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 $Y_l^m(\theta,0)$?
[Math] spherical harmonics for the two dimensional case
spherical harmonics
Best Answer
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.