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Exchanging the factors in the Clebsch-Gordan decomposition of two equal spins gives the phase $(-1{)}^{2s-j}$. This follows from the exchange symmetry of the Clebsch-Gordan coefficients. The total exchange parity includes the spatial state and any other internal factors: the spin parity alone does not impose the full symmetrization postulate. For two half-integer spins, even $j$ is spin-antisymmetric and odd $j$ spin-symmetric.

Use the uncoupled angular momentum eigenstates $|m_L,\uparrow ⟩$ and $|m_L,\downarrow ⟩$, with $m_L=-1,0,1$. For the Hamiltonian operator actually written in the paper, these are the energy eigenstates, withThe initial $J=3/2$ states are not all energy eigenstates of this operator. To find their energy populations, first perform the Clebsch-Gordan decomposition. The extremal states are $|3/2,3/2⟩=|1,\uparrow ⟩$ and $|3/2,-3/2⟩=|-1,\downarrow ⟩$. Applying the total angular momentum lowering operator to the first givesSimilarly,The initial preparation is an equal incoherent mixture of the four coupled states. Each extremal state contributes population $1/4$ to its one energy eigenvalue, while each nonextremal state contributes $2/3$ and $1/3$ of its own $1/4$ population to the two displayed components. The resulting Zeeman splitting is thereforefor nonzero $B_z$. The relative populations are $\boxed{3:2:2:2:3}$ in this order; the two zero-energy states each have population $1/12$. These Zeeman populations from an unpolarized angular-momentum multiplet remain constant under subsequent evolution by the stated static Hamiltonian operator, although off-diagonal quantum coherence can evolve.

There is a distinct weak-field model in which a retained spin-orbit interaction keeps $J=3/2$ isolated and one projects the magnetic perturbation into that quartet. The Landé g-factor is then $4/3$, giving four shifts ${\mu }_B{B}_z(\pm 2,\pm 2/3)$ with equal populations. That calculation requires the additional projection or dominant spin-orbit interaction; it is not diagonalization of the full magnetic Hamiltonian operator specified here, which also mixes the $J=1/2$ sector.

On the tensor product of spins $j_1$ and $j_2$, the combined angular momentum operator is $\mathbf{J}={\mathbf{S}}^{(1)}\otimes I+I\otimes {\mathbf{S}}^{(2)}$. The Clebsch-Gordan decomposition iswhere $J$ increases by one and each irreducible spin representation occurs once. The dimensions agree: ${\sum }_J(2J+1)=(2j_1+1)(2j_2+1)$.

The uncoupled orthonormal basis diagonalizes $S_3^{(1)},S_3^{(2)}$; the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2,J_3$. The Clebsch-Gordan coefficients are their change-of-basis entries, with conventionThey vanish unless $m_1+m_2=M$. Since both bases are orthonormal, this change-of-basis matrix obeys unitarity. Overall phases of the coupled basis vectors are fixed by a convention and do not change the energy eigenvalues.

Use $|m,m^{\prime }⟩$ as shorthand for the tensor product of two total-isospin-one states with third components $m,m^{\prime }$. A total-isospin-zero state must have total third component zero, so write $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$. Applying the total raising operator givesAn isospin singlet must be annihilated by this operator; hence $b=-a$ and $c=a$. Normalization gives the two-isovector singletIt is also annihilated by the total lowering operator. A raising or lowering action changes the third-component label by one; the unchanged output label in the PDF's preliminary ladder formula is a typographical omission, not the action used here.

Exchange commutes with the total isospin operators. The displayed singlet is symmetric. The $I=2$ multiplet has symmetric highest state $|1,1⟩$; applying the symmetric total lowering operator preserves that exchange symmetry. The symmetric square of a three-dimensional space has dimension six, exactly $1+5$, while its antisymmetric square has dimension three. The remaining $I=1$ multiplet is therefore antisymmetric. Thus the Clebsch-Gordan decomposition isPions have zero intrinsic spin and form an isospin triplet. Bose-Einstein statistics requires the complete two-pion state to be symmetric under exchange. Its relative orbital state changes by $(-1{)}^{\ell }$, so odd orbital angular momentum requires antisymmetric isospin: an odd-$\ell$ two-pion state has $I=1$. This statement treats the charge components as states of the same isospin multiplet, rather than ignoring exchange when their charges differ.

For the decay, let $|\chi ⟩=H_I|K^0⟩$. The stated commutator with $I_3$ gives $I_3|\chi ⟩=0$. More decisively, the commuting raising operator givesThe kaon doublet cannot be raised twice. An $I=2,M=0$ component would survive two raisings, so it is excluded. Meanwhile a spinless kaon decaying into two spinless pions has $\ell =0$ by conservation of total angular momentum, independently of whether parity is conserved by the decay. The exchange condition excludes $I=1$. This proves the highest-weight weak-isospin selection in kaon decay: the two-pion final state has $I=0$.

