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Finding eigenfunctions of the linear momentum operator

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    I'm just totally lost with this question. The theory just eludes me totally. Just how do you determine whether it is/isn't an eighenfunction of the linear momentum operator?
     
  2. jcsd
  3. Mar 8, 2008 #2

    kdv

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    A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction).

    So apply the operator [tex] p_x [/tex] on that wavefunction and see if the result is a constant times the initial wavefunction
     
  4. Mar 9, 2008 #3
    In other words, one of the postulates of QM relates observables by eigenfunctions and eigenvectors. For example

    [tex]\hat{H}\psi = E \psi[/tex]

    would be the hamiltonian operator giving energy eigenvalues. You could also have

    [tex]\hat{x_o}\psi = x_n \psi[/tex]

    or more relevantly

    [tex]\hat{p} \psi = p_n \psi[/tex]

    In the end, do exactly what kdv told you, see if when you apply the operator you get the function back with a constant.
     
  5. Mar 9, 2008 #4
    pardon my lack of knowledge, but what is the linear momentum operator?
     
  6. Mar 9, 2008 #5

    Hootenanny

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    The linear momentum operator is a partial differential operator and in three dimensions has the form,

    [tex]\hat{p} = -i\hbar\nabla[/tex]

    And in one dimension,

    [tex]\hat{p} = -i\hbar\frac{d}{d x}[/tex]

    So for example, in 3D

    [tex]\hat{p}\left(xyz\right) = -i\hbar\left(yz,xz,xy\right)[/tex]

    So as both Mindscrape & kdv have said, all you need to do is apply the operator to the wave function and see if you get the wave function (eigenfunction) multiplied by a constant (eigenvalue).
     
  7. Mar 9, 2008 #6
    Ok, so I want to be using the one dimensional form. Pardon my lack of knowledge again, but what do you mean by apply?
     
  8. Mar 9, 2008 #7

    kdv

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    it means take the derivative with respect to x of the wavefunction then multiply by [itex] -i
    hbar [/itex]
     
  9. Mar 10, 2008 #8
    oh no, I havn't done calculus for a while...

    Do I just treat A and i as constants?

    If yes, by chain rule I get

    [​IMG]

    Then I multiple by i*h/2pi (although it's probably wrong, worth a try though)...?
     
    Last edited: Mar 10, 2008
  10. Mar 10, 2008 #9

    Hootenanny

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    Yes, you have taken the derivative correctly, now just multiply it by [itex]-i\hbar[/itex] and see if you can arrange it into the form,

    [tex]\hat{p}\psi = p\psi[/tex]

    Where p is constant. If you can, then [itex]\psi[/itex] is a linear eigenfunction of [itex]\hat{p}[/itex]; if not, then [itex]\psi[/itex] is not a linear eigenfunction of [itex]\hat{p}[/itex].
     
  11. Mar 11, 2008 #10
    Just one question, is i the imaginary number? i.e. i^2=-1?
     
  12. Mar 11, 2008 #11

    malawi_glenn

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    that is correct! :-)
     
  13. Mar 11, 2008 #12
    Sorry, I was looking at my help and I shouldn't have put [itex]p_n[/itex] as eigenvalues because that would imply they are discrete, but really the momentum eigenvalues are continuous, as well as the position eigenvalues. I should have distinguished with maybe capitals as operators and lower case as eigenvalues.

    [tex]
    \hat{X}\psi = x \psi
    [/tex]

    [tex]
    \hat{P}\psi = p \psi
    [/tex]

    We should really be talking about an eigenket, or state vector, that lives in Hilbert space (or its suburbs), but it would be bad to burden you with technicalities, so don't worry about it unless you want to.
     
    Last edited: Mar 11, 2008
  14. Mar 11, 2008 #13
    Ok, so after all that I end up with...

    [​IMG]

    So my constant is i*(hbar)*x + (hbar) ?
     
  15. Mar 11, 2008 #14
    Hold on there, is x a constant?
     
  16. Mar 11, 2008 #15
    In that case, it isn't an eigenfunction of the linear momentum operator?
     
  17. Mar 11, 2008 #16
    Does it satisfy the postulate?
     
  18. Mar 12, 2008 #17
    "A function psi is an eigenfunction of an operator A if applying A on psi gives the function back times some constant (and the constant is then the correspodning eigenvalue of that eigenfunction)."

    x is not a constant, therefore applying the linear momentum operator on psi does not give the wavelength function back times some constant.

    Therefore, psi is not an eigenfuction of the linear momentum operator.

    (Well, that's what I think anyway...)
     
  19. Mar 12, 2008 #18

    Hootenanny

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    Sounds good to me :approve:
     
  20. Mar 14, 2008 #19
    Excellent! Thanks
     
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