1. Not finding help here? Sign up for a free 30min 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!

Commutation Relations and Ehrenfest

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\psi(\vec{r},t)[/tex] be the wavefunction for a free particle of mass m obeying Schrodinger equation with V=0 in 3 dimensions. At t=0, the particle is in a known initial state [tex]\psi_0(\vec{r})[/tex]. Using Ehrenfest's theorem, show that the expectation value [tex]<x^2>[/tex] in the state [tex]\psi(\vec{r},t)[/tex] is a quadratic function of t:
    [tex]<x^2>_t=a+bt+ct^2[/tex]
    where a, b and c are constants. Relate the constants b and c to the expectation values of operators at t=0. Show that these lead to the expression for [tex]<(\delta x)^2>_t[/tex] quoted in class.


    2. Relevant equations
    we have that [tex]<(\delta x)^2>=<x^2>-<x>^2=<(\delta x)^2>_0+\frac{1}{m}(<xp_x+p_xx>_0-2<x>_0<p_x>_0)+\frac{<(\delta p_x)^2>_0}{m^2}[/tex]
    where
    [tex]<x>=<x>_0+\frac{<px>_0}{m}t[/tex]


    3. The attempt at a solution
    [tex]\frac{d}{dt}<x^2>=\frac{1}{i\hbar}<[x^2,H]>=\frac{1}{m}<xp_x+p_xx>[/tex]

    [tex]\frac{d}{dt}\frac{1}{m}<xp_x+p_xx>=\frac{1}{i\hbar m}<[xp_x,H]+[p_xx,H]>=\frac{1}{i\hbar m}<x[p_x,H]+[x,H]p_x+p_x[x,H]+[p_x,H]x>=\frac{1}{i\hbar m}<\frac{i\hbar}{m}p_xp_x+p_x\frac{i\hbar}{m}p_x>=\frac{i\hbar}{i\hbar m^2}<2p_{x}^2>[/tex]

    [tex]\frac{d}{dt}\frac{1}{m^2}<p_{x}^2>=0[/tex]

    Taking three integrals of the last equation from 0 to t and substituting the values we got in the previous equations we get the equation for <x^2> as

    [tex]<x^2>=<x^2>_0+\frac{1}{m}<xp_x+p_xx>_0t+\frac{2<p_{x}^2>_0}{m^2}t^2[/tex]

    The problem is if I try to subtract [tex]<x>^2[/tex] from this, the 2 in the last term ([tex]\frac{2<p_{x}^2>_0}{m^2}t^2[/tex]) causes problems. I feel there should have been a 1/2 pop out somewhere in here that will cancel with that 2 to make this work. But i can't see where it is. Does anyone see where I am going wrong that is giving me this extra factor of 2?

    Thanks for any input.
     
  2. jcsd
  3. Oct 7, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You're gunna kick yourself:

    [tex]\int_0^t \frac{2\langle p_x^2\rangle_0}{m^2}t'dt'=\frac{\langle p_x^2\rangle_0}{m^2}t^2[/tex]

    :wink:
     
  4. Oct 7, 2009 #3
    HAHAHA oh... that hurts....

    Thank you =)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Commutation Relations and Ehrenfest
  1. Commutator relations (Replies: 4)

  2. Commutation relations (Replies: 6)

  3. Commutator relations (Replies: 19)

  4. Commutator relations (Replies: 3)

  5. Commutator relations (Replies: 1)

Loading...