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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.

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2 Answers 2

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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

  1. satisfy the same commutation relationship, but
  2. have half-integer quantum numbers, and therefore
  3. cannot be represented in the form $\mathbf S= \mathbf x \times \mathbf p$.

That concludes the proof, really.

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  • $\begingroup$ Yes, I thought about this possibility and mentioned "Assume one could add some condition specifying it happens in coordinate, not spin space." in the question. Not sure if this condition would be legitimate. (Sakurai's Modern quantum mechanics rev edi 3.6 discusses relation between generator and $\mathbf{L} = \mathbf{x} \times \mathbf{p}$ in the context when spin-angular momentum is zero or can be ignored) $\endgroup$
    – RandomUser
    Jul 17, 2022 at 21:35
  • $\begingroup$ It is very hard to make that condition rigorous while keeping the problem meaningful. And, in addition, even when there is in fact a coordinate space representation, you can't know for sure whether a radial variable is present, or whether it is only angle variables over a sphere - in which case the commutator holds but the cross product is not definable. $\endgroup$ Jul 19, 2022 at 0:01
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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 $L_0^2=L_{0,x}^2+L_{0,y}^2+L_{0,z}^2$.

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  • $\begingroup$ Well, there must be identical infinite degeneration for every couple $(n,m)$, that is $s= 1,2, ...$. Under these requirements (including the other you wrote), there is in fact a unitary isomorphism as wanted essentially as a consequence of the Peter-Weyl theorem. These hypotheses are a bit brute force, in my view, in comparison with the Stone von Neumann theorem however. $\endgroup$ Jul 20, 2022 at 15:36
  • $\begingroup$ However a gap in your proof is the existence of the said basis. It does not arise from pure algebraic manipulations, one should assume that at least one of the vectors of the basis exists. However, using the fact that SU(2) is compact and the theorem of Peter-Weyl it is easy to prove everything you wrote with few differnces. $\endgroup$ Jul 20, 2022 at 16:06

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