I've been thinking about this for quite some time, and from looking online I haven't found a satisfying answer.

Lots of photons, such as visible-light photons have very small wavelength (which from my understanding of basic physics is the distance between two crests/troughs), but I also know that some EM waves have wavelengths a few metres or even kilometres long e.g radio waves.

What keeps me up at night is the question "How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?"

Does this mean that one individual photon is several kilometres long? If so, wouldn't it be subject to so many variations between the beginning of the wave and its end?

I realise that matter is also wave-like, where it's uncertainty in position is given by its De Broglie wavelength. Does this apply to the photon?

In other words, is the wavelength of a photon simply the uncertainty in its position?

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    $\begingroup$ I deleted some comments; please keep in mind that comments are not to be used for answering the question. $\endgroup$
    – David Z
    Jul 8, 2016 at 21:43
  • $\begingroup$ A photon of a kilometre wavelength you hardly will produce. For radio waves - which are a modulated EM radiation and contain zillions of photons of different wavelengths - this is possible. Photons exist in the range from infrared over visible light to X-Ray and gamma only. $\endgroup$ Jul 9, 2016 at 9:15
  • $\begingroup$ @HolgerFiedler Nonesense. See the article I discusses in another question for an example of a process generating discrete photn emmisions in band close to the one used by your FM radio. There is no hard distinction between the classical and quantum descriptions of light. There are ranges where itis generally more convenient to use one description than the other, but that is a different thing. $\endgroup$ Oct 10, 2019 at 19:37
  • $\begingroup$ See this reference arguing "There are no particles, there are only fields". arxiv.org/ftp/arxiv/papers/1204/1204.4616.pdf $\endgroup$
    – Jagerber48
    Nov 8, 2022 at 15:32

5 Answers 5


The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by $E=h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy.

This is where the confusion comes. The wavelength and frequency characterize the emergent electromagnetic wave from very many photons. How the classical wave emerges can be seen here although it needs a quantum field theory background to understand it. The photon, as a quantum mechanical entity, has a quantum mechanical wavefunction. This wavefunction complex conjugate squared gives the probability density for the specific photon to be at $(x,y,z,t)$. The frequency in the wavefunction is the frequency of the possible emergent classical wave, but for the individual photon it is only connected with probability of manifestation, as for example in the single photon double slit experiments.


single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames

you ask:

"How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?

It does not. It takes zillions of photons to build up the classical electromagnetic wave. In the photos above each individual photon gives a little dot. The build up gives the probability density distribution for photons, and lo, there is a frequency associated with the interference pattern, even though the photon manifests individually as a dot at the $(x,y)$ of the screen.

That is why we need quantum mechanics.

Edit after this question became the main duplicate of another one, where I have a long answer/comment that might be of interest to readers.

Does a single photon have a wavelength or not? [duplicate]

  • $\begingroup$ Thank you and to the others who answered and commented. I like this answer simply because at the end it directly answers my main question in an easy-to-understand way (I've only just finished first year of engineering in college). I hadn't realised that people also differentiate between photons and EM waves, one being part of QM and the other being part of Classical Physics. This clears it up a lot more so thank you. $\endgroup$
    – hopper19
    Jul 9, 2016 at 22:42
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    $\begingroup$ Anna, you say that the photon does not have a wavelength. However, if we say that coherent states have a wavelength, and if we consider coherent states a simple vector sum of particle number states, then surely the single particle states must have a wavelength? Because they must be the same type of object to admit a simple sum. (I am assuming single particle states must be plane wave if they have any spatial character) Just learning QFT so if you can point my error that would be very helpful to me. $\endgroup$
    – user183966
    May 16, 2018 at 10:17
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    $\begingroup$ @user183966 The coherence of photons comes in their wavefunction, i.e. it is probabilities that have phases seemy answer here physics.stackexchange.com/questions/403412/… . the wave is a qm probability wave, not a classical wave of energy intensity. $\endgroup$
    – anna v
    May 16, 2018 at 12:11
  • $\begingroup$ Ok, so if I understand your intention this answer is an explanation of traditional wave particle duality and not something beyond, thank you for clarifying. $\endgroup$
    – user183966
    May 16, 2018 at 19:07
  • $\begingroup$ -1 A single photon has a wavelength that is measurable in principle. You don't have to measure in the position basis. It's not true that photons are spatially localized "pieces" of the classical EM wave as you seem to be suggesting. $\endgroup$
    – benrg
    Feb 8, 2021 at 4:12

My answer is close to that of @AnnaV but there is a subtle difference. The wave function is not the result of many photons, but rather gives the expectation value of a measurement. Maxwell's equations are to photons what the Schrödinger and Dirac equation are to electrons. Their solutions predict statistical observations of photons. Electrons do not have a wavelength, only electron wave functions have. The same is true for photons.

