Scalar, Vector, and Tensor Operators

Scalar, vector, and tensor operators are classified by how their components transform under rotations. This tells us which components mix under a rotation and which transitions the operator can connect:

$$U(\mathcal R)^\dagger O_a U(\mathcal R) = \sum_b D_{ab}(\mathcal R)O_b,$$

where $U(\mathcal R)$ is the quantum rotation operator and $D(\mathcal R)$ is the finite-dimensional rotation matrix acting on the component label $a$.

This page explains the Cartesian classification. The chapter guide maps the full route from transformation laws to spectroscopy. Commutators with Angular Momentum gives the infinitesimal tests for these transformation laws. Irreducible Spherical Tensors rewrites the same idea in the spherical basis used by the Wigner–Eckart theorem.

Rotating Operator Components

Use the active convention fixed in Rotations Preview. A state rotates as

$$|\psi\rangle \mapsto U(\mathcal R)|\psi\rangle.$$

An operator is classified by its adjoint transformation law:

$$O \mapsto U(\mathcal R)^\dagger O U(\mathcal R).$$

For one operator $O$, scalar behavior means

$$U(\mathcal R)^\dagger O U(\mathcal R) = O$$

for every proper rotation $\mathcal R$. For a collection of three operators $V_i$, vector behavior means

$$U(\mathcal R)^\dagger V_i U(\mathcal R) = \sum_j \mathcal R_{ij}V_j.$$

For a two-index Cartesian tensor operator $T_{ij}$, tensor behavior means

$$U(\mathcal R)^\dagger T_{ij} U(\mathcal R) = \sum_{a,b} \mathcal R_{ia} \mathcal R_{jb} T_{ab}.$$

The same pattern extends to higher-index tensors: every Cartesian index is rotated by a copy of the ordinary three-dimensional rotation matrix.

Infinitesimal Version

Let $\mathbf J=(J_x,J_y,J_z)$ be the generator of rotations on the Hilbert space being considered. The finite transformation laws above have infinitesimal versions in terms of commutators.

A rotational scalar satisfies

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

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

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

These commutators test the transformation law without calculating every finite rotation. Their derivation is given in Commutators with Angular Momentum.

For example, the position and momentum operators obey

$$[L_i,R_j] = i\hbar \sum_k\epsilon_{ijk}R_k, \qquad [L_i,P_j] = i\hbar \sum_k\epsilon_{ijk}P_k$$

for orbital rotations. Therefore $\mathbf R$ and $\mathbf P$ are vector operators under orbital angular momentum.

Scalars

A scalar operator is unchanged by rotations. Important examples include:

• $I$, the identity operator;
• $\mathbf R^2$, $\mathbf P^2$, $\mathbf L^2$, and $\mathbf S^2$;
• scalar products such as $\mathbf L\cdot\mathbf S$ when $\mathbf L$ and $\mathbf S$ are both rotated by the same total generator;
• rotationally invariant Hamiltonians such as

$$H = \frac{\mathbf P^2}{2m} + V(\mathbf R^2).$$

If an operator is a rotational scalar, it commutes with the rotation generators:

$$[J_i,O]=0.$$

Consequently it cannot change $j$ or $m$, even when several copies of the same angular-momentum representation occur. By the rank-$0$ case of the Wigner–Eckart theorem,

$$\langle \alpha',j',m'|O|\alpha,j,m\rangle =0$$

unless $j'=j$ and $m'=m$, though the operator may still act nontrivially on additional labels $\alpha$.

For example, in an arbitrary radial basis a central Hamiltonian can have off-diagonal radial matrix elements at fixed $j$. Within each angular multiplet its matrix elements are independent of $m$. Diagonalizing the radial part finds the energy eigenstates; rotational invariance alone does not do that calculation.

Vector Operators

A vector operator has three components that rotate into one another. Standard examples include:

• the position operator $\mathbf R$;
• the momentum operator $\mathbf P$;
• angular momentum $\mathbf J$ itself under rotations generated by $\mathbf J$;
• the electric dipole moment $\mathbf d=\sum_a q_a\mathbf r_a$;
• magnetic moment operators, under proper rotations.

