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Everyone is always talking about photon's wavelength. But what about its dimensions? What is length and width of it? And does it even have a point to think about such things? Or those dimensions are non-existent in such cases?

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  • $\begingroup$ related: physics.stackexchange.com/q/103904/58382 $\endgroup$
    – glS
    Oct 27, 2016 at 9:44
  • $\begingroup$ well, the wavelength is easily considered its length. But there is no width or depth. What's the width or depth of a wave in the ocean? What's the width of sound? A photon is a wave that sometimes can behave in ways we normally attribute to particles, but a wave nonetheless. There's no need for it to have a volume of any kind $\endgroup$
    – Jim
    Oct 27, 2016 at 12:47
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    $\begingroup$ Although there have been arguments (including some of the accepted answers') against it, the description of a photon as a wave usally (i.e., "classically") seems to employ the shape, size, and velocities of the particle in terms of the probabilities of it being found at particular points on that portion of its trajectory through spacetime which the wave's outline represents, $\endgroup$
    – Edouard
    Sep 8, 2020 at 22:36
  • $\begingroup$ A photon does not have wavelength or frequency. Its wave function has. A photon has energy and momentum. $\endgroup$
    – my2cts
    Jan 13 at 9:07

7 Answers 7

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The fundamental particles we know today (of which the photon is one) are called fundamental exactly because they have no substructure, or indeed, spatial extent, we know of. They are point-like when localized.

Note that these "particles" are quantum objects, not classical particles, so you should not imagine them as points whizzing about in space - they possess delocalized states where they take no definite shape at all (for example, the "electron cloud" around atoms is such a delocalized state).


The above is a short, non-relativistic view of "particles". When going to the relativistic description that is actually needed for the full description of fundamental particles, things get considerably more murky. For one, we lose the naive position operators, and the notion of "localization" becomes a bit ill-defined because the new "position operator", the Newton-Wigner operators, do not allow to speak of localization in an observer-independent way. The generic particle state that is scattered in QFT calculations is usually a sharp momentum state, and therefore strongly delocalized, so any notion of "point-like" can't really rely on the localization of a particle state.

In this picture, the proper notion of a "point-like" particle is one whose scattering behaviour indicates no substructure or spatial extent. For extended objects consisting of subobjects, their scattering behaviour will typically change when the energies/length scales of the scattering process reach their size, because then their internals get resolved and the individual subobjects start participating in the scattering. So then our notion of size becomes that the scattering behaviour is scale-independent. For more on this notion of size in QFT, see e.g. this answer by Bosoneando.

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  • $\begingroup$ Does it mean that a photon is infinitely wide and long, because there's density of probability of finding it asymptotically tending to zero? Like in that well known example of an electron in finite potential well? $\endgroup$
    – user46147
    Jan 23, 2015 at 17:23
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    $\begingroup$ @user46147: Though the probability may behave as you say, you should not identify the extent of the wavefunction with the size of the object. Quantum mechanically, it is quite unclear what one would mean by "length" or "width" of a quantum object. $\endgroup$
    – ACuriousMind
    Jan 23, 2015 at 17:27
  • $\begingroup$ @ACuriousMind : what you say?! And what about coherence length? This is our practical order of magnitude of the linear dimension of the wave-packet. $\endgroup$
    – Sofia
    Jan 23, 2015 at 17:34
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    $\begingroup$ @Sofia: Though coherence length is an important concept, it gives the scale at which we see quantum behaviour of the object, rather than the size of the object, it seems to me. $\endgroup$
    – ACuriousMind
    Jan 23, 2015 at 17:46
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    $\begingroup$ I think it can be confusing to acknowledge the wavelength of a photon while simultaneously saying they have no "width". Mathematically the photon (or electron for that matter) as a building block in QED can be treated pointlike and without any wavelength, but real world photons are more complicated quantum states with both "width" and wavelength. I think you're implying this in your second paragraph. In other words, a real photon with a given wavelength also probably has a spatial extent by a similar definition. It is unfortunate that "photon" is used for both definitions.. $\endgroup$
    – BjornW
    Jan 23, 2015 at 17:58
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The photon can be experimentally shown to not be point-like. The Young's slits experiment involves the interference of a photon with itself (the photon behaves in some ways like a particle, in some ways like a wave and in some ways like a probability distribution. in reality these are convenient models that we apply to it - in reality it is none of these - it is a photon). The interference patterns shown in the Young's slit experiment remain even if the stream of photons is reduced to the extent that photons go through 1 at a time. The length difference in the 2 paths can be varied to determine the coherence length. I believe this turns out to be in the order of 1m. Further experimental proof that the photon must have a significant length is that the spread in its frequency is minimal. if it reduced to 0 amplitude over (e.g.) only 3 wavelengths then this would give it a significant spread in frequencies

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  • $\begingroup$ The wave function of the photon is not point like, but the photon is. $\endgroup$
    – my2cts
    Jan 11 at 21:15
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For me I prefer an experimental definition of the size of a photon. If you pass light through an aperture you start to see interference effects when the aperture approaches the wavelength of the photon, as if you're clipping the edges. Why do we need to make it more complicated than this?

