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Energy eigenvalues and momentum distributions

  1. Aug 2, 2007 #1
    In the quantum version of the symmetric infinite well, the energy eigenvalues are, in principle, well-determined. Why would the momentum then have a spread or distribution for a given energy eigenvalue i.e.

    [tex] \phi(p) = 1/(2\pi\hbar) \int_{-a}^{a}dx u_n (x) e^{-ipx/\hbar}[/tex]
    where u_n is the solution with wavenumber n = 1,3,5... or n = 2,4,6...


    when there is the relationship E = p^2/2m?

    So transforming the solutions into momentum space gives a distribution, while transforming the solution into energy space give the Dirac delta function I think is another way to put it?

    Is it because of the time-dependence of momentum?
     
  2. jcsd
  3. Aug 2, 2007 #2
    This is because the energy operator is now E = p^2/2m + V(x), where V(x) is the potential energy. So E and p do not commute anymore, and p is time-dependent as a result of that.

    Eugene.
     
  4. Aug 2, 2007 #3

    Hurkyl

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    I'm not convinced by that explanation. For the infinite square well, you are obliged to restrict yourself to a subset of wavefunctions upon which V(x) is the zero operator. Upon that subset, p and V(x) do commute.

    Of course, the same isn't true for a finite square well, because you have,
    [tex][V(x), p] = -i \hbar V ( \delta(x - a) - \delta(x + a) )[/tex]

    (or something like that)


    As I puzzle through the situation, I strongly suspect that you cannot expect p to resemble a well-behaved operator on this restricted space of wavefunctions.
     
    Last edited: Aug 2, 2007
  5. Aug 2, 2007 #4
    This would be completely unphysical. [p,H]=0 would mean that momentum is conserved. The same would be true in the classical limit as well. But classical momentum is definitely not conserved, because the particle is bouncing off the walls.

    Eugene.
     
  6. Aug 2, 2007 #5
    Why does that mean momentum would be conserved? I think commutation of two operators simply implied that they could not have a simultaneous eigenstate?
     
  7. Aug 2, 2007 #6

    Hurkyl

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    Well, the infinite square well has some poor mathematical properties (the set of wave functions I mentioned don't even form a Hilbert space!); it wouldn't surprise me to see other complications.

    If there is a poor classical limit, I wouldn't think that a problem -- I would simply expect it to be an issue of conflicting approximations. Note that as [itex]\hbar \to 0[/itex], the energy eigenvalues all become zero.
     
    Last edited: Aug 2, 2007
  8. Aug 2, 2007 #7
    Non-zero commutators are not simply signs of simultaneous non-measurability of observables. Commutator of an observable with the Hamiltonian fully determines its (observable's) time evolution. This follows from the Heisenberg equation

    [tex] -i \hbar \frac{\partial p}{\partial t} = [H, p] [/tex]

    Which, in turn, follows from general formula for the time evolution of observables

    [tex] p(t) = e^{\frac{i}{\hbar}Ht} p(0) e^{-\frac{i}{\hbar}Ht} [/tex]

    Eugene.
     
    Last edited: Aug 2, 2007
  9. Aug 2, 2007 #8

    Dick

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    If the walls were not there, then the E eigenfunction would be a combination of plane waves and the position space to momentum space transform would be a combination of delta functions telling you E=p^2/2m. If the walls are there just because the potential is zero in the support of the wavefunction, it still restricts the region of integration. I'm with Eugene on this. The hamiltonian doesn't commute with p.
     
  10. Aug 2, 2007 #9
    Probability

    My book states this as obvious, but what are the intermediate steps between the second equality?

    < (x - <x>)^2> = < x^2 - 2x<x>+<x>^2> = <x^2> - <x>^2
     
  11. Aug 3, 2007 #10

    Dick

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    There is an implicit sum or integration in there. <x^2-2x<x>+<x>^2>>=<x^2>-2<x><x>+<x>^2. Hence?
     
  12. Aug 3, 2007 #11
    I see. Thanks. :biggrin:
     
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