1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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!

  2. jcsd
  3. Jan 23, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    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.

  5. Jan 24, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook