# Why is a photon's position not considered part of its quantum state?

I'm confused why the quantum state of a photon is typically regarded as its momentum (which presumably is described by 4 numbers---wavelength, direction, and angular momentum) and not by its position (which seems to be described by 3 numbers).

I understand that position and momentum are complementary measurements, so as one is known more accurately, the other is known less accurately.

So perhaps this question should read, "what is the full quantum wave equation for a photon" ? Does this question even make sense? Or does it only make sense to describe photons as a transfer of energy from one atom to another atom?

• Jan 25, 2020 at 4:51

If a photon has a well-defined momentum and energy, then it has an ill-defined position and time. A wave $$e^{ik\cdot x}$$ with one particular four-momentum, $$\hbar k$$, is a plane wave extending throughout all of spacetime.

You can consider a photon that has a well-defined position at a well-defined time, but then it doesn’t have well-defined momentum or energy.

You can’t have it both ways. This is the essence of the uncertainty principle, and one way to understood it is using basic Fourier analysis of wave packets. The spread in space, and the spread in wavelength, are inversely related.

• This is nice, and seems like the right level for the OP. It might be improved by adding a sentence or two describing the ideas in physics.stackexchange.com/questions/492711/…
– user4552
Jan 25, 2020 at 14:21
• I understand that Δx · Δp ~ ℏ, but I think that there's still a lot of specificity that can be had by specifying Δx and Δp. I can have it both ways, because ℏ is really, really small. I just can't have it precisely both ways. So should x and p be fixable within a certain measure of uncertainity?
– vy32
Jan 26, 2020 at 2:33
• Yes. Perhaps there is a formalism of quantum field theory that does it that way. But nobody uses it. Jan 26, 2020 at 2:54