The Wigner-Eckart Theorem

Derivation

The Wigner-Eckart Theorem

D-354 Home PU-401 Threads symmetry · matter Depends on Addition of Angular Momenta

Statement

For a spherical tensor operator \(\hat{T}^{(k)}_q\) of rank \(k\), the matrix element between angular-momentum eigenstates \(\lvert \alpha\, j\, m\rangle\) and \(\lvert \alpha'\, j'\, m'\rangle\) factors into a purely geometric Clebsch–Gordan coefficient carrying all the \(m,q,m'\) dependence, times a single reduced matrix element that depends only on \(\alpha,\alpha',j,j',k\): \[ \langle \alpha'\, j'\, m' \lvert \hat{T}^{(k)}_q \rvert \alpha\, j\, m\rangle = \langle j\, m;\, k\, q \,\vert\, j'\, m'\rangle \,\frac{\langle \alpha'\, j' \,\Vert\, \hat{T}^{(k)} \,\Vert\, \alpha\, j\rangle}{\sqrt{2j'+1}}. \]

Why it matters

The theorem is the workhorse of every calculation involving rotational symmetry in quantum mechanics. It collapses a table of \((2j'+1)(2k+1)(2j+1)\) apparently independent matrix elements onto a single physical number — the reduced matrix element — with everything else fixed by group theory. Selection rules (dipole transitions, forbidden lines, magnetic-resonance intensities) fall out immediately from when the Clebsch–Gordan coefficient vanishes.

Physically it separates geometry from dynamics: how a transition is oriented in space (the CG factor) is universal, while how strong it intrinsically is (the reduced element) encodes the specific interaction. This is why one measured line intensity predicts all its Zeeman components, and why the same reduced element governs both absorption and emission.

Assumptions

The states carry a definite total angular momentum \(j\) and project onto \(SU(2)\) irreducible representations.If \(\lvert\alpha j m\rangle\) is not a rotation eigenstate, \(\hat{J}_\pm\) does not act with the standard ladder coefficients and the recursion that pins the matrix elements to CG coefficients never closes. \(\hat{T}^{(k)}_q\) is an irreducible spherical tensor, i.e. it satisfies \([\hat{J}_z,\hat{T}^{(k)}_q]=\hbar q\,\hat{T}^{(k)}_q\) and \([\hat{J}_\pm,\hat{T}^{(k)}_q]=\hbar\sqrt{k(k+1)-q(q\pm1)}\,\hat{T}^{(k)}_{q\pm1}\).A Cartesian or reducible operator mixes ranks; the factorization then holds only after decomposing it into irreducible pieces, one reduced element per rank. Rotational invariance of the underlying theory (Wigner's theorem): rotations are represented by unitary operators \(\hat{U}(R)=e^{-i\theta\hat{n}\cdot\hat{\mathbf J}/\hbar}\).Without a unitary rotation representation the transformation law \(\hat{U}\hat{T}^{(k)}_q\hat{U}^\dagger=\sum_{q'}\mathcal{D}^{(k)}_{q'q}\hat{T}^{(k)}_{q'}\) fails and Schur's lemma cannot be invoked.

Derivation

\[ \hat{T}^{(k)}_q\lvert \alpha\, j\, m\rangle \;\equiv\; \lvert \psi\rangle \]

Define the state obtained by acting with one tensor component. The whole strategy is to show \(\lvert\psi\rangle\) transforms under rotations exactly like a coupled state \(\lvert k q\rangle\otimes\lvert j m\rangle\). A

\[ \hat{J}_z\,\hat{T}^{(k)}_q\lvert j m\rangle =\big([\hat{J}_z,\hat{T}^{(k)}_q]+\hat{T}^{(k)}_q\hat{J}_z\big)\lvert j m\rangle =\hbar(q+m)\,\hat{T}^{(k)}_q\lvert j m\rangle \]

Insert the commutator \([\hat{J}_z,\hat{T}^{(k)}_q]=\hbar q\,\hat{T}^{(k)}_q\) and use \(\hat{J}_z\lvert jm\rangle=\hbar m\lvert jm\rangle\). (Suppressing \(\alpha\) for brevity.) A

\[ \langle \alpha'\, j'\, m'\lvert \hat{T}^{(k)}_q\rvert \alpha\, j\, m\rangle = 0 \quad\text{unless}\quad m'=q+m \]

