advanced30 min studyLesson 405 in Physics

The Wigner–Eckart theorem and selection rules

Separate angular geometry from reduced dynamics and distinguish an allowed transition from a guaranteed transition.

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What you’ll understand

  • State a reduced-matrix-element convention explicitly.
  • Derive magnetic and triangle selection rules for a spherical tensor.
  • Identify dynamical zeros that remain possible after angular tests pass.

Many transition amplitudes differ only because the initial state, final state, or probing field points in a different direction. The Wigner–Eckart theorem organizes these relationships by separating a known angular coefficient from an unknown reduced matrix element. It can eliminate a forbidden transition before any radial integral is calculated. It cannot usually prove that a transition occurs, since the remaining dynamical factor may vanish or the required field component may be absent.

Operators can carry angular momentum labels

A spherical tensor of rank kk consists of 2k+12k+1 operator components that transform among themselves like a spin-kk multiplet. A scalar has rank zero, and an ordinary vector has rank one. Tensor rank describes transformation behavior, not the number of spatial indices in an arbitrary notation. A Cartesian array must be decomposed into irreducible parts before a single-rank theorem can be applied. State labels can include additional quantum numbers, collectively denoted by α\alpha, which rotations leave untouched.

The commutator with the z generator says that component qq changes a magnetic label by qq. The raising and lowering commutators relate all components and force their matrix elements into the same angular pattern. For a vector, adopt T0(1)=TzT^{(1)}_0=T_z and T+1(1)=−(Tx+iTy)/2T^{(1)}_{+1}=-(T_x+iT_y)/\sqrt2, with the corresponding positive combination for the negative component. Declaring this convention prevents the signs of circular components from being silently reversed.

[Jz,Tq(k)]=ℏqTq(k),[J±,Tq(k)]=ℏk(k+1)−q(q±1) Tq±1(k)[J_z,T^{(k)}_q]=\hbar qT^{(k)}_q,\qquad [J_\pm,T^{(k)}_q]=\hbar\sqrt{k(k+1)-q(q\pm1)}\,T^{(k)}_{q\pm1}
The tensor components have the units of the physical operator. The dimensionless indices q run from minus k to k, and the equations define an irreducible spherical tensor.↔ Scroll wide equations horizontally.

One reduced number determines an angular family

Use the convention in which a matrix element equals a Clebsch–Gordan coefficient times a reduced matrix element, with no additional square-root factor. Other references place a square root of a multiplet dimension in the formula; their reduced matrix elements therefore have a different normalization. Either convention predicts the same actual amplitudes when used consistently. The reduced factor retains the physical units of the tensor operator because the angular coefficient is dimensionless.

The theorem applies between states belonging to specified irreducible rotation representations and to an operator with the stated tensor transformation law. An isotropic Hamiltonian supplies natural energy states with these labels, but the algebraic theorem is about representations, not a numerical approximation to the interaction. If an external field mixes different angular-momentum sectors, its actual eigenstates must first be expanded into the relevant basis. In that situation, directly assigning one unmixed label to the state can produce false selection rules.

⟨α′j′m′∣Tq(k)∣αjm⟩=⟨jm;kq∣j′m′⟩ Rα′j′,αj(k)\langle\alpha^{\prime}j^{\prime}m^{\prime}|T^{(k)}_q|\alpha jm\rangle=\langle jm;kq|j^{\prime}m^{\prime}\rangle\,\mathcal R_{\alpha^{\prime}j^{\prime},\alpha j}^{(k)}
This defines the reduced quantity R with no extra dimension factor, matching the declared Clebsch–Gordan convention. R is independent of m, m prime, and q.↔ Scroll wide equations horizontally.

Allowed means that symmetry has not ruled it out

A Clebsch–Gordan coefficient vanishes unless the final magnetic label equals the initial label plus the component index, and the final angular momentum satisfies the triangle inequalities. For a vector operator, zero-to-zero is forbidden even though a superficial rule saying the change is at most one might suggest otherwise. There can also be zeros of a particular angular coefficient inside the triangle range. Checking the actual coefficient is stronger than checking the broad necessary conditions.

