Angular Momentum Commutation Relations

Every qubit gate is this algebra cashed in. A two-level system is a $j = \tfrac12$ representation; universal single-qubit gates are exponentials of the three generators, and gate calibration is a daily fight with equation (1): rotations about $x$ and $z$ do not commute, so pulse order is physics, and compilers track it the way the book-rotation figure does. NMR has run on the same grammar since the 1940s. The vector-operator grammar of (5) is equally alive: selection rules in spectroscopy, tensor interactions in nuclei, and the angular analysis of particle decays all descend from asking "how does this operator commute with $\hat{\mathbf{J}}$?" — the question Spherical Tensor Operators and The Wigner-Eckart Theorem industrialize downstream.

Hold a book flat. Rotate it $90°$ about $x$, then $90°$ about $y$: the cover faces you. Reverse the order: the spine does. Finite rotations about different axes do not commute — a fact about the geometry of space, older than quantum mechanics. The generators inherit it: since $\hat L_x$ generates $x$-rotations and $\hat L_y$ generates $y$-rotations, the mismatch between orders must live in their commutator, and $\varepsilon_{ijk}$ is the right-hand rule doing the bookkeeping.

What does a state with definite angular momentum look like, then? Not an arrow — a cone. Sharpen $\hat J^2$ (it commutes with everything) and one projection, say $\hat J_z$. The other two components are forced to smear: definite length, definite $z$-shadow, while $J_x$ and $J_y$ average to zero with irreducible spread — the arrow replaced by its cone of possibilities. The classical arrow re-emerges only at large quantum numbers, where the transverse smear falls as $1/\sqrt{2j}$ — 50% of the length at $j = 1$, 7% at $j = 100$, 0.7% at $j = 10^4$ (machine-computed). Flywheels look like arrows; electrons do not.

The derivation needs only the canonical relations $[\hat x_a, \hat p_b] = i\hbar\,\delta_{ab}$ — owned by Canonical Commutation Relation — and the algebra pack of The Commutator and Compatible Observables (bilinearity, product rule; consumed, never re-proved). Everything angular follows from what position and momentum already knew.

where: $\varepsilon_{ijk}$ is the totally antisymmetric symbol ($\varepsilon_{xyz} = +1$), and the sum over $k$ is implied. Read it as a closure statement: the friction between two components is not some new operator — it is the third component. The algebra feeds itself, which is exactly what lets it be promoted to a definition.

Here is the move that separates this chapter from classical mechanics. Notice what the coming constructions will actually use: never the cross product, only equation (1). So quantum mechanics flips the logic and defines:

where: the neutral letter $\hat{\mathbf{J}}$ signals the generality. Orbital $\hat{\mathbf{L}}$ satisfies (2) by the derivation above — one representation of the algebra. But the definition asks for no wavefunction, no angles: any matrices obeying (2) qualify. The generosity is not idle. The electron carries intrinsic angular momentum with no spatial winding whatsoever — Spin Angular Momentum introduces it, and Spin-1/2 and the Pauli Matrices exhibits the $2\times2$ matrices satisfying (2) perfectly. Sums of angular momenta of independent subsystems obey (2) again (the seed of Addition of Angular Momenta). One algebra, many bodies. Mathematically, (2) says the generators span the Lie algebra $\mathfrak{su}(2)$ — the structure of rotations, with $\hbar$ as the scale. The Commutator and Compatible Observables introduced the bracket as an algebraic product; here is its most consequential instance. Exponentiating into finite rotations, and the $2\pi$ surprise awaiting half-integer representations, belong to Rotation Operators and SU(2).

Every component quarrels with every other. Remarkably, one object gets along with all of them — the squared magnitude $\hat J^2 = \hat J_x^2 + \hat J_y^2 + \hat J_z^2$:

where: the statement holds because of, not despite, the component algebra: the cross terms generated by each $\hat J_j^2$ cancel in the sum by antisymmetry. An operator commuting with every generator of an algebra is called a Casimir invariant — the naming honors the 1931 thesis of the previous article's portrait, which first exploited such invariants systematically (his band lives there; we only borrow the name).

The payoff is the labeling scheme all of physics uses. By the compatibility theorem of The Commutator and Compatible Observables (both directions proved there, consumed here), $[\hat J^2, \hat J_z] = 0$ guarantees a common eigenbasis — the states $|j, m\rangle$, forming the complete commuting set $\{\hat J^2, \hat J_z\}$ in the sense of Simultaneous Diagonalization and Complete Sets of Commuting Observables. A third label would have to commute with both, and no other function of the components does. Two labels is the algebra's allowance. What values $j$ and $m$ take — $j(j+1)\hbar^2$, $m = -j \dots j$, and the half-integer surprise — is derived purely from the algebra in Ladder Operators and the Angular Momentum Spectrum, next; this article stops at the licence.

Generalized Uncertainty Principle owns the Robertson bound $\sigma_A\sigma_B \geq \tfrac12|\langle[\hat A,\hat B]\rangle|$ — consumed here, never re-derived. Feed it the algebra:

where: the floor is state-dependent — the more angular momentum points along $z$, the more the transverse components must fluctuate. On the stretched state $|j, j\rangle$ the verifier finds $\sigma_{J_x}^2 = \sigma_{J_y}^2 = j\hbar^2/2$ exactly: the product equals the floor — saturated, for $j = \tfrac12, 1, \tfrac32$; stretched states are the algebra's minimum-uncertainty states. Two fine-print lines. On $|j, 0\rangle$ the floor collapses to zero yet $\sigma_{J_x} = \sigma_{J_y} \neq 0$ ($\hbar$ at $j{=}1$, machine-checked) — a zero Robertson floor never promises sharpness. And the sole exception to "no two components sharp" is $j = 0$: every component annihilates the state — a one-dimensional loophole, not a counterexample. One more piece of grammar the algebra exports. An operator triple $\hat{\mathbf{V}}$ is a vector operator if it transforms under rotations the way $\hat{\mathbf{J}}$ itself does:

where: $\hat{\mathbf{r}}$, $\hat{\mathbf{p}}$, and $\hat{\mathbf{L}}$ all qualify (machine-checked); equation (1) says $\hat{\mathbf{J}}$ is a vector operator with respect to itself. This grammar — operators, not just states, carrying rotation labels — grows into Spherical Tensor Operators and The Wigner-Eckart Theorem, far downstream.

