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Use flavor hypercharge $Y=B+S$, where $B$ is baryon number and $S$ is strangeness. The isospin coordinate is $I_3$. The baryon octet has coordinatesHere the nucleons have $S=0$, the Sigma baryons and Lambda baryon have $S=-1$, and the Xi baryons have $S=-2$. The central Sigma baryon belongs to an isospin triplet while the central Lambda baryon is an isospin singlet; equal coordinates do not identify the states.

The pseudoscalar meson octet isThe pions have $S=0$, the upper kaons have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral pion and the Eta octet state. This is the octet basis of flavor symmetry; the physical eta can also mix with the flavor-singlet state.

For the flavor SU(3) Cartan generators, choose the Hermitian physics convention and the inner product $⟨A,B⟩=\mathrm{tr}(AB)$. Then an orthonormal diagonal basis isStrictly, $i{h}_1,i{h}_2$ are elements of the anti-Hermitian SU(3) Lie algebra; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down quarks form an isospin doublet and the strange quark is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each quark has baryon number $1/3$, and their strangeness values are $(0,0,-1)$. It follows thatThese flavor hypercharge conventions differ from the electroweak hypercharge convention. If instead the inner product is $2\mathrm{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt{2}$, $t_8=h_2/\sqrt{2}$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt{3}$.

In the ordinary quark model, the proton has valence content $uud$ and charge $+1$, while the neutron has content $udd$ and charge zero. Additivity of electric charge gives $2q_u+q_d=1$ and $q_u+2q_d=0$, hence $q_u=2/3$, $q_d=-1/3$. The Sigma baryon ${\Sigma }^{-}$ has content $dds$ and charge $-1$, giving $2q_d+q_s=-1$. Thus the quark triplet's electric charges, in units of the positive elementary charge, areThe Gell-Mann--Nishijima formula is consequentlyIt also reproduces every baryon octet charge from the first diagram. On antiquarks the additive quantum numbers reverse sign, and combining a quark with an antiquark reproduces the meson octet charges.

Because the down and strange quarks have identical electric charge, $Q$ commutes with the U-spin generatorsThey satisfy $[U_a,U_b]=i{ϵ}_{abc}{U}_c$; equivalently the anti-Hermitian matrices $i{U}_a$ span an $\mathfrak{su}(2)$ subalgebra. The entries on the $d,s$ block of $Q$ are equal, so $[Q,U_a]=0$ for all three U-spin generators. The electric charge is therefore constant within each irreducible U-spin multiplet. For example, U-spin relates ${\pi }^{+}$ and $K^{+}$, and relates $p$ and ${\Sigma }^{+}$, without changing their charge. It does not imply exact mass degeneracy: unequal down- and strange-quark masses break U-spin.

For pion-nucleon octet channels, assume the collision is governed by the strong interaction. The initial baryon number is one and strangeness is zero, so an outgoing meson-baryon pair must preserve $B=1$, $S=0$, and electric charge. Thus its total flavor hypercharge is $Y=1$. The allowed types areA kaon of $S=+1$ can accompany a Lambda baryon or Sigma baryon of $S=-1$. An antikaon cannot balance the nonpositive strangeness of an octet baryon. A Xi baryon would require a meson of $S=+2$, which the meson octet does not contain.

Resolving these types by electric charge gives all possible pairs:In the isospin-symmetric approximation, total isospin is conserved as well: the incoming $1\otimes \frac{1}{2}$ contains $I=\frac{1}{2},\frac{3}{2}$. The $\pi N$ and $K\Sigma$ channels contain both values, while ${\eta }_{8}N$ and $K\Lambda$ contain only $I=\frac{1}{2}$. The extreme-charge initial states are pure $I=\frac{3}{2}$, consistently excluding ${\eta }_{8}N$ and $K\Lambda$. Clebsch-Gordan coefficients relate amplitudes in different charge channels; the table establishes permission, not equal probabilities. Electromagnetism and unequal up- and down-quark masses introduce small violations of isospin symmetry.

Finally, energy and momentum conservation require $\sqrt{s}\ge m_M+m_B$ for a particular pair, where $s$ is the squared total four-momentum. Only channels above their own threshold can occur. Total angular momentum and parity symmetry in quantum field theory constrain the partial waves of the meson-baryon scattering: a pseudoscalar meson and a positive-parity spin-one-half baryon have pair parity $-(-1{)}^L$ and total angular momentum $J=L\pm \frac{1}{2}$ (only $J=\frac{1}{2}$ for $L=0$). Initial and final partial waves must have matching $J$ and parity. Sufficient energy can open a channel, but cannot remove the electric charge conservation, baryon number, or strangeness constraints.

