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A symmetrized form of the Clebsch-Gordan coefficients used for coupling three angular momenta. The relation is $⟨j_1{m}_1,j_2{m}_2|j_3{m}_3⟩=(-1{)}^{j_1-j_2+m_3}\sqrt{2j_3+1}\left(\begin{array}{ccc}{j}_1& j_2& j_3\\ m_1& m_2& -m_3\end{array}\right)$. The symbol vanishes unless the magnetic indices sum to zero and the angular momenta obey the triangle and integrality conditions. Products of two such symbols express the Gaunt integral; the zero-magnetic-index symbol enforces even total multipole for integer spherical harmonics.

The operator $J_{-}=J_x-i{J}_y$ acts by $J_{-}|j,m⟩=\hslash \sqrt{(j+m)(j-m+1)}|j,m-1⟩$. It generates lower magnetic states and determines Clebsch-Gordan coefficients.

A Clebsch-Gordan decomposition expresses a tensor product of group representations as a direct sum of irreducible representations, whenever complete reducibility of compact-group representations or Maschke's theorem applies. For a finite group over the complex numbers, the multiplicity of $V_{\lambda }$ is the character inner product $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V{\chi }_W⟩$. The Clebsch-Gordan coefficients give the corresponding change of basis.

A complete set of commuting observables consists of mutually commuting quantum observables whose joint eigenvalues distinguish all vectors of a simultaneous eigenbasis, up to phase. For a finite-dimensional Hilbert space, commuting self-adjoint operators admit a simultaneous orthonormal basis of eigenvectors; completeness means that every joint eigenspace is one-dimensional. An additional quantum observable that is already a function of the others supplies no independent quantum number.

For the addition of angular momentum, the uncoupled labels $(j_1,j_2,m_1,m_2)$ correspond to $\{J_1^2,J_2^2,J_{1z},J_{2z}\}$, whereas the coupled labels $(j_1,j_2,j,m)$ correspond to $\{J_1^2,J_2^2,J^2,J_z\}$. Within a fixed pair of irreducible angular momentum spaces each choice gives a natural eigenbasis. The Clebsch-Gordan coefficients implement the change between them.

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.

An isospin-invariant decay amplitude is an intertwining map from the initial isospin multiplet into the final tensor product. The Clebsch-Gordan decomposition for SU2 contains each allowed final isospin once. Commutation with the third generator and the quadratic Casimir operator places the image in the matching coupled state. Commutation with the ladder operators makes its scalar coefficient independent of the magnetic component. Orthogonality of the Clebsch-Gordan coefficients then gives a summed squared amplitude $|a{|}^2$. A physical decay rate additionally includes phase space and spin sums, which can be absorbed into a common reduced coefficient when exact isospin symmetry makes them identical between channels.

Apply a total ladder operator to a coupled state and evaluate against an uncoupled product state. The total operator is the sum of the two factor operators, so its one coupled ladder amplitude times the corresponding one of the Clebsch-Gordan coefficients equals the sum of two factor ladder amplitudes times shifted coefficients. An overall common rescaling of the ladder operators cancels from the recurrence. Nonzero coefficients obey $m_1+m_2=m$, and the positive ladder amplitudes require a consistent phase convention within each irreducible summand.

// Target: lie-theory.bigb

Put $s=j_1+j_2$, with both $j_i>0$. At magnetic quantum number $-s+1$, let $|a⟩$ raise only the first factor of the lowest product state and $|b⟩$ raise only the second. Applying total $J_{+}$ to the lowest product gives the normalized $J=s$ combination $\sqrt{j_1/s}|a⟩+\sqrt{j_2/s}|b⟩$. Its orthogonal combination is annihilated by total $J_{-}$ and therefore has $J=s-1$. This obtains a useful Clebsch-Gordan coefficients without a table.

A spherical tensor operator of rank $k$ transforms under rotation like an angular momentum $k$. The Wigner-Eckart theorem writes its matrix elements asThe reduced matrix element contains the radial and other dynamical information; the Clebsch-Gordan coefficients determine the dependence on the projections. Ratios within a fixed pair of multiplets therefore follow without a radial integral. The angular selection rules include $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$.

For the position vector, choose the rank-one componentsLet $R=⟨{\lambda }^{\prime },3/2\Vert T^{(1)}\Vert \lambda ,1/2⟩$. The stretched Clebsch-Gordan coefficients and one lowering step giveFor example, apply the total angular momentum ladder operator to the stretched state. The result has components with coefficients $1/\sqrt{3}$ and $\sqrt{2/3}$, corresponding to lowering the initial angular momentum and the vector component, respectively.

Only the $q=+1$ component of $x$ contributes to $M_1$. The $q=0$ operator $z$ cannot connect the projections in $M_2$, so that matrix element vanishes. With the convention above,Thus the requested relative magnitudes and their squared values areThese ratios presume a common nonzero reduced matrix element. The position operator has odd parity, so an electric dipole transition requires opposite-parity multiplets; for one electron the electric-dipole selection rule also gives $\Delta \ell =\pm 1$. The displayed $j,m$ labels alone do not establish that condition. If the common reduced matrix element vanishes, all three matrix elements vanish and no nonzero ratio can be inferred.

