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Hamiltonian math problem

  1. Oct 8, 2006 #1
    One more question. Sorry!

    Here's the problem:

    An electron moves in a straight line under the influence of a conservative force so that the Hamiltonian is [tex]H = \frac{p\wedge^2}{2m} + V(x)[/tex], where [tex]p\wedge[/tex] means the momentum operator and I think V(x) is the potential energy. I need to find an expression for [tex]\frac{d}{dt} <\frac{p^2}{2m}>[/tex].

    Sigh. Anyone have any idea how to do this? I wish I could show you something that I've done to get started but I really don't have a clue. I do have an expression for the time-derivative of an operator.. but plugging this into it involves time derivatives of integrals over x and figuring out the commutator of H and p^2 / 2m, which involves doing calculations with the potential energy, and I have no idea how to deal with that. Thanks.
     
    Last edited: Oct 9, 2006
  2. jcsd
  3. Oct 8, 2006 #2
    I don't really know the meaning of "I need to find an expression for..."
    If you want to calculate something you need to know the wave function, they probably mean that you should do it to an arbitrary one (?)

    Note that you have to calculate the time derivative of the kinetic energy expectation value. Calculate the total energy (you have to know the wave function for that), calculate the <V(x)> and the difference is the desired k.e exp. val.
    I'm not sure though, sorry :blushing:
     
  4. Oct 8, 2006 #3
    The wave function isn't given, so I'm not sure what to do :( oh well, thanks!
     
  5. Oct 8, 2006 #4

    George Jones

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    This is a nice question that helps to show the relationship between classical mechanics and quantum mechanics.

    Have you covered the expression for the time derivative of expectation value of an operator?

    As you say, you have to calculate [H, p^2/2m], but this isn't so bad.
     
  6. Oct 8, 2006 #5
    OK, I think I know what you're talking about. I did the commutator you were talking about and got [H, p^2/2m] = [tex]\frac{\hbar^2}{2m}\frac{\delta^2H}{\delta x^2}[/tex]

    sound right?

    So now I need to do the second derivative with respect to x of H.. and I'm not sure how to go about that, since I'm going to be taking the second derivative of a second derivative?
     
  7. Oct 8, 2006 #6
    k, i'm really stuck on this one. based on what i'm trying to prove (i realize i left this off, but i thought it would be elementary once i got the solution - stupid, i know), which is that in the classical limit [tex]\hbar \rightarrow 0[/tex], the expression reduces to [tex]\triangle K = \int Fdx[/tex], the work-energy theorem, I gather that my expression for d/dt <p^2/2m> must reduce to a double time integral integral of dV(x)/dx, which corresponds to the integral of force... but i have no idea how to get there. anybody have any hints or suggestions? :)
     
  8. Oct 8, 2006 #7
    Changed title of post to match what is going on here, because the original one wasn't very accurate.

    Update before I go to bed: as I said in the last post, and as alluded to by George, I'm trying to show the relationship between the result and the one given by classical mechanics. This is where I'm stuck now:

    I have [tex]\frac{d}{dt} <\frac{p^2}{2m}> = \frac{\hbar i}{2m}<\frac{\delta^2H}{\delta x^2}>[/tex]

    This is after the commutator and setting the time derivative of p^2/2m to 0.

    I have in my notes that you can set dH/dx equal to dV/dx.. but I still don't see how this is going to lead to the work-energy theorem. I'm also confused because my hbar is being multiplied by my expectation value.. so if I let it go to 0, the whole equaltion should go to zero, which doesnt make much sense at all. I could really use some help here:)
     
  9. Oct 9, 2006 #8

    George Jones

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    This isn't the expression that I get.

    [H , P^2] = [P^2/2m + V(x) , P^2] = ?
     
  10. Oct 9, 2006 #9
    Ah, in class we did it the way I was describing, just leaving H in there without explicity expressing it. I have tried it your way as well and gotten V(x) p^2/2m - p^2/2m V(x) for the commutator.. I still fail to see how that's going to lead to the answer, though, since the hbar is still outside... Unless I'm still doing something wrong, because my end result [tex]\frac{d}{dt} <\frac{p^2}{2m}> = \frac{\hbar i}{2m}<\frac{\delta^2V(x)}{\delta x^2}>[/tex] is still the same.
     