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.

Direct formation in $e^{+}{e}^{-}$ through one virtual photon selects the vector meson quantum numbers $J^{PC}=1^{--}$. The nonzero $K_S^0{K}_L^0$ mode and absence of $K_S^0{K}_S^0$ are consistent with this assignment: the two spin-zero kaons require odd orbital angular momentum for negative parity, whereas identical $K_S^0$ bosons cannot have an odd relative wave. This argument neglects the tiny CP violation in neutral kaons.

Assuming the $\rho \pi$ decays conserve isospin, their equal rates indicate $I=0$. The Clebsch-Gordan coefficients giveAn $I=1,I_3=0$ state lacks ${\rho }^0{\pi }^0$, while $I=2,I_3=0$ has squared coefficients $1:4:1$ rather than $1:1:1$. The branching fractions alone determine neither $J$ nor these relative phases; the single-photon formation and approximate strong interaction isospin conservation are essential additional inputs.

The isospin coupling $1\otimes 1\to 1$, with third component $-1$, has the normalized stateThe two Clebsch-Gordan coefficients have equal squared magnitude. Neglecting electromagnetic effects and charged-neutral mass differences, the two rho meson charge channels have equal rates. Their subsequent strong decays are ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ and ${\rho }^{-}\to {\pi }^{-}{\pi }^0$, with nearly unit probabilities. Consequentlyin the ideal isospin model.

Each final state contains one pair of identical pions. Summing the alternative resonant-pair assignments coherently, with the corresponding identical-particle phase space factors, does not generate an extra factor of two for just one channel. In the narrow-rho meson approximation the equality is already evident from equal $\rho \pi$ rates; the consistent isospin amplitudes extend it to the integrated three-body distributions.

The tabulated ratio $9.02/9.30=0.970$ is close to this expectation but is not exactly one. Treating the printed errors as independent gives an uncertainty about $0.012$, so a percent-level comparison requires isospin breaking, radiative corrections, realistic resonant dynamics and experimental correlations. The idealized argument predicts the leading pattern rather than an exact identity for the rounded measurements.

For independent angular momenta ${\mathbf{J}}_1,{\mathbf{J}}_2$, set $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Since different subsystems commute, $\mathbf{J}$ satisfies the same angular-momentum algebra. The product basis $|j_1{m}_1⟩|j_2{m}_2⟩$ diagonalizes $J_1^2,J_2^2,J_{1z},J_{2z}$; the coupled basis diagonalizes $J_1^2,J_2^2,J^2,J_z$. Its labels obeyThe Clebsch-Gordan coefficients are the unitary change of basis between them. Start from the maximum-weight state $j=j_1+j_2,m=j$, apply $J_{-}=J_{1-}+J_{2-}$ with wavefunction normalization, and construct the remaining multiplets by orthogonality.

For two spin-one-half systems, the symmetric triplet state is $|\uparrow \uparrow ⟩$, $(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, $|\downarrow \downarrow ⟩$, with $j=1$; the antisymmetric singlet state $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ has $j=0$. The dimension identity ${\sum }_j(2j+1)=(2j_1+1)(2j_2+1)$ checks completeness. Coupling orbital and intrinsic spin similarly gives total atomic angular momentum and organizes spin-orbit energy splittings. Exchange symmetry is an additional restriction when the constituents are identical.

angular momentum generates rotations. Its components obey $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$, and $J^2$ commutes with each component. A compatible basis has $J^2={\hslash }^{2}j(j+1)$ and $J_z=m\hslash$, where $m=-j,-j+1,\dots ,j$. ladder operators change $m$ by one, terminate at $\pm j$, and give $2j+1$ magnetic substates. Different Cartesian components generally cannot all be sharp simultaneously.

orbital angular momentum is $\mathbf{L}=\mathbf{r}\times \mathbf{p}=-i\hslash \mathbf{r}\times \nabla$. Its angular eigenfunctions are spherical harmonics, with integer $\ell$ arising from their spatial rotation properties and single-valuedness. spin is intrinsic angular momentum, not literal rotation of a finite charged sphere. It may be half-integer; spin $1/2$ is represented by $\mathbf{S}=\hslash \mathbf{\sigma }/2$. For several contributions, $\mathbf{J}=\mathbf{L}+\mathbf{S}$ obeys the same algebra. Coupling angular momenta $j_1,j_2$ permits $J=|j_1-j_2|,\dots ,j_1+j_2$, with Clebsch-Gordan coefficients relating product and coupled bases. spin-orbit coupling can distinguish these total-angular-momentum sectors.

