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Laser beam uncertainty

  1. Dec 3, 2008 #1
    hello forum.

    a question about light and uncertainty:

    If a laser beam has a 2 cm diameter when it passes through an aperture that is 6 cm wide, the beam is not affected by the aperture.
    Why? What happens to the uncertainty principle there?
    Isn't the probability wavefunction affected by the edges of the aperture causing a larger momentum uncertainty?

    For a laser beam all the photons belong to the same quantum state, or almost, and so the diffraction spread of a single photon represents the spread of the whole beam.


    thanks!
    fisico30
     
  2. jcsd
  3. Dec 3, 2008 #2

    ZapperZ

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    The HUP has nothing to do with the size of the laser beam. It has to do with how well you can determine the uncertainty in the location of the photon that passed through the aperture, and in this case, the characteristic length is the wavelength of that laser. At 6 cm wide, this uncertainty is HUGE when compared to the optical wavelength, meaning [itex]\Delta x[/itex] is large. Then [itex]\Delta p_x[/itex] is miniscule, and you see practically no diffraction.

    Now, if you use radio waves, that's a different matter.

    Zz.
     
  4. Dec 4, 2008 #3
    Here's a possibly awful analogy: spray water from a 3/4 inch hose thru an adjacent hole, say 6 inches in diameter, and the flow is smooth and normal....now begin the make the hole smaller and smaller...with water the spray will likely begin to be disrupted at maybe an inch or maybe 7/8 or definitely at 3/4"....but the optical wavelength has little resemblence to errant water molecules on the edge of a spray pattern....
     
    Last edited: Dec 4, 2008
  5. Dec 4, 2008 #4
    Thank you ZapperZ and Naty1

    ZapperZ: A regular laser beam can be expanded or made smaller in diameter using system of lenses. Is that possible also with only one photon?

    also, forgive me, but I am not sure I fully get your comments: the position uncertainty in one photon is proportional to the wavelength, which is very small at optical frequencies, correct? So the momentum uncertainty is very large..... which means large diffraction.....
    Don't we have to consider the smallest uncertainty, instead of the size of the aperture?
    thanks!
     
  6. Dec 4, 2008 #5

    ZapperZ

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    That doesn't make any sense. The size of the beam simply is the size of the aperture. It isn't the size of the photon, a quality that also do not have any definition.

    Where did this come from? If this is true, then the position uncertainty for any photon of a particular frequency would be FIXED to the wavelength. This is certainly not the case.

    The single-slit diffraction is the clearest manifestation of the uncertainty principle. Think about it. If the slit width represents the uncertainty in position, and the spread of the diffraction pattern represents the spread in the lateral momentum AFTER it passes through the slit, then guess what happens to the spread of the diffraction patterns and the position uncertainty becomes smaller (i.e. the slit becomes smaller)?

    Zz.
     
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