    Last edited: Oct 9, 2006
  11. Oct 9, 2006 #10

    George Jones

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    You need to do a little more work on the second term.
     
  12. Oct 9, 2006 #11
    Well, if you meant doing the product rule, I've tried that now and gotten a slightly different result (I messed up the sign before):

    [tex][H,\frac{p^2}{2m}] = \frac{\hbar^2}{2m} ( 2V(x) \frac{\delta^2}{\delta x^2} + \frac{\delta^2 V(x)}{\delta x^2})[/tex]

    but I fail to see how this is any more useful :(
     
    Last edited: Oct 9, 2006
  13. Oct 9, 2006 #12

    George Jones

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    Yes.

    Now the first term isn' right.

    Can you change the integration variable to time?
     
  14. Oct 9, 2006 #13
    It isn't? I've done it three times now and gotten the same result each time..

    And as far as changing the integration variable to time.. you mean, by Fourier transform? I'm lost.
     
  15. Oct 9, 2006 #14

    George Jones

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    Without seeing your work, I can't tell where you went wrong.

    No, I just mean write F(x) as F(x(t)), where x(t) is the path of the particle, and write dx = (dx/dt) dt.

    Then, what is

    [tex]\frac{d}{dt} \left( \Delta K \right)?[/tex]
     
  16. Oct 9, 2006 #15
    ok, here's what I did:

    [tex][\frac{p^2}{2m} + V(x)](\frac{p^2}{2m}) - \frac{p^2}{2m}[\frac{p^2}{2m}+V(x)][/tex]
    .
    .
    [tex]\frac{p^4}{4m^2} + V(x)\frac{p^2}{2m} - \frac{p^4}{4m^2} - \frac{p^2}{2m}V(x)[/tex]
    .
    .
    [tex]\frac{\hbar^2}{2m}V(x)\frac{\delta^2}{\delta x^2} + \frac{\hbar^2}{2m}[V(x)\frac{\delta^2}{\delta x^2} +\frac{\delta^2 V(x)}{\delta x^2}][/tex]
    .
    .
    [tex]\frac{\hbar^2}{2m}[2V(x)\frac{\delta^2}{\delta x^2} + \frac{\delta^2 V(x)}{\delta x^2}][/tex]


    and... would [tex]\frac{d}{dt} \left( \Delta K \right)[/tex] just be F(x(t))?
     
    Last edited: Oct 9, 2006
  17. Oct 9, 2006 #16

    George Jones

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    Good!

    When expanding this term, a little care is needed. Do it one step at a time, and, if it helps, put in a [itex]\psi[/itex].

    [tex]\frac{p^2}{2m}V\psi = \frac{p}{2m} \left{ p \left( V \psi \right) \right}[/tex]

    Also, I think there is a sign mistake in the first term of the third line.
     
  18. Oct 9, 2006 #17
    Ahhhh, I see. Now I'm getting [tex]\frac{\hbar^2}{2m}[\frac{\delta^2 V}{\delta x^2} + 2\frac{\delta V}{\delta x}\frac{\delta}{\delta x}][/tex]

    Still not seeing it :(
     
  19. Oct 9, 2006 #18

    George Jones

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    What is the time derivative of the expectation value of the first term?

    Write the last term in terms of p.
     
  20. Oct 9, 2006 #19
    You mean actually go through and calculate [H, d^2V/dx^2]? Or is there something I'm supposed to be seeing here?

    Write in terms of p? I'm not sure what you mean, unless it's replace d/dx with ip/hbar.
     
  21. Oct 9, 2006 #20

    George Jones

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    No, it's much simpler. The expectation value is an an integral, and since V doesn't depend on time, the only time dependence is in the [itex]\Psi[/itex] parts. But since V'' is a scalar it can be move throught a [itex]\Psi[/itex], giving [itex]V'' \Psi \Psi *[/itex] under the integral.

    What is [itex] \Psi \Psi *[/itex]? Does the integral have any time dependence?

    Yes.
     
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