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Expectation momentem

  1. Sep 12, 2006 #1
    Problem 1.17 in griffiths gives, at time t = 0, the state psi =A(a^2-x^2) for -a to a, and 0 otherwise. It asks then to find the expected value of momentum p at 0 and also the uncertainty in p. How do I do this? The only way momentum is defined is md<x>/dt, and since the state is only for time t, there seems to be no way to do this.

    thanks
     
  2. jcsd
  3. Sep 17, 2006 #2
    To determine the expectation value of the momentum you may use
    [tex] \langle p \rangle = m\frac{d}{dt} \langle x \rangle [/tex]
    or
    [tex] \langle p \rangle = \langle \Psi \mid p \mid \Psi \rangle [/tex]
    Note that these two are equivalent statements. In either case, when doing the integrals, notice that you always obtain an odd integrand over over symmetric limits about the origin, what does that mean? Can you guess the solution from the initial problem statement. Give it a try.
     
    Last edited: Sep 17, 2006
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