Expect Momentum Problem 1.17 Griffiths: Find Expected Value & Uncertainty

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In Problem 1.17 of Griffiths, the state function is defined as psi = A(a^2 - x^2) for the interval -a to a, and zero otherwise. To find the expected value of momentum, p, at time t = 0, utilize the formula <p> = m(d/dt)<x> or <p> = <Ψ|p|Ψ>. Both methods yield equivalent results, but integrals over symmetric limits result in odd integrands, indicating that the expected value of momentum is zero.

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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
 
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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.
 
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