Additional symmetries impose independent constraints. Parity can kill an electric-dipole amplitude whose angular coefficient is nonzero. A radial integral can vanish through cancellation, and a probing field might have no component that drives the needed tensor index. Transition rates also depend on field strength, final-state density, and dynamical assumptions. Thus the theorem organizes amplitudes and relative strengths, while a complete spectroscopy prediction needs the interaction Hamiltonian and the physical preparation of the radiation.

m′=m+q,∣j−k∣≤j′≤j+k,∣m′∣≤j′m^{\prime}=m+q,\qquad |j-k|\leq j^{\prime}\leq j+k,\qquad |m^{\prime}|\leq j^{\prime}
These are necessary angular conditions, not a guarantee of a nonzero amplitude. In a pure rank-one transition they exclude j=0 to j prime=0.↔ Scroll wide equations horizontally.

A scalar initial state and a vector probe

An initial state has j=m=0j=m=0. A rank-one tensor connects it to a j′=1j^{\prime}=1 multiplet with reduced matrix element R=3d0\mathcal R=3d_0, where d0d_0 has dipole units. Apply T+1(1)T^{(1)}_{+1} and find the amplitudes to final magnetic labels +1 and 0.

  1. The magnetic condition gives m′=0+1=1m^{\prime}=0+1=1. The final label zero fails this condition, so its matrix element vanishes independently of the reduced factor.

  2. Coupling angular momentum zero to rank one leaves only total label one. The coefficient ⟨0,0;1,1∣1,1⟩=1\langle0,0;1,1|1,1\rangle=1 in the chosen basis convention, because there is only one normalized product state for that label.

  3. Multiply the nonzero coefficient by the reduced factor to obtain 3d03d_0. Its squared magnitude is 9∣d0∣29|d_0|^2, while the other squared matrix element is zero. These are strengths, not normalized detection probabilities or transition rates by themselves.

The result

The +1 amplitude is 3d0; the zero-label amplitude vanishes by the magnetic selection rule.

Explore the model

Interactive lab

Interactive lab · vectors

A probe component can vanish geometrically

Identify the plotted x projection with a chosen longitudinal probe component. In fresh runs Reset and set angle and magnitude to (0∘,1)(0^\circ,1), then (90∘,1)(90^\circ,1). Predict x readouts 1 and 0. This planar vector projection illustrates how a driving component disappears; it does not compute spherical-tensor coefficients, polarization helicity, or a transition rate.

A probe component can vanish geometricallyPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn angularly permitted matrix element still needs the appropriate nonzero driving component.

Check your understanding

A nonzero angular coefficient has been found for a transition. Which conclusion is justified?

Try it yourself

For j=1j=1, take the vector operator T=J\mathbf T=\mathbf J and initial state ∣1,0⟩|1,0\rangle. Compute the matrix elements of T+1(1)T^{(1)}_{+1} to ∣1,1⟩|1,1\rangle and of T0(1)T^{(1)}_0 to ∣1,0⟩|1,0\rangle. Explain why one broad allowed channel has zero strength.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use T+1(1)=−J+/2T^{(1)}_{+1}=-J_+/\sqrt2 and T0(1)=JzT^{(1)}_0=J_z, together with the ladder coefficient for the initial state.

Compare with the solution

The raising coefficient is 2ℏ\sqrt2\hbar, so the first matrix element is −ℏ-\hbar. Its strength is ℏ2\hbar^2. The second is ℏm=0\hbar m=0 even though the magnetic and triangle inequalities permit j=1,m=0j=1,m=0 to itself with q=0q=0. Necessary inequalities do not exclude zeros of the specific angular coefficient.

Take it with you

  • The Wigner–Eckart theorem factors angular geometry from a reduced matrix element.
  • Reduced-matrix-element normalizations vary, so the convention must be declared.
  • Selection rules identify required zeros; passing them does not guarantee a nonzero transition.

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