Is the Heisenberg Uncertainty Principle Limited to Momentum?

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Sorry for this first year level question but something is really not clear in my head concerning the Heisenberg's uncertainty principle. First remark: my question here is neither a critic of this principle nor a tentative to collapse it. No; I consider the propagation of the light in the air or in vacuum. I stay at the origin of an inertial frame. I know via experiments two things: 1) the speed of light with a hight precision (c); 2) the trajectory of the light: it is automatically going straightforward as long as the beam does not encounter a mirror, a prism, a big concentration of matter, ... So: I can predict the position and the speed simultaneously ! What is wrong in this manner to present the reality ? Thanks for the help because I get some panic with this. Must I think that the trajectory only is an average one? I know: it must be unpleasant to always repeat the same things ...
 
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Do you know diffraction? If yes how can you predict with certainty the position of a photon within a beam of light?

Seratend.
 
The uncertainty principle deals with momentum, not velocity. The momentum of a photon is related to its wavelength by p = h / \lambda. So, even though the speed of a light beam is exactly known (in vacuum), its momentum can still be uncertain. This corresponds to an uncertainty in the wavelength. No light source is completely monochromatic. Even a laser has a small but finite spread in wavelengths.
 
jtbell said:
The uncertainty principle deals with momentum, not velocity.
Of course ! Oh I am so sorry: I think I need holydays... Thanks for the rapid answer; to seratend too.
 
jtbell said:
The uncertainty principle deals with momentum, not velocity. The momentum of a photon is related to its wavelength by p = h / \lambda. So, even though the speed of a light beam is exactly known (in vacuum), its momentum can still be uncertain. This corresponds to an uncertainty in the wavelength. No light source is completely monochromatic. Even a laser has a small but finite spread in wavelengths.

Just to add to this excellent answer...

Momentum commutes with position within the context of the HUP. Momentum, of course, having 2 components, one of which is the velocity which can be considered constant and known. The other component therefore exhibits the expected uncertainty as described above.

Funny that no matter how you approach it, the HUP alway comes out intact.
 
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