Angular Momentum Operators and Commutation Relations

Every discussion of atomic and molecular orbital shape rests on this algebra. The reason chemists label orbitals by $l$ (s, p, d, f) and $m$ (orientation within a shell, e.g. $p_x,p_y,p_z$) rather than by three independent numbers is a direct consequence of $[\hat L^2,\hat L_z]=0$ paired with $[\hat L_x,\hat L_z]\neq 0$: nature permits a shared eigenbasis for magnitude and one projection, and no more. This shows up immediately in Eigenvalues of L-squared and Lz — the quantized values $\hbar^2 l(l+1)$ and $m\hbar$ are extracted using exactly the algebra derived here, typically via the Ladder Operators and the Angular Momentum Algebra technique that constructs raising and lowering operators $\hat L_\pm = \hat L_x \pm i\hat L_y$ directly from these commutators. In molecular spectroscopy, selection rules for electronic and rotational transitions ($\Delta l = \pm 1$, $\Delta m = 0,\pm 1$) trace back to how the dipole operator transforms under the same rotation algebra. In magnetic resonance (NMR, EPR), the nuclear and electron spin operators obey an identical commutation algebra $[\hat S_i,\hat S_j]=i\hbar\varepsilon_{ijk}\hat S_k$, even though spin has no classical position–momentum origin at all — the entire vector model of NMR pulse sequences (90° and 180° pulses, precession, spin echoes) is built on this non-commuting operator structure. In coordination chemistry and crystal-field theory, the splitting of $d$-orbitals in an octahedral or tetrahedral field is organized by how $\hat L^2$ and $\hat L_z$ eigenstates transform under the point group of the ligand environment. In modern quantum computing with spin qubits, single-qubit gates are literally implemented as controlled non-commuting rotations generated by these same operators, and the impossibility of simultaneously fixing all three spin projections is what necessitates careful pulse-sequence design. And in astrophysical and atmospheric spectroscopy, the fine structure of atomic emission lines (used to determine stellar and interstellar compositions) is a direct fingerprint of angular momentum coupling built from exactly this non-commuting algebra.

:::quiz[What is $[\hat L_y,\hat L_z]$?]

• $i\hbar \hat L_x$ ✓
• $i\hbar \hat L_y$
• $-i\hbar \hat L_z$
• $0$

:::quiz[Which combination of operators is used to prove $[\hat L^2,\hat L_z]=0$ most efficiently?]

• Direct substitution of Cartesian coordinates for every term
• The identity $[\hat A^2,\hat B]=\hat A[\hat A,\hat B]+[\hat A,\hat B]\hat A$ applied to $\hat L_x^2$ and $\hat L_y^2$ ✓
• Fourier transforming into momentum space first
• Assuming $\hat L^2$ is a scalar and therefore automatically commutes

:::quiz[A Hamiltonian has potential $V=\tfrac12k(x^2+y^2)$ with a single, equal force constant $k$. What does the algebra predict about $[\hat H,\hat L_z]$?]

• It is always nonzero because $V$ depends on two coordinates
• It vanishes, since $V$ is invariant under rotation about $z$ ✓
• It equals $i\hbar\hat L_x$
• It cannot be determined without knowing the mass

Ask the tutor: Can you show me how the ladder operators $\hat L_\pm = \hat L_x \pm i\hat L_y$ are built from this algebra and used to derive the quantized eigenvalues $\hbar^2l(l+1)$ and $m\hbar$?

