Wigner–Eckart Theorem

Purpose

The Wigner–Eckart theorem separates every matrix element of an irreducible spherical tensor into

• a universal angular coefficient fixed by $j,m,k,q,j',m'$;
• one reduced matrix element containing radial, dynamical, and internal-state information.

Let $\alpha$ and $\alpha'$ collect all labels other than angular momentum. In the convention used throughout this card,

$$\begin{aligned} &\langle\alpha',j',m'\rvert T_q^{(k)} \lvert\alpha,j,m\rangle \\ &\qquad= (-1)^{j'-m'} \begin{pmatrix} j'&k&j\\ -m'&q&m \end{pmatrix} \langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle. \end{aligned}$$

Equivalently,

$$\begin{aligned} &\langle\alpha',j',m'\rvert T_q^{(k)} \lvert\alpha,j,m\rangle \\ &\qquad= \frac{ \langle j,m;k,q\vert j',m'\rangle }{ \sqrt{2j'+1} } \langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle. \end{aligned}$$

The theorem and its representation-theoretic meaning are developed at Wigner–Eckart Theorem. This card fixes conventions and organizes direct calculations.

Convention block

The reduced matrix element in this card is defined by the two displayed equations. Other books may absorb $\sqrt{2j+1}$ or $\sqrt{2j'+1}$ into the double-bar quantity. Never compare reduced matrix elements by symbol alone; compare the defining equation.

An irreducible spherical tensor of rank $k$ has

$$q=-k,-k+1,\ldots,k$$

and satisfies

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

For a Cartesian vector $\mathbf V$, the spherical-component convention is

$$V_0=V_z, \qquad V_{\pm1} = \mp\frac{V_x\pm iV_y}{\sqrt2}.$$

Changing the spherical-component phases requires corresponding changes in coefficient tables and reduced matrix elements.

At a glance

Task	Formula or rule
Factor a matrix element	angular coefficient times reduced matrix element
Magnetic rule	$m'=m+q$
Triangle rule	$\lvert j-k\rvert\le j'\le j+k$
Tensor components	$q=-k,\ldots,k$
Extract the reduced element	multiply by $\sqrt{2j'+1}$ and divide by a nonzero Clebsch–Gordan coefficient
Scalar specialization	$k=0$ implies $j'=j$ and $m'=m$
Vector specialization	$k=1$ permits $j'=j-1,j,j+1$ subject to valid labels

Rotational permission is necessary, not sufficient. Parity, exchange symmetry, additional conserved charges, and zeros of the reduced matrix element can still eliminate an angularly allowed amplitude.

Angular selection rules

The Clebsch–Gordan or $3j$ coefficient vanishes unless

$$m'=m+q$$

and

$$\lvert j-k\rvert \le j' \le j+k.$$

The labels must also satisfy

$$\lvert m\rvert\le j, \qquad \lvert m'\rvert\le j', \qquad \lvert q\rvert\le k,$$

and the angular momenta must have consistent integer or half-integer parity so that $j+k+j'$ is an integer.

The full decision sequence is:

1. identify the irreducible rank $k$ and component $q$ of the physical operator;
2. apply the magnetic and triangle rules;
3. apply parity and other discrete-symmetry rules;
4. check spectator labels and identical-particle constraints;
5. evaluate or import the reduced matrix element;
6. remember that an allowed reduced matrix element may vanish dynamically.

An “allowed transition” means not forced to zero by the stated rules. It does not mean large, observable, or resonant.

Reduced matrix element

The double-bar quantity

$$\langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle$$

is independent of $m,m'$, and $q$. It can still depend on

• $j,j'$ and the additional labels $\alpha,\alpha'$;
• the normalization and physical definition of $T^{(k)}$;
• radial wavefunctions, coupling constants, charges, and model parameters;
• phase and reduced-matrix-element conventions.

If one nonzero ordinary matrix element is known, then

$$\begin{aligned} &\langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle \\ &\qquad= \sqrt{2j'+1} \frac{ \langle\alpha',j',m'\rvert T_q^{(k)} \lvert\alpha,j,m\rangle }{ \langle j,m;k,q\vert j',m'\rangle }, \end{aligned}$$

provided the denominator is nonzero. Once extracted, the same reduced element generates every magnetic-sublevel amplitude between those two multiplets.

For two allowed amplitudes sharing the same reduced element,

$$\frac{ \langle j'm_1'\rvert T_{q_1}^{(k)}\lvert jm_1\rangle }{ \langle j'm_2'\rvert T_{q_2}^{(k)}\lvert jm_2\rangle } = \frac{ \langle j,m_1;k,q_1\vert j',m_1'\rangle }{ \langle j,m_2;k,q_2\vert j',m_2'\rangle }.$$

Thus relative sublevel amplitudes and line strengths can often be obtained without computing any radial integral.