For proper rotations, polar vectors and axial vectors both transform with the same $3\times3$ matrix $\mathcal R$. They differ under parity. For example, $\mathbf R$ is parity odd, while $\mathbf L=\mathbf R\times\mathbf P$ is parity even. That parity distinction matters for selection rules but is separate from the SO(3) vector classification.

A vector operator does not mean “any operator with three entries.” The three entries must transform as a vector under the same rotation generator. If $\mathbf A$ is assembled from three unrelated observables, it is not a vector operator just because it has three components.

External Fields and Symmetry Breaking

External fields are a common source of confusion. A coupling such as

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

is a scalar if both $\boldsymbol\mu$ and the background $\mathbf B$ are rotated together. But in a laboratory Hamiltonian, $\mathbf B$ is usually a fixed external direction. For nonzero $\mathbf B$ and a nontrivial vector coupling, this term generally preserves only rotations that leave $\mathbf B$ fixed.

For $\mathbf B=B\hat{\mathbf z}$, the coupling becomes

$$H_B=-B\mu_z.$$

This preserves rotations about the $z$ axis but breaks the full rotational symmetry. Therefore $J_z$ may remain a good label while $J^2$ need not be protected by the same argument. Always distinguish transformation covariance of an operator from invariance of a particular Hamiltonian with fixed backgrounds.

Two-Index Tensor Operators

A Cartesian two-index tensor operator $T_{ij}$ transforms with one rotation matrix for each index:

$$U^\dagger T_{ij}U = \sum_{a,b} \mathcal R_{ia} \mathcal R_{jb} T_{ab}.$$

Such a tensor is usually reducible under rotations. It can be separated into three pieces:

$$T_{ij} = \frac{1}{3}\delta_{ij}T_{aa} + \frac{1}{2}(T_{ij}-T_{ji}) + \left[ \frac{1}{2}(T_{ij}+T_{ji}) - \frac{1}{3}\delta_{ij}T_{aa} \right],$$

where repeated Cartesian indices are summed. The three terms are:

Cartesian piece	Rotational content	Dimension
trace $T_{aa}$	scalar	$1$
antisymmetric part	vector-like	$3$
symmetric traceless part	rank-$2$ tensor	$5$

The dimensions add as

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

This is the Cartesian version of angular-momentum addition:

$$1\otimes1 = 0\oplus1\oplus2.$$

The symmetric traceless piece is the part usually meant by a quadrupole tensor. For a charge distribution, a common Cartesian quadrupole operator is

$$Q_{ij} = \sum_a q_a \left( 3r_{a,i}r_{a,j} - \delta_{ij}r_a^2 \right),$$

which is symmetric and traceless:

$$Q_{ij}=Q_{ji}, \qquad \sum_i Q_{ii}=0.$$

The precise numerical factor is conventional; the transformation type is not.

Reducible Versus Irreducible

“Tensor operator” is sometimes used in two different ways:

• a Cartesian tensor with one or more indices;
• an irreducible spherical tensor $T_q^{(k)}$ of definite rank $k$.

These are related but not identical. A general Cartesian two-index tensor contains rank-$0$, rank-$1$, and rank-$2$ rotational pieces. An irreducible spherical tensor contains only one rank.

For selection-rule work, the irreducible form is usually the useful one. A rank-$k$ spherical tensor component $T_q^{(k)}$ carries a definite angular-momentum label $k$ and component $q$. The Wigner–Eckart theorem can then read off angular dependence from Clebsch–Gordan coefficients.

A quadrupole calculation therefore starts by removing the trace, then converting the remaining five components into a rank-$2$ spherical tensor. Applying a rank-$2$ rule to the unreduced nine components would incorrectly constrain its scalar and vector parts.

Matrix-Element Consequences

The classification controls which matrix elements can be nonzero before any radial integral is evaluated.