Classically the orbital angular momentum of a particle is $\mathbf{L} = \mathbf{r}\times\mathbf{p}$, a vector perpendicular to the plane of motion whose three Cartesian components,

are all simultaneously well-defined numbers for any trajectory. To go quantum mechanical we do exactly what was done for the Hamiltonian: replace each classical variable by its operator, $x\to\hat x$ (multiplication by $x$) and $p_x \to \hat p_x = -i\hbar\,\partial/\partial x$, and similarly for $y,z$. Because position and momentum operators along different axes commute (only $\hat x$ and $\hat p_x$ fail to commute with each other; $\hat x$ and $\hat p_y$ commute freely), there is no ordering ambiguity in writing

The surprise appears only when you ask whether these three new operators commute with each other. They do not, and the failure is not a minor technical defect — it is the entire reason atomic orbitals have the shapes and quantum numbers they do. Physically, $\hat L_z$ measures the tendency of the electron's probability current to circulate around the $z$-axis; $\hat L_x$ measures circulation around the $x$-axis. Rotating an electron's motion to have well-defined circulation about $z$ necessarily randomizes any well-defined sense of circulation about $x$: you cannot simultaneously pin down "how fast it spins around this axis" and "how fast it spins around that axis" for axes that are not parallel. What does survive is the total magnitude squared of angular momentum, $\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2$. This operator is rotationally invariant — it doesn't care which axis you call $z$ — and that invariance is exactly why it commutes with every individual component. The practical upshot, verified rigorously below, is that we can find wavefunctions that are simultaneous eigenfunctions of $\hat L^2$ and $\hat L_z$ (these become the spherical harmonics $Y_l^m(\theta,\phi)$ underlying every $s$, $p$, $d$, $f$ orbital), but never simultaneous eigenfunctions of $\hat L_z$ and $\hat L_x$ (except in the trivial case $l=0$).

The foundational commutators, from which everything else in angular momentum theory (ladder operators, quantized eigenvalues, spin algebra) is built:

where: $\hat L_x,\hat L_y,\hat L_z$ are the Cartesian orbital angular momentum operators; $\hbar$ is the reduced Planck constant; the three relations are cyclic permutations of $x\to y\to z\to x$.

Reading. None of the three components commutes with either of the other two. The commutator of two components is proportional to the third, with the proportionality constant $i\hbar$ — structurally identical to the classical Poisson-bracket relation $\{L_x,L_y\}=L_z$, except quantum mechanically the bracket is an operator identity, not a numerical statement about a single trajectory. Compactly, using the Levi-Civita symbol $\varepsilon_{ijk}$ (which is $+1$ for $(i,j,k)=(x,y,z)$ and cyclic permutations, $-1$ for anticyclic permutations, and $0$ if any two indices repeat):

where: the sum runs over the third Cartesian index $k$; this single line reproduces all three relations above.

Reading. This is the defining algebra of angular momentum in quantum mechanics — the same algebra (with $\hat L\to\hat S$ or $\hat J$) governs electron spin, nuclear spin, and total angular momentum, even for particles like the electron where no classical $\mathbf r\times\mathbf p$ construction applies.

where: $\hat L^2$ is the (rotationally invariant) total angular-momentum-squared operator; the identity holds with $\hat L_z$ replaced by $\hat L_x$ or $\hat L_y$ as well, i.e. $\hat L^2$ commutes with every Cartesian component.

Reading. $\hat L^2$ and one chosen component (by convention $\hat L_z$) form a complete commuting set: they can be diagonalized simultaneously, giving wavefunctions labeled by two quantum numbers, $l$ (magnitude) and $m$ (projection). This is precisely the origin of the $(l,m)$ labeling of atomic orbitals.

:::derivation[Deriving $[\hat L_x,\hat L_y]=i\hbar\hat L_z$ from canonical commutators] Goal. Show that $[\hat L_x,\hat L_y]=i\hbar\hat L_z$ follows purely from the position–momentum commutation relations $[\hat x_i,\hat p_j]=i\hbar\delta_{ij}$ and $[\hat x_i,\hat x_j]=[\hat p_i,\hat p_j]=0$.

