# Parity of spherical harmonics

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 $$$$Y_{\ell m}(\theta, \varphi) = \sqrt{\frac{2\ell + 1}{4\pi}\frac{(\ell-|m|)!}{(\ell+|m|)!}}\, P_\ell^m(\cos\theta)e^{\mathrm{i}m\varphi}.$$$$ For the associated Legendre polynomials, I follow the convention $$$$P_\ell^m(x) = \frac{(-1)^m}{2^\ell \ell!} (1-x^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}x^{\ell+m}}(x^2-1)^\ell.$$$$

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 $$(-x)^2=x^2$$.

Remember you are also taking the derivative, so you must apply the change rule, e.g, call $$z = -x$$
$$\begin{eqnarray} P_\ell^m(z) &=& P_\ell^m(-x) \\ &=& \frac{(-1)^m}{2^\ell \ell!} (1-z^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}z^{\ell+m}}(z^2-1)^\ell \\ &=& \frac{(-1)^m}{2^\ell \ell!} (1-(-x)^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}(-x)^{\ell+m}}((-x)^2-1) \\ &=& (-1)^{\ell + m} P_\ell^m(x) \end{eqnarray}$$
• Thanks! Why can you not just replace $x$ with $-x$ behind the derivative? Could you give me a more detailed explanation? Oct 30 '20 at 17:41
• @physicist23 Do you mean something like $$\frac{{\rm d}}{-{\rm d} x^{\ell + m}}$$ ? Oct 30 '20 at 17:44
• 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 $f'=g'$. Oct 30 '20 at 17:47
• @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 $-x$ Nov 1 '20 at 14:37
• @ryan1 Indeed, you can get that from applying the chain rule $l + m$ times Mar 16 at 12:04