Sandwich Step 2 between \(\langle\alpha'j'm'\rvert\) and use \(\langle\alpha'j'm'\rvert\hat{J}_z=\hbar m'\langle\alpha'j'm'\rvert\). Both sides give \(\hbar m'\) versus \(\hbar(q+m)\) times the matrix element, forcing it to vanish off the diagonal. This is the first CG selection rule. A

\[ \hat{J}_\pm\,\hat{T}^{(k)}_q\lvert j m\rangle =[\hat{J}_\pm,\hat{T}^{(k)}_q]\lvert j m\rangle + \hat{T}^{(k)}_q\hat{J}_\pm\lvert j m\rangle \]

Apply the ladder operator and split with the Leibniz rule for commutators — the key move that generates a recursion in \((q,m)\). B

\[ \hat{J}_\pm\lvert\psi\rangle =\hbar\sqrt{k(k{+}1){-}q(q{\pm}1)}\;\hat{T}^{(k)}_{q\pm1}\lvert j m\rangle +\hbar\sqrt{j(j{+}1){-}m(m{\pm}1)}\;\hat{T}^{(k)}_{q}\lvert j,m{\pm}1\rangle \]

Substitute the tensor ladder commutator and the state ladder relation \(\hat{J}_\pm\lvert jm\rangle=\hbar\sqrt{j(j{+}1){-}m(m{\pm}1)}\lvert j,m{\pm}1\rangle\). B

\[ \hat{J}_\pm\big(\lvert k q\rangle\otimes\lvert j m\rangle\big) =\hbar\sqrt{k(k{+}1){-}q(q{\pm}1)}\,\lvert k,q{\pm}1\rangle\otimes\lvert j m\rangle +\hbar\sqrt{j(j{+}1){-}m(m{\pm}1)}\,\lvert k q\rangle\otimes\lvert j,m{\pm}1\rangle \]

Write the ladder action on the tensor-product basis of two angular momenta \(k\) and \(j\). Compare with Step 5: the coefficients are identical. Hence \(\hat{T}^{(k)}_q\lvert jm\rangle\) obeys the same \(\hat{\mathbf J}\)-algebra as \(\lvert kq\rangle\otimes\lvert jm\rangle\). C

\[ \lvert \Psi^{J M}\rangle \equiv \sum_{q,m}\langle k\,q;\, j\,m\,\vert\, J\,M\rangle\;\hat{T}^{(k)}_q\lvert j m\rangle \]

Because the set \(\{\hat{T}^{(k)}_q\lvert jm\rangle\}\) carries the SAME representation of \(\hat{\mathbf J}\) as the product basis, projecting it with Clebsch–Gordan coefficients builds states of definite total \((J,M)\): \(\hat J^2\lvert\Psi^{JM}\rangle=\hbar^2 J(J{+}1)\lvert\Psi^{JM}\rangle\), \(\hat J_z\lvert\Psi^{JM}\rangle=\hbar M\lvert\Psi^{JM}\rangle\). This uses only that both sides satisfy the identical ladder recursions, so the same linear combination diagonalizes \(\hat J^2\). C

\[ \langle \alpha' j' m' \,\vert\, \Psi^{J M}\rangle = \delta_{j' J}\,\delta_{m' M}\; C(\alpha',j';\alpha,j,k) \]

\(\lvert\Psi^{JM}\rangle\) is a genuine angular-momentum eigenstate, and \(\langle\alpha'j'm'\rvert\) is another; by orthogonality of irreducible representations (Schur's lemma) their overlap vanishes unless \(j'=J,\;m'=M\), and when they match it is a single \(M\)-independent constant \(C\). \(M\)-independence follows because \(\hat J_\pm\) relates different \(M\) with fixed coefficients on both states. C

\[ \langle \alpha' j' m'\lvert \hat{T}^{(k)}_q\rvert \alpha j m\rangle =\sum_{J,M}\langle k q; j m\,\vert\, J M\rangle\,\langle\alpha'j'm'\vert\Psi^{JM}\rangle =\langle k q; j m\,\vert\, j' m'\rangle\, C \]

