I've read several threads on what exactly a photon is and how it can't be modeled with a wavefunction the same way electrons can. In particular, this answer to my question about whether photons are actually particles says explicitly that what makes photons different is that they're massless, and I read a few other posts mentioning that photons don't have a standard position operator because the position operator comes from the non-relativistic limit, which massless particles don't have.

What I don't get is why? Why does being massless mean a particle can't be localized? This answer to a related question says it's because photons don't preserve causality because they can go back in time, which doesn't make any sense to me -- isn't SR supposed to preserve causality? And even if it didn't, what does that have to do with position?

  • $\begingroup$ We have no trouble, beyond the Uncertainty Principle, localizing photons in experiments. Just like other particles. $\endgroup$
    – John Doty
    Commented May 2, 2023 at 21:36
  • $\begingroup$ @johndoty, so then why don't they have a normal position operator? $\endgroup$ Commented May 2, 2023 at 22:49
  • $\begingroup$ It is localized by the position of the detector that has absorbed it, i.e., it has disappeared by the act of localization. $\endgroup$
    – hyportnex
    Commented May 3, 2023 at 1:16
  • $\begingroup$ One of the reasons the photon is a particle is from the standard model ..... but the standard model just says we have fields and excitations of fields .... and the excitations are particles ... all excitations are assumed to be a point .... for the photon its probably an erroneous assumption. $\endgroup$ Commented May 3, 2023 at 2:27
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    $\begingroup$ @hyportnex A double Compton telescope is an example of a detector that localizes photons without causing them to disappear. Indeed, in such a machine, the photon from the first scattering site propagates like a particle to the second site. $\endgroup$
    – John Doty
    Commented May 3, 2023 at 13:38

1 Answer 1


I will only make an answer about the position operator.

The most well-defined things that we may reliably say that we can measure, are Hamiltonian eigenvalues. It should be good on hindsight---you can keep measuring in succession, over a region of time, and keep getting the same value. This is the most certain we may become of anything in quantum theory.

This means that, really, momentum eigenstates are more fundamentally understandable than position eigenstates, because at least the free particle Hamiltonian, or the asymptotically free states of interacting Hamiltonians, corresponds to momentum eigenstates. Position representations are really just for us to make sure that we write down spacetime localised field interactions. (Remember that, as long as we want the field commutators to vanish for ``macroscopically space-like", meaning that fields do not interact with each other if they are sufficiently far apart, then they must also vanish microscopically. The proof of this is in "PCT, Spin and Statistics, and All That".)

The offending reason why position eigenstates are non-relativistic, is that $$ \hat\phi_{\text{NR}}(\vec x,t)=\int\frac{\mathrm d^3p}{(2\pi)^3}\hat{\tilde\phi}(\vec p,t)e^{i\vec p\cdot\vec x} \qquad \text{but} \qquad \hat\phi_{\text{SR}}(\vec x,t)=\int\frac{\mathrm d^3p}{(2\pi)^32E_p}\hat{\tilde\phi}(\vec p,t)e^{i\vec p\cdot\vec x} $$ notice that it is just a $2E_p$ in the denominator that is all the difference. The non-relativistic field has a sensible interpretation as a probability amplitude in the position representation, but it is manifestly non-relativistic because $\hat{\tilde\phi}(\vec p,t)e^{i\vec p\cdot\vec x}$ is manifestly relativistically invariant while $\frac{\mathrm d^3p}{(2\pi)^3}$ is missing the $2E_p$ in the denominator away from being relativistically invariant. This is contrasted with the relativistic version, which is manifestly relativistically invariant but has no nice interpretation as a probability amplitude in the position representation.

This is also why you will not find QFT authors discussing position representation of wavefunctions. They know this is a pain point. It is not just for massless particles; it affects massive particles too, but for the case of massless particles, it is particularly bad, because then we cannot pretend that there would be a sensible $2E_p\to2m_0c^2$ region, which can then be extracted out of the integral as a constant, and thus approximately considered as a position representation probability amplitude.

  • $\begingroup$ Why does having a factor of E in the denominator mean we can't treat it as a probability amplitude? $\endgroup$ Commented May 3, 2023 at 8:13
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    $\begingroup$ I am talking about the trivial issue that it causes us to not be able to get the Dirac delta distribution $\delta^3(\vec x-\vec y)$, amongst other things. physicsforums.com/threads/position-in-qft.974556 but you might also want to look up Newton-Wigner, because those are more thorough studies of the matter. $\endgroup$ Commented May 3, 2023 at 10:07
  • $\begingroup$ This is mathematics, not physics. In real life, we have no difficulty measuring the probability as a function of position. And, with less sophisticated models, we can predict it, too. $\endgroup$
    – John Doty
    Commented May 3, 2023 at 13:45

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