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Understanding the uncertainty principle

  1. Aug 15, 2010 #1
    I've been trying to understand the uncertainty principle and so far the Wikipedia definition has been very good explainig position and momentum relationship, but the aperture size explanation below is beyond me. Any help?

    "If a large aperture is used for the microscope, the electron's location can be well resolved (see Rayleigh criterion); but by the principle of conservation of momentum, the transverse momentum of the incoming photon and hence the new momentum of the electron will be poorly resolved. "
  2. jcsd
  3. Aug 15, 2010 #2
    Yeah, don't worry too much about that one--its a bad way to think about it.

    The smaller (or farther away) something is that you want to see with a microscope (or telescope) the bigger the lens/aperture has to be.

    Then as a separate thing--unrelated to lenses--the only way you could 'see' an electron, is if the light (photons) you bounce off of the electron have a wavelength the-same-size or smaller than an electron. A photon with such a small wavelength has a large energy (and momentum) and therefore bouncing it off of the electron means you are going to impart a large (unknown) velocity to the electron --> thereby not knowing much about its momentum.

    Does that help?
  4. Aug 15, 2010 #3
    Why the wavelength has to be smaller than an electron? I have had this question for a long time.
  5. Aug 16, 2010 #4
    Why the photons wavelength has to be the-same-size or smaller than an electron?
  6. Aug 16, 2010 #5
    The way I look at it is if you're on boat and there are waves with long wavelengths relative to the boat, you don't feel them so much, perhaps a gentle rocking, but if they are the size of the boat or smaller you would crash into them.

    Note: All the waves have equal amplitude.
  7. Aug 16, 2010 #6
    Think about an antenna: if an antenna is shorter than the wavelength it is trying to receive, the electrons in the antenna don't have enough space to respond. If the wavelength is shorter than the length of the antenna, the electrons can't respond as coherently (that's a different subject though).

    Its the same idea with interaction in general.
  8. Aug 16, 2010 #7
    Another one: Think about a (dark) room with a hole in the wall. You keep on throwing balls against the wall. If the balls are small enough they go through the hole. If the balls are too big, they just bounce back, and you won't notice wether there is a hole in the wall.

    Same idea goes for the wavelength of the photon and the electron.
  9. Aug 17, 2010 #8
    Bla bla bla.... this analogies are incorrect. I would like to see some explanation with maths involved, not analogies. Thnx!
  10. Aug 20, 2010 #9
    I doubt you'll get them. The reason is that you need concepts to apply maths to physics.

    That being said, I'm not really a fan of "explaining" the uncertainty principle in terms of these thought experiments. Heisenberg didn't seek to explain the HUP with reference to them; rather, he invented them as a construction to show that it couldn't be violated. The real point is that the position and momentum do not have well defined values prior to measurement. You can explain/prove the HUP without reference to measurement at all. The wave mechanical formulation is probably the most intuitive- you do it by considering how many plane waves (i.e. momentum eigenstates) you'd need to add together to get a position eigenstate (answer- infinitely many).
  11. Aug 24, 2010 #10


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    Just like the finite value of the speed of light implies the impossibility of a sharp seperation between time and space (relativity), the finite value of the quantum of action implies the impossibility of a sharp seperation between the behavior of a 'system' and its interaction with the measuring instruments (complementarity). This leads to the well-known difficulties with the concept of 'state' in quantum theory and the uncertainty relation for the variables p and q is a consequence of this.
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