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How to prove that for any bound electronic state, < p > = 0

  1. Aug 9, 2010 #1
    Hey all. So, I understand that every bound electronic state will have zero average electronic momentum, because otherwise the electron will fly off the atom. But how do I show mathematically that < p > = 0 for any bound state. Any help or reference greatly appreciated. Thanks.
     
  2. jcsd
  3. Aug 9, 2010 #2
    Try relating [tex]\langle p \rangle[/tex] to [tex] \frac{d}{dt} \langle x \rangle [/tex] (Ehrenfest's Theorem). If "bound state" means "stationary state in a time-independent potential", then use the fact that [tex] \frac{d}{dt} \langle x \rangle = 0 [/tex].
     
    Last edited: Aug 9, 2010
  4. Aug 9, 2010 #3
    If d<x>/dt = 0 then that means <p> is a constant. but how I prove that <p> = 0?
     
  5. Aug 10, 2010 #4
    The atom's hamiltonian commutes with the parity operator P. Hence any energy eigenstate [tex]|\psi>}[/tex] must have definite parity:

    [tex]P|\psi>=\pm|\psi>[/tex]

    Now, the momentum operator changes sign under parity

    [tex]PpP^\dagger=-p[/tex]

    So, using [tex]P^\dagger P = I[/tex], we find

    [tex]<\psi|p|\psi>=(<\psi|P^\dagger)( P p P^\dagger) (P|\psi>)=-<\psi|p|\psi>[/tex]

    hence

    [tex]<\psi|p|\psi>=0[/tex]
     
  6. Aug 10, 2010 #5
    This is not properly justified. You did not use the knowledge that the state is "bounded" (whatever it means rigorously...?)

    For example set [itex]V=0[/itex]. The Hamiltonian of a free particle commutes with the parity operator too, but still its eigenstates [itex]\psi(x)=e^{ikx}[/itex] don't satisfy [itex]P|\psi\rangle = \pm|\psi\rangle[/itex].

    How do you prove, that bounded states are non-degenerate? That would help. In the previous counter example the eigenstates [itex]\sin(kx)[/itex] and [itex]\cos(kx)[/itex] would be eigenstates of parity operator, but their linear combinations are not.
     
  7. Aug 10, 2010 #6
    For a time independent potential:

    [tex] \langle p \rangle = m \frac{d}{dt} \langle x \rangle [/tex]

    which is zero for a bound state.
     
  8. Aug 10, 2010 #7
    ok someone please tell me if this is right:

    d<p>/dt = <F> : this is always valid. so if I want to calculate the average force on a particle in any stationary state / non stationary state, I can just calculate the average momentum and take the time derivative.

    For example, if I have an atom in presence of a time-dependent potential, I can calculate the perturbed wave function using time-dependent perturbation theory, then take the average of p over the perturbed wave function and then take its time derivative to get the average force. Is that right?
     
  9. Aug 10, 2010 #8

    dextercioby

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    You can easily show that it's time independent, but showing it equal to 0 is a much stronger statement. Compute the integral in the case of the infinite square potential in 1 dimension. Is the integral = 0 ?
     
  10. Aug 10, 2010 #9
    yeah I can show that it's time independent. For example, consider the ground electronic state of an atom in absence of any external field. then in ground state,

    <p> = <psi * Exp[i * w * t] ] p [psi * Exp [ -i * w *t]>

    = <psi] p [psi>

    = time independent.

    but, is <p> = 0? someone told me it's = 0 but I can't understand why.

    Also, someone please tell me if my previous post is right or wrong.
     
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