Orthogonality and line-strength checks

Clebsch–Gordan orthogonality implies, for fixed $j',m'$,

$$\begin{aligned} &\sum_{m,q} \left\lvert \langle\alpha',j',m'\rvert T_q^{(k)} \lvert\alpha,j,m\rangle \right\rvert^2 \\ &\qquad= \frac{ \left\lvert \langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle \right\rvert^2 }{2j'+1}. \end{aligned}$$

Summing over the final magnetic sublevel as well gives

$$\begin{aligned} &\sum_{m,m',q} \left\lvert \langle\alpha',j',m'\rvert T_q^{(k)} \lvert\alpha,j,m\rangle \right\rvert^2 \\ &\qquad= \left\lvert \langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle \right\rvert^2. \end{aligned}$$

Therefore an average over the $2j+1$ initial magnetic substates, with a sum over all final substates and tensor components, is

$$\frac{1}{2j+1} \left\lvert \langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle \right\rvert^2.$$

These identities are convention dependent through the normalization of the reduced element, but they are exact for the convention fixed here. Real experimental line strengths may include field polarization, populations, frequency factors, density of states, and unresolved degeneracies in addition to this angular sum.

Hermitian tensors

For a spherical tensor built from a Hermitian physical observable, the components commonly satisfy

$$\left(T_q^{(k)}\right)^\dagger = (-1)^qT_{-q}^{(k)}.$$

In the present convention, the reduced matrix elements then obey

$$\begin{aligned} &\langle\alpha',j'\lVert T^{(k)}\rVert\alpha,j\rangle \\ &\qquad= (-1)^{j'-j} \left[ \langle\alpha,j\lVert T^{(k)}\rVert\alpha',j'\rangle \right]^*. \end{aligned}$$

This is a reciprocity check, not permission to discard complex conjugation. If the tensor is not Hermitian or the component convention differs, use the adjoint tensor explicitly.

Scalar operators

For $k=0$, only $q=0$ exists. The triangle and magnetic rules force

$$j'=j, \qquad m'=m.$$

The Clebsch–Gordan coefficient is

$$\langle j,m;0,0\vert j,m\rangle=1,$$

so

$$\langle\alpha',j,m\rvert T_0^{(0)} \lvert\alpha,j,m\rangle = \frac{ \langle\alpha',j\lVert T^{(0)}\rVert\alpha,j\rangle }{ \sqrt{2j+1} }.$$

The ordinary matrix element is independent of $m$. A scalar may still mix additional labels $\alpha$ and $\alpha'$ when no other symmetry forbids it.

For the identity operator within one $j$ multiplet,

$$\langle j\lVert I^{(0)}\rVert j\rangle = \sqrt{2j+1}.$$

Vector operators

A vector is a rank-$1$ tensor with $q=-1,0,1$. Rotations permit

$$j'=j-1, j, j+1$$

when those values are nonnegative and satisfy the triangle rule. In particular, a rank-$1$ operator cannot connect $j=0$ to $j'=0$.

For the angular-momentum operator itself,

$$\langle j\lVert J^{(1)}\rVert j\rangle = \hbar \sqrt{j(j+1)(2j+1)}.$$

Using

$$\langle j,m;1,0\vert j,m\rangle = \frac{m}{\sqrt{j(j+1)}},$$

the theorem recovers

$$\langle j,m\rvert J_0\lvert j,m\rangle = \hbar m.$$

For a transition from $j=0,m=0$ through a vector component,

$$\begin{aligned} &\langle\alpha',1,m'\rvert T_q^{(1)} \lvert\alpha,0,0\rangle \\ &\qquad= \frac{\delta_{m'q}}{\sqrt3} \langle\alpha',1\lVert T^{(1)}\rVert\alpha,0\rangle. \end{aligned}$$

Each spherical component feeds the matching final magnetic sublevel.

Parity and multipole rules

Rotational rank does not determine parity. If the initial and final states have parities $\pi$ and $\pi'$ and the tensor has parity $\pi_T$, a nonzero matrix element requires

$$\pi'=\pi_T\pi.$$

For electric multipoles of order $L$, the operator parity is $(-1)^L$; for magnetic multipoles it is $(-1)^{L+1}$. These statements assume the standard electromagnetic multipole definitions.

For an electric dipole, $k=1$ and the operator is odd. Applied to central- potential orbital states, rotations alone permit $\ell' = \ell-1,\ell,\ell+1$, while parity removes $\ell'=\ell$. The combined rule is

$$\Delta\ell=\pm1.$$

Use Parity Selection Rules and Dipole Transitions for the full physical assumptions.

Composite systems and recoupling

When $T^{(k)}$ acts on one factor of a coupled state, the reduced matrix element between total-$J$ states can be expressed through a subsystem reduced matrix element and a Wigner $6j$ symbol. The exact phase and square-root factors depend on which factor is acted on and on the coupling order.

