If an EM wave only gives us a probability of where a photon may be at a given moment, and the HUP tells us that we can't know the exact location of the photon. Then would it be correct to say that a photon does not travel in a straight line?

If this is true, wouldn't the photon's crooked path mean that it must travel faster than $c$ for us to measure its straight line speed at $c$?

Here's a comparison to explain my question further: If two sprinters run a 100 yard dash in 10 seconds, but one sprinter is required to run in a zigzag manner wouldn't that runner need to run faster than the straight line sprinter for both runners to run the race in 10 seconds. Does the crooked traveling photon need to exceed $c$ for us to measure its straight line speed at $c$?

  • $\begingroup$ the HUP tells us that we can't know the exact location of the photon. No, that is not true. Also, the photon does have not any definite path, crooked or straight. $\endgroup$
    – user140606
    Feb 8, 2017 at 16:59
  • $\begingroup$ When we measure c don't we assume it is traveling in a straight line? $\endgroup$
    – Lambda
    Feb 8, 2017 at 17:04
  • $\begingroup$ We assume it, in order for it to correspond with what we expect classically, but in Q.M, the laws are different. The second answer on this question says it better than I can: physics.stackexchange.com/q/186170 $\endgroup$
    – user140606
    Feb 8, 2017 at 17:11
  • $\begingroup$ Great photo of the interferogram by the way. Lovely fringe visibility: I could look at interferograms all day, especially if they are taken with my favorite colors. $\endgroup$ Feb 9, 2017 at 0:01
  • $\begingroup$ WetSavannaAnimal aka Rod Vance: Thank you. I have some shots where I put a paper cone under the image. It produces a cool 3d look. The interferometer is mounted onto steel tubing and is rock solid. I can carry it around and the fringes for the most part stay stationary. $\endgroup$
    – Lambda
    Feb 9, 2017 at 1:15

4 Answers 4


Your crooked path argument holds for a classical electromagnetic wave going through a medium. That is why the speed of light in a medium can be less than c, the speed of light in vacuum. At the photon level it is that photons effectively go a larger distance at velocity c, as they interact individually with the lattice of the material, and a lower group velocity is measured.

In vacuum there are no interactions for the photon, which is a point particle, and its velocity is c and the same is true for the light beam. The Heisenberg uncertainty does not enter.

The validation of this is the validation of the special theory of relativity which has been tested by the myriads of experiments in physics.


From the point of view of worry that the HUP might contradict the existence of the speed limit $c$, one can assuage your worries by pointing out that:

  1. $c$ does not primarily mean the speed of light, rather it is a geometric constant that defines our spacetime and experimentally light is observed to move at that speed. See more about this in my answer here and
  2. The ban on speeds amounts to a limit to the speed at which information, or a cause-effect relationship can propagate. So an inferred speed exceeding $c$ is not needfully a problem: it only violates known physics if the speed can be interpreted as a propagation speed for a cause-effect relationship. Thus, for example, phase speeds of waves can happily exceed $c$, e.g. at anomalous dispersion wavelengths for light propagation in dielectric materials. Moreover, gainsaying many texts, even group velocities can exceed $c$ in some wavelength bands and for slightly weird materials: that's not even a problem because, although the group velocity approximates the signal speed propagation, it's not exact and the step response of the medium still propagates at less than $c$.

Any speed inferred from the kind of reasoning you are making falls into the category in my point (2) above. There are two reasons for this:

  1. Lone photons without quantum measurement cannot be used to transmit information. In particle physics, the propagation of the electromagnetic field is calculated from a superposition of paths of photons - and so called "off-shell" photons (those with speeds other than $c$ - above or below)- enter this calculation. This is not a problem, because information-bearing aspects of the field are still calculated to propagate at $c$;
  2. I don't think you have a sound conception of a photon. Please don't take that as an attack as there are a great many education resources out there that give people these kinds of ideas (I was 35 before I shifted on from these ideas). The modern idea is subtler, but much simpler and much sounder logically than the 1920s Bohrian ideas that are still taught. I recommend you begin with Daniel Sank's glorious answer here. Inspired by his answer, I followed up with this one (which may also be useful to you). Indeed, for massless particles like the photon, one cannot even define a sound notion of "position" as one can for nonrelativistic electrons following the nonrelativistic Schrödinger equation.

Light does travel in a straight line, or else there wouldn't be shades!! lol

The EM Wave

The EM wave is used to describe the possibilities of finding a particle at a certain location. But we can't say that the particle is at some particular point in the wave because then we lose the wave analogy. We can only say that the particle exists somewhere in the wave, and by that, we mean the particle is simultaneously everywhere on the wave.

So if the wave continues throughout the universe, that means at the next point in time, the particle could appear from anywhere throughout the universe.

So does that mean an electron near you could end up being at the edge of the universe at the very next moment? YES!!

But basically, the consequences of particles ending up at a very far distance is extremely low that it's considered never to happen.

The Action

In the Uncertainty Principle, there is this thing called action.

The action calculates the crest and troughs in a wave using the mass, time, and distance between the starting and final position of a particle or an object.

If say the final position of a particle is 10 units away(large distance for the particle say), the action of this would correspond to the crest of the wave(skipping the math); simultaneously, the same particle would also appear somewhere on that original wave, say, its 10.1 units away from the final position, the action would correspond to the trough of the wave. Lastly, we only have to add these waves together in order to get the final wave function for the final position of the wave.

The crest and the troughs cancel each other out, and so we get a final wave function of nearly zero possibility of finding the particle at such a large distance.

On the other hand, the possibilities of a particle ending up at a close location are huge. Because instead of canceling each other out, the waves stacks themselves up into a larger wave with much more amplitude, which means a higher chance of finding the particle at that location.

That's why the particles can't get far from their original position, and therefore you shouldn't be worried about its speed exceeding the speed of light.


First of all, as Countto10 pointed out, in this other question you can find a nice discussion on the meaning of trajectories in quantum mechanics.

Heisenberg's uncertainty principle tells you that when you measure the photon's position or momentum (or any other pair of complementary variables), you will find the results to always affect each other. In other words, a photon cannot have a definite value of position and momentum at the same time.

This does not really come into play if you are trying to measure the speed of light using single photons. Let us imagine for the purpose an experiment in which a single photon is emitted at point $A$ at a known time $t$, and travels towards the point $B$. After the photon is emitted, its wavefunction will expand and evolve according to a variety of factors. For example, the smaller the error in the transverse position when it is emitted, the faster its wavefunction will disperse and so the harder it will be to detect it at $B$. This kind of things can however be taken care of relatively easily, so to make the probability of detecting the photon in $B$ high enough. When this detection event happens, we can measure the time it took and derive the speed of light accordingly.

The photon did not follow a zig zag path going from $A$ to $B$. It didn't simply because that is not how a photon usually travels, and nothing in Heisenberg's uncertainty principle says it should. Its wavefunction did evolve in the process, but it is wrong to think of this in terms of a point particle jiggling around. It means instead that the probabilities of detecting the photon in the various transverse points vary with the longitudinal position.

As a final note, it is worth noting that there should be a lot of buts and ifs in the above argument. For example, I assumed that the emitted photon can be more or less be thought of as a particle-like thing, in the sense of it being relatively well localized. A photon can however also be highly delocalized, or have a complex inner structure. This kind of things can matter: an interesting example is a recent experiment by Bareza and Hermosa in which it was shown that photons carrying an orbital angular momentum can have (in the vacuum) a group velocity smaller than $c$. This is an interesting reminder that the speed of light being $c$ only strictly holds for plane waves in the vacuum, not really for light with finite extent and complex inner structure.