Commutators with Angular Momentum

Angular momentum classifies operators by commutators. Instead of checking every finite rotation, one can often test how an operator responds to the generators $J_x,J_y,J_z$.

For rotations, the central idea is:

$$\text{transformation law} \quad \Longleftrightarrow \quad \text{commutators with }\mathbf J.$$

This page is the angular-momentum specialization of the general commutator action developed in Infinitesimal Transformations. The classification of scalar, vector, and tensor operators is introduced in Scalar, Vector, and Tensor Operators. The irreducible spherical form used in matrix elements is developed in Irreducible Spherical Tensors.

Convention

Let

$$U(\hat{\mathbf n},\theta) = \exp\left( -\frac{i}{\hbar}\theta\,\hat{\mathbf n}\cdot\mathbf J \right)$$

be the unitary rotation operator. With the state-rotation convention used in this volume, the operator appearing in expectation values transforms as

$$O \mapsto U^\dagger O U.$$

For a small rotation about the $i$ axis,

$$U_i(\delta\theta) = I-\frac{i}{\hbar}\delta\theta J_i +O(\delta\theta^2),$$

so

$$U_i^\dagger O U_i = O + \frac{i}{\hbar}\delta\theta [J_i,O] +O(\delta\theta^2).$$

This sign convention matches the commutators used in the angular-momentum and spherical-tensor pages. Some books transform operators as $UOU^\dagger$ instead; that reverses the first-order sign.

Scalar Operators

A rotational scalar is unchanged by every proper rotation:

$$U^\dagger O U=O.$$

Infinitesimally, this is equivalent to

$$[J_i,O]=0, \qquad i=x,y,z.$$

Examples include $\mathbf R^2$, $\mathbf P^2$, $\mathbf L^2$, $\mathbf S^2$, central Hamiltonians, and scalar products such as $\mathbf L\cdot\mathbf S$ when all relevant degrees of freedom are rotated by the same total angular momentum.

Do not confuse this with conservation. The commutator

$$[J_i,O]=0$$

says that $O$ is invariant under rotations generated by $J_i$. The conservation test for time evolution is instead a commutator with the Hamiltonian, such as $[H,O]=0$.

Vector Operators

A vector operator $\mathbf V=(V_x,V_y,V_z)$ satisfies

$$U^\dagger V_j U = \sum_k \mathcal R_{jk}V_k.$$

For a small rotation by $\delta\theta$ about the $i$ axis,

$$\mathcal R_{jk} = \delta_{jk} - \delta\theta\,\epsilon_{ijk} +O(\delta\theta^2).$$

Therefore

$$U_i^\dagger V_j U_i = V_j - \delta\theta \sum_k\epsilon_{ijk}V_k +O(\delta\theta^2).$$

Comparing with the commutator expansion gives the vector-operator commutator:

$$[J_i,V_j] = i\hbar \sum_k\epsilon_{ijk}V_k.$$

For example, with $i=z$,

$$[J_z,V_x]=i\hbar V_y, \qquad [J_z,V_y]=-i\hbar V_x, \qquad [J_z,V_z]=0.$$

The corresponding infinitesimal transformed components are

$$U_z^\dagger V_xU_z = V_x-\delta\theta V_y, \qquad U_z^\dagger V_yU_z = V_y+\delta\theta V_x,$$

which is the ordinary small-angle rotation of the component labels.

Angular Momentum Itself

Angular momentum is a vector operator under rotations generated by itself:

$$[J_i,J_j] = i\hbar \sum_k\epsilon_{ijk}J_k.$$

This is both the angular-momentum algebra and a transformation statement. It says that the three components of $\mathbf J$ rotate into one another.

If a system has orbital and spin angular momentum, the relevant generator must be chosen carefully. Under total rotations,

$$\mathbf J=\mathbf L+\mathbf S,$$

and both $\mathbf L$ and $\mathbf S$ transform as vectors:

$$[J_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k, \qquad [J_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k.$$

Under orbital rotations generated only by $\mathbf L$, spin components commute with $\mathbf L$ in the usual product Hilbert space model:

$$[L_i,S_j]=0.$$

This distinction is one reason total angular momentum is the natural generator for spin–orbital systems.

