Does the uncertainty principle apply at all to photons?

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ChrisVer

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I will repeat myself...
Once you measure the position with infinite precision , you have absolutely no clue what the momentum of the measured photon is...
If you tell me that the photon starts with some momentum [itex]p[/itex] and I measure it at point [itex]x_{0}[/itex] then you'll probably never know its momentum again- not until you measure it. It can still be [itex]p[/itex] of course, but it can as well be [itex]p+Δp[/itex]... In fact the momentum is totally undetermined, it spans the whole p-space.
Why does this happen?
Because the position and momentum quantum operators don't commute. And also because of that, if you measure the [itex]x[/itex] component of the photon's position, you will lose information about the [itex]p_{x}[/itex] component of momentum, not necessarily [itex]p_{y,z}[/itex] since they indeed commute with the operator you measure. Of course that's nonsence, because what you measure is a position vector, so you lose momentum's vector.
As I said this problem corresponds to finding the delta function's Fourier Transform.

Also before the photon collides on your measuring machine, you have absolutely no idea what its position is... In fact the [itex]Δx=L[/itex] where L is the distance between your source and your measuring machine (where the photon scatters or is absorbed to be measured). That's the natural way of choosing the x uncertainty in that thought experiment. What about p? If p is fully determined (that means you can somehow measure it before you measure the photon's position), then the photon's propagation is space is going to be like a wave - you won't know anything about the x.
 
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DrChinese

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...With an uncertainty of infinity? Really?
Sure, as the other approaches zero. That's the experiment! In other words, the HUP is confirmed in practical situation, including observations of entangled particles.

As others say, the HUP itself is not fundamental, it is a consequence of other assumptions.
 

jtbell

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consider the case of spontaneous emission. That is, radiative transition of an electron in an atom from a higher energy level to a lower. In the textbooks they usually just note that a photon is created with definite energy hf = E2-E1.
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.
 
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?
 

ChrisVer

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knowing the momentum p, the probability to find the particle is propagating as a normal wave with momentum p...
 

DrChinese

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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.
 
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Nugatory

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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.
 
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vanhees71

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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 [itex]\hat{\vec{x}}[/itex] that have the Heisenberg-Born commutation relations with momentum, i.e., fulfill
[tex][\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}.[/tex]

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

http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html
 
Not while it's in space, but upon collision with a screen, perhaps?
 
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.
 

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