Invert Step 7 (the CG matrix is orthogonal/unitary, so \(\hat T^{(k)}_q\lvert jm\rangle=\sum_{JM}\langle kq;jm\vert JM\rangle\lvert\Psi^{JM}\rangle\)), then insert Step 8. Only \(J=j',M=m'\) survives, leaving one CG coefficient times the constant \(C\). C

\[ C \equiv \frac{\langle \alpha' j'\Vert \hat{T}^{(k)}\Vert \alpha j\rangle}{\sqrt{2j'+1}} \]

Define the reduced matrix element by absorbing the constant \(C\) and the conventional normalization \(\sqrt{2j'+1}\) (Sakurai convention). The reduced element is what remains once all \(m,q,m'\) dependence is stripped out. A

Result

\[ \boxed{\;\langle \alpha'\, j'\, m'\lvert \hat{T}^{(k)}_q\rvert \alpha\, j\, m\rangle = \langle j\, m;\, k\, q \,\vert\, j'\, m'\rangle\;\frac{\langle \alpha'\, j'\Vert \hat{T}^{(k)}\Vert \alpha\, j\rangle}{\sqrt{2j'+1}}\;} \]

Reading. Every dependence on the magnetic quantum numbers \(m,q,m'\) — that is, on the orientation of the states and operator in space — lives entirely in the Clebsch–Gordan coefficient \(\langle jm;kq\vert j'm'\rangle\), a pure number fixed by group theory. All the physics of the specific operator and states is compressed into one orientation-independent scalar, the reduced matrix element \(\langle\alpha'j'\Vert\hat T^{(k)}\Vert\alpha j\rangle\). Two selection rules are immediate: \(m'=m+q\) and the triangle condition \(\lvert j-k\rvert\le j'\le j+k\).

Units check. The CG coefficient and the factor \(1/\sqrt{2j'+1}\) are dimensionless, so both sides carry the units of the operator \(\hat T^{(k)}_q\). For an electric-dipole operator \(\hat T^{(1)}_q\sim e\hat r\) both the left side and the reduced element have units of C·m; for a dimensionless spherical harmonic \(\hat T^{(k)}_q = Y_{kq}\) all quantities are dimensionless. Consistency is automatic because only dimensionless geometric factors are peeled off.

Limiting cases

Scalar operator, \(k=0\): \(\langle jm;00\vert j'm'\rangle=\delta_{jj'}\delta_{mm'}\), so \(\langle\alpha'j'm'\vert\hat T^{(0)}\vert\alpha jm\rangle=\delta_{jj'}\delta_{mm'}\,\langle\alpha'j\Vert\hat T^{(0)}\Vert\alpha j\rangle/\sqrt{2j+1}\) — diagonal and \(m\)-independent, as any rotational invariant must be.
Vector operator, \(k=1\) (projection theorem): the reduced element of any vector \(\hat{\mathbf V}\) is proportional to that of \(\hat{\mathbf J}\) within a fixed \(j\), giving \(\langle jm'\vert\hat V_q\vert jm\rangle=\frac{\langle\hat{\mathbf J}\cdot\hat{\mathbf V}\rangle}{\hbar^2 j(j+1)}\langle jm'\vert\hat J_q\vert jm\rangle\) — the basis of the Landé \(g\)-factor.
Stretched transition, \(j'=j+k\), \(m'=j'\): the CG coefficient equals \(1\), so the matrix element equals the reduced element (up to \(1/\sqrt{2j'+1}\)) — the cleanest way to extract a reduced element from experiment.
Identical states, \(\alpha'=\alpha,\,j'=j\): only tensors of rank \(k\le 2j\) can have nonzero expectation, since the triangle rule needs \(j'=j\) reachable — why a spin-\(\tfrac12\) has no permanent quadrupole moment.

Breaks when

Rotational symmetry is broken at the operator level. If the "operator" is not a well-defined spherical tensor (e.g. a term that mixes ranks under rotation, or a symmetry-violating interaction), no single reduced element exists and the factorization fails — you must first decompose into irreducible tensors, one reduced element per rank.
States are not true \(\hat J^2\) eigenstates. Near-degenerate levels mixed by an external perturbation (strong-field Paschen–Back regime, configuration interaction across \(j\)) carry no sharp \(j\); the CG labels lose meaning and one must work in the correct diagonal basis before applying the theorem.
Relativistic / continuous or non-compact groups without the same Schur structure. For the Lorentz group (non-compact) or when the relevant symmetry is only approximate (isospin breaking), the "reduced matrix element" is no longer a single exact constant; symmetry-breaking corrections split what would be one number into several.
Half-integer total with time-reversal subtleties. Naïve application to matrix elements between Kramers-degenerate doublets can give phase/sign errors unless the antiunitary structure is tracked; the modulus factorizes but relative phases need care.

