Commutation Relations and Ehrenfest

  • Thread starter keniwas
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  • #1
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Homework Statement


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.


Homework 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]


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.
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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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:
 
  • #3
59
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HAHAHA oh... that hurts....

Thank you =)
 

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