Cartesian Tensor Operators

A two-index Cartesian tensor operator $T_{jk}$ transforms with one rotation matrix for each index. Its infinitesimal commutator is

$$[J_i,T_{jk}] = i\hbar \sum_a \left( \epsilon_{ija}T_{ak} + \epsilon_{ika}T_{ja} \right).$$

The formula says that each index transforms like a vector index. For an $n$-index tensor, the commutator is a sum of $n$ terms, one for each index.

If $T_{jk}=A_jB_k$ is a product of two vector operators, the result follows from the Leibniz rule:

$$[J_i,A_jB_k] = [J_i,A_j]B_k + A_j[J_i,B_k].$$

Substituting the vector commutators gives the two-index formula.

The trace

$$T_{aa}$$

is scalar-like, the antisymmetric part is vector-like, and the symmetric traceless part is the Cartesian rank-$2$ piece. This is the commutator version of

$$3\otimes3=1\oplus3\oplus5.$$

Spherical Tensor Commutators

For angular-momentum calculations, Cartesian components are usually converted to irreducible spherical components

$$T_q^{(k)}, \qquad q=-k,-k+1,\ldots,k.$$

The defining commutators are

$$[J_z,T_q^{(k)}] = \hbar q\,T_q^{(k)}$$

and

$$[J_\pm,T_q^{(k)}] = \hbar \sqrt{k(k+1)-q(q\pm1)} \,T_{q\pm1}^{(k)}.$$

The second formula is understood to give zero when $q\pm1$ lies outside the allowed range. Thus $T_k^{(k)}$ is killed by $J_+$ and $T_{-k}^{(k)}$ is killed by $J_-$.

These are exactly the same ladder relations obeyed by a spin-$k$ multiplet, except that the rotation generator acts by commutator rather than by ordinary left multiplication on a state.

Why the Ladder Form Is Forced

The $J_z$ commutator says that $T_q^{(k)}$ has magnetic component label $q$. Apply the Jacobi identity to $J_z$, $J_\pm$, and $T_q^{(k)}$:

$$[J_z,[J_\pm,T_q^{(k)}]] = [[J_z,J_\pm],T_q^{(k)}] + [J_\pm,[J_z,T_q^{(k)}]].$$

Using

$$[J_z,J_\pm]=\pm\hbar J_\pm, \qquad [J_z,T_q^{(k)}]=\hbar q T_q^{(k)},$$

one obtains

$$[J_z,[J_\pm,T_q^{(k)}]] = \hbar(q\pm1)[J_\pm,T_q^{(k)}].$$

Thus $[J_\pm,T_q^{(k)}]$ must be proportional to a component with label $q\pm1$. The square-root coefficient is the standard angular-momentum normalization for a rank-$k$ multiplet.

Magnetic Selection Rule from a Commutator

The simplest selection rule follows directly from the $J_z$ commutator. Let

$$J_z|\alpha,j,m\rangle = \hbar m|\alpha,j,m\rangle.$$

Taking the matrix element of

$$[J_z,T_q^{(k)}] = \hbar qT_q^{(k)}$$

between $|\alpha,j,m\rangle$ and $|\alpha',j',m'\rangle$ gives

$$\hbar(m'-m) \langle\alpha',j',m'|T_q^{(k)}|\alpha,j,m\rangle = \hbar q \langle\alpha',j',m'|T_q^{(k)}|\alpha,j,m\rangle.$$

Therefore a nonzero matrix element requires

$$m'=m+q.$$

The triangle rule for $j,j',k$ needs the full irreducible-tensor and Wigner–Eckart theorem machinery. The broader list of symmetry-enforced zeros is collected in Selection Rules.

Fixed Backgrounds

Commutator tests must be applied to the actual operator in the actual symmetry setting. If

$$H_B=-B\mu_z$$

with $B$ a fixed laboratory field, then $H_B$ is not a scalar under all rotations. It may commute with $J_z$, but generally not with $J_x$ and $J_y$.

