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Average electronic momentum in bound state: please see this

  1. Aug 18, 2010 #1
    Someone please tell me if I am thinking right:

    Let's consider an unperturbed electronic state of an atom/molecule. If we denote it by [a>, then the average electronic momentum in state [a> is,

    <p> = <a]p[a> = (<a]p<a])* (because p is hermitian)
    = (<a]*p*[a>*)
    = (<a]p*[a>) (because [a> is real.)
    = - (<a]p[a>) ( because p is purely imaginary. so p* = -p)
    so, <p> = 0

    so, the average momentum in any unperturbed eigenstate = 0

    Now, the atom/molecule is being perturbed by a time-dependent scalar potential of the form V(t) = V Cos(wt). So we can write the time-dependent wave function as a perturbation series. Since there is no magnetic field, the wave function will be real and <p> = 0.

    Is it true? Please correct me if I am doing it wrong. Thanks.
     
  2. jcsd
  3. Aug 18, 2010 #2
    not quite
    (<a|p <a|)* is nonsensical.
    you could do <a| (p |a>) = <a| (<a| p^dagger)^dagger = <a| (<a| p)^dagger
    or (<a|p ) |a> = (p^dagger |a>)^dagger |a> = (p |a>)^dagger |a>.

    what do you mean |a> is real? |a> is a STATE, not a number.

    ----
    here's what i think you mean.

    notice p is a hermitian operator so that for any state |a> we have
    (<a|p|a>)* = <a|p|a>^dagger = <a| p^dagger |a> = <a|p|a>.

    suppose |a> is a state such that <x|a> = a(x) is a real valued function.
    [i.e. the wave-function is real (in the x - representation)]

    then <a|p|a>= [itex] -i \hbar \int d^3x ( a^*(x) \nabla a(x) ) = -i \hbar \int d^3x ( a(x) \nabla a(x) ) [/itex]
    first equality is definition, and second is assumption that a(x) is real valued.

    since the integral of a real valued function (a a') is real, we have
    (<a|p|a>)* = - (<a|p|a>).

    so that <a|p|a> = 0.
    -------

    this doesn't follow. What makes you think the time-dependent wave function
    will be real? what does the perturbation series have to do with anything?
     
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