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Problem in Quantum Mechanics

  1. Jul 16, 2003 #1

    pmb

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    There's a problem in Liboff's text "Introductory Quantum Mechanics - 3rd Ed."

    On page 176 problem 6.12 states

    "A particle moving in one dimension interacts with a potential V(x). In a stationary state of this system show that

    (1/2) <x dV/dx > = <T>

    where T = p^2/2m is the kinetic energy of the particle."

    Liboff gives the answer but starts off with

    "In a stationary state,

    d<xp>/dt = (i/hbar)<[H,xp]> = 0
    ..."

    Why? I.e. why is d<xp>/dt = (i/hbar)<[H,xp]> = 0 for a stationary state?

    Pete
     
  2. jcsd
  3. Jul 16, 2003 #2

    Tom Mattson

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    Look up the Heisenberg equation of motion for operators. The equation is:

    dO/dt=(i/hbar)[H,O]+&part;O/&part;t

    for any operator O. Evidently, x and p have no explicit time dependence in your problem so the partial with respect to t is zero. The derivation should be in your book, but the basic reason is that the Hamiltonian is the generator of time translations, and so you would expect it to be closely associated with the time evolution of operators.
     
  4. Jul 16, 2003 #3

    pmb

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    Re: Re: Problem in Quantum Mechanics

    {Note: Liboff is is a quick review for me for the summer so I've bneen through this before - but 10 years ago. We used Cohen-Tannoudji in grad school - both semesters - so I'm brushing up to jump into that}

    What you've said is in a way related to this section in a certain sense - this was a section on the relation

    d<A>/dt = <i/hbar [H,A] +&part;A/&part;t>

    In this case A = xp. Th partial drops out and we're left with


    d<ap>/dt = i/hbar <[H,xp]>

    But Liboff sets that to zero - why?

    Pete
     
  5. Jul 16, 2003 #4

    Tom Mattson

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    Re: Re: Re: Problem in Quantum Mechanics

    OK, now I understand your question. He sets it to zero because you are looking at an expecation value, which for stationary states does not evolve in time (by definition of "stationary state"). Take away the < > brackets, and you do not necessarily get zero.
     
  6. Jul 16, 2003 #5

    pmb

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    Re: Re: Re: Re: Problem in Quantum Mechanics

    '

    Ahhh! The expectation for any operator for a stationary state is a constant in time!

    Okay - Thanks. I get it now. Duh! :-) I can't see why I missed that now. Thanks Tom

    Pete
     
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