I have a very basic understanding of quantum physics, but as I understand it the uncertainty principle says that the more precisely you know a particle momentum and the less you know the particle's position.

But I wonder with the photon: given that the velocity is a constant $c$ so there is no uncertainty at all in the speed (and so in the momentum), does that mean for a photon that the uncertainty of the position is "infinite"?

  • 1
    $\begingroup$ You're confusing velocity and speed. While there is no uncertainty in speed, there is uncertainty in velocity (which is a vector, with a direction). $\endgroup$ Aug 9, 2018 at 9:38
  • $\begingroup$ Please refer to "The first ever photograph of light as both a particle and wave" for your interest. $\endgroup$ Aug 9, 2018 at 9:46
  • $\begingroup$ @JohnReenie So you are saying the HUP only involves uncertainty in the direction of the velocity? $\endgroup$ Aug 9, 2018 at 10:22
  • $\begingroup$ @AaronStevens I don't know about that. But perfect knowledge of speed does not mean perfect knowledge of momentum, unlike what is claimed in this question. Therefore everything that follows is not necessarily true, thanks to logic. $\endgroup$ Aug 9, 2018 at 11:13
  • 1
    $\begingroup$ @AaronStevens Yes, this is what I meant. $\endgroup$ Aug 9, 2018 at 11:17

2 Answers 2


As explained in If photons have no mass, how can they have momentum?, it is impossible to assign photons a classical momentum $p=mv$, because their mass is zero.

Instead, the photon momentum is determined by its wavelength $\lambda$ via $$ p = \frac h\lambda, $$ where $h$ is Planck's constant. This means that the only way to have a completely determined momentum (i.e. $\Delta p=0$) is to have a completely determined wavelength, and that can only be done if the wavepacket has infinite extent (because, if it doesn't, what's the wavelength at the edge of the wavepacket). Thus, the photon momentum is fully compatible with the Heisenberg uncertainty principle.


There is an uncertainty in momentum! Because, for a photon,

$$p = \frac{h}{\lambda} = \frac{h \nu}{c}$$

where $p$ is the magnitude of the momentum, $\lambda$ is the wavelength of the photon and equivalently $\nu$ is the frequency.

So even though photons travel at $c$, their momentum can be uncertain if their frequency is uncertain.

And this ties in exactly with the uncertainty principle: when looking at a wave, be it light, sound, or whatever, how accurately you can know the wavelength depends on how distributed in space (equivalently, in time) the wave is: a brief pulse of sound, say, doesn't have a well-defined frequency at all, while a sound that goes on for a very long time does (or may do). So a photon has a well-defined frequency, and hence a well-defined momentum, only if it is very spread out in space, while a photon which is localised has an ill-defined frequency and hence an ill-defined momentum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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