  • $\begingroup$ This is related to a question about photon reflection between mirror spaced about one wavelength. There should be a wavefunction for the photon itself, in terms of probability at least. I do see that the two things are erroneously mixed, and even textbook about dual slit cause confusion (the electron wf is compared to light em waves, inducing to treat a photon like that em wave with the little possible intensity). $\endgroup$
    – Alchimista
    May 8, 2021 at 12:46

Concerning massless particles, don't forget that the spacetime of their lightlike worldline is empty (= zero). That means that the point of emission and of absorption are adjacent in spacetime, even if the space interval between them measures billions of lightyears. By consequence, there is no problem for the transmission of particle characteristics for massless particles.

The wave of a photon is propagating through space with velocity c, and the lenght of a wave is what we can measure in space (with a meterstick), even if the spacetime interval is zero.

This rule does not apply to photons moving at speed v < c through matter. The particle characteristics are transmitted, but the spacetime interval of the worldline of their timelike movement with speed v < c is not empty. This is one of the phenomena of quantum nonlocality, and we can only describe and calculate it, but we haven't got an explanation yet.


It had been known for a long time that light showed interference effects, just as in other waves such as sound and water waves. So in the double slit experiment with monochromatic light you get the light and dark bands on the screen, from which you can work out a wavelength for the light. It was thus assumed that light being a wave there must be a vibrating medium to transmit it, which they called the ether. The big difference between light and other known waves was that for light there was no alternating physical phenomenon such as the height of water or pressure of air which for classical waves could be directly measured. The ether theory was knocked on the head by Planck and Einstein when light became a particle, and Max Born gave the only possible feasible explanation of the interference effect, that the wave property (the square of the modulus of the complex number obtained by adding up the various possible paths) determines the probability of the photon landing at that point on the screen. Its as if nature had been fooling us into believing that light must be a classical wave, when all along the meaning of 'wavelength' and 'interference' are quite different to a classical wave. In Feynman's book 'QED' he talks about 'arrows' which are complex numbers represented in the complex plane which are rotating according to the frequency of the photon, thus describing a spiral as the photon moves along. The wavelength is the distance for which the arrow goes once around. Its a mathematical device (the complex plane does not exist as a real object) that however gives us the results of real experiments.


A photon is a measurement on a quantum field. It's a "one time deal", if you like. Each photon has an energy and a helicity (sometimes confused with "spin"), but that's not enough to produce a "wavelength", which is a property of a classical electromagnetic wave. We only recover the wave by measuring many photons, which then approximate the classical wave shape. In order to have a sensible definition of wavelength, these photons all have to have a similar energy, so that the coherence length of the resulting wave is long (enough). Strictly speaking one would not assign that wavelength to the single photon since the single photon measurement can't tell us that the wave is sufficiently coherent.

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    $\begingroup$ I'm understanding from your answer(comment) that measuring the energy of single photons in a monochromatic beam can lead to several possible energy values? $\endgroup$
    – Alexander
    Jul 8, 2016 at 23:00
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    $\begingroup$ I didn't talk about measuring the monochromacity of the beam or coherence. @Matija Milenovic asked abouth wavelength of a single photon. In your answer you said "one would not assign that wavelength to the single photon", talking about coherence. I commented that in fact you can measure the wavelength of a single photon (i.e. you can assign a wavelength to a single photon) - nothing about coherence here. In other words, I think your answer needs modification, distincting coherence (which is many photon phenomena) and single photon wavelength, which is deduced from its energy. $\endgroup$
    – Alexander
    Jul 8, 2016 at 23:20
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    $\begingroup$ @Alexander: That question is clearly answered by "No, as single photon doesn't have a wavelength.". It has an energy. A wavelength is a property of a coherent wave. It's not even a property of a classical wave with a mixed spectrum. Just because you "assign" a property to something because of a mathematical relation that is actually an expression of temporal behavior ($\omega$ is a frequency, not a wavelength) doesn't make it so for spatial behavior. $\endgroup$
    – CuriousOne
    Jul 8, 2016 at 23:26
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    $\begingroup$ @CuriousOne would you mind providing some authoritative sources to back up your statement? $\endgroup$
    – Previous
    Jul 9, 2016 at 6:26
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    $\begingroup$ As anna v mentioned in many answers the photon is part of the Standard model of particle physics and photons are real particles. A point of view like CuriousOne 's is possible but has to led to the same results as the classical point of view. If not, one of the two views or better one of the two interpretations has to be wrong. $\endgroup$ Jul 9, 2016 at 9:10

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