The Stern-Gerlach apparatus sends a collimated neutral atomic beam through an inhomogeneous magnetic field and records its deflection. A magnetic moment experiences $F_z\simeq {\mu }_z\partial B_z/\partial z$; field uniformity would produce precession but no analogous net separating force. A classically continuous distribution of moment orientations would broaden the beam continuously. Gerlach and Stern instead observed separated silver-beam components. Reversing or removing the field and comparing the undeflected profile controls the magnetic origin; collimation, velocity spread and finite resolution affect the spots. The original experiment demonstrated space quantization; its modern silver interpretation uses the unpaired electron spin with nearly zero orbital angular momentum, rather than a direct measurement of a free electron's trajectory.^[ref]

For two spin-$1/2$ particles, the triplet state sector has $S=1$ and symmetric spin statesThe spin singlet state has $S=0$ and antisymmetric spin stateTheir total $S^2$ values are $2{\hslash }^2$ and zero, while ${\mathbf{S}}_1\cdot {\mathbf{S}}_2$ has values ${\hslash }^2/4$ and $-3{\hslash }^2/4$. The singlet is rotationally invariant and gives anticorrelated spin measurements along a common axis. For identical fermions the complete state is antisymmetric: a symmetric spatial state pairs with a singlet spin state, whereas an antisymmetric spatial state pairs with triplet spin. Orbital, spin and coupled angular momenta share an algebra, but their physical origins and allowed representations differ.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

A tensor operator transforming in an irreducible representation $R$ between states in $V$ defines an intertwining operator $R\otimes V\to V$. In a basis of these intertwining maps, its scalar coefficients $g_{\nu }$ are reduced matrix elements, independent of the component indices. Schur lemma makes the number of independent coefficients the multiplicity of $V$ in $R\otimes V$. The component dependence is fixed by the associated Clebsch-Gordan coefficients.

Finite optical pumping time lets an atom climb a light-shift hill before transfer into a lower magnetic sublevel. Spontaneous emission dissipates the gained potential energy; climbing alone is conservative.^[ref]

To make the geometry and signs explicit, take an original choice of phase origin for equal opposing waves:The real electric field is $\mathrm{Re}[\mathcal{E}(z)e^{-i{\omega }_{L}t}]$; $\lambda$ is wavelength. The fractions $w_{\pm }$ are squared projections onto the defined circular basis vectors, divided by total intensity. Their sum is one: interference changes electromagnetic polarization despite constant total intensity. At $z=0,\lambda /4,\lambda /2$ the field is respectively ${\mathbf{ϵ}}_{+},{\mathbf{ϵ}}_{-},{\mathbf{ϵ}}_{+}$; halfway between these positions it is linearly polarized. Specifying the vectors avoids an observer-dependent verbal handedness convention.

For an illustrative $J_g=1/2\to J_e=3/2$ angular momentum transition, use the squared Clebsch-Gordan coefficients $1$ and $1/3$ for the strong and weak circular couplings from each lower magnetic sublevel. Interchanging circular components interchanges these strengths. The weighted sums are $w_{+}+w_{-}/3=(2+\cos (2kz))/3$ and $w_{-}+w_{+}/3=(2-\cos (2kz))/3$. Rescaling their common negative shift givesBoth potential energies are negative relative to the unilluminated level. Each hill has height $U_0$, and opposite hills are displaced by $\lambda /4$. An atom starting at the $U_{+}$ minimum at $z=0$ loses $U_0$ of kinetic energy if it reaches $z=\lambda /4$ on that branch. A pump transfer there lowers its potential energy by $U_0$, leaving it at a $U_{-}$ minimum. The next uphill segment ends at $z=\lambda /2$, where the favoured transfer is reversed. Actual transfers are stochastic and need not occur exactly at these extrema.

For one completed climb and pump, conservation of energy gives $\Delta K=-U_0$, while the ground-state potential energy returns to its starting value. Neglecting recoil and treating the scattered photon as the sole outgoing radiation change, $\hslash ({\omega }_{\mathrm{out}}-{\omega }_L)=U_0$. A potential drop is an internal-state reset, not an automatic equal upward kick in kinetic energy. Random emission nevertheless adds momentum fluctuations and recoil energy, so useful cooling requires the dissipated energy to compete with that heating.

