Does the uncertainty principle apply at all to photons?

[DrChinese]

If I know its momentum with exact precision, given to me by the frequency of the source, and the direction of emission, will my very calculation will make me infinitely uncertain about its position? That's impossible. Will the calculation of momentum send the photon out into infinity?

You may be being a little bit dramatic by talking about the calculation thing. But the fact is, as you localize a quantum particle, its momentum does spread wider.

It still has an average value for momentum. But that average value becomes less and less likely to be observed in an experiment.

If you repeat the experiment many times, the predicted distribution will appear. If you repeated enough times, a few extreme values would result as well.

 

[Nugatory]

But that doesn't make sense. You are trying to imply that if I know a source which emits photons in a certain direction and with a certain frequency, I can never know its momentum and position with infinite certainty even if I know when it was emitted! If I know its momentum with exact precision, given to me by the frequency of the source, and the direction of emission, will my very calculation will make me infinitely uncertain about its position? That's impossible. Will the calculation of momentum send the photon out into infinity?

You're still making the same mistake... Photons are not what you think they are, and therefore your entire model of the thought experiment is misleading you.

Photons are always fuzzy and spread out through space so that you can no more specify an exact position for them than you could specify the edge of a cloud of smoke. That's the uncertainty in spatial position (and equivalently, time of emission/detection - when exactly does a cloud of smoke start to make your eyes water?). This is one of the ways that a photon does not behave like a little tiny grain of sand, despite our using the word "particle" to describe it.

The more tightly you constrain the spatial extent of the cloud, the greater the spread of momentum (and equivalently, frequency, energy and wavelength). That's an inherent property of a photon, and another of the ways that a photon does not behave the way the word "particle" suggests that it might.

The calculation itself has nothing to do with the uncertainty and the way that we can trade uncertainty in position against uncertainty in momentum. That's always there, and the calculation just tells us how that tradeoff is working for us in a particular setup.

 

[vanhees71]

Photons are indeed far from anything you associate with "particle". Already massive "particles" are different from miniature billard balls when it comes to situations where you have to use quantum theory to adequately describe them. At least massive particles always have a proper position observable, i.e., there exist self-adjoint operators \hat{\vec{x}} that have the Heisenberg-Born commutation relations with momentum, i.e., fulfill
[\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk}.

This is not the case for massless particles with a spin \geq 1. There exists a well-defined momentum operator but not a position operator. For details, see

http://arnold-neumaier.at/physfaq/topics/position.html

 

[Reallyfat]

Not while it's in space, but upon collision with a screen, perhaps?

 

[QuasiParticle]

Textbooks generally assume that the atom is at rest initially, for simplicity. However, the atom cannot be definitely at rest, nor in any other exact state of motion. The "fuzziness" in the atom's initial state leads to "fuzziness" in the emitted photon's energy.

I like this answer. Often, though, the time-energy uncertainty is invoked in this context, which might be the source of my confusion. If I'm not mistaken, spontaneous emission is not allowed in the old quantum theory. But still it is claimed that some of the broadening of the spectrum is due to the time-energy uncertainty.