Addition of Angular Momenta

Derivation

Addition of Angular Momenta

D-353 Home PU-401 Threads symmetry · matter Depends on angular-momentum-ladder-operators

Statement

Given two independent angular momenta \(\hat{\vec{J}}_1\) and \(\hat{\vec{J}}_2\) with fixed quantum numbers \(j_1,j_2\), the tensor-product Hilbert space \(\mathcal{H}_{j_1}\otimes\mathcal{H}_{j_2}\), of dimension \((2j_1+1)(2j_2+1)\), decomposes into a direct sum of irreducible eigenspaces of the total angular momentum \(\hat{\vec{J}}=\hat{\vec{J}}_1+\hat{\vec{J}}_2\), \[ \mathcal{H}_{j_1}\otimes\mathcal{H}_{j_2}\;=\;\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}\mathcal{H}_j , \] and the change of basis from the uncoupled basis \(|j_1\,m_1\rangle|j_2\,m_2\rangle\) to the coupled basis \(|j\,m\rangle\) is effected by the Clebsch–Gordan coefficients \(\langle j_1 m_1\, j_2 m_2 | j\, m\rangle\), which we construct via the highest-weight state, the lowering operator \(\hat{J}_-\), and orthogonality.

Why it matters

Combining angular momenta is the arithmetic of quantum symmetry. Any time two rotational degrees of freedom couple — orbital and spin (spin–orbit coupling), two electron spins (singlet/triplet), nuclear spins in hyperfine structure, or isospin in nuclear and particle physics — the physically meaningful states are eigenstates of the total angular momentum, because it is \(\hat{\vec{J}}\), not \(\hat{\vec{J}}_1\) or \(\hat{\vec{J}}_2\) separately, that commutes with a rotationally invariant Hamiltonian.

The decomposition tells you exactly which total-\(j\) sectors appear and with what multiplicity (here, once each), fixing spectroscopic term symbols, selection rules, and the block structure that makes otherwise intractable Hamiltonians diagonal by symmetry alone.

Assumptions

The two angular momenta act on distinct factors and commute: \([\hat{J}_{1i},\hat{J}_{2k}]=0\). If dropped, \(\hat{\vec{J}}_1\) and \(\hat{\vec{J}}_2\) are not independent observables, \(\hat{\vec{J}}^2\) need not commute with each \(\hat{\vec{J}}_a^2\), and the product space does not factor as assumed.

\(j_1\) and \(j_2\) are fixed (sharp): we work inside a single \((2j_1+1)(2j_2+1)\)-dimensional block. If dropped, \(\hat{J}_1^2\) or \(\hat{J}_2^2\) are not good quantum numbers and the neat finite decomposition is replaced by a sum over their spectra.

Each factor carries an irreducible representation of \(\mathfrak{su}(2)\): the ladder operators connect all \(2j_a+1\) states of the factor. If dropped (a reducible factor), the multiplicity of each total \(j\) is no longer one and the Clebsch–Gordan series acquires multiplicity labels.

Derivation

\[ \hat{\vec{J}}=\hat{\vec{J}}_1\otimes\hat{\mathbb{1}}+\hat{\mathbb{1}}\otimes\hat{\vec{J}}_2,\qquad \hat{J}_z=\hat{J}_{1z}+\hat{J}_{2z} \]

Total angular momentum is the generator of simultaneous rotations of both factors; additivity of the generators is the statement that the product representation is \(D^{(j_1)}\otimes D^{(j_2)}\). A

\[ [\hat{J}_i,\hat{J}_k]=\varepsilon_{ikl}\,i\hbar\,\hat{J}_l \]

Adding two commuting copies of the \(\mathfrak{su}(2)\) algebra gives a third copy, because \([\hat{J}_{1i}+\hat{J}_{2i},\hat{J}_{1k}+\hat{J}_{2k}]=\varepsilon_{ikl}i\hbar(\hat{J}_{1l}+\hat{J}_{2l})\) using the commuting-factor assumption. So \(\hat{\vec{J}}\) is a bona fide angular momentum. A

\[ [\hat{J}^2,\hat{J}_1^2]=[\hat{J}^2,\hat{J}_2^2]=[\hat{J}^2,\hat{J}_z]=0 \]