For $5s5d$, addition of angular momentum gives $L=2$ and $S=0,1$. The even-parity atomic terms and atomic fine-structure levels areThe electric-dipole selection rules exclude all connections to the even-parity $5s^2$ and $5s6s$ electron configurations. To the odd-parity $5s5p$ electron configuration, all seven allowed connections in pure LS coupling areHere a connection is listed independently of whether the $5s5d$ energy level is above or below its partner.

In a weak magnetic field, use the Landé g-factorFor the three ${}^3{D}_J$ energy levels, this gives $1/2,7/6,4/3$; for ${}^3{P}_1,{}^3{P}_2$ it gives $3/2$. The six-component spectral line is thereforeA component has frequency displacement $({\mu }_{B}B/h)(m_D/2-3m_P/2)$. The electric dipole transition with $m_D=m_P=0$ has zero matrix element, since the relevant $J=1\leftrightarrow 1$ rank-one Clebsch-Gordan coefficients vanish there. The remaining allowed pairs giveThus all six frequency positions $\pm 1/2,\pm 1,\pm 3/2$ are distinct. Perpendicular to the magnetic field, both the $\pi$ and $\sigma$ components are visible, with mutually perpendicular linear polarizations. Along the magnetic field the $\pi$ components disappear. This is the weak-field Zeeman effect, before the Paschen-Back effect changes the coupling scheme.

An electric dipole transition has amplitude proportional to $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$. The electric dipole operator is odd under parity, so the initial and final states must have opposite parity. As a rank-one spherical tensor it also requires $\Delta J=0,\pm 1$, excluding $J=0\leftrightarrow 0$, and $\Delta m_J=0,\pm 1$. These follow from angular-momentum addition and the Wigner-Eckart theorem; vanishing Clebsch-Gordan coefficients can impose further restrictions. Relative to the quantization axis, $\pi$ polarization drives $\Delta m=0$, while ${\sigma }^{\pm }$ drive $\Delta m=\pm 1$.

For a single active electron, parity and the orbital triangle rule combine to give $\Delta l=\pm 1$. In LS coupling, the spin-independent electric dipole operator gives $\Delta S=0$ and $\Delta L=0,\pm 1$, excluding $L=0\leftrightarrow 0$; total parity must still change. Spin-orbit mixing can weakly relax the approximate spin rule. For resolved hyperfine states the corresponding triangle rule is $\Delta F=0,\pm 1$, excluding $F=0\leftrightarrow 0$, with $\Delta m_F=0,\pm 1$. These labels require the appropriate coupling regime. There is no universal E1 rule demanding a particular change in principal quantum number. An E1-forbidden line may occur through another multipole or a multiphoton process.

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.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

Choose the doublet bases so that the column of barred fields $B=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $R=({\psi }_R^1,{\psi }_R^2{)}^T$ both transform by the doublet matrices in d. This choice follows the source's assignment of the barred doublet itself. Since $H$ transforms in $\epsilon$, its Yukawa coupling must multiply a fermion bilinear in $\epsilon$.

There is exactly one such bilinear in $2\otimes 2$: the Clebsch-Gordan coefficients in d give $B_1{R}_2-B_2{R}_1$. Equivalently, writing the upper block of the Yukawa matrix as $A$, invariance requiresThe first equation sets $A_{11}=A_{22}=0$, because ${\alpha }^2\ne 1$; the second sets $A_{21}=-A_{12}$.

A mixed doublet-singlet bilinear remains a doublet after multiplication by $H$, so all four mixed entries vanish. Finally ${\bar{\psi }}_L^3{\psi }_R^3$ transforms in $\epsilon \otimes 1=\epsilon$, and multiplication by $H$ makes it invariant. ThereforeThe allowed Yukawa interaction is $H\{a({\bar{\psi }}_L^1{\psi }_R^2-{\bar{\psi }}_L^2{\psi }_R^1)+b{\bar{\psi }}_L^3{\psi }_R^3\}$, together with its Hermitian conjugate when forming a real Lagrangian density.

The displayed texture depends on basis. If instead the unbarred left doublet is assigned $D$ in the same ordering as the right doublet, its barred doublet has the dual representation $D^{\ast }$ and the equivalent upper block is $a\mathrm{diag}(1,-1)$. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex Yukawa couplings.

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.

The Clebsch-Gordan decomposition for SU2 iswhere $j$ increases by one and every summand occurs once. The tensor product of Lie algebra representations is antihermitian for the product inner product, so an invariant subspace has an invariant orthogonal complement; this gives complete reducibility. To see the particular summands, suppose $j_1\ge j_2$. The product weight diagram has one state of weight $j_1+j_2$, two of weight $j_1+j_2-1$, and increasing multiplicities down to the plateau determined by $j_1-j_2$. Subtracting the weight string of $V_{j_1+j_2}$ leaves a single highest state of weight $j_1+j_2-1$. Repeating highest-weight character subtraction yields exactly the displayed sequence, down to $j_1-j_2$. Its dimensions exhaust the tensor product becauseThe case $j_2\ge j_1$ is identical with the factors exchanged.

