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Spherical harmonics and P operator

  1. Dec 14, 2009 #1
    Let's define operator P:
    P \phi(\vec{r})=\phi(-\vec{r})

    Does anyone know simple and elegant prove that [tex]P|lm\rangle = (-1)^l |lm\rangle[/tex]
    ([tex]|lm\rangle[/tex] is spherical harmonic).
  2. jcsd
  3. Dec 14, 2009 #2
    I don't think this is true.

    |1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\theta)e^{i\phi}
    P|1 1\rangle = (3/(8\pi))^{(1/2)}sin(\theta)e^{-i\phi}

    The [tex]\phi[/tex] component screws up the relation.
  4. Dec 14, 2009 #3
    The spherical harmonics are a product of the associated Legendre function [tex] P_{l}^{m}(x)[/tex] and the azimuthal exponential. Under parity [tex](r, \theta, \phi) [/tex] changes to [tex](r,\pi-\theta, \phi\pm\pi)[/tex], which means that [tex]cos \theta[/tex] changes to [tex]-cos \theta[/tex]

    From Rodrigos formula [tex]P_{l}^{m}(x) \mbox{ }\alpha \mbox{ } (1-x^2)^{\frac{m}{2}} \frac{d^{l+m}}{dx^{l+m}}(1-x^2)^l [/tex]

    where x is [tex]cos \theta[/tex]. Under x-->-x this changes by [tex](-1)^{m+l}[/tex]. However, the azimuthal exponential changes by: [tex]e^{i m (\phi \pm \pi)}=(-1)^m e^{i m \phi}[/tex].

    Multiplying the azimuthal and polar parts together, the result changes by [tex](-1)^{l+m}(-1)^m=(-1)^l[/tex]
  5. Dec 14, 2009 #4
    In spherical coordinates P acts as follows:
    P \psi(r,\theta,\phi)=\psi(r,\pi-\theta,\phi+\pi)
    (Am I right?)
    P|1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\pi-\theta)e^{-i(\phi+\pi)} =- |1 1\rangle
  6. Dec 14, 2009 #5
    Ah, OK. Thanks for the correction. I was going outside the domain of [tex]\theta[/tex] and [tex]\phi[/tex] and the [tex]\phi[/tex] transformation was just wrong.
    Last edited: Dec 14, 2009
  7. Dec 14, 2009 #6
    Maybe somone knows more "physical" prove??? I mean prove which doesn't use of explicite formlua for spherical harmonics.
  8. Dec 14, 2009 #7
    Maybe somone knows more "physical" prove??? I mean prove which doesn't use explicite formlua for spherical harmonics.
  9. Dec 14, 2009 #8
    Parity and rotations commute (why?). Since parity and rotations commute, by Schur's lemma that means all states of total angular momentum [tex]\ell [/tex] have the same parity. So parity does not depend on the quantum number 'm', but only '[tex]\ell [/tex]' (if you're not familiar with Schur's lemma then the fact that rotations commute with parity implies that angular momentum commutes with parity, so the raising and lowering operators that are built from the angular momentum operators commute with parity - therefore states with different m but same [tex]\ell [/tex] have the same parity). An easy test for parity is to test the m=0 or m=[tex]\ell [/tex] spherical harmonic. The m=0 spherical harmonic coincides exactly with the Legendre function, which is odd if [tex]\ell[/tex] is odd and even if [tex]\ell[/tex] is even.

    The important point is that you can test m=anything to determine the parity for all [tex]\ell[/tex]. The spherical harmonics for m=[tex]\ell[/tex] takes on the simple form: [tex] sin^{\ell} \theta e^{i\ell \phi}[/tex] and that's all you have to test for parity.
  10. Dec 15, 2009 #9


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    All spherical harmonics with m ne 0 can be obtained by rotation and linear combination from the one with m=0 so it is sufficient to look at the function with m=0 (thats the Schur's lemma stuff of RedX). The harmonic with m=0 and l can be obtained from z^l by orthogonalization to the functions with l'<l. As under parity z->-z, the function transforms as z^l ->(-z)^l.
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