01 Jun 2026
Angular Momentum Operators
Commutation relations, ladder operators, and the algebraic structure of angular momentum in quantum mechanics.
In classical mechanics, angular momentum is defined as $\mathbf L=\mathbf r\times\mathbf p$. In quantum mechanics, position and momentum become operators, so angular momentum also becomes an operator. Because the basic operators do not all commute, the components of angular momentum do not commute with one another.
This non-commutativity is the central point. It explains why $J^2$ and one component, usually $J_z$, can be specified together, but $J_x$, $J_y$, and $J_z$ cannot all have sharp values in the same state.
Orbital angular momentum
For a particle with position operator $\mathbf r$ and momentum operator $\mathbf p$, the orbital angular momentum operator is
\[\mathbf L=\mathbf r\times \mathbf p.\]In Cartesian components,
\[L_x=yp_z-zp_y,\qquad L_y=zp_x-xp_z,\qquad L_z=xp_y-yp_x.\]The canonical commutation relations are
\[[x_i,p_j]=i\hbar\delta_{ij},\qquad [x_i,x_j]=0,\qquad [p_i,p_j]=0.\]Using these, one obtains the angular momentum algebra
\[\boxed{ [L_x,L_y]=i\hbar L_z,\qquad [L_y,L_z]=i\hbar L_x,\qquad [L_z,L_x]=i\hbar L_y. }\]In compact notation,
\[\boxed{ [L_i,L_j]=i\hbar\epsilon_{ijk}L_k. }\]Here $\epsilon_{ijk}$ is the Levi-Civita symbol.
The result follows directly from the canonical commutation relations. Thus angular momentum algebra is not an extra assumption for orbital motion; it is a consequence of the quantum relation between position and momentum.
Total angular momentum algebra
The same algebra is satisfied by any angular momentum operator $\mathbf J$:
\[\boxed{ [J_i,J_j]=i\hbar\epsilon_{ijk}J_k. }\]This includes orbital angular momentum $\mathbf L$, spin angular momentum $\mathbf S$, and total angular momentum
\[\mathbf J=\mathbf L+\mathbf S.\]Commutation with $J^2$
Define
\[J^2=J_x^2+J_y^2+J_z^2.\]Although the components of $\mathbf J$ do not commute with each other, $J^2$ commutes with every component:
\[\boxed{ [J^2,J_x]=[J^2,J_y]=[J^2,J_z]=0. }\]Therefore $J^2$ and one chosen component, conventionally $J_z$, can have simultaneous eigenstates.
Ladder operators
Define the raising and lowering operators
\[J_+=J_x+iJ_y,\qquad J_-=J_x-iJ_y.\]They satisfy
\[\boxed{ [J_z,J_\pm]=\pm\hbar J_\pm }\]and
\[\boxed{ [J_+,J_-]=2\hbar J_z. }\]The operators $J_+$ and $J_-$ change the magnetic quantum number $m$ without changing $j$.
Useful identities
The ladder operators are related to $J^2$ by
\[J^2=J_-J_+ + J_z^2+\hbar J_z\]and
\[J^2=J_+J_- + J_z^2-\hbar J_z.\]Equivalently,
\[J_+J_-=J^2-J_z^2+\hbar J_z,\] \[J_-J_+=J^2-J_z^2-\hbar J_z.\]These identities are used to derive the allowed eigenvalues of angular momentum.
Derivation of one commutator
Using
\[L_x=yp_z-zp_y, \qquad L_y=zp_x-xp_z,\]and the canonical commutators, one obtains
\[[L_x,L_y] = [yp_z-zp_y,\; zp_x-xp_z].\]Terms containing only commuting coordinates or commuting momenta vanish. The non-zero terms reduce to
\[[L_x,L_y] =i\hbar(xp_y-yp_x).\]Therefore
\[\boxed{ [L_x,L_y]=i\hbar L_z. }\]The other two relations follow by cyclic interchange of $x,y,z$.
Ladder action
If $\lvert j,m\rangle$ is an eigenstate of $J^2$ and $J_z$, then
\[J_z(J_\pm\lvert j,m\rangle) =([J_z,J_\pm]+J_\pm J_z)\lvert j,m\rangle.\]Using $[J_z,J_\pm]=\pm\hbar J_\pm$,
\[J_z(J_\pm\lvert j,m\rangle) =\hbar(m\pm1)(J_\pm\lvert j,m\rangle).\]Thus $J_+$ raises $m$ by one unit and $J_-$ lowers $m$ by one unit.
Main points
- Angular momentum components do not commute.
- The operator $J^2$ commutes with all components of $\mathbf J$.
- The pair $(J^2,J_z)$ can be diagonalized simultaneously.
- Ladder operators move between states with different $m$ values.
- The algebra is the same for orbital, spin, and total angular momentum.
Practice questions
- Starting from $L_x=yp_z-zp_y$, derive $[L_x,L_y]=i\hbar L_z$.
- Prove that $[J^2,J_z]=0$.
- Derive $[J_z,J_\pm]=\pm\hbar J_\pm$.
- Show that $[J_+,J_-]=2\hbar J_z$.
- Explain why $J_x$, $J_y$, and $J_z$ cannot all be measured sharply in the same state.
Discussion