How does the uncertainity principle apply in this situation?

Question

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:

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):

$$x=ct$$

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?

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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?

Answer 1

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.

Answer 2

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.