Step 1 — Write out the operators explicitly. Using $\hat L_x = \hat y\hat p_z - \hat z\hat p_y$ and $\hat L_y = \hat z\hat p_x - \hat x\hat p_z$, expand the commutator by linearity:

Distribute over the four cross terms:

Step 2 — Eliminate the terms that vanish outright. Any commutator $[\hat A\hat B,\hat C\hat D]$ where $\hat A,\hat B,\hat C,\hat D$ pairwise commute except for at most one canonically conjugate pair reduces quickly. The second term, $[\hat y\hat p_z,\hat x\hat p_z]$: here $\hat y$ commutes with $\hat x$ and $\hat p_z$, and $\hat x$ commutes with $\hat p_z$ — every operator in one factor commutes with every operator in the other, so

Likewise the third term, $[\hat z\hat p_y,\hat z\hat p_x]$: $\hat z$ commutes with itself and with $\hat p_x$, and $\hat p_y$ commutes with both $\hat z$ and $\hat p_x$, so

Step 3 — Evaluate the surviving first term. Use the identity $[\hat A\hat B,\hat C\hat D] = \hat A[\hat B,\hat C]\hat D + \hat A\hat C[\hat B,\hat D] + [\hat A,\hat C]\hat D\hat B + \hat C[\hat A,\hat D]\hat B$, or more directly expand with the product rule $[\hat A\hat B,\hat C] = \hat A[\hat B,\hat C]+[\hat A,\hat C]\hat B$ applied twice. For $[\hat y\hat p_z,\hat z\hat p_x]$, only the pair $(\hat p_z,\hat z)$ is canonically conjugate ($[\hat p_z,\hat z] = -i\hbar$); every other pair among $\{\hat y,\hat p_z\}$ and $\{\hat z,\hat p_x\}$ commutes. Peel off the commuting spectators $\hat y$ (left) and $\hat p_x$ (right):

Step 4 — Evaluate the fourth term the same way. For $[\hat z\hat p_y,\hat x\hat p_z]$, the only conjugate pair is $(\hat z,\hat p_z)$ with $[\hat z,\hat p_z]=i\hbar$; the spectators $\hat p_y$ (left) and $\hat x$ (right) commute with everything else present:

Step 5 — Assemble. Adding the two nonzero contributions from Steps 3 and 4 (the Step 2 terms are zero):

The combination in parentheses is exactly the definition of $\hat L_z$, so

The other two relations, $[\hat L_y,\hat L_z]=i\hbar\hat L_x$ and $[\hat L_z,\hat L_x]=i\hbar\hat L_y$, follow by the same steps under the cyclic relabeling $x\to y\to z\to x$ — the algebra is identical because the original definitions of $\hat L_x,\hat L_y,\hat L_z$ are themselves cyclic permutations of one another.

:::derivation[Proving $[\hat L^2,\hat L_z]=0$] Goal. Show that the total-angular-momentum-squared operator commutes with each Cartesian component, using only the results just derived, so that $\hat L^2$ and $\hat L_z$ can share eigenfunctions.

Step 1 — Split $\hat L^2$ into pieces. Write $\hat L^2 = \hat L_x^2 + \hat L_y^2 + \hat L_z^2$, and use linearity of the commutator:

The last term vanishes trivially since any operator commutes with itself: $[\hat L_z^2,\hat L_z]=0$.

Step 2 — Reduce $[\hat L_x^2,\hat L_z]$ using the product rule. For any operators, $[\hat A^2,\hat B] = \hat A[\hat A,\hat B] + [\hat A,\hat B]\hat A$. Apply this with $\hat A=\hat L_x$, $\hat B=\hat L_z$:

From the cyclic relation derived above, $[\hat L_z,\hat L_x]=i\hbar\hat L_y$, so $[\hat L_x,\hat L_z] = -i\hbar\hat L_y$. Substituting:

Step 3 — Reduce $[\hat L_y^2,\hat L_z]$ the same way. With $\hat A=\hat L_y$, $\hat B=\hat L_z$:

Since $[\hat L_y,\hat L_z]=i\hbar\hat L_x$ directly from the first derivation's cyclic partner:

Step 4 — Add the two nonvanishing pieces. The terms $-i\hbar(\hat L_x\hat L_y+\hat L_y\hat L_x)$ from Step 2 and $+i\hbar(\hat L_y\hat L_x+\hat L_x\hat L_y)$ from Step 3 are identical in operator content and opposite in sign, since $\hat L_x\hat L_y+\hat L_y\hat L_x = \hat L_y\hat L_x+\hat L_x\hat L_y$ trivially:

