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Heisenberg uncertainty principle

  1. Sep 2, 2013 #1
    I read somewhere that Heisenberg described his uncertainty principle by saying that you can't measure position more accurately than the wavelength of light (which makes sense), so Δx > λ.

    This is what I don't get. He then says that p=h/λ, so Δp > h/λ2 Δλ. He the multiplies and sets Δλ ≈ λ to get:

    ΔxΔp > h

    Why does the initial momenta of the photon, p=h/λ, determine the uncertainty of the momentum in the object scattered by light? And what if you knew the momentum of the photon exactly, then Δp=0?
  2. jcsd
  3. Sep 2, 2013 #2
    Hmmm... This is one of those fishy explanations that people give to *sort of* explain the Heisenberg uncertainty principle. The argument works better in words, which is clearly evidenced by the unusual algebra you've used to write this proof. Moreover, you get the wrong answer! The correct HUP is ΔxΔp ≥ h/4π

    Here is the intuitive argument in words. Let's say we're trying to locate the position of a particle by shooting a photon at it. In other words, we will shoot a photon at the particle, and then we might determine where the particle is by seeing whether the photon misses it (the particle wasn't there) or the photon interacts with it (maybe it is absorbed or it bounces off).

    According to you, you can't locate the photon any more precisely than its wavelength λ. So if we really want to determine the exact location of the particle, we want λ to be small. But if λ is small, then the momentum p=h/λ is big. If the photon interacts with the particle, then it could transfer a momentum as big as 2p (if it bounces off). So then we've messed up the momentum of the particle. On the other hand, if we wanted to not change the momentum very much, we would need a big λ and then we would have an imprecise idea of where the particle is. So the reason the photon's initial momentum is important is because it sets a bound on how much the photon can change the particle's momentum.

    Anyway this is a very fishy argument and it's certainly not rigorous. The better way to understand the HUP is by solving the Schrodinger equation for minimum uncertainty wavepackets. If you go through the formalism, you can easily answer your other questions: you'll find that if Δp=0, then Δx=∞. If you want a *sort of* explanation for why that is so, you might think about the energy-time version of the HUP which is ΔEΔt ≥ h/4π. If we knew the photon's momentum p exactly, then we know the energy E=pc exactly, and so if we were trying to use a photon with an exactly-known momentum to measure the position of a particle, we would have no idea about when the photon was shot out. If we waited long enough and a photon came back, signalling the particle was detected, we wouldn't know whether the photon was travelling for a nanosecond or a gigayear, so the point in space where it bounced off the particle would have huge uncertainty.
    Last edited: Sep 2, 2013
  4. Sep 2, 2013 #3

    Vanadium 50

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    It's very important that you tell us where. Otherwise we're just guessing what you read and why it confused you. Apart from wasting time, we might end up simply confusing you more.
  5. Sep 3, 2013 #4
    I forget where I read the explanation from. Sorry.

    But is the change in the particle's momentum the uncertainty in the particle's momentum?

    It's very enticing, to derive quantum mechanics from classical concepts: you can only observe the size of something with the wavelength of light smaller than the object (classical optics). If you can reason your way classically with the momentum, then you have just derived quantum mechanics: ΔxΔp ≠ 0

    That's why I'm interested in the argument.

    Again, sorry for not providing a reference. I actually thought it would be an argument common to a lot of textbooks so it would be well-known.
  6. Sep 6, 2013 #5

    Claude Bile

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    The most convincing justification of the HUP is in the fact that certain properties we derive from wavefunctions are Fourier transforms of one another. It is easy to demonstrate that the width of a function is inversly proportional to the width of its Fourier transform, hence the HUP.

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