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I know there exists the property:

$$Y_\ell^m(0,\phi)=\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m0}$$

but can it be applied equally as simply to

$$Y_\ell^{m-1}(0,\phi)\stackrel{?}{=}\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m1}$$

or

$$Y_\ell^{m+1}(0,\phi)\stackrel{?}{=}\sqrt{\frac{2\ell+1}{4\pi}}\delta_{m(-1)}$$

or am I missing something?

Thanks in advance

Karl
  • 217

1 Answers1

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You can definitely do so. You've only changed $m$ to $m\pm1$, shifting the value of $m$ that yields a zero upper index. The statement that

$$\begin{cases}Y_{\ell}^{0}\left(0,\varphi\right)=\sqrt{\frac{2\ell+1}{4\pi}}\\Y_{\ell}^{\rm non-zero\:number}\left(0,\varphi\right)=0\end{cases}$$

is kept unchanged.

eranreches
  • 5,903