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How do I determine if it's a Hermitian Operator or not

  1. Jan 25, 2006 #1
    First post so please go easy on me, here goes:

    I have looked over the basic definition of what is a Hermitian operator such as: <f|Qf> = <Qf|f> but I still am unclear what to do with this definition if I am asked prove whether or not i(d/dx) or (d^2)/(dx^2) for example are Hermitian operators. I am using Griffiths' "Intro to Quantum Mechanics" and it really seems like he skips a lot of steps. Steps I need to make sure I understand. Can someone help enlighten me.
     
  2. jcsd
  3. Jan 26, 2006 #2

    Meir Achuz

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    You have to put the operator in the integral \int\psi*[id/dx]\psi dx.
    Then integrate by parts, using the BC at the endpoints. If you get the same result after integratilng by part, the operator is hermitian.
     
  4. Jan 26, 2006 #3
    This is exactly how it's done in Griffiths too. Maybe Mixil had some specific part of this in mind?
     
  5. Jan 26, 2006 #4
    I think I am a little hazy on integration by parts, so that apeears to be where my problem lies. Can someone recommend a good refresher on the matter?
     
  6. Jan 26, 2006 #5
    http://archives.math.utk.edu/visual.calculus/4/int_by_parts.3/
    I just found this on google, and it appears to be a nice introduction, and takes you through each step. Perhaps a little basic for what you want though?
    Hope it helps,
    ~Gareth
     
  7. Jan 26, 2006 #6
    OK, that somewhat helps but if I have [tex]\int \frac{d^2}{d \phi^2}d \phi[/tex] how do I evaluate that? It's been a while since calc obviously. It's the 2nd derivative thing that's throwing me. Hopefully I am not tiring you guys, cause I am certainly getting bleary. Thanks for your help this far. I am thankful for your continued assistance.
     
  8. Jan 26, 2006 #7

    George Jones

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    You can't evaluate that. On what is [itex]\frac{d^2}{d \phi^2}[/itex] operating?

    Regards,
    George
     
  9. Jan 26, 2006 #8
    I am given [tex]Q = \frac{d^2}{d \phi^2} [/tex] And am asked to prove if eigenvalues are real. In order to do that I need to determine if Q is in fact a Hermitian operator. So I am at: [tex] \int f*(\frac{d^2}{d \phi^2})g d\phi[/tex] This is integrated over 0 to 2 pi.
     
    Last edited: Jan 26, 2006
  10. Jan 26, 2006 #9

    George Jones

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    You need to use integration by parts twice. For the first integration by parts, it may be helpful to think of

    [tex]\left( \frac{d^2}{d \phi^2} \right)g[/tex]

    as

    [tex]\frac{dg'}{d \phi}[/tex]

    where

    [tex]g' = \frac{dg}{d \phi}.[/tex]

    Regards,
    George
     
  11. Jan 26, 2006 #10
    For the first integration then would [tex]v = \frac{dg}{d \phi} [/tex] and [tex] du = (\frac{df}{d\phi})^*[/tex]?

    [edit] Well I think I solved it, but it feels wrong. I got [tex]-\int f^*(\frac{d^2}{d \phi^2}) g d\phi[/tex] It the initial integral but minus. Can that be right?
     
    Last edited: Jan 26, 2006
  12. Jan 27, 2006 #11

    dextercioby

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    Assuming the operator is densly defined, can u at least identify the operator's adjoint...? Next thing to worry you is to find whether your operator is bounded or not...

    Daniel.
     
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