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Hermitian Operator in Inner Product

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int d^{3} \vec{r} ψ_{1} \hat{A} ψ_{2}[/itex] = [itex]\int d^{3} \vec{r} ψ_{2} \hat{A}* ψ_{1}[/itex]


    Hermitian operator A, show that this condition is equivalent to requiring [itex] <v|\hat{A}u>[/itex] = [itex]< \hat{A}v|u>[/itex]


    2. Relevant equations

    I changed the definitions of ψ into their bra-ket forms first of all.

    Hints say something about the Identity operator, but I don't have any bra's in my equation, what do I do?

    3. The attempt at a solution

    After changing the ψ into their bra-ket forms and substituting, I am stumped. Any pointers please?

    I have introduction into adjoint hermitian operators but I cannot see how this would fit into it.

    I would write more about what I did, but your equation editor is very difficult to use, is it possible that you could make a code that mirror's the equation editor on Microsoft Word 2007 - 2011?


    Please help me, I'm doing this problem WAY ahead of time and I just want to be good at this stuff. Please don't ignore me, just give me a push in the right direction and I promise I will do all the rest and show you what I did.
     
  2. jcsd
  3. Oct 16, 2011 #2
    Help?
     
  4. Oct 17, 2011 #3

    vela

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    It would help if you showed what you did rather than just describe it.
     
  5. Oct 17, 2011 #4
    Latex takes forever to use... Well here goes:

    We are supposed to assume that [itex]ψ_{1}*[/itex] is a wavefunction, so we use a trick to show just that, whilst the conjugate is now something wierd

    [itex]ψ_{1}* = < \vec{r}|ψ_{1} >[/itex]

    [itex]ψ_{1} = < ψ_{1}|\vec{r} >[/itex]

    So subst into

    [itex] \int d^{3} \vec{r} < ψ_{1}|\vec{r} > \hat{A} < \vec{r}|ψ_{2} >[/itex] = [itex]\int d^{3} \vec{r} < \vec{r}|ψ_{2} > \hat{A}* < ψ_{1}|\vec{r} >[/itex]

    ([itex] | \vec{r} > < \vec{r} | [/itex]

    Left side can have the r thingy rearranged in terms of identity bra-ket which is 1, so we get:

    [itex]< ψ_{1}|\hat{A}|ψ_{2} >[/itex] = [itex]\int d^{3} \vec{r} < \vec{r}|ψ_{2} > \hat{A}* < ψ_{1}|\vec{r} >[/itex]

    Right side becomes:

    [itex]| ψ_{2}>\hat{A}*< ψ_{1}|[/itex]

    Swap around the functions on the right side to get:

    [itex]< ψ_{1}|\hat{A}|ψ_{2} >[/itex] = [itex]< ψ_{2}|\hat{A} *|ψ_{1} >[/itex]

    Remembering that [itex] \hat{A} = \hat{A}*[/itex] for an hermitian function. Dunno whether this is correct.

    Latex is so hard to use, anything easier?
     
    Last edited: Oct 17, 2011
  6. Oct 17, 2011 #5

    vela

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    Can you explain how you went from
    [tex]\int d^3\vec{r}\,\langle\psi_1\vert\vec{r}\rangle\hat{A}\langle\psi_2\vert\vec{r}\rangle[/tex]to [itex]\langle\psi_1\vert\hat{A}\vert\psi_2\rangle[/itex]?
     
  7. Oct 17, 2011 #6
    That's not what I did..

    I went from [itex] \int d^{3} \vec{r} < ψ_{1}|\vec{r} > \hat{A} < \vec{r}|ψ_{2} > [/itex] to [itex]< ψ_{1}|\hat{A}|ψ_{2} > [/itex]

    I used that identity relation with the bra of the r facing the ket of the r... it equals 1.
     
  8. Oct 17, 2011 #7

    vela

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    Sorry, that was a typo. The thing I don't like is you have the operator sandwiched between the [itex]|\vec{r}\rangle[/itex] and [itex]\langle \vec{r}|[/itex]. It's not clear that you can just then eliminate the pair from the expression. It would be better if you wrote it as follows:
    \begin{align*}
    \int d^3\vec{r}\, \psi_1^* \hat{A} \psi_2 &= \int d^3\vec{r}\, \psi_1^* (\hat{A} \psi_2) \\
    &= \int d^3\vec{r}\, \langle\psi_1 | \,\vec{r}\rangle\langle\vec{r}\, |\, \hat{A}\psi_2\rangle \\
    &= \langle \psi_1 |\, \hat{A}\psi_2 \rangle
    \end{align*}
     
  9. Oct 18, 2011 #8
    Nicely done broslice, I assume I do the same thing for the right side?

    EDIT:

    Done, thanks alot, since it's hermitian, the conjugate remains the same as normal and the condition [itex] <v|\hat{A}u> [/itex] = [itex] < \hat{A}v|u> [/itex] is met.

    Can you help me on my other topic? Just a point in what to do with the probabilities, thank you in advance.
     
    Last edited: Oct 18, 2011
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