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I would like to proof $Y_{\ell m}(-\mathbf{r}) = (-1)^\ell\, Y_{\ell m}(\mathbf{r})$. In this formula, $Y_{\ell m}$ are the spherical harmonics given by \begin{equation} Y_{\ell m}(\theta, \varphi) = \sqrt{\frac{2\ell + 1}{4\pi}\frac{(\ell-|m|)!}{(\ell+|m|)!}}\, P_\ell^m(\cos\theta)e^{\mathrm{i}m\varphi}. \end{equation} For the associated Legendre polynomials, I follow the convention \begin{equation} P_\ell^m(x) = \frac{(-1)^m}{2^\ell \ell!} (1-x^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}x^{\ell+m}}(x^2-1)^\ell. \end{equation}

I already figured out that $\mathbf{r}\to -\mathbf{r}$ corresponds to $\theta\to \pi-\theta$ and $\varphi\to \pi+\varphi$ in spherical coordinates. This yields \begin{align} \cos\theta&\to -\cos\theta\\ e^{\mathrm{i}m\varphi}&\to (-1)^m\, e^{\mathrm{i}m\varphi}, \end{align} but I do not see how $P_\ell^m(-x) = (-1)^{\ell+m}\, P_\ell^m(x)$. To me, it seems like it should not change anything as $(-x)^2=x^2$.

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Remember you are also taking the derivative, so you must apply the change rule, e.g, call $z = -x$

\begin{eqnarray} P_\ell^m(z) &=& P_\ell^m(-x) \\ &=& \frac{(-1)^m}{2^\ell \ell!} (1-z^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}z^{\ell+m}}(z^2-1)^\ell \\ &=& \frac{(-1)^m}{2^\ell \ell!} (1-(-x)^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}(-x)^{\ell+m}}((-x)^2-1) \\ &=& (-1)^{\ell + m} P_\ell^m(x) \end{eqnarray}

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  • $\begingroup$ Thanks! Why can you not just replace $x$ with $-x$ behind the derivative? Could you give me a more detailed explanation? $\endgroup$ Oct 30 '20 at 17:41
  • $\begingroup$ @physicist23 Do you mean something like $$ \frac{{\rm d}}{-{\rm d} x^{\ell + m}} $$ ? $\endgroup$
    – caverac
    Oct 30 '20 at 17:44
  • $\begingroup$ No, I mean $((-x)^2-1)^\ell$ is just the same as $(x^2-1)^\ell$ and I thought I do not need to care for the derivative as $f=g$ implies $f'=g'$. $\endgroup$ Oct 30 '20 at 17:47
  • $\begingroup$ @physicist23 You're correct, $f=g$ implies $f'=g'$ but this is not what you're doing, you're calculating the derivative w.r.t another variable, namely $-x$ $\endgroup$
    – caverac
    Nov 1 '20 at 14:37
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    $\begingroup$ @ryan1 Indeed, you can get that from applying the chain rule $l + m$ times $\endgroup$
    – caverac
    Mar 16 at 12:04

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