Expanding \(\hat{J}^2=\hat{J}_1^2+\hat{J}_2^2+2\hat{\vec{J}}_1\cdot\hat{\vec{J}}_2\), each term commutes with \(\hat{J}_1^2\) and \(\hat{J}_2^2\) (scalars under each rotation) and with \(\hat{J}_z\). Hence \(\{\hat{J}_1^2,\hat{J}_2^2,\hat{J}^2,\hat{J}_z\}\) is a complete set of commuting observables — the coupled basis. B

\[ \hat{J}_z\,|j_1 m_1\rangle|j_2 m_2\rangle=\hbar(m_1+m_2)\,|j_1 m_1\rangle|j_2 m_2\rangle \]

The uncoupled product states already diagonalize \(\hat{J}_1^2,\hat{J}_2^2,\hat{J}_z\). Their \(\hat{J}_z\) eigenvalue is \(m=m_1+m_2\); only \(\hat{J}^2\) remains to be diagonalized. A

\[ N(m)=\#\{(m_1,m_2):m_1+m_2=m\},\qquad n_j=N(m{=}j)-N(m{=}j+1) \]

Count product states in each \(m\)-ladder. Because each irreducible \(D^{(j)}\) contributes exactly one state at every \(m\in\{-j,\dots,j\}\), the multiplicity \(n_j\) of total \(j\) is the drop in the degeneracy \(N(m)\) between \(m=j\) and \(m=j+1\). This yields \(j=|j_1-j_2|,\dots,j_1+j_2\), each once. B

\[ \sum_{j=|j_1-j_2|}^{j_1+j_2}(2j+1)=(2j_1+1)(2j_2+1) \]

Dimension check confirming completeness of the series: summing the arithmetic run (take \(j_1\ge j_2\)) gives \(\sum_{j=j_1-j_2}^{j_1+j_2}(2j+1)=(2j_2+1)(2j_1+1)\), matching the product-space dimension. No sector is missing or double-counted. C

\[ |j_1+j_2,\,j_1+j_2\rangle=|j_1\,j_1\rangle|j_2\,j_2\rangle \]

The highest-weight state. The maximal \(m=j_1+j_2\) subspace is one-dimensional, so this single product state must be the top of the \(j=j_1+j_2\) multiplet. The Condon–Shortley phase convention fixes its coefficient to \(+1\). A

\[ \hat{J}_-\,|j\,m\rangle=\hbar\sqrt{j(j+1)-m(m-1)}\;|j\,m-1\rangle,\qquad \hat{J}_-=\hat{J}_{1-}+\hat{J}_{2-} \]

Apply the total lowering operator (from the assumed ladder-operator result) to generate the whole \(j=j_1+j_2\) multiplet. Because \(\hat{J}_-=\hat{J}_{1-}+\hat{J}_{2-}\), acting on a product state distributes over both factors, producing a definite linear combination of product states at each \(m\). B

\[ |j_1+j_2,\,j_1+j_2-1\rangle=\sqrt{\tfrac{j_1}{j_1+j_2}}\,|j_1\,j_1-1\rangle|j_2\,j_2\rangle+\sqrt{\tfrac{j_2}{j_1+j_2}}\,|j_1\,j_1\rangle|j_2\,j_2-1\rangle \]

Explicit first descent: equate \(\hat{J}_-\) applied to the coupled state (LHS coefficient \(\hbar\sqrt{2(j_1+j_2)}\)) with \((\hat{J}_{1-}+\hat{J}_{2-})\) applied to the product state, then normalize. This is the algorithm; iterate to fill the multiplet. B

\[ |j_1+j_2-1,\,j_1+j_2-1\rangle\perp|j_1+j_2,\,j_1+j_2-1\rangle \]

