# Using parity properties to evaluate the inner product of spherical harmonics

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?

• It is not necessarily zero (consider $l=j$, $m=k$.) May 17, 2021 at 17:11
• 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. May 17, 2021 at 17:26

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.