Question
Let $\vec J=(J_x,J_y,J_z)$ denote the angular momentum operator, $\hat n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ a unit vector.
What is the eigenstates of $\hat n\cdot\vec J$？
As is often the case, we can just write down the matrix element of $\hat n\cdot\vec J$ and figure out its eigenvectors. But the workload increases quickly with dimesion and there isn't a closed form to express these eigenstates.
My attempt
if $\hat n =\hat z$, then $\hat n\cdot\vec J=J_z$, whose eigenstates are $|\hat z,{jm}\rangle, m=-j,\cdots,j$ in some irreducible subspace, and $$J_z|\hat z,jm\rangle=m\hbar|\hat z,jm\rangle.$$
There must be a rotation operation $R$, which rotates $\hat z$ axis to $\hat n$ axis, and the corresponding operator is denoted by $\hat U(R)$. So $$ |\hat n,jm\rangle=\hat U(R)|\hat z,jm\rangle $$ is an eigenstate of $\hat n\cdot\vec J$ with eigenvalue $m\hbar$. Above equation can be expanded as $$ |\hat n,jm\rangle= \sum_{m'}|\hat z,jm\rangle \langle \hat z,jm|\hat U(R)|\hat z,jm\rangle =\sum_{m'}D^j_{m'm}(R) |\hat z,jm\rangle $$ where $$ D^j_{m'm}(R)= \langle \hat z,jm'|\hat U(R)|\hat z,jm\rangle=e^{-\mathrm i m'\alpha}d^j_{m'm}(\beta)e^{\mathrm im\gamma}, $$ where $\alpha,\beta,\gamma$ are Euler angles and $d^j_{m'm}$ is Wigner's (small) d-matrix.
In my case, $\gamma$ can be chosen to be $0$ because it leads to only a phase shift to $|\hat n,jm\rangle$. And $$ \beta = \arccos(\hat z\cdot\hat n), \quad \alpha = \mathrm{sgn}(\hat n\cdot\hat y) \arccos[(\hat n\times\hat z)\cdot\hat x]. $$
The expression of $d^j_{m'm}$ is quite complicated, too. Is there any simpler way to solve this problem? For example, in the case of dimension 2, 3 or 4.
