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Proof that the parity operator is hermitian

  1. Mar 9, 2006 #1
    hey, it's good to be back at pf. :cool:

    anyway, today i had an exam in my honors modern course, and one of the questions was a proof that the parity operator is hermitian. i don't think i got it right. :/

    here's what i did:

    1:
    [tex]
    \int(P_(op) \psi_2(x))^* \psi_1(x) dx

    = \int \psi_2^*(-x) \psi_1(x) dx. [/tex]

    and
    2:

    [tex]
    \int \psi_2^*(x) P_(op) \psi_1(x) dx

    = \int \psi_2^*(x) \psi_1(-x)dx. [/tex]


    But... that doesn't really get me anywhere. if (1) equaled (2), then i'd be satisfied, but... it doesn't appear that this is the case.


    so how would you have gone about this?
     
    Last edited: Mar 9, 2006
  2. jcsd
  3. Mar 9, 2006 #2
    What do you know about the eigenvalues of a Hermitian operator?

    -Dan
     
  4. Mar 9, 2006 #3
    they're real. (that was the next question on the exam!)

    but psi1 and psi2 needn't be eigenfunctions of the parity operator.
     
  5. Mar 9, 2006 #4

    Physics Monkey

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    Hi Brad,

    The point is that your choice of psi1 and psi2 have nothing to do with it. The easiest approach is just to show directly that the eigenvalues are real. Hint: think about what [tex] P^2 [/tex] is. Another approach is to look at things like [tex] P | x \rangle [/tex] where [tex] | x \rangle [/tex] are the position eigenstates.
     
  6. Mar 9, 2006 #5
    ah, i forgot that eigenvalues real => operator hermitian, and not just vice-versa. sucks that i forgot that direction; i showed my professor a proof of it after class on tuesday! :redface:

    oh well.
     
  7. Mar 10, 2006 #6

    Galileo

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    Is that true? What about the matrix:
    [tex]\left( \begin{array}{cc}1 & 4\\1 & 1\end{array}\right)[/tex].
    It's not Hermitian, but it has real eigenvalues 3 and -1.
     
  8. Mar 10, 2006 #7
    :/

    well, i know there's a reason why i found it tough to swallow. but that fact(?) is in my prof's lecture notes. i'll email him.
     
  9. Mar 10, 2006 #8
    I would have to look it up in my notes as well, but I suspect that the actual theorem is that any Unitary Hermitian operator must have real eigenvalues and that if you have only real eigenvalues you may construct a Unitary Hermitian matrix from them.

    -Dan
     
  10. Mar 10, 2006 #9

    Galileo

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    Ah, but if you can show P is unitary, then the result follows immediately from P^2=1.
     
  11. Mar 12, 2006 #10

    Meir Achuz

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    Just make the variable substitution x'=-x.
     
  12. Mar 20, 2006 #11
    yep, that's the way to do it, it turns out!


    also, the theorem is that operator is hermitian <=> expectation values are real.

    this, i guess, is not quite the same as "eigenvalues are real."
     
  13. Mar 21, 2006 #12

    Meir Achuz

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    [QUOTEalso, the theorem is that operator is hermitian <=> expectation values are real.
    This, I guess, is not quite the same as "eigenvalues are real."[/QUOTE]
    Proving the evs are real is a bit longer, but still simple. An expectation value can be real even if the evs are not all real.
     
  14. Nov 16, 2007 #13
    I have to do this as a homework problem now and I'm just not seeing it. I can get <Pf(x)|g(x)> = <f(x')|Pg(x')> where x' = -x, but I don't see how that's a proof, since it's comparing two different variables.
     
    Last edited: Nov 16, 2007
  15. Nov 19, 2007 #14
    I still haven't figured this one out. I tried <Pf(x)|g(x')> = <f(x')|Pg(x)>, but that's still not the same thing has having f(x) and g(x), so I don't know what else to do.
     
  16. Nov 19, 2007 #15

    Galileo

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    That P is unitary and P^2=1 is all you need. so just prove those properties.
     
  17. Nov 19, 2007 #16
    P^2 = 1 was given to me.

    I don't understand how I'd prove that it's unitary, though.

    Is it that since <Pf(x)|g(x)> makes <f(x)| -> <f(-x)| and the same for |g(x)> -> |g(-x)>, so I can multiply P* and P and get 1?
     
  18. Nov 19, 2007 #17
    well, i think it would be most illustrative to write the brackets out as integrals in position space. then you can see how the variable substitution will get you what you need.
     
  19. Nov 19, 2007 #18
    I did, got what you got in the first post, and no amount of switching around variables has helped me see the answer.
     
  20. Nov 19, 2007 #19
    ok, my tex'ing is kind of rusty...

    [tex]\int_{-\infty}^\infty(P\psi(x))^*\psi(x)dx = \int_{-\infty}^\infty\psi^*(-x)\psi(x)dx.[/tex]

    let y = -x.

    then the above becomes...

    [tex]-\int_{\infty}^{-\infty}\psi^*(y)\psi(-y)dy = \int_{-\infty}^\infty\psi^*(y)P\psi(y)dy.[/tex]

    note that the variable of integration is a dummy variable. you can return it all to the dirac bra-ket notation, if you would like. in any case, it's clear now that we proved what we wanted.
     
  21. Nov 20, 2007 #20
    Okay, but now you have a function of y's instead of x's. If I were to take it back into x's, I'd get something totally different. How can you justify setting x = y if you state that x = -y before that?
     
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