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Heisenberg UP and space homogeneity - an intuitive reasoning

  1. Nov 18, 2006 #1
    Let's consider the position x and momentum p of a free particle.

    [tex]\Delta\ x\Delta\ p\geq \hbar/2[/tex] so, if [tex]\Delta \ x[/tex] is little enough, [tex]\Delta \ p[/tex] is big enough.

    The fact [tex]\Delta \ p[/tex] is big implies that we cannot say the momentum conservation law is valid anylonger.


    space is homogeneus --> momentum conservation,
    non conservation of momentum --> space is not homogeneus

    So, for [tex]\Delta \ x[/tex] very little, that is, considering space at very small distances, space itself must be non-homogeneus.

    Consider that this is a very rough reasoning with no pretence at all.

    Can it have any meaning?
    Last edited: Nov 18, 2006
  2. jcsd
  3. Nov 18, 2006 #2
    It will be valid, it's just you weren't sort of momentum before or after an interaction (within certain limits obviously).

    Giving wildly unrealistic numbers here consider particles A and B interacting. A has momentum somewhere in the range [5,6] and B is somewhere in [6,7]. If after interactioning A has momentum somewhere in [7,8] then you'd expect B to be in [4,5].

    This is totally unrigorous and I half expect someone much more versed in QM to come along and correct me, but that's how I would reconcile the HUP and conservation of momentum.
  4. Nov 19, 2006 #3


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    Yes, that's right. There is in principle exact momentum conservation, but it are the uncertainties on the initial conditions which allow for the uncertainties on the final states. This can in fact be illustrated by using Bohmian mechanics where this is explicit.
  5. Nov 21, 2006 #4


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    In QFT this issue deals with real/virtual particles versus conservation/nonconservation of 4-momentum.


    P.S. Nowadays we prefer the interpretation: there are virtual particles because at any point in spacetime the 4 momentum is conserved.
  6. Nov 21, 2006 #5
    You mean this is assumed as a postulate? And, of course, we are talking about SR only (not GR).
  7. Nov 21, 2006 #6
    Does that mean that during an interaction (a) "virtual particle(s)" is created which carries off the missing momentum to maintain the conservation law?

    I was under the impression that a virtual particle was just a tool to make it easier to understand Feynman diagrams and path integrals.
    Last edited: Nov 21, 2006
  8. Nov 22, 2006 #7


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    In QFT on Minkowski background, Poincare' symmetry is a must, so you can take it an an axiom.

    That's right.

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