By contrast, the formal expression

$$-\boldsymbol\mu\cdot\mathbf B$$

is a scalar only if both $\boldsymbol\mu$ and the background vector $\mathbf B$ are transformed. This distinction between covariance and invariance is essential in spectroscopy, perturbation theory, and selection-rule arguments.

Common Mistakes

• Using the wrong sign because one switched between $U^\dagger O U$ and $UOU^\dagger$ conventions.
• Checking only $[J_z,O]$ and concluding that $O$ is a rotational scalar. Full rotational scalar behavior requires commutation with all three components of $\mathbf J$.
• Confusing $[J_i,O]=0$ with $[H,O]=0$. The first is rotational invariance of $O$; the second is conservation under time evolution.
• Calling a three-component object a vector without checking the vector commutator.
• Treating a two-index Cartesian tensor as if every part were irreducible rank $2$.
• Forgetting that fixed external fields reduce the symmetry group.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• M. E. Rose, Elementary Theory of Angular Momentum, Wiley, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.

Exercises

1. Derive the vector commutator from the finite vector transformation law.

Solution

For a small rotation about the $i$ axis,

$$U_i^\dagger V_j U_i = V_j + \frac{i}{\hbar}\delta\theta[J_i,V_j] +O(\delta\theta^2).$$

The vector transformation law gives

$$U_i^\dagger V_j U_i = V_j - \delta\theta \sum_k\epsilon_{ijk}V_k +O(\delta\theta^2).$$

Equating first-order terms,

$$\frac{i}{\hbar}[J_i,V_j] = - \sum_k\epsilon_{ijk}V_k.$$

Multiplying by $\hbar/i=-i\hbar$ gives

$$[J_i,V_j] = i\hbar \sum_k\epsilon_{ijk}V_k.$$

2. Suppose $\mathbf A$ and $\mathbf B$ are vector operators. Show that $\mathbf A\cdot\mathbf B=\sum_j A_jB_j$ is a scalar under simultaneous rotations.

Solution

Use the Leibniz rule:

$$[J_i,\mathbf A\cdot\mathbf B] = \sum_j \left( [J_i,A_j]B_j + A_j[J_i,B_j] \right).$$

Substitute the vector commutators:

$$[J_i,\mathbf A\cdot\mathbf B] = i\hbar \sum_{j,k} \epsilon_{ijk} \left( A_kB_j + A_jB_k \right).$$

In the first term, interchange the dummy labels $j$ and $k$. Since $\epsilon_{ikj}=-\epsilon_{ijk}$, the first term cancels the second. Therefore

$$[J_i,\mathbf A\cdot\mathbf B]=0.$$

3. Use $[J_z,T_q^{(k)}]=\hbar qT_q^{(k)}$ to derive the rule $m'=m+q$ for a nonzero matrix element.

Solution

Insert angular-momentum eigenstates:

$$\langle\alpha',j',m'| [J_z,T_q^{(k)}] |\alpha,j,m\rangle = \hbar q \langle\alpha',j',m'| T_q^{(k)} |\alpha,j,m\rangle.$$

The left side is

$$\hbar(m'-m) \langle\alpha',j',m'| T_q^{(k)} |\alpha,j,m\rangle.$$

If the matrix element is nonzero, the coefficients must agree:

$$m'-m=q.$$

4. Let $T_{jk}=A_jB_k$ with $\mathbf A$ and $\mathbf B$ vector operators. Derive the commutator of $T_{jk}$ with $J_i$.

Solution

Start with

$$[J_i,A_jB_k] = [J_i,A_j]B_k + A_j[J_i,B_k].$$

Using

$$[J_i,A_j] = i\hbar\sum_a\epsilon_{ija}A_a, \qquad [J_i,B_k] = i\hbar\sum_a\epsilon_{ika}B_a,$$

gives

$$[J_i,T_{jk}] = i\hbar \sum_a \left( \epsilon_{ija}T_{ak} + \epsilon_{ika}T_{ja} \right).$$