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Probability of Momentum in Infinite Well

  1. Oct 28, 2007 #1
    1. The problem statement, all variables and given/known data

    Need to find the probability density of a ground state in an infinite square well.

    2. Relevant equations

    Ground State psi(x) = Asin(Bx), don't want to look up the constants, don't think they are relevant anyway.

    3. The attempt at a solution

    Took the derivative of psi(x) with respect to x to get momentum, times h/(2*pi*i).

    Then tried to integrate psi*psi, but before I did that I stopped myself, because taking the limits of integration to be infinity and negative infinity like for the position probability, I'd end up with 0. Since I get cos(x)^2 = 1 - sin(x)^2

    Sin(x)^2 integrated from -inf. to inf. = 1, so this would be 0. That doesn't make sense, so it's clearly not the way to go about it.

    Should I go a step deeper and just try to solve Schrodinger's equation for momentum? Would that mean I have to take d/dx for the whole thing and then solve it? This problem is harder than I thought it would be.
     
  2. jcsd
  3. Oct 29, 2007 #2
    probability density function is the integrand! since the problem is asking you to find this integrand, you don't need to integrate!

    Also, when you do integrate the probability density function of momentum, you find that <p>= 0 (as you claimed) and this is perfectly fine. Since, in infinite square well potential, you would expect <x> to be constant! (to wit: <p> = m d<x>/dt)
     
  4. Oct 29, 2007 #3
    I thought it was only constant in the classical case? Since the ground state is an arc (half a sine wave), there's more probability in the middle, so wouldn't the momentum be smaller in the middle?
     
  5. Oct 29, 2007 #4
    Wait, you're just using the expectation value, which isn't what I am trying to find at all.
     
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