Failure modes

Swapping CG argument order/convention. \(\langle jm;kq\vert j'm'\rangle\ne\langle kq;jm\vert j'm'\rangle\) in general (they differ by \((-1)^{j+k-j'}\)); mixing Condon–Shortley, Sakurai, and 3\(j\)-symbol conventions silently corrupts signs.
Forgetting the \(\sqrt{2j'+1}\) normalization. The reduced element's numerical value is convention-dependent; students quote a value from a table using a different normalization and get intensities wrong by a factor \(\sqrt{2j'+1}\).
Confusing rank \(k\) with a magnetic index. Treating a Cartesian vector component \(\hat V_z\) as a spherical component \(\hat V_0\) directly, without the \(\hat V_0=\hat V_z\), \(\hat V_{\pm1}=\mp(\hat V_x\pm i\hat V_y)/\sqrt2\) map, drops the \(\mp1/\sqrt2\) factors.
Assuming the reduced element is real or positive. It is generally complex; only \(\lvert\langle j'\Vert T\Vert j\rangle\rvert^2\) (line strength) is physically constrained.
Applying it across different \(j\) with a projection-theorem shortcut. The \(\hat{\mathbf V}\propto\hat{\mathbf J}\) replacement is valid only within a single \(j\)-multiplet; using it for \(j'\ne j\) is wrong.

Discussion

The deep content of the theorem is a statement about representation theory dressed in physics language: the tensor operator \(\hat T^{(k)}_q\) is an intertwiner between the representation \(D^{(j)}\) carried by the ket and \(D^{(j')}\) carried by the bra. Schur's lemma says that a map between irreducible representations, commuting appropriately with the group action, is either zero or unique up to a scalar. That scalar is precisely the reduced matrix element. Everything else — the full \(m,q,m'\) structure — is the fixed Clebsch–Gordan machinery of how \(D^{(k)}\otimes D^{(j)}\) decomposes into \(\bigoplus_{j'} D^{(j')}\).

This is why the theorem generalizes far beyond \(SU(2)\). Any compact group with a Clebsch–Gordan (Wigner) decomposition has its own Wigner–Eckart theorem: \(SU(3)\) flavor symmetry in particle physics (isospin/hypercharge multiplets and the Gell-Mann–Okubo mass formula), point groups in crystal-field theory, and \(SO(3)\) for classical multipole radiation. The reduced element is the invariant content; the group's CG coefficients are the geometry.

Experimentally, the theorem underlies the additivity and ratios of spectral line intensities. In atomic and nuclear spectroscopy, the relative intensities of all Zeeman/hyperfine components of a line are pure squared CG coefficients (the reduced element cancels in ratios), so measuring one component predicts the rest. It is also the reason polarization and angular-distribution measurements — which probe the \(q\) dependence — directly read out geometry, decoupled from the dynamical strength.

A subtle rigor point: the proof establishes existence and uniqueness of the reduced element but says nothing about its value, which requires the actual dynamics (wavefunctions, radial integrals). The theorem's power is precisely this separation — it is a kinematic constraint from symmetry, not a dynamical calculation. In the language of the Wigner–Eckart / 3\(j\)-symbol formulation, \(\langle j'm'\vert\hat T^{(k)}_q\vert jm\rangle=(-1)^{j'-m'}\begin{pmatrix}j'&k&j\\-m'&q&m\end{pmatrix}\langle j'\Vert\hat T^{(k)}\Vert j\rangle\), which makes the symmetry under interchange of the three angular momenta manifest and is the form used in modern computation.

Common misconceptions. The theorem does not tell you the transition strength — it factors it. It does not require the perturbation causing the transition to be weak; it is exact whenever the states are genuine angular-momentum eigenstates. And the reduced element is not a matrix element of any particular \(q\) — it is a derived normalization constant whose value depends on the chosen convention.