Step 5 — Conclude. Combining with the trivial Step 1 result:

Because $\hat L^2$ is built symmetrically from $\hat L_x^2,\hat L_y^2,\hat L_z^2$, the identical argument (relabeling which component plays the role of "$\hat L_z$") shows $[\hat L^2,\hat L_x]=0$ and $[\hat L^2,\hat L_y]=0$ as well: $\hat L^2$ commutes with every Cartesian component, even though the components fail to commute with one another. This is exactly what permits a common eigenbasis for $\hat L^2$ and any single chosen component — conventionally $\hat L_z$.

Example 1 — Foundational: verifying $[\hat L_z,\hat L_x]=i\hbar\hat L_y$ by direct operator action on a test function. Rather than trust the general derivation blindly, apply both sides of $[\hat L_z,\hat L_x]=i\hbar\hat L_y$ to the concrete (non-symmetric, asymmetric-in-all-three-coordinates) test function

chosen with unequal, non-round powers on each coordinate precisely so that no accidental symmetry could mask an error in the general formula. First compute $\hat L_x f = (\hat y\hat p_z - \hat z\hat p_y)f$. With $\hat p_z = -i\hbar\partial_z$, $\hat p_y=-i\hbar\partial_y$:

Now apply $\hat L_z = \hat x\hat p_y - \hat y\hat p_x$ to this result. Let $g \equiv \hat L_x f = -i\hbar(3x^2y^2z^2 - x^2z^4)$. Compute

More carefully, using $\hat p_y = -i\hbar\partial_y$ and $\hat p_x=-i\hbar\partial_x$ directly on $g$:

So

Next compute $\hat L_x\hat L_z f$. First $\hat L_z f = \hat x\hat p_y f - \hat y\hat p_x f$, with $\partial_y f = x^2z^3$, $\partial_x f = 2xyz^3$:

Let $h\equiv \hat L_z f = -i\hbar(x^3z^3-2xy^2z^3)$. Apply $\hat L_x = \hat y\hat p_z-\hat z\hat p_y$:

Wait — recompute cleanly: $\hat L_x h = y(-i\hbar)\cdot\partial_z h - z(-i\hbar)\cdot \partial_y h$, and since $\partial_z h$ and $\partial_y h$ already carry a factor $-i\hbar$, each term picks up $(-i\hbar)(-i\hbar)=-\hbar^2$:

So

Now form the commutator:

Compare this to $i\hbar\hat L_y f$. Compute $\hat L_y f = (\hat z\hat p_x-\hat x\hat p_z)f$ with $\partial_x f = 2xyz^3$, $\partial_z f=3x^2yz^2$:

Then

This matches $[\hat L_z,\hat L_x]f$ term for term, confirming $[\hat L_z,\hat L_x] = i\hbar\hat L_y$ on this deliberately asymmetric test function.

Limiting-case check. If the test function depended only on $z$ (no $x,y$ dependence at all, e.g. $f=z^3$), every angular momentum operator acting on it would vanish, since each $\hat L_i$ contains at least one derivative with respect to a coordinate other than the one paired in its own definition combined with multiplication by that same excluded coordinate — concretely $\hat L_x z^3 = (\hat y\hat p_z-\hat z\hat p_y)z^3 = y(-i\hbar)(3z^2) - 0 \neq 0$ actually, so the truly trivial check is $l=0$: for a spherically symmetric function like $f=e^{-r^2}$ (function of $r=\sqrt{x^2+y^2+z^2}$ only), all three $\hat L_i f = 0$, so every commutator trivially annihilates it — consistent with $s$-orbitals carrying zero angular momentum entirely.