The recoil energy $E_R=(\hslash k{)}^2/(2m)$ sets a different scale from $k_B{T}_D=\hslash \Gamma /2$. Conventional Sisyphus cooling reaches the recoil regime when reducing the hill height also weakens cooling; it is not a claim of arbitrarily small temperature at vanishing intensity. Subrecoil techniques need further physics.^[ref]

A rank-$k$ spherical tensor operator has components $T_q^{(k)}$, $-k\le q\le k$, that transform under rotation like an angular momentum $k$. Equivalently, their commutators with the angular momentum operator satisfyFor a vector operator, the rank is one and a conventional choice is $T_0=V_z$, $T_{\pm 1}=\mp (V_x\pm i{V}_y)/\sqrt{2}$. The Wigner-Eckart theorem separates the matrix elements of a spherical tensor operator into Clebsch-Gordan coefficients and a reduced matrix element.

Consider an original hypothetical atom with a lower angular momentum multiplet $F=1$, $m=-1,0,+1$, and one opposite-parity excited level $F^{\prime }=0,m^{\prime }=0$. Stipulate that all spontaneous emission returns to these three lower quantum states, with no other decay channel. This is an isolated four-state model, not the actual sodium or rubidium D-line structure. In particular, the physical rubidium cell above contains multiple hyperfine structure levels.

The electric-dipole selection rule permits $\Delta F=-1$ here; ${\sigma }^{+}$ absorption with $\Delta m=+1$ excites only $m=-1$. Decay from $F^{\prime }=0$ has three equal integrated branching probabilities $b_{-}=b_0=b_{+}=1/3$: rotational symmetry gives equal squared Clebsch-Gordan coefficients. The emitted photons carry the three permitted angular momentum projections; equal integrated branching does not imply equal intensity in every observation direction. Thus $m=0$ and $m=+1$ are both dark. The chosen beam cannot prepare a unique pure $m=+1$ state.

Let $R\ge 0$ be the absorption rate of an atom in $m=-1$, and let $\Gamma$ be the excited-state spontaneous emission rate, both in inverse seconds. Assume $R,\gamma \ll \Gamma$, negligible stimulated emission, and no maintained quantum coherence between sublevels. Eliminating the short-lived excited population gives normalized lower populations $p_{-}+p_0+p_{+}=1$ to this approximation. At each scattering event the atom leaves $m=-1$ with probability $2/3$; the remaining $1/3$ simply returns it there.

Model depolarizing collisions and other population relaxation by resets to the unpolarized distribution at rate $\gamma \ge 0$. The rate equations areThis relaxation prescription assumes that Zeeman splitting is small compared with thermal energy, so the unpolarized distribution approximates thermal equilibrium. It is not an energy-independent relaxation law at arbitrarily low temperature.

Every transfer has a nonnegative rate: optical transfers $-\to 0$ and $-\to +$ each have rate $R/3$, and relaxation transfers between each distinct pair have rate $\gamma /3$. These define a continuous-time Markov chain. The three derivatives sum to zero, and at $p_i=0$ the corresponding derivative cannot be negative. Hence the model conserves population and preserves nonnegative populations. Optical cycles that return to $m=-1$ contribute to scattering even though they cause no net population transfer.

Two pions, each of isospin one, have a nine-dimensional charge-state space. Clebsch-Gordan coefficients decompose it into an isospin-two quintet, an isospin-one triplet and an isospin-zero singlet. Under exchange of the two particle slots, the $I=2,0$ states are symmetric and the $I=1$ states are antisymmetric. Because pions have zero spin and obey bosonic exchange symmetry, the $I=1$ two-pion channel requires odd relative orbital angular momentum, while $I=0,2$ require even orbital angular momentum.

The Wigner-Eckart theorem separates the dependence of a spherical tensor operator on angular momentum projections from its dynamical content. In one standard convention,The reduced matrix element on the right is independent of $m,m^{\prime },q$. The Clebsch-Gordan coefficients impose $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$; additional selection rules, including parity, can make the reduced matrix element vanish.

To see why the factorization holds, combine the rank-$k$ transformation law of the operator with the spin-$j$ transformation law of the initial state. This gives an intertwining operator from the tensor product representation $j\otimes k$ to the final representation $j^{\prime }$. The Clebsch-Gordan decomposition contains each allowed $j^{\prime }$ once. Therefore the corresponding intertwining operator is unique up to an overall scalar, which is the reduced matrix element after the chosen normalization. Ratios within the same pair of multiplets depend only on Clebsch-Gordan coefficients, whenever the common reduced matrix element is nonzero.