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Matter Waves

  1. Aug 11, 2003 #1
    I was told that for an electron, the wave length lamda calculated from De Broglie's equation ( = h/p ) refers to the wave length of its probability function (as solution to Schroedinger's wave equation ), and is not its physical wave length as exhibited in the electron's double slit diffraction nor diffraction through crystals.

    At the same time, for photons we take lamda= h/p as the actual/physical wave length of the light.

    Can someone shed some light on this apparent inconsistency?
  2. jcsd
  3. Aug 11, 2003 #2
    I don't see any inconsistency. If you perform an electron diffraction experiment (for example, Bragg reflection in a crystal lattice) diffraction will occur for specific values of d (lattice spacing), theta (angle of incidence) and lambda. If you can get values for d and theta, and calculate the wavelength, it should correspond (within experimental error) to the theoretical value given by the de Broglie relation.
  4. Aug 12, 2003 #3
    Speaking of the de Broglie equation I noticed that if it were possible to know that a particles momentum was zero then it's wavelength would be infinitely long? So does the uncertainity principle prevent you from making a measurement of zero for the momentum of a particle since then you would be able to know it's position?
  5. Aug 12, 2003 #4
    The HUP says that dx*dp = h. That is, the product of the uncertainty in x (position) and p (momentum) is equal to a constant h (Planck's constant). So, the smaller one of these gets, the larger the other one does to keep this relation. Or, the more precisely we know one value, the less precisely we know the other.

    So, if we could measure the momentum of a particle to be exactly zero, then technically it would have infinite wavelength, which is of course not physically meaningful. The HUP prevents this from happening. If we measured exactly 0 for p, then dp would also be 0, meaning dx would have to be infinite -- clearly, we could not tell anything useful about its position with an infinite uncertainty (not to mention the above equality would be violated).
  6. Aug 13, 2003 #5
    The Uncertainty Principle works with ... uncertainties. Nothing can stop you from measuring p=0. But are you ABSOLUTELY SURE you measured 0? Or you measured 0 &plusmn something? That something is your uncertainty. If you're absolutely sure (which you can't be because you measure with something that makes some aproximations) then you get an infinite uncertainty for the position. This doesn't stop you though to measure the position. You'll get a value, but you'll have to mean value &plusmnn infinity.
  7. Aug 13, 2003 #6
    thank you everyone for your reply!I see what you both mean but I'm pretty sure that my lamp is at rest relative to me so it has zero momentum relative to me. So does that mean that the de broglie equation is not always true for macroscopic objects? Is there an uncertainity I'm not aware of?
  8. Aug 13, 2003 #7
    The de Broglie relation can be applied to macroscopic objects, but it is generally not meaningful. Take your lamp for example. Let's say the mass was 1 kg, and we threw it through the air at some velocity, say 10 m/s. If you calculcate it's wavelength, you will get something on the order of 10^-35 m, which is far too small to measure experimentally. For comparison, the diameter of a typical atom is on the order of 10^-10 m, or 25 orders of magnitude greater than your lamp's wavelength. Basically, this means that we cannot observe the "wave" nature of your lamp, and we would treat it as a pure "particle".
  9. Sep 25, 2003 #8
    To Futz:

    Thanks for your reply.

    So what you are saying is that the "physical" wavelength is also the wavelength of the wave function, whose modulus calculated at a location represents the probability of finding the electron at that position?

    Can we apply this to photons and say that their wavelengths are the same as the wavelengths of their wave function?
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