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Hermitian Operators

  1. Feb 7, 2007 #1
    1. The problem statement, all variables and given/known data
    Within the framework of quantum mechanics, show that the following are Hermitian operators:

    a) [tex]p=-i\hbar\bigtriangledown[/tex]

    b) [tex]L=-i\hbar r\times\bigtriangledown[/tex]

    Hint: In Cartesian form L is a linear combination of noncommuting Hermitian operators.
    2. Relevant equations

    [tex]\int\psi_{1}^{*}\L\psi_{2}d\tau=\int(\L\psi_{1})^{*}\psi_{2}d\tau[/tex]

    3. The attempt at a solution
    I understand that a Hermitian operator is self-adjoint, and that it's eigenvalues are real, but as far as proving it, I'm not exactly sure how to use the formula above to do that.
     
    Last edited: Feb 7, 2007
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  3. Feb 7, 2007 #2

    Tom Mattson

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    You need to insert the operator into each side of the identity separately and show that they both work out to the same quantity. That means you'll be manipulating one integrand until it looks like the other one.
     
  4. Feb 7, 2007 #3

    Dick

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    And consider integration by parts to move the derivative operators.
     
  5. Feb 7, 2007 #4
    [tex]\int\psi^{*}(\frac{\delta\psi_{2}}{\delta x}+\frac{\delta\psi_{2}}{\delta y}+\frac{\delta\psi_{2}}{\delta z})d\tau[/tex]

    So do I start off like this? If [tex]\psi[/tex] is a function of x, y and z then how do I handle the integration by parts? (The textbook does it, but only dependent on one variable)
     
  6. Feb 7, 2007 #5

    Dick

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    You do it just the same way as with one variable. (?) I'm not sure what is confusing you. Split the integral into three separate parts if you need to.
     
  7. Feb 7, 2007 #6

    Tom Mattson

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    That expression in parentheses in the integrand doesn't look much like [itex]\nabla(\psi_2)[/itex] to me.

    You're working in Cartesian coordinates, which all appear symmetrically in your integrand. So you have the luxury of working out the 1D problem (say, in x) and then saying "Similarly, we have for y and z..."
     
  8. Feb 7, 2007 #7
    I think I got it,

    [tex]\int\psi_1^*(-i\hbar\bigtriangledown\psi_2)d\tau=-i\hbar\psi_1^*\psi_2+\int i\hbar\bigtriangledown\psi_1^*\psi_2 d\tau[/tex]

    Since [tex]\psi_1^*\psi_2=0[/tex] (eigenfunctions are orthogonal) then

    [tex]\int\psi_1^* (-i\hbar\bigtriangledown\psi_2)d\tau=\int i\hbar\bigtriangledown\psi_1^*\psi_2 d\tau[/tex]

    Which is the same as above equation. Is this right?
     
  9. Feb 7, 2007 #8

    Dick

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    I'm afraid it's not orthogonality. The expression without an integral sign on it should be a difference betwen its values at + and - infinity. What is the the answer?
     
  10. Feb 7, 2007 #9
    So it's zero because it's basically [tex]-\infty+\infty[/tex]?
     
  11. Feb 7, 2007 #10

    Dick

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    Nooooo. You usually assume wave functions vanish at infinity or have some other similar boundary condition.
     
  12. Feb 7, 2007 #11

    Tom Mattson

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    Also, here's another nitpick.

    This:

    Should be carried one step further to this:

    [tex]\int\psi_1^* (-i\hbar\bigtriangledown\psi_2)d\tau=\int (-i\hbar\bigtriangledown\psi_1)^*\psi_2 d\tau[/tex]
     
  13. Feb 7, 2007 #12
    I did part b) using the same method, but only the x component. (Since from cross product I get a vector.) Am I required to prove for each component, or is the linear combination hint enough to extend this proof to all components?
     
  14. Feb 7, 2007 #13

    Tom Mattson

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    Like I said, the symmetry of Cartesian coordinates affords you the luxury of generalizing your argument for one component to the others with a minimum of trouble. You don't have to do it all again.
     
  15. Feb 7, 2007 #14
    Thank you!
     
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