Let $C_m^j(m_1,m_2)=⟨m_1:m_2|j,m⟩$ be the Clebsch-Gordan coefficients. Since the total $J_3$ is $J_3^{(1)}\otimes 1+1\otimes J_3^{(2)}$, orthogonality of its eigenspaces gives the selection rule $C_m^j(m_1,m_2)=0$ unless $m_1+m_2=m$. Normalize the coupled states by lowering from a unit highest-weight vector, as in part (i); a common phase for an entire summand is immaterial. ThenTake the inner product with $⟨m_1:m_2|$. On the other hand, the total ladder operator is $J_{\mp }^{(1)}\otimes 1+1\otimes J_{\mp }^{(2)}$, and $J_{\mp }^{†}=J_{\pm }$. Its action on the product bra is thereforeEquating the two evaluations and multiplying by $\sqrt{2}$ proves both signs of the ladder recurrence for Clebsch-Gordan coefficients:Terms with an index outside its allowed weight range are zero. Merely specifying eigenvectors of $J_3$ would allow unrelated phases for different $m$; the coherent positive-ladder convention above is what makes these recurrences hold with the displayed positive square roots.

A highest-weight vector is a nonzero vector $v_0$ satisfying $J_3{v}_0=j{v}_0$ and $J_{+}{v}_0=0$; its eigenvalue $j$ is the highest weight. Existence does not require the Lie algebra representation to be antihermitian. Since $V$ is a nonzero finite-dimensional complex vector space, $J_3$ has an eigenvector $v$ with some eigenvalue $\lambda$. The commutator with $J_{+}$ shows that each nonzero $J_{+}^{k}v$ has eigenvalue $\lambda +k$. Infinitely many such vectors would have distinct eigenvalues and be linearly independent, which is impossible. The last nonzero vector in this raising chain is therefore a highest-weight vector $v_0$.

Put $v_{\ell }=J_{-}^{\ell }{v}_0$. The ladder operator relations give $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. The raising coefficient follows by expanding the commutator of $J_{+}$ with a power of $J_{-}$:The factor $1/2$ is essential: these ladder operators have $[J_{+},J_{-}]=J_3$, rather than the convention with $2J_3$. The same finite-dimensionality argument gives a last nonzero lowered vector $v_r$, with $J_{-}{v}_r=v_{r+1}=0$. Applying $J_{+}$ to $v_{r+1}=0$ yieldsso $2j=r\in {\mathbb{Z}}_{\ge 0}$. All $v_0,\dots ,v_r$ are nonzero and have distinct eigenvalues. Their span is invariant under $J_3,J_{+},J_{-}$ and therefore under the original three generators. Irreducibility of the Lie algebra representation makes this span equal to $V$. HenceIn particular, the initial construction produces the largest eigenvalue of $J_3$, since its full spectrum is now $j,j-1,\dots ,-j$. This establishes the highest weight without assuming beforehand that all eigenvalues are real or that $J_3$ is diagonalizable.

For the additional antihermitian Lie algebra representation, the given definitions imply $J_3^{†}=J_3$ and $J_{+}^{†}=J_{-}$. Thus distinct $J_3$ eigenvalues give orthogonal weight vectors. With $\Vert v_0\Vert =1$, write $N_{\ell }=\Vert v_{\ell }{\Vert }^2$. The adjoint relation and the raising coefficient giveMultiplying this recursion proves the normalized highest-weight lowering formula in the present normalization:Every factorial argument is an integer, even for half-integral $j$. The displayed states have unit norm, mutually distinct eigenvalues, and span $V$, so they form an orthonormal basis. Their phase convention also fixes the positive ladder amplitudeswith the endpoint actions interpreted as zero. These amplitudes will determine the Clebsch-Gordan coefficients.

Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\dots ,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in \{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.

For $j=1/2$, abbreviate the product state $|1,q⟩|1/2,m⟩$ by $|q,\uparrow ⟩$ or $|q,\downarrow ⟩$. The highest-weight state is $|3/2,3/2⟩=|1,\uparrow ⟩$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it givesThese six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding Clebsch-Gordan coefficients.

The unique maximum-$M$ product state isApply the total lowering operator $J_{-}=J_{1-}+J_{2-}$ and normalize to obtain that multiplet. At the next highest remaining $M$, choose a normalized vector orthogonal to all previously constructed multiplets and annihilated by $J_{+}$. It is the next highest-weight state; lower it in the same way. Iterating produces every allowed total angular momentum and the Clebsch-Gordan coefficients.

For equal constituent spins, a total-spin singlet has $M=0$, so only the product basis vectors with opposite magnetic quantum numbers can occur. Acting with $J_{+}$ on the expansion and examining the coefficient of $|j,m+1{⟩}_1|j,-m{⟩}_2$ givesThus ${\alpha }_{m+1}=-{\alpha }_m$. Normalization requires $(2j+1)|{\alpha }_m{|}^2=1$. Choosing a conventional overall phase givesThe exponent is an integer even for half-integral $j$. Projecting the specified extreme product state onto this unique singlet gives the measurement probability