The algebra (2) predates quantum mechanics by eighty-two years — carved, in different clothing, into the stone of Broome Bridge.

Example 1 — the uncertainty budget of a stretched p-state (confidence builder). An electron occupies $|\ell{=}1, m{=}1\rangle$ (labels licensed by (3); values from the next article, taken as given). Audit the budget: length, shadow, transverse spread, and the cone angle.

Step 1. Length and shadow: $\langle\hat L^2\rangle = 2\hbar^2$ ($|\mathbf{L}| = \hbar\sqrt2$), $\langle\hat L_z\rangle = \hbar$ — the lean is already visible.

Step 2. Transverse budget: $\langle\hat L_x^2 + \hat L_y^2\rangle = \langle\hat L^2\rangle - \langle\hat L_z^2\rangle = 2\hbar^2 - \hbar^2 = \hbar^2$. By symmetry each takes half: $\sigma_{L_x} = \sigma_{L_y} = \hbar/\sqrt2 \approx \boxed{0.707\,\hbar}$.

Step 3. Check against the floor (4): $\sigma_{L_x}\sigma_{L_y} = \hbar^2/2$, and $\tfrac{\hbar}{2}|\langle\hat L_z\rangle| = \hbar^2/2$ — saturated: the stretched state is as arrow-like as the algebra permits, and still smeared by 71% of $\hbar$ per transverse component.

Step 4. Cone angle: $\cos\theta = m/\sqrt{\ell(\ell+1)} = 1/\sqrt2$, so $\theta = \boxed{45°}$. (For $\ell = 2$: $35.3°$ — all numbers machine-verified.)

Example 2 — measuring the wrong component (exam level). The same electron in $|1, 1\rangle$ (sharp $L_z = +\hbar$) is sent through an analyzer that measures $\hat L_x$. What outcomes appear, with what probabilities? If $L_z$ is then measured again, what is the chance of recovering $+\hbar$?

Step 1. Diagonalize $\hat L_x$ in the three-state multiplet and expand $|1,1\rangle$ in its eigenbasis. The machine-checked amplitudes give $P(L_x = +\hbar) = \boxed{1/4}$, $P(L_x = 0) = \boxed{1/2}$, $P(L_x = -\hbar) = \boxed{1/4}$.

Step 2. Sanity-check with moments: $\langle\hat L_x\rangle = \tfrac14\hbar + 0 - \tfrac14\hbar = 0$ and $\langle\hat L_x^2\rangle = \tfrac14\hbar^2 + 0 + \tfrac14\hbar^2 = \hbar^2/2$ — exactly Example 1's transverse budget, arrived at from the distribution.

Step 3. Now remeasure $\hat L_z$. The $L_x$ measurement collapsed the state (the update rule of Measurement and Wavefunction Collapse); summing the two-step paths, $P(\text{return to } m{=}+1) = \boxed{3/8}$ (machine-checked). Measuring an incompatible component scrambled five-eighths of the original certainty — incompatibility is a physical cost, paid in probability.

References. Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 3 · Griffiths, Introduction to Quantum Mechanics, 3rd ed., Ch. 4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 12 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. 1, Ch. VI · Townsend, A Modern Approach to Quantum Mechanics, Ch. 3

Provenance. All commutators and matrix elements are standard. A verifier should: (1) reproduce $[L_x,L_y]=i\hbar L_z$ from $[x_i,p_j]=i\hbar\delta_{ij}$ (Griffiths §4.3); (2) confirm $[\mathbf{J}^2,J_z]=0$ by the cancellation shown (Sakurai §3.5); (3) check the spectrum derivation $2j\in\mathbb{Z}_{\geq0}$ and ladder coefficients $\hbar\sqrt{j(j+1)-m(m\pm1)}$ (Shankar §12.5, Cohen-Tannoudji Vol.1 Ch.VI); (4) verify $\mathbf{S}^2=\tfrac34\hbar^2$ for $s=\tfrac12$ from the Pauli matrices. Numerical constant $\hbar\approx1.055\times10^{-34}\,\mathrm{J\,s}$ is CODATA-standard. For this edition, items (1), (2), (4) plus every number above were machine-verified (sympy/numpy, 55 checks): the algebra by two routes (a symbolic normal-ordering engine imposing only the canonical relations, all nine pairs; differential operators on test functions), the Casimir property abstractly and in three matrix reps, Robertson saturation for $j = \tfrac12, 1, \tfrac32$, the quaternion realization, the vector-operator instances, both worked examples, and the classical-limit smears. Item (3) — the spectrum and ladder coefficients — is deliberately NOT reproduced: per the chapter ownership map it belongs to Ladder Operators and the Angular Momentum Spectrum; the Pauli matrices' anatomy belongs to Spin-1/2 and the Pauli Matrices.

Builds on: Orbital Angular Momentum Operators · Eigenfunctions of L_z and the Azimuthal Quantization

Leads to: Ladder Operators and the Angular Momentum Spectrum · Spin Angular Momentum · Rotation Operators and SU(2) · Spin-1/2 and the Pauli Matrices