I would like to know how to argue if the inner product of two spherical harmonics is zero using symmetry arguments. If the inner product is given by the following integral,

$$\left\langle Y_{\ell}^{m},Y_{j}^{k}\right\rangle=\int_0^{2\pi}\int_0^{\pi} \left(Y_{\ell}^{m}(\theta, \varphi)\right)^*Y_{j}^{k}(\theta, \varphi)\sin\theta\,\mathrm d\theta\,\mathrm d \phi$$

and keeping in mind that theire parity is such that $$ Y_{\ell}^{m}(\pi-\theta, \pi+\phi)=(-1)^{\ell} Y_{\ell}^{m}(\theta, \phi) $$

How could this feature be used?

  • $\begingroup$ It is not necessarily zero (consider $l=j$, $m=k$.) $\endgroup$
    – K.defaoite
    May 17, 2021 at 17:11
  • $\begingroup$ Ok, but I mean that there may be some values for $\ell,m,k$ and $j$ such that we know a priori if the integral is going to be zero because of the parity properties, without having to compute it. $\endgroup$
    – Invenietis
    May 17, 2021 at 17:26

1 Answer 1


It can be shown that the inner product of two spherical harmonics $\left\langle Y_{\ell_1}^{m_1},Y_{\ell_2}^{m_2}\right\rangle$ cancels out whether

  1. if $m_2-m_1 \in \mathbb{Z}\setminus \{0\}$ or
  2. if $\ell_1+\ell_2$ is odd.

The first condition, $m_2 -m_1\in \mathbb{Z}\setminus \{0\}$, arises from the fact that $Y_{\ell}^{m}=\Phi_m(\varphi)\Theta_{l,m}(\theta)$, with $\Phi_m(\varphi)=\frac{1}{\sqrt{2\pi}}e^{im\varphi}$, so

$$\int_0^{2\pi} \left( \Phi_{m_1}(\varphi) \right)^* \Phi_{m_2}(\varphi) \mathrm d \phi= \frac{1}{{2\pi}} \int_0^{2\pi} e^{i(m_2-m_1)\varphi} \mathrm d \phi$$

which is zero if $m_2 -m_1$ is a non-zero integer.

You can get to the second one by making a change of variable $\tilde\theta = \pi-\theta$ and $\tilde\varphi = \pi+\varphi$ in the integral and applying the parity property of the spherical harmonics.


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