Can the quantum angular momentum operator be derived from its commutation relations with position and momentum?

Question

Exercise 12.2.2 in Shankar's Principles of Quantum Mechanics asks to derive the expression for the angular momentum operator $L_z$ \begin{equation} L_z = XP_y-YP_x \end{equation} using its commutation relations with the coordinate and momentum operators \begin{align} [X,L_{z}]&=-i\hbar Y\\ [Y,L_{z}]&=i\hbar X \\ [P_{x},L_{z}]&=-i\hbar P_{y}\\ [P_{y},L_{z}]&=i\hbar P_{x} \end{align} I have seen a solution which uses the coordinate representations of the momenta.

However, I would like to find one that relies purely on the abstract relations.

All I have been able to prove so far is that $L_z$ and $XP_y-YP_x$ commute. Multiply the first relation on the right by $P_{y}$ and the third relation \begin{align} [X,L_{z}]P_{y}&=-i\hbar YP_{y}\\ Y[P_{x},L_{z}]&=-i\hbar YP_{y} \end{align} Equating the LHS's we obtain \begin{align} (XL_{z}-L_{z}X)P_{y}&=Y(P_{x}L_{z}-L_{z}P_{x})\\ XL_{z}P_{y}-L_{z}XP_{y}&=YP_{x}L_{z}-YL_{z}P_{x} \end{align} Now subtract $XP_{y}L_{z}$ from both sides \begin{align} XL_{z}P_{y}-XP_{y}L_{z}-L_{z}XP_{y}&=YP_{x}L_{z}-XP_{y}L_{z}-YL_{z}P_{x}\\ X[L_{z},P_{y}]-L_{z}XP_{y}&=(YP_{x}-XP_{y})L_{z}-YL_{z}P_{x}\\ \end{align} Add $L_{z}YP_{x}$ to both sides \begin{align} X[L_{z},P_{y}]+L_{z}YP_{x}-L_{z}XP_{y}&=(YP_{x}-XP_{y})L_{z}+L_{z}YP_{x}-YL_{z}P_{x}\\ X[L_{z},P_{y}]+L_{z}(YP_{x}-XP_{y})&=(YP_{x}-XP_{y})L_{z}+[L_{z},Y]P_{x}\\ [L_{z},XP_{y}-YP_{x}]&=X[P_{y},L_{z}]-[Y,L_{z}]P_{x} \end{align} However, using the commutation relations we deduce that $X[P_{y},L_{z}]=[Y,L_{z}]P_{x}=i\hbar XP_{x}$, therefore \begin{equation} [L_{z},XP_{y}-YP_{x}]=0 \end{equation}

Any other manipulations left me spinning in logical circles. I would appreciate any help in completing the argument.

Answer 1

What you want is not really possible. The reason is that the angular momentum of a particle may have a spin component, or there may be other particles for which you must also include the angular momentum. More specifically, all you may conclude from symmetry arguments is that in a rotationally invariant theory there exists a pseudovector operator $\hat{\mathbf{J}}$ whose commutation relations with the position and momentum components of any one particle in the system are the ones you posted. It will typically include the orbital angular momentum of the particle, $\hat{\mathbf{L}}$, as well as other operators such as spin which will commute with all position and momentum operators.

This, of course, obviates the fact that if you only do have the orbital degrees of freedom of a single particle, then there is nothing there that will have more angular momentum, and the equality you want should follow.

The way to go about proving that is the following. You begin with a total angular momentum operator $\hat{\mathbf{J}}$ about which you know only its commutation relations with $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$. You then construct the orbital angular momentum operator $\hat{\mathbf{L}}$ and show that it has the same commutation relations with $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$ as $\hat{\mathbf{J }}$. This means, then that $\hat{\mathbf{J }}-\hat{\mathbf{L}}$ commutes with all components of $\hat{\mathbf{r}}$ and $\hat{\mathbf{ p}}$.

Now, if your system really does consist of a single particle, then it carries the direct sum of three irreducible representations of the Heisenberg group, the algebra of which is of course spanned by the components of $\hat{\mathbf{r}}$ and $\hat{\mathbf{p}}$. This means that you can apply Schur's lemma to conclude that each component of $\hat{\mathbf{J }}-\hat{\mathbf{L}}$ must be a multiple of the unit operator.

Finally, in an isotropic system, such a vector would break the global rotational symmetry, and must therefore be zero.