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Free particle, classical vs quantum mechanics

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider first a free particle (Potential energy zero everywhere). When the particle at a given time is prepared in a state ## \psi (x) ## it has <x> = 0 and <p> = 0.
    The particle is now prepared in
    ## \Psi (x, t=0) = \psi (x) e^{ikx} ##

    1. Give <p> at time t=0.

    2. It can be shown in quantum mechanics that <p> is independent of time for a free particle, in anology to Newton's first law in classical mechanics. Give <x> as a function of time.


    2. Relevant equations
    ## \displaystyle <p> = \int ^{ \infty} _{- \infty} \Psi (x,t)^* \left( \dfrac{\hbar}{i} \dfrac{ \partial}{\partial x} \right) \Psi (x,t) dx ##


    3. The attempt at a solution

    So I put Psi into the integral and get:

    ## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* e^{-ikx} \left( \dfrac{ \partial}{\partial x} \right) \psi (x) e^{ikx} dx ##

    ## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* e^{-ikx} \left[ \dfrac{d \psi (x)}{dx} e^{ikx} + ik \psi (x) e^{ikx} \right] dx ##

    The exponentials cancel out and I get:

    ## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \dfrac{d \psi (x)}{dx} dx + \dfrac{ik \hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \psi (x) dx ##


    So, for a normalised distribution, the area under the curve wil be 1, and so the second integral will give me ## \hbar k##
    What about the first integral?
     
  2. jcsd
  3. May 4, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Can you relate the first integral to some property of the state ##\psi (x)## that was specified at the beginning of the problem statement?
     
  4. May 5, 2013 #3
    Oh right silly me, the first integral is just the expectation value of the momentum of the function ## \psi (x) ## which is given to us as 0.
     
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