In a consistent pion phase convention, the singlet's charge components have the structureThe normalized symmetric charged channel has coefficient $\sqrt{2/3}$ and the neutral channel coefficient $-1/\sqrt{3}$. Thus, in the isospin limit with equal pion masses and matching two-body phase space,Equivalently, ordered charged and neutral amplitudes have equal magnitudes, but neutral identical particles contribute the phase-space factor $1/2!$. This is the same counting expressed in another basis, and must not be included a second time after using the normalized charge-channel coefficients. Small pion-mass and electromagnetic effects can modify the ideal ratio.

For the addition of angular momentum, the uncoupled orthonormal basis is $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$, whereas the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2$ and $J_3$ for $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Define the Clebsch-Gordan coefficients byThe Clebsch-Gordan decomposition allows $|j_1-j_2|\le J\le j_1+j_2$ in unit steps, and a coefficient vanishes unless $m_1+m_2=M$. Taking the inner product of the expansion with itself provesWith the customary real phase convention the modulus signs can be omitted, giving the sum of squares used in the question. In an arbitrary phase convention the modulus-square identity is the invariant statement.

Treat the strong interaction as exactly isospin-invariant for this calculation. Its transition operator $T$ commutes with the three isospin generators and hence with their quadratic Casimir operator. Acting on the initial state $|IM⟩$, it can therefore produce only final states with the same $I$ and $M$. The tensor product representation of final multiplets $I_1,I_2$ contains one copy of each permitted total isospin, so $T|IM⟩=a_M|(I_1{I}_2)IM⟩$. Commuting $T$ with the lowering operator givesEvery connecting ladder factor is nonzero, so all $a_M$ coincide with one reduced coefficient $a$. Expanding the final coupled state now gives the isospin-invariant two-body decay amplitudeThe reduced coefficient may depend on the particle species, momenta and spin channel, but not on the three isospin projections. Summing the squared amplitude over the orthogonal charge channels givesA physical decay rate also includes phase space, angular integration and any spin sums. In the isospin-degenerate limit these factors are common to the charge channels; absorbing their square root into $a$ gives ${\Gamma }_{IM}=|a{|}^2$ for this specified two-body multiplet channel. If several distinct species or independent spin channels are included in “anything”, their reduced rates must additionally be summed. There is no extra factor $2I+1$ when the initial magnetic component is fixed.

The Delta baryon states have $I=3/2$ and projections $3/2,1/2,-1/2,-3/2$ for ${\Delta }^{++},{\Delta }^{+},{\Delta }^0,{\Delta }^{-}$. The proton and neutron form $I=1/2$, with projections $+1/2,-1/2$, and the pion triplet has $I=1$ with projections $+1,0,-1$. The highest coupled state isApply the total lowering operator $I_{-}=I_{-}^{(N)}+I_{-}^{(\pi )}$. Its coefficient on the left is $\sqrt{3}$, while on the right the nucleon and pion coefficients are $1$ and $\sqrt{2}$:The squared Clebsch-Gordan coefficients are consequently $1/3$ and $2/3$, compared with coefficient one for the ${\Delta }^{++}$ channel. Thus the Delta baryon pion branching ratios areAt the opposite end of the same isospin multiplet, $|3/2,-3/2⟩=|n{\pi }^{-}⟩$, so the strong decay is ${\Delta }^{-}\to n{\pi }^{-}$, with the same reduced rate in this symmetry limit. Physical mass splittings and electromagnetic effects perturb these idealized equal-kinematics ratios.

We consider finite-dimensional continuous complex representations of SU(2). Its maximal torus is $\mathrm{diag}(e^{i\theta },e^{-i\theta })$. The irreducible representations areThey may be realized as homogeneous polynomials of degree $n$ in two variables, with the induced linear substitution action. Their torus weights are $n,n-2,\dots ,-n$, each once. The centre $-I$ acts by $(-1{)}^n$, so precisely the even-$n$ representations descend to SO(3). In physics $n=2j$ labels integral or half-integral spin $j$.

Here is a classification argument. Average any positive definite Hermitian form over the compact group using normalized Haar measure. The averaged form is invariant, so the orthogonal complement of an invariant subspace is invariant. Repeatedly splitting gives complete reducibility. Differentiate an irreducible representation and complexify to the Lie algebra ${\mathfrak{sl}}_2$, with generators $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Torus periodicity gives integer $H$-weights. Choose a vector $v$ of maximal weight $n$; then $Ev=0$. Induction on $k$ using the commutators givesThe chain terminates: if $F^{p}v\ne 0$ and $F^{p+1}v=0$, the displayed formula forces $n=p\ge 0$. The vectors $v,Fv,\dots ,F^{n}v$ have distinct weights, span an invariant subrepresentation, and hence span the irreducible representation. Their generator action agrees with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Invariance under the Lie algebra is equivalent to invariance under the connected group, completing the classification. Conversely, a nonzero invariant subspace of a symmetric power contains a weight vector by projection onto torus weights. Raising it to the highest weight and then lowering generates every monomial, proving that the symmetric power is irreducible.