Worked examples

Ratio of dipole line intensities in a \(j=1\to j'=0\) transition

Electric-dipole operator is a rank-1 spherical tensor \(\hat T^{(1)}_q\). We want the relative strengths of the three \(m=+1,0,-1\to m'=0\) components. Set up symbolically first. A

\[ \langle 0\,0\vert\hat T^{(1)}_q\vert 1\,m\rangle=\langle 1\,m;1\,q\vert 0\,0\rangle\,\frac{\langle 0\Vert\hat T^{(1)}\Vert 1\rangle}{\sqrt{2\cdot0+1}} \]

Apply Wigner–Eckart; the reduced element and \(\sqrt{2j'+1}=1\) are common to all three components. A

\[ \langle 1\,m;1\,q\vert 0\,0\rangle=\frac{(-1)^{1-m}}{\sqrt{3}}\,\delta_{q,-m},\qquad m=+1,0,-1 \]

Insert the known CG coefficients for \(1\otimes1\to0\): each equals \(\pm1/\sqrt3\) in magnitude, nonzero only when \(q=-m\). B

\[ \frac{\lvert\langle0\,0\vert\hat T^{(1)}_{-1}\vert1\,1\rangle\rvert^2:\lvert\langle0\,0\vert\hat T^{(1)}_{0}\vert1\,0\rangle\rvert^2:\lvert\langle0\,0\vert\hat T^{(1)}_{+1}\vert1\,{-}1\rangle\rvert^2}{} =\tfrac13:\tfrac13:\tfrac13 \]

Square each; the reduced element cancels in the ratio, leaving equal magnitudes \(1/3\). B

\[ I_{+1}:I_0:I_{-1}=1:1:1 \]

Reading. All three Zeeman components carry equal intensity, fixed entirely by geometry; the reduced element sets the overall scale but drops out of the ratio.

Units check. Ratios are dimensionless; individual \(I\propto\lvert\text{C·m}\rvert^2\) for a real dipole.

Landé projection: \(g\)-factor of a \({}^2P_{3/2}\) level

The magnetic moment operator \(\hat{\boldsymbol\mu}=-\mu_B(g_L\hat{\mathbf L}+g_S\hat{\mathbf S})/\hbar\) is a vector (rank-1). Use the projection theorem (\(k=1\), \(j'=j\)) to get its diagonal matrix elements. Take \(L=1,S=\tfrac12,J=\tfrac32\), \(g_L=1,g_S=2\). A

\[ \langle J m'\vert \hat V_q\vert J m\rangle=\frac{\langle\hat{\mathbf J}\cdot\hat{\mathbf V}\rangle}{\hbar^2 J(J+1)}\,\langle J m'\vert\hat J_q\vert J m\rangle \]

Projection theorem — a corollary of Wigner–Eckart for a vector within one \(J\)-multiplet, since both \(\hat{\mathbf V}\) and \(\hat{\mathbf J}\) share the same rank-1 CG structure. B

\[ g_J=g_L\frac{\langle\hat{\mathbf J}\cdot\hat{\mathbf L}\rangle}{\hbar^2 J(J{+}1)}+g_S\frac{\langle\hat{\mathbf J}\cdot\hat{\mathbf S}\rangle}{\hbar^2 J(J{+}1)} \]

Apply the projection to each of \(\hat{\mathbf L}\) and \(\hat{\mathbf S}\). Use \(\hat{\mathbf J}\cdot\hat{\mathbf S}=\tfrac12(J(J{+}1){+}S(S{+}1){-}L(L{+}1))\hbar^2\) and similarly for \(\hat{\mathbf J}\cdot\hat{\mathbf L}\). B

\[ g_J=1+\frac{J(J{+}1)+S(S{+}1)-L(L{+}1)}{2J(J{+}1)} =1+\frac{\tfrac{15}{4}+\tfrac34-2}{2\cdot\tfrac{15}{4}} \]

Insert \(g_L=1,g_S=2\) to get the standard Landé formula, then substitute \(J=\tfrac32,S=\tfrac12,L=1\): \(J(J{+}1)=\tfrac{15}{4}\), \(S(S{+}1)=\tfrac34\), \(L(L{+}1)=2\). B

\[ g_J=1+\frac{15/4+3/4-2}{15/2}=1+\frac{10/4}{15/2}=1+\frac13=\frac{4}{3}\approx1.333 \]

Reading. The whole vector operator collapses onto \(\hat{\mathbf J}\); the single number \(g_J=4/3\) then fixes every Zeeman sublevel splitting \(\Delta E=g_J\mu_B B\,m_J\). Measured value for \({}^2P_{3/2}\) is indeed \(4/3\).