Do not improvise this recoupling formula from memory. Fix the basis as $\lvert(j_1j_2)JM\rangle$, state which subsystem carries the tensor, and use a convention-matched $6j$ formula from Recoupling and Wigner Symbols or the Wigner Symbols Table.

Calculation workflow

1. Verify that the operator has been decomposed into irreducible spherical tensors and record the $V_{\pm1}$ phase convention.
2. Record all state labels, including spectator labels $\alpha,\alpha'$.
3. Apply $m'=m+q$ and the triangle rule before looking up any coefficient.
4. Apply parity, exchange, and additional symmetry rules separately.
5. Choose either the $3j$ or Clebsch–Gordan form and stay in one convention.
6. Obtain one reduced matrix element from dynamics, a radial integral, data, or a convention-matched table.
7. Generate the remaining magnetic-sublevel amplitudes algebraically.
8. Check a normalization or line-strength sum when many components are used.

Common mistakes

• Applying the theorem before showing that the operator is an irreducible spherical tensor.
• Mixing reduced-matrix-element normalizations from different books.
• Forgetting the factor $1/\sqrt{2j'+1}$ in this convention.
• Treating the reduced matrix element as independent of $j,j'$ or of the physical operator.
• Forgetting $m'=m+q$ or the triangle condition.
• Assuming a rotationally allowed amplitude must be nonzero or large.
• Treating parity selection rules as part of the theorem rather than a separate symmetry constraint.
• Confusing Cartesian vector components with spherical components.
• Omitting complex conjugation in Hermitian reciprocity.
• Using a subsystem recoupling formula without fixing factor order and $6j$ convention.

Canonical links

• Irreducible Spherical Tensors defines rank and component conventions.
• Selection Rules separates exact symmetry zeros from weak mechanisms.
• Clebsch–Gordan Coefficients supplies the basis coefficients in the theorem.
• Wigner Symbols Quick Reference provides convention-aware angular coefficients.
• Atomic Selection Rules applies the theorem to spectroscopy.
• Selection Rules in Transition Rates embeds amplitudes in rates and experimental observables.

References

• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.
• D. M. Brink and G. R. Satchler, Angular Momentum, 3rd ed., Oxford University Press, 1993.
• R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.

Exercises

1. Specialize the theorem to a scalar operator and recover its reduced matrix element from one diagonal magnetic-sublevel matrix element.

Solution

For $k=0$, $q=0$, the triangle rule gives $j'=j$ and the magnetic rule gives $m'=m$. Since

$$\langle j,m;0,0\vert j,m\rangle=1,$$

the theorem becomes

$$\langle\alpha',j,m\rvert T_0^{(0)}\lvert\alpha,j,m\rangle = \frac{ \langle\alpha',j\lVert T^{(0)}\rVert\alpha,j\rangle }{\sqrt{2j+1}}.$$

Therefore

$$\langle\alpha',j\lVert T^{(0)}\rVert\alpha,j\rangle = \sqrt{2j+1} \langle\alpha',j,m\rvert T_0^{(0)}\lvert\alpha,j,m\rangle.$$

The right-hand side is independent of the chosen $m$.

2. A rank-$2$ component has $q=-1$ and acts on an initial $j=1/2$ state. List the rotationally allowed final $j'$ values and the magnetic change.

Solution

The triangle rule gives

$$\left\lvert\frac12-2\right\rvert \le j'\le \frac12+2,$$

so

$$j'=\frac32,\frac52.$$

The component label gives

$$m'=m-1.$$

For a specific initial $m$, one must still check that the resulting $m'$ lies inside the selected final multiplet.

3. A vector operator connects $j=1/2,m=1/2$ to the $j'=3/2$ multiplet. Compare the squared angular factors for $q=+1$ and $q=0$.

Solution

The relevant coefficients are

$$\left\langle \frac12,\frac12;1,1 \middle\vert \frac32,\frac32 \right\rangle =1$$

and

$$\left\langle \frac12,\frac12;1,0 \middle\vert \frac32,\frac12 \right\rangle = \sqrt{\frac23}.$$

The common reduced element and the common factor $1/\sqrt{2j'+1}$ cancel in the ratio. Therefore the squared angular factors are in the ratio

$$1:\frac23 = 3:2.$$

This ratio is purely angular; an observed intensity ratio may contain additional polarization and population factors.

4. Explain why the electric-dipole orbital rule is $\Delta\ell=\pm1$, even though rank one alone also permits $\Delta\ell=0$.

Solution

Rank one gives the triangle possibilities

$$\ell'=\ell-1,\ell,\ell+1.$$

The electric-dipole operator is odd under parity, while a central-potential orbital state has parity $(-1)^\ell$. A nonzero dipole matrix element therefore requires opposite initial and final parity. The case $\ell'=\ell$ has the same parity and is removed. The remaining possibilities are

$$\Delta\ell=\pm1.$$