Eigenfunctions of angular momentum operators in momentum representation I want to know the eigenfunctions of $L^2$ and $L_z$ in the momentum representation. Do I need to Fourier transform the spherical harmonics? 
 A: Following up on yellowquark's comment in the OP, we can explicitly show this by computing the Fourier transform of a 3D wave function.
$$\renewcommand{\vec}[1]{\mathbf{#1}}$$
We assume a wave function of the form
$$
\psi\left(  \vec{r}\right)  =f\left(  r\right)  Y_{\alpha,\beta}\left(
\Omega_{\vec{r}}\right)  .
$$
The Fourier transform of this wave function is given by
$$
\tilde{\psi}\left(  \vec{k}\right)  =\frac{1}{(2\pi)^{3/2}}\int d^{3}r\left[ e^{-i\vec{k}\cdot\vec
{r}} f\left(  r\right)  Y_{\alpha,\beta}\left(
\Omega_{\vec{r}}\right) \right] .
$$
We expand the complex exponential in spherical harmonics as
$$
e^{-i\vec{k}\cdot\vec{r}}=4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}e^{-i\pi
l/2}j_{l}\left(  kr\right)  Y_{l,m}\left(  \Omega_{\vec{k}}\right)
Y_{l,m}^{\ast}\left(  \Omega_{\vec{r}}\right)  ,
$$
where $j_{l}\left(  kr\right)  $ is a spherical Bessel function, and write
\begin{align*}
\tilde{\psi}\left(  \vec{k}\right)    & =\frac{1}{(2\pi)^{3/2}} \int r^{2}dr\int d\Omega_{\vec{r}
}\left\{
%replacing the expansion
\left[
4\pi\sum_{l=0}^{\infty}\sum
_{m=-l}^{l}e^{-i\pi l/2}j_{l}\left(  kr\right)  Y_{l,m}\left(  \Omega_{\vec{k}
}\right)  Y_{l,m}^{\ast}\left(  \Omega_{\vec{r}}\right) 
\right]
%the functiont we want to find the Fourier transform of
f(r) Y_{\alpha,\beta}\left(
\Omega_{\vec{r}}\right)  \right\}\\
%%next line
& =\frac{4\pi}{(2\pi)^{3/2}}\sum_{l=0}^{\infty}\sum
_{m=-l}^{l}\left\{e^{-i\pi l/2}Y_{l,m}\left(  \Omega_{\vec{k}}\right)  \int dr \left[r^{2} f(r)
j_{l}\left(  kr\right) \right] \int d\Omega_{\vec{r}}\left[Y_{l,m}^{\ast}\left(
\Omega_{\vec{r}}\right)  Y_{\alpha,\beta}\left(  \Omega_{\vec{r}}\right) \right]\right\} .
\end{align*}
Using the orthogonality relation for the spherical harmonics, this becomes
\begin{align*}
\tilde{\psi}\left(  \vec{k}\right)    & =\frac{4\pi}{(2\pi)^{3/2}}\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\left\{e^{-i\pi l/2}
Y_{l,m}\left(  \Omega_{\vec{k}}\right) \delta_{\alpha,l}\delta_{\beta,m} 
\int dr \left[r^{2}f(r)j_{l}\left(  kr\right)\right]
\right\}\\
& =Y_{\alpha,\beta}\left(  \Omega_{\vec{k}}\right)  e^{-i\pi\alpha/2}\sqrt{\frac{2}{\pi}}\int dr \left[r^{2}f(r)j_{\alpha}\left(  kr\right)\right].
\end{align*}
We can see directly that we get the same spherical harmonic out, now just a function of the angular variables of the momentum rather than position.
A: Yes. The Fourier transforms of the spatial representations are the momentum space representations. Don't forget to include the radial wave function when doing do.
