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Expectation Value of Momentum Squared

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is in the state

    Psi(x,t) = Ae^(-a[(mx^2)+it])

    where A and a are positive real constants.

    a) Find A

    b) For what potential energy function V(x) does Psi satisfy the Shrodinger equation?

    c) Calculate the expectation values of x, x^2, p, and p^2

    d) Find sigma_x and sigma_p. Is their product consistent with the uncertainty principle?


    2. Relevant equations
    Intetgral of p(x)dx = 1 (from negative infinity to infinity)
    Shrodinger equation
    <x> = Integral of Psi* (x) Psi dx
    <p> = Integral of Psi* (hbar/i partial derivative with respect to position) Psi dx
    sigma_x times sigma_p > or = hbar/2


    3. The attempt at a solution
    I managed to get solutions for parts a and b, and most of part c. Using the equations listed above, I got 0 for <x> and <p> and 1/4 hbar/(am). The problem I had was figuring out how to solve <p^2>. I tried squaring hbar/i and the partial derivative term but that lead to some problems.

    The partial derivative is A^2 (-2amx/hbar) e^(-amx^2/hbar - iat).

    When I squared only the (-2amx/hbar) part, I managed to get a solution of hbar*am, and when I solved for the standard deviations and put them into the Heisenberg Uncertainty Principle I got exactly hbar/2. However, a friend of mine informed me that e^(-amx^x/hbar - iat) is also part of the derivative, and when I tried squaring that as well I ran into problems with simplification.

    Is hbar*am the correct answer? And if so, why don't I have to square e^(-amx^2/hbar - iat)? Or if I'm completely wrong, what am I supposed to do to calculate <p^2>?

    Thanks for your help!

    -Pie
     
  2. jcsd
  3. Jan 23, 2010 #2

    vela

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    Your friend is correct, and you could just grind it out.

    Here's a trick that might help you. The time-independent Schrodinger equation is

    [tex]\frac{p^2}{2m}\psi(x)+V(x)\psi(x) = E\psi(x)[/tex]

    Since you already found V(x) earlier, you can solve for [itex]p^2\psi(x)[/itex] without having to differentiate twice, and use the result when evaluating the integral.
     
  4. Jan 24, 2010 #3
    Hi

    Could you please show me how did the OP get [tex]<p>= 1/4 \hbar/(am)[/tex]? I'm a little bit new to this stuff.

    AB
     
  5. Jan 24, 2010 #4

    vela

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    I think the OP made a typo and meant that result was [itex]\langle x^2\rangle[/itex] and that both [itex]\langle x\rangle[/itex] and [itex]\langle p\rangle[/itex] were 0.
     
  6. Nov 13, 2011 #5
    Hello,

    I'm really new to this whole thing, would you be able to explain how you got those answers, i'm getting confussed when i try it myself
     
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