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Quantum operators: Where to start?

  1. Feb 9, 2010 #1
    Hello!

    I have a task to do where I do not know where to start or where to find more information.

    At first, this is just the problem statement:

    Velocity operator [tex]\mathbf{\hat{v}}[/tex] is defined by the following equations:

    [tex]\frac{d}{dt} \mathbf{\bar{r}} = \left< \psi | \mathbf{\hat{v}} | \psi \right>[/tex]

    [tex]\mathbf{\bar{r}} = \left< \psi | \mathbf{\hat{r}} | \psi \right>[/tex]

    where [tex]\mathbf{\hat{r}}[/tex] is the position operator (just the coordinate [tex]\mathbf{r}[/tex] itself). Show that the following relation holds between the operators:

    [tex]\mathbf{\hat{v}} = \frac{\mathbf{\hat{p}}}{m}[/tex]

    (momentum operator [tex]\mathbf{\hat{p}}[/tex] is given by [tex]\mathbf{\hat{p}} = -i \hbar \nabla[/tex])

    I do not understand what the momentum has to do with that. I tried around a while with the integral-definitions of the operators but was not successful. At the moment I have absolutely no idea where to start.

    Can anybody give me a hint?

    Also some literature which might help me would be greatly appreciated!

    Thank you very much and Regards,
    divB
     
  2. jcsd
  3. Feb 9, 2010 #2

    Hurkyl

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    Two operators are equal iff they have the same effect on all kets.
     
  4. Feb 9, 2010 #3
    Hi,

    Thanks for your reply.

    Anyway I think I found a source where it is derived:

    * http://de.wikibooks.org/wiki/Quantenmechanik#Der_Impulsoperator

    But - for sure - I do not really understand the derivation :-( Maybe somebody can help me with that?

    First question: Why is

    [tex]\frac{d}{dt}\int \psi^* r \psi\,dr = \int \psi^* r \frac{d}{dt}\psi + \psi r \frac{d}{dt}\psi^*\,d r[/tex]

    ?

    (For the sake of simplicity I take only scalars - I hope this does not harm the generalization)

    divB
     
  5. Feb 9, 2010 #4
    That's your old buddy, the chain rule...
    Hurky is telling you what to do. Try applying the operators to the psi ket vector to show that they give the same result.
     
  6. Feb 9, 2010 #5
    Hi,

    Thank you very much! I just wanted to write that I got it, really dumb easy :-) Thank you anyway.

    About Hurky's hint: Aaah, now understand what he meant. But I think this is exactly what I am doing now: I start with the derivation of the [tex]\mathbf{\hat{r}}[/tex] (as this is v!) and the result should and will be something proportional to p with factor m. Of course this is true if I replace

    [tex]-i\hbar\nabla = \mathbf{\hat{p}}[/tex]

    in the final equation :-) So I have shown that both are the same.

    But I think there is just one piece of mathematics missing for me to understand:

    How to come from

    [tex]\frac{\hbar}{2mi} \left[-\int d^3r \psi^*(\vec{r},t) \vec{r} \Delta\psi(\vec{r},t)+\psi(\vec{r},t) \vec{r} \Delta\psi^*(\vec{r},t)\right][/tex]

    to

    [tex]\frac{\hbar}{2mi} \left[-\int \vec{f}\left(\psi^*\vec{r}\vec\nabla\psi-\psi\vec{r}\vec\nabla\psi^*\right) + \int d^3r \left(\vec\nabla\psi^*\vec{r}\right)\vec\nabla\psi - \int d^3r\left(\vec\nabla\psi\vec{r}\right)\vec\nabla\psi^* \right][/tex]

    Thank you again,
    divB
     
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