Example 2 — Applied: what the algebra says about a real $p$-orbital. Consider an electron in a hydrogen $2p_z$ orbital, described (angular part) by the spherical harmonic $Y_1^0(\theta,\phi) \propto \cos\theta$. This state is an eigenfunction of both $\hat L_z$ and $\hat L^2$:

(The eigenvalue structure $\hbar^2 l(l+1)$ for $\hat L^2$ and $m\hbar$ for $\hat L_z$ is derived rigorously elsewhere from this same algebra; here we use the $2p_z$ values $l=1,m=0$ as an illustration.) The physically important question this article's algebra answers: does the electron in this state have a definite value of $L_x$ or $L_y$? Since $[\hat L_x,\hat L_z]=-i\hbar\hat L_y \neq 0$, and $Y_1^0$ is an eigenfunction of $\hat L_z$ with a nonzero commutator between $\hat L_x$ and $\hat L_z$, the generalized uncertainty relation built from any commutator,

tells us that if $\Delta L_z = 0$ exactly (a genuine eigenstate), the uncertainty in $L_x$ cannot also vanish unless $\langle \hat L_y\rangle=0$ trivially satisfies the inequality — but direct calculation (using ladder operators, foreshadowed here) shows $L_x$ and $L_y$ each have expectation value zero and nonzero variance in a $Y_1^0$ state: $\langle \hat L_x^2\rangle = \langle \hat L_y^2\rangle = \hbar^2$ (since $\langle \hat L^2\rangle = 2\hbar^2 = \langle \hat L_z^2\rangle + \langle \hat L_x^2\rangle + \langle \hat L_y^2\rangle = 0+\hbar^2+\hbar^2$). So the electron in a $2p_z$ orbital has a perfectly sharp $L_z=0$ and $L^2=2\hbar^2$, but $L_x$ and $L_y$ are each spread with root-mean-square value $\hbar$ — never zero, never sharp. This is not experimental ignorance; it is what the operator algebra forces, and it is the rigorous quantum-mechanical statement behind the classical picture of a vector "precessing" indeterminately about the $z$-axis.

Example 3 — Exam/competition level: commutator of $\hat L_z$ with a non-trivial combination, and consistency with rotation. A common exam trap is to compute $[\hat L_z, \hat x^2+2y^2]$ for a potential-like operator (relevant, for instance, to whether angular momentum is conserved in an anisotropic 2D oscillator potential $V=\tfrac12 k_1x^2+\tfrac12 k_2y^2$ with $k_1\neq k_2$, chosen non-symmetric on purpose). Use $\hat L_z=\hat x\hat p_y-\hat y\hat p_x$ and the elementary commutators $[\hat p_y,y^2]=-i\hbar\cdot 2y$ (i.e. $[\hat p_y,f(y)] = -i\hbar\,\partial f/\partial y$) and $[\hat p_x,x^2]=-i\hbar\cdot 2x$:

Since $\hat p_y$ commutes with any function of $x$ alone, $[\hat p_y,\hat x^2]=0$. Since $[\hat p_x,\hat x^2] = -i\hbar\,\partial(x^2)/\partial x = -2i\hbar x$:

Similarly,

Adding these unequal-coefficient contributions (the "$2$" on $y^2$ was chosen precisely so the two pieces do not cancel by accident):

This is manifestly nonzero whenever $\hat x\hat y \neq 0$ (i.e., generically, off the coordinate axes) — confirming that $\hat L_z$ does not commute with an anisotropic potential $\hat x^2+2\hat y^2$, exactly as expected: angular momentum about $z$ is conserved only when the potential is invariant under rotation about $z$, i.e., when the coefficients of $x^2$ and $y^2$ are equal. As a limiting-case check, setting the coefficient of $y^2$ to match that of $x^2$ (isotropic case, coefficient $1$ instead of $2$) gives

correctly recovering the textbook fact that $\hat L_z$ commutes with any function of $r=\sqrt{x^2+y^2+z^2}$ alone (here $x^2+y^2$), since a rotationally symmetric potential conserves angular momentum about every axis. This exercise shows how the raw commutation algebra directly encodes the quantum version of Noether's theorem: symmetry of the Hamiltonian under rotation about an axis $\Leftrightarrow$ that axis's angular momentum component commutes with $\hat H$ and is conserved.