Build the next multiplet's highest-weight state. The \(m=j_1+j_2-1\) subspace is two-dimensional; one direction is already used by the \(j=j_1+j_2\) ladder. The orthogonal unit vector (Condon–Shortley: coefficient of the highest-\(m_1\) product state chosen real positive) is the top of \(j=j_1+j_2-1\). Lower it, repeat orthogonalization at each \(m\), and descend \(j\) by one until \(j=|j_1-j_2|\). C

\[ |j\,m\rangle=\sum_{m_1+m_2=m}\langle j_1 m_1\,j_2 m_2|j\,m\rangle\,|j_1 m_1\rangle|j_2 m_2\rangle \]

Collecting the coefficients generated in steps 7–10 defines the Clebsch–Gordan coefficients. The sum is restricted to \(m_1+m_2=m\) because \(\hat{J}_z\) is diagonal; the coefficients are real in the Condon–Shortley convention. A

\[ \sum_{m_1,m_2}\langle j_1 m_1\,j_2 m_2|j\,m\rangle\langle j_1 m_1\,j_2 m_2|j'\,m'\rangle=\delta_{jj'}\delta_{mm'} \]

Orthonormality/unitarity of the transformation. Since both bases are orthonormal and the CG matrix is real, it is orthogonal; the inverse relation \(|j_1 m_1\rangle|j_2 m_2\rangle=\sum_{j}\langle j_1 m_1\,j_2 m_2|j\,m\rangle|j\,m\rangle\) follows by transposition. This closes the construction. C

Result

\[ D^{(j_1)}\otimes D^{(j_2)}=\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}D^{(j)},\qquad |j\,m\rangle=\sum_{m_1+m_2=m}\langle j_1 m_1\,j_2 m_2|j\,m\rangle\,|j_1 m_1\rangle|j_2 m_2\rangle \]

Reading. Two angular momenta of size \(j_1\) and \(j_2\) combine into total angular momenta running from \(|j_1-j_2|\) to \(j_1+j_2\) in integer steps, each value occurring exactly once. The Clebsch–Gordan coefficients are the fixed, convention-independent-up-to-phase amplitudes that rotate the "each part sharp" basis into the "total sharp" basis; \(\hat{J}_z\) conservation forces \(m=m_1+m_2\).

Units check. Clebsch–Gordan coefficients are pure numbers (dimensionless), being inner products of normalized states. Dimensions live only in the eigenvalues: \(\hat{J}^2\to\hbar^2 j(j+1)\) and \(\hat{J}_z\to\hbar m\), so both sides of each ladder relation carry a single power of \(\hbar\) per lowering, consistent throughout. The dimension count \(\sum(2j+1)=(2j_1+1)(2j_2+1)\) is an integer identity.

Limiting cases

\(j_2=0\): the series collapses to the single value \(j=j_1\); coupling to a scalar does nothing, and every CG coefficient is \(1\).
\(j_1=j_2=\tfrac12\): \(\tfrac12\otimes\tfrac12=1\oplus 0\), the triplet plus singlet; the singlet CG coefficients are \(\pm 1/\sqrt2\).
Stretched states \(m=\pm(j_1+j_2)\): unique product states, CG coefficient \(=1\); no superposition needed.
\(j_1\gg j_2\) (or \(j_2\) fixed, \(j_1\to\infty\)): the \(2j_2+1\) allowed \(j\) values cluster around \(j_1\), the semiclassical regime where \(\hat{\vec{J}}_2\) precesses about a nearly-fixed \(\hat{\vec{J}}_1\).

Breaks when

The two angular momenta do not commute (share degrees of freedom, e.g. attempting to "add" \(\hat{L}\) to \(\hat{L}\) of the same particle as independent): the tensor-product structure fails and the multiplicity-one series is wrong.
A factor is reducible or \(j_1,j_2\) are not sharp: the decomposition acquires nontrivial multiplicities \(n_j>1\), requiring an extra label, and the simple triangle rule no longer gives one copy each.
Relativistic / field-theoretic regimes where particle number is not fixed: coupling angular momenta of a fixed pair is ill-defined when creation and annihilation mix sectors; one must couple within the full Poincaré representation instead.
Coupling three or more momenta without care: the pairwise decomposition is basis-dependent (recoupling), and \(6j\)/\(9j\) symbols, not a single CG series, govern the result.

