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Is (d^2/dx^2) a Hermitian Operator?

  1. Oct 15, 2006 #1
    Hi, I'm doing a Quantum mechanics and one of my question is to determine if [tex]\frac{d^2}{dx^2}[/tex] (a second derivative wrt to x) is a Hermitian Operator or not.

    An operator is Hermitian if it satisfies the following:
    [tex]\int_{-\infty}^{\infty}\Psi^{*}A\Psi = \int_{-\infty}^{\infty}\left(A\Psi\right)^{*}\Psi[/tex]
    where [tex] \Psi^{*} [/tex] is a complex conjugate of the wavefunction psi and A is the operator.

    I do the LHS of the equation with A= [tex]\frac{d^2}{dx^2}[/tex]. I get:
    [tex]\left[\Psi^{*} \frac{d}{dx}\Psi - \frac{d}{dx}\Psi^{*}\Psi\right]_{-\infty}^{\infty} + \int_{-\infty}^{\infty}\left(A\Psi\right)^{*}\Psi [/tex]
    using integration by parts twice.

    The second term obviously is the RHS, and in order for [tex]\frac{d^2}{dx^2}[/tex] to be Hermitian (and I'm pretty sure it is), the first term has to be equal to zero. Can anyone justify?

    Thanks a bunch!
    -Rick
     
    Last edited: Oct 16, 2006
  2. jcsd
  3. Oct 16, 2006 #2

    StatusX

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    Psi has to vanish at infinity to be square integrable.
     
  4. Oct 16, 2006 #3
    kcirick, you might be interested in this thread: https://www.physicsforums.com/showthread.php?t=120888
     
  5. Oct 16, 2006 #4
    OK, Thanks to both of you.

    I wish I bought the Griffiths textbook. Our prof's using "Quantum Mechanics" by B.H. Bransden & C.J. Joachain. I find it not too helpful and I've heard many good things about Griffiths' texbook.
     
  6. Oct 16, 2006 #5

    Hurkyl

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    Shortcut: A is a real multiple of P².
     
  7. Oct 18, 2009 #6
    Here is an easier procedure for proving that the second derivative (wrt to x) is Hermitian. And I just discovered this!
    1) Prove that the momentum operator is Hermitian. (it involves first derivative)
    2) Prove that the operator aA (where a is some number and A is a hermitian operator) is Hermitian only when a is .......... (I will not tell you what that condition is, it isn't that difficult to prove)
    3) Relate the momentum squared operator to the second derivative operator and use the condition in step 2 to show that second derivative operator is Hermitian.

    Edit: Hurky! already mentioned it, which I didn't see :)
     
  8. Nov 30, 2009 #7
    Hello, I have a question about the definition of Hermtian operator.

    Which one is correct?
    This:
    [itex]
    \langle Ax|y\rangle\ =\ \langle x|Ay\rangle
    [/itex]
    or this:
    [tex]
    \int_{-\infty}^{\infty}\Psi^{*}A\Psi = \int_{-\infty}^{\infty}\left(A\Psi\right)^{*}\Psi
    [/tex]

    Everytime I see the definition of Hermitian I see the first one (where they use x and y, different to each other). But when someone wants to prove if an operator is Hermitian (just like they did in this post) they use the second one (where they use only one Psi, not two different functions).

    Thanks
     
  9. Nov 30, 2009 #8

    Hurkyl

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    For operators on the "usual" Hilbert space of wavefunctions, L2(R), those are exactly the same thing.

    The first one applies to any Hilbert space, though.
     
  10. Dec 1, 2009 #9
    mhm I still don't get why they use different functions on the first definition but the same function on the second one.

    Thanks
     
  11. Dec 1, 2009 #10

    Hurkyl

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    Oh! I didn't notice that your second expression was [itex]\langle A \Psi \mid \Psi \rangle = \langle \Psi \mid A \Psi\rangle[/itex]; I thought you just wrote [itex]\langle A \Psi \mid \Phi \rangle = \langle \Psi \mid A \Phi \rangle[/itex] in integral form.


    There's a neat trick for converting between quadratic and bilinear forms:
    [tex](A(x+y), x+y) = (Ax, x) + (Ay, y) + (Ax, y) + (Ay, x)[/tex]

    [tex](x+y, A(x+y)) = (x, Ax) + (y, Ay) + (x, Ay) + (y, Ax)[/tex]​

    Equating the two gives
    [tex](Ay, x) + (Ax, y) = (x, Ay) + (y, Ax)[/tex]​


    And then we apply tricks for splitting complexes apart. Substituting x -> ix:
    [tex]i (Ay, x) - i (Ax, y) = -i (x, Ay) + i (y, Ax)[/tex]

    [tex](Ay, x) - (Ax, y) = -(x, Ay) + (y, Ax)[/tex]​

    and adding to the original:
    [tex]2(Ay, x) = 2(y, Ax)[/tex]​

    Assuming I haven't made any mistakes.



    Note that we are dealing with complex inner products is essential. The identity
    <x,Ax> = <Ax,x>​
    holds for all operators on any finite-dimensional real inner product space. (Maybe for infinite-dimensional too?)
     
  12. Dec 1, 2009 #11
    Now I get it. Thank you very much for your help.
     
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