If the photon is point-like then the energy density would be infinite which seems unrealistic.

It must be localized in space as photons can be detected from the other end of the universe. If photons are spreading out then their energy density would tend to zero over these distances.

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  • $\begingroup$ The wave function of the photon is not point like, but the photon is. $\endgroup$
    – my2cts
    Jan 11 at 21:16
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This is an excellent question. Following is an abbreviated version of the answer that I wrote in the physics textbook "Light and Waves: A Conceptual Exploration of Physics" (2023).

A photon is typically described as a point particle, with no size whatsoever. This makes sense when considering that there is no limit to the number of photons that can exist in a fixed volume of space, the fact that individual photons can interact with very tiny objects, and the fact that photons are absorbed in their entirety and all at once. In this point particle explanation, the photon is just a point but its position is some random location that has a probability that is given by the corresponding wave function.

On the other hand, it’s often helpful to think of photons as having larger sizes. For example, when a molecule absorbs a photon, the photon is not absorbed at a tiny spot within the molecule, but by the entire molecule at once. Likewise, when a radio antenna absorbs a radio wave photon, it gets absorbed over the entire length of the antenna. Thus, it’s reasonable to say that when individual photons interact with objects, then the size of the photon must be the same as the size of the object.

Yet larger sizes arise when considering diffraction and interference experiments. Light (like all types of waves) goes through holes that are bigger than a wavelength with low power loss and is essentially blocked by holes that are smaller than a wavelength, suggesting a photon size of about one light wavelength. In two-slit interference, single photons interfere with themselves, implying that each individual photon goes through both slits at once; this suggests a size of a millimeter or more. In an interferometer, a single photon gets split into two pieces that might be meters, or even kilometers, apart from each other, now suggesting that a single photon’s size could be reasonably considered to be multiple kilometers across.

As another answer, one can imagine measuring the electromagnetic field in the light beam, where one would see regular peaks and troughs over short distances, but different sets of waves over longer distances. The length scale over which the waves are reasonably regular is called the coherence length of the light source. Lasers, and light from point-sources such as distant stars, have large coherence lengths, often of meters or larger; however, light from large sources, such as the sun or light bulbs, have short coherence lengths that are in the micron range. Coherence lengths set the limit on the sizes of interference phenomena, such as the maximum separations of two-slit interference experiments. Thus, another useful measure for the size of a photon is the light’s coherence length.

Taking all of these arguments together, there is no good single answer for the size of a photon. Instead, there are situations where photons are best considered as point particles, and other situations where larger sizes, even up to kilometers or longer, are more meaningful.

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  • $\begingroup$ 'Thus, it’s reasonable to say that when individual photons interact with objects, then the size of the photon must be the same as the size of the object.' I don't think it is. Electrons are point particles. They do not have the size of the wavelength or of the orbital they are in. The same is true for a photon. $\endgroup$
    – my2cts
    Jan 11 at 21:19
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Fundamental particles have no dimensions, they are point like. At times it is convenient to associate some finite, non-zero value though.

For example, the classical radius of the electron helps with some electromagnetic problems. $r_e=\frac{e^2}{4\pi \epsilon_0m_ec^2}$.

One can also associate the Reduced Compton Wavelength of the electron: $\lambda_e=\frac{\hbar}{m_ec}\implies r_e=\lambda_e\alpha = a_0\alpha^2$ where $\alpha_0$ is the Bohr Radius and $\alpha$ is the Fine Structure Constant, $\alpha \approx 1/137 $.

Something similar applies to photons. $\lambda_p=\frac{hc}{E}$ where E is the energy of the photon.

In general, the uncertainty is position has a minimum of half the Compton wavelength as mentioned here.

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It’s a great question. In college, my professor was rather unclear on this topic.

Electrons are experimentally point sources. No matter how high an energy is used to smash them together no size is evident in the result.

Photons can’t collide so this experiment doesn’t help us.

It is interesting that in NSOM photons can pass through extremely tiny apertures, much, much smaller than a wavelength of light.

Experimentally photons seem to behave as objects that have a size on the order of a few wavelengths. On the other hand, photons of a wavelength of roughly $500$ - $1,000\ nm$ can be absorbed and emitted by atoms with dimensions of about $0.3\ nm$.

I guess if you think in terms of string theory you would say that the inherent size of a photon is on the order of the Planck length but it has a range of action on the order of a few wavelengths.

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    $\begingroup$ Photons can collide with electrons as in compton scattering, or create point on a detector screen. 'Experimentally photons seem to behave as objects that have a size on the order of a few wavelengths. ' This is not true. $\endgroup$
    – my2cts
    Jan 11 at 21:17
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IF d is set as the distance light travels in one quantum of time, t; and c is set at exactly 3 d/t, then light will encircle its “diameter” exactly 3 times in the one time-unit (not pi times). One of those 3 revolutions can be thought to represent the x direction, with the other two revolutions corresponding to y and z… A “Planck’s constant” = 3^.5 then becomes applicable. This indicates a photon “sphere” with radius= 3^.5 (diameter=3) and volume = 4/3 * 3 * r^3 = 4r^3 = 4*3^(3/2)

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