Is that possible to from the commutator relation of angular momentum to derive the coordinate representation? From
$[x,p]=i$, one could somehow show the coordinate representation of the momentum operator, e.g., in Dirac's principles of quantum mechanics section (22), as $p_x = i \frac{ \partial }{\partial x} + f(x) $.
The Stone von-Neumann theorem confirms such representation is unique up to unitary equivalence. Hence, the commutator defines everything.
Could one do similar for angular momentum? Namely entirely from the commutation relation, E.g., $[L_x,L_y] = i L_z$, without relying on neither $\mathbf{L}:= \mathbf{x} \times \mathbf{p}$ nor the rotation generator? Assume one could add some condition specifying it happens in coordinate, not spin space.
I naively guess, one may try something similar in Dirac's book, say, guess  $\mathbf{L}:= \mathbf{x} \times \mathbf{p}$ would work, and $L_x = L_x + f(z)$ would also work.
 A: No, this cannot be done.
The simple answer as to why is that the angular-momentum commutation relation, $[L_i,L_j] = i\epsilon_{ijk}L_k$, is also compatible with spin angular momentum operators $S_i$, which

*

*satisfy the same commutation relationship, but

*have half-integer quantum numbers, and therefore

*cannot be represented in the form $\mathbf S= \mathbf x \times \mathbf p$.

That concludes the proof, really.
A: I have a different opinion.
Actually commutation relationship tells you a lot about eigenstates and eigenvalues. Consider the operator $L^2=L_x^2+L_y^2+L_z^2$. Since $[L_z,L^2]=0$ (one can check this with commutation relationship), we can find a basis $\{|\psi^s_{n,m}\rangle\}$ such that
\begin{align}
L^2|\psi^s_{n,m}\rangle & = n(n+1)|\psi^s_{n,m}\rangle \\
L_z|\psi^s_{n,m}\rangle & = m|\psi^s_{n,m}\rangle
\end{align}
which is to say, $\{|\psi^s_{n,m}\rangle\}$ is simultaneously the basis of $L^2$ and of $L_z$, and note that we do not restrict any value of $n$ and $m$ here. The additional label $s$ represents possible degeneracy since there can be $\psi^0_{n,m}, \ \psi^1_{n,m}, \ \psi^2_{n,m}, \cdots$ all satisfying the above equations.
Then, consider the operators $L_+=L_x+iL_y$ and $L_-=L_x-iL_y$. Since
\begin{align}
L_zL_+|\psi^s_{n,m}\rangle & = \big((L_xL_z+iL_y)+(iL_yL_z+L_x)\big)|\psi^s_{n,m}\rangle \\
& = (L_x+iL_y)(L_z+1)|\psi^s_{n,m}\rangle \\
& = (m+1)L_+|\psi^s_{n,m}\rangle
\end{align}
and in the same way
\begin{align}
L_zL_-|\psi^s_{n,m}\rangle & = \big((L_xL_z+iL_y)+(-iL_yL_z-L_x)\big)|\psi^s_{n,m}\rangle \\
& = (L_x-iL_y)(L_z-1)|\psi^s_{n,m}\rangle \\
& = (m-1)L_-|\psi^s_{n,m}\rangle
\end{align}
we know $L_+|\psi^s_{n,m}\rangle$ and $L_-|\psi^s_{n,m}\rangle$ (if there are non-zero) are eigenstates of $L_z$ with eigenvalues $(m+1)$ and $(m-1)$.
In the end, let us determine the possible values of $n$ and $m$. Since $L_x^2+L_y^2=L_-L_++L_z=L_+L_--L_z$, if $m<n$ and $L_+|\psi^s_{n,m}\rangle=0$,
\begin{align}
\langle\psi^s_{n,m}|L^2|\psi^s_{n,m}\rangle & = \langle\psi^s_{n,m}|(L_-L_++L_z+L_z^2)|\psi^s_{n,m}\rangle \\
& = m(m+1) \neq n(n+1)
\end{align}
we have contradiction. That is to say, when $m<n$, $L_+|\psi^s_{n,m}\rangle$ must be non-zero (it is in parallel with $|\psi^s_{n,m+1}\rangle$). We can also show, in a similar way, when $m>-n$, $L_-|\psi^s_{n,m}\rangle$ is non-zero (parallel with $|\psi^s_{n,m-1}\rangle$). One can also see that since the eigenvalue of $L_+|\psi^s_{n,m}\rangle$ under the action of $L_z$, which is $(m+1)$, must not exceed $n$, the increasing process with serial actions of $L_+$ must end at $n$ with $L_+|\psi^s_{n,n}\rangle=0$. And similarly, the decreasing process must end at $-n$ with $L_-|\psi^s_{n,-n}\rangle=0$. Therefore, $n-(-n)=2n=k$ where $k$ is some non-negative integer, and
$$n \in \big\{0,{1 \over 2},1,{3 \over 2},2,\cdots\big\} \ \ \ \text{and} \ -n \leq m \leq n$$
To answer your problem, one can just pick up some orthonormal countable basis (like the basis of simple harmonic oscillator), and assign each wavefunction to $|\psi^s_{n,m}\rangle$ as he or she wishes. The operators $L_x$, $L_y$, $L_z$, by construction, exist although it may in principle very hard to express them in other frequently-used operators including position and momentum operators. Also, they mostly are not isomorphic to the usual angular momentum operators $L_{0,x}$, $L_{0,y}$ and $L_{0,z}$ up to unitary transformation. If we want the operators to be isomorphic to usual angular momentum operators, it requires $n=0,1,2,3,\cdots$ and no degeneracy simultaneously in $n$ and $m$ ($s$ can only be $0$). Given the above criteria met by $L_x$, $L_y$, and $L_z$, there would be some unitary operator $U$ such that $U|\psi^0_{n,m}\rangle=|n,m\rangle$ where $|n,m\rangle$ is the eigenstate of $L_{0,z}$ and $L_0^2=L_{0,x}^2+L_{0,y}^2+L_{0,z}^2$.