Units check. \(g_J\) dimensionless; \(\Delta E=g_J\mu_B B\,m_J\) has units \([\text{J/T}][\text{T}]=\text{J}\), correct.

Problems

Show that a scalar operator (\(k=0\)) can only connect states with \(j'=j\) and \(m'=m\), and that its matrix element is independent of \(m\).
Solution
The CG coefficient \(\langle jm;00\vert j'm'\rangle=\delta_{jj'}\delta_{mm'}\). Wigner–Eckart gives \(\langle\alpha'j'm'\vert\hat T^{(0)}\vert\alpha jm\rangle=\delta_{jj'}\delta_{mm'}\,\langle\alpha'j\Vert\hat T^{(0)}\Vert\alpha j\rangle/\sqrt{2j+1}\). The right side has no residual \(m\) dependence, so all \(2j+1\) diagonal elements are equal — as required since a scalar commutes with all \(\hat J_i\) and cannot distinguish orientations.
For an electric-quadrupole transition (\(k=2\)) from \(j=\tfrac12\) to \(j'=\tfrac12\), determine whether it is allowed.
Solution
The triangle condition requires \(\lvert j-j'\rvert\le k\le j+j'\), i.e. \(0\le 2\le 1\). Since \(k=2>j+j'=1\), the CG coefficient vanishes: the transition is forbidden. A spin-\(\tfrac12\) system cannot support a rank-2 (quadrupole) matrix element — hence no static quadrupole moment for \(j=\tfrac12\).
Given \(\langle j'\Vert\hat T^{(1)}\Vert j\rangle=A\) for a \(j=2\to j'=2\) transition, find the ratio \(\langle 2\,2\vert\hat T^{(1)}_0\vert2\,2\rangle/\langle2\,1\vert\hat T^{(1)}_0\vert2\,1\rangle\).
Solution
Both share the reduced element and \(\sqrt{2j'+1}=\sqrt5\), so the ratio is \(\langle2\,2;1\,0\vert2\,2\rangle/\langle2\,1;1\,0\vert2\,1\rangle\). Using \(\langle j m;1\,0\vert j m\rangle=m/\sqrt{j(j+1)}\): numerator \(=2/\sqrt6\), denominator \(=1/\sqrt6\). Ratio \(=2\). This is the projection-theorem statement that \(\langle jm\vert\hat V_0\vert jm\rangle\propto m\).
A rank-1 tensor connects \(j=1\to j'=2\). List all nonzero components \((m,q,m')\) and state the selection rules used.
Solution
Selection rules: \(m'=m+q\), \(q\in\{-1,0,1\}\), \(m\in\{-1,0,1\}\), \(m'\in\{-2,\dots,2\}\), triangle \(\lvert1-2\rvert\le1\le1+2\) satisfied. Nonzero \((m,q,m')\): \((1,1,2),(1,0,1),(1,-1,0),(0,1,1),(0,0,0),(0,-1,-1),(-1,1,0),(-1,0,-1),(-1,-1,-2)\) — nine components, each a distinct CG coefficient times the same reduced element. Note \(m'=\pm2\) is reached only via \(m=\pm1,q=\pm1\).
Two vector operators \(\hat{\mathbf V}\) and \(\hat{\mathbf W}\) act within a \(j=1\) multiplet with reduced elements \(\langle1\Vert V\Vert1\rangle=a\), \(\langle1\Vert W\Vert1\rangle=b\). Show \(\langle 1m'\vert\hat V_q\vert1m\rangle/\langle1m'\vert\hat W_q\vert1m\rangle=a/b\) for every \((m,q,m')\), and explain the physical meaning.
Solution
Wigner–Eckart gives \(\langle1m'\vert\hat V_q\vert1m\rangle=\langle1m;1q\vert1m'\rangle\,a/\sqrt3\) and \(\langle1m'\vert\hat W_q\vert1m\rangle=\langle1m;1q\vert1m'\rangle\,b/\sqrt3\). The identical CG factor and \(\sqrt3\) cancel, leaving \(a/b\) independent of \((m,q,m')\). Physical meaning: within a single \(j\)-multiplet all vector operators are proportional to \(\hat{\mathbf J}\) (projection theorem), so any two are proportional to each other — the geometry is universal, only the scalar strength differs.