A common (but, as I think, incomplete) description of the uncertainity principle is the following:

You cannot determine a particle's momentum and position at high accuracy at the same time

It could also be other properties, but those two are the most commonly used to introduce the uncertainity principle. As fas as I understand, this is due to the measuring devices interacting with the particle, i.e. when you measure the momentum, you change the position and vice-versa.

Now consider the following situation:

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Some Source (e.g. a laser) emits a photon at some time $t_0$. The photon travels with velocity $v=c$ (Since every photon travels with the speed of light) and hits a wall at time $t_1$ (Let's assume the wall is made in such a way that it lights up when hit by a photon) Since we know that the distance light source - wall is equal to $d=\frac{t_1 - t_0}{c}$, we can calculate the photon's position at any point in time (Let's assume for simplicity that the photon is moving along one axis of our coordinate system):


where $t$ is the time that has elapsed since the photon was emitted.

We now know the particle's velocity ($v=c$) and position ($x=ct$), both to (theoretically) infinite accuracy. But this contradicts the uncertainity principle. How is this possible?

Here are some thoughts of mine:

  1. Uncertainity principle does not apply to photons because they are always travelling with $v=c$. For any other particle, such as an electron, there is no definite speed (i.e. you have to measure it). But the uncertainity principle does apply to photons, as far as I know.
  2. We don't measure the position and momentum of the photon, but calculate it. Maybe this is some sort of trick to "escape" the uncertainity principle?

Here's an addition: Suppose we had a light source that only emits one specific wavelength. As stated in the existing answer, the momentum is dependant on wavelength, so momentum would be the same for every emitted photon. We then would only have to worry about position and could measure it with high accuracy. How does that not violate the uncertainity principle?

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    $\begingroup$ This is not a complete answer: Uncertainty deals with components of momentum, not the absolute value. So the direction of the photon may have some uncertainty. $\endgroup$ Commented Oct 19, 2020 at 12:52
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    $\begingroup$ The uncertainty relation would also apply for the canonic conjugate observables energy and time, which means knowing exactly the time at which the photon hits the screen results in a high uncertainty about the energy of the photon $\endgroup$
    – Milarepa
    Commented Oct 19, 2020 at 13:24
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    $\begingroup$ You might also notice that we can't determine the time $\tau_0$ of emission to arbitrary accuracy either $\endgroup$ Commented Oct 19, 2020 at 13:40
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    $\begingroup$ Knowing the velocity is not the same as knowing the momentum of a photon. $\endgroup$ Commented Oct 19, 2020 at 15:37
  • $\begingroup$ @BrainStrokePatient You're right, for some reason I always think that UP refers to velocity. It now makes sense that the momentum and position cannot be both known accurately, because, as Milarepa mentioned, the more accurate we know the emission time, the less accurate its energy, which again determines momentum (for photons, at least). $\endgroup$
    – jng224
    Commented Oct 19, 2020 at 15:54

2 Answers 2


The uncertainty principle deals with position and momentum, not velocity. The momentum of a photon is not given by $p=mv$ (which vanishes) but rather by $$p=\frac{h}{\lambda}$$ where $p$ is momentum, $h$ is Plank's constant, and $\lambda$ is it's wavelength. One can derive an expression relating uncertainty in position and wavelength (as I do here), getting $$(\Delta x)(\Delta \lambda)\geq\frac{\lambda}{4\pi}$$ where little $\lambda$ is the mean value of possible wavelength values. As you can see, you can't get away from Heisenberg; now, if you try to exactly determine position, you'll have no idea what the wavelength (and therefore the momentum is), while if you try to determine wavelength, you'll have no idea what the position is.

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    $\begingroup$ As described above, a photon is associated with a wave. All we can say about the position of a photon is that its energy can be absorbed somewhere within the duration of that wave. $\endgroup$
    – R.W. Bird
    Commented Oct 28, 2020 at 18:46
  • $\begingroup$ That makes sense. But I wonder how it would be if we had a light source that only emits one specific wavelength? That way, we wouldn't have to worry about wavelength (and thus momentum) and could measure position with high accuracy. Wouldn't this "overcome" the uncertainity principle? Or would measuring the positions alter the momentum? (Although that would mean changing the wavelength; I don't know if that is possible) $\endgroup$
    – jng224
    Commented Nov 3, 2020 at 21:32

This is only an addition to the other excellent answer.

One of my misunderstandings when asking this question was that I thought that the photon does have a specified position at any point in time. However, since the uncertainty principle is not only a measurement problem, the photon simply does not have a well-defined position until we measure it. I will quote from Manisheart's answer here:

The particle has all (possible) velocities at once;it is in a wavefunction, a superposition of all of these states. This can actually be verified by stuff like the double-slit experiment with one photon--we cannot explain single-photon-fringes unless we accept the fact that the photon is in "both slits at once".

So, it's not a knowledge limit. The particle really has no definite position/whatever.


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