometry is given by the maxwell equations as usual. And for light it's just an EM field shaped like a sine wave (or a combination of sines) $\endgroup$ Apr 30, 2021 at 8:08
  • $\begingroup$ in other words, electric fields do work on charged particles, but without knowing the location of the charged particle you can't say the energy is here or there. And even if you knew where the charged particle is, where would be that energy "located"? In the field, or on the particle, or between the particles that originated the light and the particle that light acts upon ??? Energy involves spacial positions (ex gravity potential energy) but there is no specific location for the enrgy itself $\endgroup$ Apr 30, 2021 at 8:19
  • $\begingroup$ Sure, we cannot directly measure the details, but it also concerns e.g. processes in the center of stars - such difficulty should not stop us: from extrapolating knowledge to self consistent models including given situation, like the cited two articles trying to predict something about EM field configuration of photon. If you don't know, what do you think about them and experimental arguments they use? $\endgroup$
    – Jarek Duda
    Apr 30, 2021 at 9:21
  • $\begingroup$ The paper is too complicated for me, I just tried to comment on your question. I do have my own issues with the photon concept that I explained above. What I get from the paper is that they try to assign a size to a photon, then get into trouble explaining double slit interference so they bring in "evanescent waves" and then claim that the photon field isnt sharply cut (and therefore the size isnt either). For me it's just simpler to consider that light isnt quantized before it's absorbed, it's just a classic EM wave, so I dont need to worry about photons size. $\endgroup$ Apr 30, 2021 at 10:47

There are really two meanings to "photon". One is the detection of a quantum of electromagnetic radiation, which always occurs at a point. The other is the probability distribution: the likelihood of detecting the quantum at each point. The probability distribution is spread out and can have pretty much any shape.

  • $\begingroup$ There is also a third one: single photon being EM field or its quantum ensemble - its dimensions, shape I am asking about. $\endgroup$
    – Jarek Duda
    Apr 30, 2021 at 18:10
  • $\begingroup$ I think, after you've digested all the answers you can get, you will conclude there are not any particular dimensions or shape. $\endgroup$
    – S. McGrew
    Apr 30, 2021 at 18:56

It makes sense to talk about the “physical nature of an object” only in the context of a defined observation, for two primary reasons. One, the “physical nature” of every object is dependent on the context of observation, or maybe better called the “context of physical manifestation,” where “physical manifestation” means an object has the capacity to interact with other ontic/physical objects, ie, it is not just an epistemic abstraction. Thus to ask, “What is the physical nature of an object without measurement?” is not an answerable question without a defined context - even analytically, there must be a defined context. Two, the condition “without measurement” is physically unrealizable anyway, as every object is always interacting with its context, or environment, and so is undergoing “measurement.” (See environmental decoherence theory.) Perfect isolation / non-measurement of an object cannot be physically realized. There is no meaning to “the physical nature of an object” without defining a context - manifest physical reality literally is context dependent. One may imagine, at least an abstract, comprehensive operator, C, that includes all elements of a defined context. This operator defines the associated eigenvalue equation possible solutions that may become physically manifest after measurement, ie, the eigenspace of C, and the observed object wave function, ψ, modulus squared, from ψ expansion in this eigenspace, defines the probability of each solution becoming physically manifest. Without a defined C, there is no eigenspace, no defined possible solution set, no ψ expansion, and no defined probabilities of outcomes - under these conditions, “the physical nature of an object” has no meaning.


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