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A Paradox?

  1. Sep 9, 2005 #1
    Let,s suppose we have a particle moving under a potential V:
    By Heisenberg picture we know that:

    [tex]\frac{dx}{dt}=p/m [/tex]

    so if we knew x(t) and x(t+h) we could calculate the expresion:

    [tex]x(t+\epsilon)-x(t)=p\epsilon/m [/tex]

    so knowing x(t) and x(t+e) we could calculate the momentum of the particle:
    Last edited: Sep 9, 2005
  2. jcsd
  3. Sep 9, 2005 #2
    Please do not mix the differential equation between operators and eigenvalues. They are completely different.

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