We know that in general spherical harmonics of a unit vector $\hat{\mathbf{r}}$ is $Y_l^m(\hat{\mathbf{r}})=Y_l^m(\theta,\phi)$. I am interested to know what happens to this sperical harmonics if the dimension of the problem is changed to two dimension. Is it effectively the same as writing $Y_l^m(\theta,0)$?


1 Answer 1


For the 1D case (a circle in 2D), it turns out to be the classical Fourier basis. More specifically, the Laplace-Beltrami operator $\Delta_{\mathbb{S}^1}$ has eigenvalues $\lambda_k=-k^2$. Each one has eigenfunctions $w_1(\theta)=\cos(k\theta)$ and $w_2(\theta)=\sin(k\theta)$ (i.e. the classic Fourier basis). These form the "circular harmonics" (see here).

Concerning your question, recall that: $$ Y_\ell^m(\theta,\phi) = \mathcal{N}(\ell,m)\exp(im\theta)\,P^m_\ell[\cos(\phi)] $$ for some normalization $\mathcal{N}$. From this you can see $\phi=0$ won't quite be as you expect.

See Notes on Spherical Harmonics and Linear Representations of Lie Groups from Jean Gallier for more info.

Note that general spherical harmonics are defined on $\mathbb{S}^n$; i.e. in $n+1$-dimensional space. See Pseudodifferential Operators and Spectral Theory by Shubin, at the end of ch. 3 for instance.


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