Failure modes

Adding the wrong quantity: writing \(j=j_1+j_2\) as the only value, forgetting the full triangle \(|j_1-j_2|\le j\le j_1+j_2\).
Adding \(m\) as if it were \(j\): \(m=m_1+m_2\) always, but \(j\ne j_1+j_2\) in general; confusing the two collapses the multiplet structure.
Dropping the \(\hat{J}_z\) constraint: including product states with \(m_1+m_2\ne m\) in the expansion of \(|j\,m\rangle\).
Sign/phase errors: ignoring the Condon–Shortley convention and getting overall signs of singlet or lower-\(j\) states wrong.
Mis-normalizing after \(\hat{J}_-\): forgetting the \(\sqrt{j(j+1)-m(m-1)}\) factor, so the descended states are not unit vectors.
Assuming CG coefficients carry \(\hbar\): they are pure numbers; only eigenvalues carry dimensions.

Discussion

The heart of the construction is representation theory: \(\mathcal{H}_{j_1}\otimes\mathcal{H}_{j_2}\) carries a generally reducible representation of \(SU(2)\), and Clebsch–Gordan decomposition is nothing but its reduction into irreducibles. The three tools — highest weight, lowering operator, orthogonalization — are the constructive proof that the reduction is the multiplicity-free series \(\bigoplus_j D^{(j)}\). Everything spectroscopic (term symbols like \(^{2S+1}L_J\), Landé factors, hyperfine splittings) is a corollary of which \(j\) sectors appear.

Physically, the coupled basis is preferred whenever the Hamiltonian is rotationally invariant, since then \([\hat{H},\hat{\vec{J}}]=0\) and \(\hat{J}^2,\hat{J}_z\) are conserved while \(\hat{J}_{1z},\hat{J}_{2z}\) individually are not (they are scrambled by any \(\hat{\vec{J}}_1\cdot\hat{\vec{J}}_2\) interaction). The Wigner–Eckart theorem elevates this to a general statement: matrix elements of any tensor operator factor into a CG coefficient times a single reduced matrix element, so the entire angular dependence of transition rates is model-independent geometry.

Group-theoretically the coefficients are the intertwiners realizing the isomorphism between \(D^{(j_1)}\otimes D^{(j_2)}\) and \(\bigoplus_j D^{(j)}\); their reality (Condon–Shortley) reflects the fact that all \(SU(2)\) irreps are (pseudo)real, and their symmetry relations under permuting \((j_1 m_1)\leftrightarrow(j_2 m_2)\) and under \(j\)-swaps are encoded compactly by the Wigner \(3j\) symbol \(\begin{pmatrix}j_1&j_2&j\\ m_1&m_2&-m\end{pmatrix}\), whose manifest symmetry makes it the natural object for recoupling three or more momenta via \(6j\) and \(9j\) symbols.

Common misconceptions. Students often think "adding angular momenta" means vector addition of definite arrows; it does not — quantum \(\hat{\vec{J}}_1\) and \(\hat{\vec{J}}_2\) cannot both point definitely, and only \(j_1,j_2,j,m\) are simultaneously sharp. Another trap is imagining the CG transformation mixes different \(j_1\) or \(j_2\); it never does — those are block-diagonal invariants, and only \(m_1,m_2\) get superposed within a fixed \((j_1,j_2)\) block.