The character on the torus iswith the endpoints interpreted by continuity. Every representation has a unique decomposition $V\cong {⨁}_{n\ge 0}{m}_n{V}_n$. Its multiplicities can be computed from character inner products,The sine orthogonality directly verifies orthonormality of the irreducible characters. Alternatively, if $w_k$ is the dimension of the weight-$k$ subspace, the weight lists give $w_n=m_n+m_{n+2}+\cdots$, so $m_n=w_n-w_{n+2}$ for $n\ge 0$. This gives a constructive decomposition from torus weights. In particular, multiplying weight characters yields the Clebsch-Gordan decomposition $V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus \cdots \oplus V_{|m-n|}$. Arbitrary unitary Hilbert-space representations of this compact group similarly decompose into Hilbert direct sums of these finite-dimensional irreducibles; the finite-dimensional statement above is the standard representation-theory setting.

In $V_n\otimes V_n$, interchange of the two factors commutes with SU2. On the highest-weight vector $D^r{x}_1^{n-r}{x}_2^{n-r}$ it acts by $(-1{)}^r$. Each irreducible summand in the Clebsch-Gordan decomposition has multiplicity one, so this sign is its scalar action on the entire summand. The symmetric square is the $+1$ eigenspace and the exterior square the $-1$ eigenspace. ThereforeFor instance, $S^2{V}_1=V_2$ and ${\Lambda }^2{V}_1=V_0$. The dimensions agree respectively with $(n+1)(n+2)/2$ and $n(n+1)/2$.

A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:

For two equal spins, exchange acts on the coupled state $|j,m⟩$ by the Clebsch-Gordan coefficient symmetry factor $(-1{)}^{2j_1-j}$. It is therefore $-1$ when $j=2j_1-1$, making every state in that multiplet antisymmetric, and $+1$ when $j=2j_1-2$, making the next multiplet symmetric.

The Clebsch-Gordan decomposition permitsThus a total-angular-momentum singlet $j=0$ occurs exactly when $j_1=j_2$.

With ${ϵ}_{0123}=+1$, define the Pauli-Lubanski pseudovector byFor a translation $a^{\mu }$ and a Lorentz transformation $\Lambda =\exp (\omega )$, a left-handed Weyl spinor transforms asEquivalently, the active field at a fixed point uses ${\psi }^{\prime }(x)=S(\Lambda )\psi ({\Lambda }^{-1}(x-a))$.

The Clebsch-Gordan decomposition isIndeed,The symmetric spinor is the $(1,0)$ representation, while its antisymmetric part is the Lorentz scalar $(0,0)$.

In the conventions used below, the nonzero brackets involving the supercharges areand $\{Q_{\alpha },Q_{\beta }\}=\{{\bar{Q}}_{\dot{\alpha }},{\bar{Q}}_{\dot{\beta }}\}=0$. These equations, together with the Poincare algebra, are the four-dimensional $\mathcal{N}=1$ Super-Poincaré algebra without central charges.

For three distinguishable spin one-half systems, the Clebsch-Gordan decomposition first combines two spins into $0\oplus 1$, then couples the third: $0\otimes 1/2=1/2$ and $1\otimes 1/2=1/2\oplus 3/2$. Thus the tensor product has one quartet and two doublets. A Hamiltonian operator depending only on the total $S^2$ and $S_z$ acts identically on the two doublet copies. In particular $H={ϵ}_t{S}^2/{\hslash }^2+{ϵ}_z{S}_z/\hslash$ has energy eigenvalues ${ϵ}_{t}S(S+1)+{ϵ}_{z}M$, generically once for each quartet projection and twice for each doublet projection. Equal numerical energies have their multiplicities added.

A total-spin-$J$ sector is an irreducible subspace on which ${\mathbf{J}}^2$ has eigenvalue $J(J+1){\hslash }^2$. For two spin-one particles, Clebsch-Gordan decomposition gives sectors $J=0,1,2$ of dimensions one, three, and five.

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.

An equal incoherent mixture within a fixed total-$J$ multiplet has density operator $\rho =P_J/(2J+1)$, where $P_J$ is its orthogonal projection. If the applied-field Hamiltonian operator is proportional to $L_z+2S_z$ and no spin-orbit interaction is retained, its energy eigenstates are the uncoupled states $|m_L,m_S⟩$. The Born rule gives their populations by the formula above, or equivalently by squared Clebsch-Gordan decomposition coefficients. Equal populations of coupled states need not give equal populations of magnetic energy eigenvalues.

For $L=1,S=1/2,J=3/2$, the nonextremal coupled states have weights $2/3$ and $1/3$ in their two uncoupled components. The five dimensionless magnetic energies $-2,-1,0,1,2$ then have populations $1/4,1/6,1/6,1/6,1/4$; the zero energy contains two independent states, each with population $1/12$. In a different, weak-field approximation with a dominant spin-orbit interaction, projecting into the isolated multiplet gives the Landé g-factor spectrum instead. The full operator and its projection must not be interchanged.