For a scalar operator:

$$\Delta j=0, \qquad \Delta m=0.$$

For a vector operator, after rewriting it in spherical components $V_q$ with $q=0,\pm1$, angular momentum allows

$$m'=m+q$$

and

$$|j-1|\le j'\le j+1,$$

with $j'$ advancing in integer steps. In particular, $j=0$ requires $j'=1$: a vector cannot connect two $j=0$ states. Particular Clebsch–Gordan coefficients can impose further zeros.

For a rank-$2$ irreducible tensor:

$$q=-2,-1,0,1,2, \qquad m'=m+q,$$

and

$$|j-2|\le j'\le j+2.$$

These are only rotational rules. Parity, exchange symmetry, charge conservation, time reversal, and dynamical details can impose additional zeros.

Common Mistakes

• Treating a fixed external field as if it rotates when testing Hamiltonian invariance.
• Calling $\mathbf B\cdot\boldsymbol\mu$ rotationally invariant without specifying whether $\mathbf B$ is a transformed background or a fixed laboratory direction.
• Forgetting that polar and axial vectors transform the same way under proper rotations but differently under parity.
• Treating every two-index tensor as an irreducible rank-$2$ tensor; the trace and antisymmetric pieces have different rotational rank.
• Assuming that a scalar operator is proportional to the identity on every label. It is identity-like only within an irreducible angular-momentum multiplet; it may still act on radial, internal, or degeneracy labels.
• Using selection rules without naming the operator whose transformation law is being used.

Exercises

1. Show that $\mathbf R^2$ is a scalar if $\mathbf R$ is a vector operator.

Solution

Use

$$[J_i,R_j] = i\hbar \sum_k\epsilon_{ijk}R_k.$$

Then

$$[J_i,\mathbf R^2] = \sum_j \left( [J_i,R_j]R_j + R_j[J_i,R_j] \right).$$

Substituting the vector commutator gives

$$[J_i,\mathbf R^2] = i\hbar \sum_{j,k} \epsilon_{ijk} \left( R_kR_j+R_jR_k \right).$$

The factor in parentheses is symmetric under $j\leftrightarrow k$, while $\epsilon_{ijk}$ is antisymmetric. The sum vanishes, so $[J_i,\mathbf R^2]=0$.

2. Take $T_{ij}=R_iR_j$. Which of the three irreducible pieces vanish, and what remains?

Solution

Position components commute, so $T_{ij}-T_{ji}=[R_i,R_j]=0$: there is no rank-$1$ part. The trace is $R^2$, and

$$R_iR_j=\frac{\delta_{ij}}3R^2+ \left(R_iR_j-\frac{\delta_{ij}}3R^2\right).$$

The second term is symmetric and traceless and supplies the five quadrupole components. Thus this particular tensor contains ranks $0$ and $2$, even though a general two-index tensor can also contain rank $1$.

3. An otherwise rotationally invariant Hamiltonian contains a nonzero coupling $H_B=-B\mu_z$. Which continuous rotations generically remain symmetries when $B\ne0$ is fixed by the laboratory?

Solution

Only rotations about the $z$ axis leave the fixed vector $B\hat{\mathbf z}$ unchanged. Because $[J_z,\mu_z]=0$, rotations about $z$ remain symmetries of this Hamiltonian. The transverse commutators are generally nonzero. If the coupling vanishes, full rotational symmetry is restored; additional accidental symmetries must be checked separately.

4. Let $O$ be a rotational scalar. Explain why $\langle \alpha',j',m'|O|\alpha,j,m\rangle$ vanishes when $j'\ne j$ or $m'\ne m$, assuming the states are organized into irreducible angular-momentum multiplets.

Solution

A scalar commutes with all components of $\mathbf J$, so it does not change the angular-momentum representation. Equivalently, it is a rank-$0$ tensor, and the rank-$0$ Wigner–Eckart selection rules give $j'=j$ and $m'=m$. The labels $\alpha$ and $\alpha'$ can still be affected if they distinguish states inside repeated copies of the same angular-momentum representation.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon Press, 1977.
• 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.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.