A photon is an excitation or a particle created in the electromagnetic field whereas an electron is an excitation or a particle created in the "electron" field, according to second-quantization.

However, it is often said in the literature that the wave function of a photon doesn't exist whereas it exists for an electron.

Why is it so?

  • $\begingroup$ There are some not-so-innocent ways in which you can defend writing down a wave function for a photon as well but the general underlying reason for the issues is that photon is no ordinary particle. It quite literally travels at the speed of light. $\endgroup$
    – user87745
    Feb 24, 2019 at 1:01
  • $\begingroup$ @DanYand have a look at the link in my answer $\endgroup$
    – anna v
    Feb 27, 2019 at 4:51

2 Answers 2


Saying that a photon doesn't have a wavefunction can be misleading. A more accurate way to say it is that a photon doesn't have a strict position observable. A photon can't be strictly localized in any finite region of space. It can be approximately localized, so that it might as well be restricted to a finite region for all practical purposes; but the language "doesn't have a wavefunction" is referring to the non-existence of a strict position observable.

An electron doesn't have a strict position observable, either, except in strictly non-relativistic models.

In relativistic quantum field theory, nothing has a strict position observable. This is a consequence of a general theorem called the Reeh-Schlieder theorem. The proof of this theorem is not trivial, but it's nicely explained in [1].

Relativistic quantum field theory doesn't have strict single-particle position observables, but it does have other kinds of strictly localized observables, such as observables corresponding to the magnitude and direction of the electric and magnetic fields inside an arbitrarily small region of space. However, those observables don't preserve the number of particles. Strictly localized observables necessarily turn single-particle states into states with an indefinite number of particles. (Actually, even ignoring the question of localization, "particle" is not easy to define in relativistic quantum field theory, but I won't go into that here.)

For example, relativistic quantum electrodynamics (QED) has observables corresponding to the amplitudes of the electric and magnetic fields. These field operators can be localized. The particle creation/annihilation operators can be expressed in terms of the field operators, and vice versa, but the relationship is non-local.

Technically, the Reeh-Schlieder theorem says that a relativistic quantum field theory can't have any strictly-localized operator that annihilates the vacuum state. Therefore, it can't have any strictly-localized operator that counts the number of particles. (The vacuum state has zero particles, so a strictly-localized particle-counting operator would annihilate the vacuum state, which is impossible according to the Reeh-Schlieder theorem.)

Strictly non-relativistic models are exempt from this theorem. To explain what "strictly non-relativistic" means, consider the relativistic relationship between energy $E$ and momentum $p$, namely $E=\sqrt{(mc^2)^2+(pc)^2}$, where $m$ is the single-particle mass. If $p\ll mc$, then we can use the approximation $E\approx mc^2+p^2/2m$. A non-relativistic model is one that treats this approximate relationship as though it were exact. The most familiar single-particle Schrödinger equation is a model of this type. Such a model does have a strict position operator, and individual particles can be strictly localized in a finite region of space in such a model.

Since photons are massles ($m=0$), we can't use a non-relativistic model for photons. We can use a hybrid model, such as non-relativistic QED (called NRQED), which includes photons but treats electrons non-relativistically. But even in that hybrid model, photons still can't be strictly localized in any finite region of space. Loosely speaking, the photons are still relativistic even though the electrons aren't. So in NRQED, we can (and do) have a single-electron position observable, but we still don't have a single-photon position observable.

"Wavefunction" is a more general concept that still applies even when strict position observables don't exist. The kind of "wavefunction" used in relativistic quantum field theory is very different than the single-particle wavefunction $\psi(x,y,z)$ familiar from strictly non-relativistic quantum mechanics. In the relativistic case, the wavefunction isn't a function of $x,y,z$. Instead, it's a function of more abstract variables, and lots of them (nominally infinitely many), and it describes the state of the whole system, which generally doesn't even have a well-defined number of particles at all. People don't use this kind of wavefunction very often, because it's very difficult, but once in a while it's used. For example, Feynman used this kind of "wavefunction" in [2] to study a relativistic quantum field theory called Yang-Mills theory, which is a simplified version of quantum chromodynamics that has gluons but not quarks.

In this generalized sense, a single photon can have a wavefunction.

In the non-relativistic case, the $x,y,z$ in $\psi(x,y,z)$ correspond to the components of the particle's position observables. When physicists say that a photon doesn't have a wavefunction, they mean that it doesn't have a wavefunction that is a function of the eigenvalues of position observables, and that's because it doesn't have any strict position observables.

Also see these very similar questions:

Can we define a wave function of photon like a wave function of an electron?

Wave function of a photon?

EM wave function & photon wavefunction


[1] Witten, "Notes on Some Entanglement Properties of Quantum Field Theory", http://arxiv.org/abs/1803.04993

[2] Feynman (1981), "The qualitative behavior of Yang-Mills theory in 2 + 1 dimensions", Nuclear Physics B 188: 479-512, https://www.sciencedirect.com/science/article/pii/0550321381900055

  • $\begingroup$ what I don't understand is that the same interference pattern is observed with photons as is observed with electrons in the double slit experiment. Except that electrons get localised after observation while photons get absorbed at the position they're detected. However, observing the interference pattern implies that photons also undergo a probabilistic wavefunction interference in the position space in the double slit experiment. Doesn't this mean that a position basis wavefunction of photons exists in some approximate sense? $\endgroup$
    – Ryder Rude
    Jul 18, 2022 at 14:28
  • $\begingroup$ @RyderRude That's correct, in some appropriate sense. I prefer to avoid giving simple yes/no answers to questions about photon wavefunctions, though, because the word "wavefunction" is overloaded, and different people might use it with different unacknowledged connotations. $\endgroup$ Jul 19, 2022 at 23:15
  • $\begingroup$ I am curious: if the wave function cannot be localized strictly, can we still give a non-perfectly localizable wave function? Because this phenomenon seems very similar to the one where that in the "particle in a box" in non-relativistic quantum mechanics, the momentum observable cannot be strictly localized, but we can still define a suitably-constrained wave function for it. Can something similar be done here that is completely mathematically analogous, or not? $\endgroup$ Feb 25 at 2:35
  • $\begingroup$ @The_Sympathizer We could define an observable that acts approximately like a position observable in the context of single-photon states, and we could use that to define a single-photon "wavefunction." These definitions aren't unique -- there is no single most-natural generalization of the single-electron-wavefunction concept to photons, because the former relies on a nonrelativistic approximation that doesn't apply for photons, but we can still invent useful definitions even though they're non-unique. $\endgroup$ Feb 28 at 3:04

Here is the wavefunction of the photon, which is a solution of a quantized maxwell's equation:


In quantum field theory it is necessary to have a plane wave function solution for the fields on which creation and annihilation operators operate.

This blog post describes how the classical fields emerge from the quantum ones of QFT.

  • $\begingroup$ This is oversimplified and basically misleading as written. QFT does not allow us to have wavefunctions like this for a relativistic particle in a state of definite particle number. Although you can cook up something like this and call it the wavefunction of a photon, it doesn't have the standard interpretation in terms of the Born rule. A long review article on this is Iwo Bialynicki-Birula, "Photon wave function," 2005, arxiv.org/abs/quant-ph/0508202 $\endgroup$
    – user4552
    Feb 28, 2019 at 22:52
  • $\begingroup$ @BenCrowell from the abstract of your link "can not have all the properties of the Schroedinger wave functions of nonrelativistic wave mechanics". Please note that QFT by construction is relativistic and needs wavefunctions to define the fields.,and that the question is about QFT. $\endgroup$
    – anna v
    Mar 1, 2019 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.