Angular momentum and rotation symmetry

Orbital angular momentum is $\mathbf L=\mathbf r\times\mathbf p$. Its components obey $[L_x,L_y]=i\hbar L_z$ and cyclic counterparts. Small rotations about different axes therefore do not generally commute. The algebra is a quantum expression of rotation geometry, not an arbitrary complication added to the classical definition. More general angular momentum $\mathbf J$, including spin, satisfies the same component commutators.

The squared magnitude $J^2=J_x^2+J_y^2+J_z^2$ commutes with each component. We conventionally choose a common basis for $J^2$ and $J_z$. The choice of z is a coordinate or experimental reference, not a privileged direction built into empty space. If a Hamiltonian is fully rotation invariant, it commutes with the rotation generators, so angular momentum organizes its states and constrains the dynamics.

A state $|j,m\rangle$ has $J^2$ eigenvalue $\hbar^2j(j+1)$ and $J_z$ eigenvalue $\hbar m$. For fixed $j$, the allowed $m$ values run from $-j$ to $j$ in unit steps, giving $2j+1$ states. General angular momentum permits integer or half-integer $j$. For orbital wavefunctions on ordinary three-dimensional space, $l$ is a nonnegative integer and the angular eigenfunctions are spherical harmonics.

Do not confuse the total magnitude $\hbar\sqrt{j(j+1)}$ with the maximum component $j\hbar$. Their difference encodes unavoidable spread in the other components. A cone drawing may suggest this distinction, but it does not prove that an individual angular-momentum vector is literally precessing with a hidden fixed direction. The state provides outcome distributions, not three simultaneously sharp classical components waiting to be uncovered.

The operators $J_\pm=J_x\pm iJ_y$ change $m$ by one while leaving $j$ fixed. Their amplitudes vanish at the top and bottom of a multiplet. This resembles the oscillator ladder, but the physical label is different: here the ladder has only $2j+1$ rungs. An angular-momentum raising operation does not necessarily raise energy; that depends on the Hamiltonian and any external fields.

In a $|j,m\rangle$ state, rotational symmetry about z gives equal x and y second moments and zero x and y means. Subtracting the known z contribution from $\langle J^2\rangle$ then determines both spreads. This is a useful strategy: combine symmetry and algebra before attempting coordinate integrals. When a magnetic field picks an axis, it can split the energies of different $m$ values without changing these basic representation rules.