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Parity of wavefunction.

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data
    The wavefunction describing state of a system is,
    [tex]\psi (r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}%
    \right) ^{3/2}\frac{r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi }[/tex]
    Find the parity of system in this state.


    3. The attempt at a solution

    [tex]\psi (-r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}%
    \right) ^{3/2}\frac{-r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi }
    =-\psi (r,\theta ,\phi )[/tex]
    odd parity.


    I'm wrong.But donno where. Pls help.
     
  2. jcsd
  3. Feb 6, 2012 #2

    vela

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    In spherical coordinates, the parity operator takes
    \begin{align*}
    r &\to r \\
    \theta &\to \pi-\theta \\
    \phi &\to \phi+\pi
    \end{align*}
     
  4. Feb 6, 2012 #3
    you mean r →-r ?

    Then that gives me the perfect answer.
    Thank you very much dear friend.
     
  5. Feb 6, 2012 #4

    vela

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    No, r goes to +r.
     
  6. Feb 6, 2012 #5
    How is that possible?
    is it bcz r^2 = x^2+y^2+z^2?
    But the reflection changes direction of vector also na?

    Can you please explain?
     
  7. Feb 6, 2012 #6

    vela

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    What?
     
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