I'm trying to figure out if it's possible to derive a general form for the linear operator $L_z$(rotational momentum in Quantum mechanics) which is a combination of the four linear operators
$$X, Y, P_X, P_Y$$
solely based on knowing its commutators. We know that all the four operators commute except for $X$ and $P_X$, and $Y$ and $P_Y$. More specifically:
$$ [X, P_X] = I = \text{the identity function} $$
$$ [Y, P_Y] = I = \text{the identity function} $$
Where $[., .]$ is the commutator of two operators such that $[A, B] = AB - BA$. (I know in quantum mechanics the right hand side is not $1$ and is $i\hbar$, but that's irrelevant to the general idea of the problem)
We know that:
$$ L_z = \sum_{\alpha,\beta,\gamma,\delta=0}^{\infty} c_{\alpha,\beta,\gamma\,\delta} X^\alpha Y^\beta P_x^\gamma P_y^\delta + d_{\alpha, \beta, \gamma, \delta} P_x^{\alpha} P_Y^\beta X^\gamma Y^\delta $$
We also know the commutators of $L_z$ with the four operators:
$$ [L_z, X] = -Y $$
$$ [L_z, Y] = X $$
$$ [L_z, P_X] = -P_Y $$
$$ [L_z, P_Y] = P_X $$
From quantum mechanics we know (one possible solution) is $L_Z = X P_Y - Y P_X$. But how can one find the general form of $L_z$?
I derived the equation below which might come in handy:(it also works for $y$)
$$ [X^{a}, P_X^{b}] = \sum_{m=0}^{a-1}\sum_{n=0}^{b-1} \begin{pmatrix}b-1 \\ n\end{pmatrix} P_X^{b-1-n} X^{m} P_X^n $$
The center of the Weyl algebra (the algebra generated by $X, Y, P_X, P_Y$) consists of scalar multiples of the identity, so the answer is that you can recover $L$ up to a scalar multiple of the identity; in other words, the general solution is $X P_Y - Y P_X + c I$ for some scalar $c$. The argument is very similar to the argument here which shows that the Weyl algebra is simple by repeatedly applying commutators.