Worked examples

Example 1 — Two spin-½ particles (\(j_1=j_2=\tfrac12\)).

\[ j\in\{|{\tfrac12-\tfrac12}|,\dots,\tfrac12+\tfrac12\}=\{0,1\},\quad \dim=(2\cdot\tfrac12+1)^2=4=3+1 \]

Triangle rule and dimension check: a triplet (\(j=1\)) and a singlet (\(j=0\)). A

\[ |1,1\rangle=|\uparrow\uparrow\rangle \]

Highest-weight state at \(m=1\), the unique product state. A

\[ \hat{J}_-|1,1\rangle=\hbar\sqrt{2}\,|1,0\rangle=(\hat{J}_{1-}+\hat{J}_{2-})|\uparrow\uparrow\rangle=\hbar(|\downarrow\uparrow\rangle+|\uparrow\downarrow\rangle) \]

Lower once; \(\sqrt{1\cdot2-1\cdot0}=\sqrt2\) on the left, unit factors on the right for spin-½. B

\[ |1,0\rangle=\tfrac{1}{\sqrt2}\big(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\big),\qquad |1,-1\rangle=|\downarrow\downarrow\rangle \]

Normalize; the bottom of the triplet is again a unique product state. A

\[ |0,0\rangle=\tfrac{1}{\sqrt2}\big(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\big) \]

The \(m=0\) singlet is the unit vector orthogonal to \(|1,0\rangle\); Condon–Shortley fixes the sign so the \(|\uparrow\downarrow\rangle\) coefficient is positive. B

\[ \tfrac12\otimes\tfrac12=\underbrace{\{|1,1\rangle,|1,0\rangle,|1,-1\rangle\}}_{\text{triplet, symmetric}}\ \oplus\ \underbrace{|0,0\rangle}_{\text{singlet, antisymmetric}} \]

Reading. CG coefficients \(\langle\tfrac12\pm\tfrac12\,\tfrac12\mp\tfrac12|1\,0\rangle=+\tfrac1{\sqrt2}\) and \(\langle\tfrac12\tfrac12\,\tfrac12-\tfrac12|0\,0\rangle=+\tfrac1{\sqrt2}\), \(\langle\tfrac12-\tfrac12\,\tfrac12\tfrac12|0\,0\rangle=-\tfrac1{\sqrt2}\). Units check. All coefficients dimensionless; states normalized (\(|{1/\sqrt2}|^2+|{1/\sqrt2}|^2=1\)).

Example 2 — Spin–orbit coupling for a p-electron (\(j_1=\ell=1,\ j_2=s=\tfrac12\)).

\[ j\in\{|1-\tfrac12|,\dots,1+\tfrac12\}=\{\tfrac12,\tfrac32\},\quad \dim=(3)(2)=6=4+2 \]

Triangle rule: a \(j=\tfrac32\) quartet and a \(j=\tfrac12\) doublet — the \(^2P_{3/2}\) and \(^2P_{1/2}\) fine-structure levels. A

\[ \big|\tfrac32,\tfrac32\big\rangle=|\ell{=}1,m_\ell{=}1\rangle\big|\tfrac12,\tfrac12\big\rangle \]

Highest-weight state at \(m=\tfrac32\). A

\[ \hat{J}_-\big|\tfrac32,\tfrac32\big\rangle=\hbar\sqrt{3}\,\big|\tfrac32,\tfrac12\big\rangle=\hbar\sqrt2\,|1,0\rangle\big|\tfrac12,\tfrac12\big\rangle+\hbar\,|1,1\rangle\big|\tfrac12,-\tfrac12\big\rangle \]

Left: \(\sqrt{\tfrac32\cdot\tfrac52-\tfrac32\cdot\tfrac12}=\sqrt3\). Right: \(\hat{L}_-\) gives \(\sqrt{1\cdot2-1\cdot0}=\sqrt2\); \(\hat{S}_-\) gives \(1\). B

\[ \big|\tfrac32,\tfrac12\big\rangle=\sqrt{\tfrac23}\,|1,0\rangle\big|\tfrac12,\tfrac12\big\rangle+\sqrt{\tfrac13}\,|1,1\rangle\big|\tfrac12,-\tfrac12\big\rangle \]

Divide by \(\sqrt3\) and read off the CG coefficients \(\sqrt{2/3},\ \sqrt{1/3}\). B

\[ \big|\tfrac12,\tfrac12\big\rangle=\sqrt{\tfrac13}\,|1,0\rangle\big|\tfrac12,\tfrac12\big\rangle-\sqrt{\tfrac23}\,|1,1\rangle\big|\tfrac12,-\tfrac12\big\rangle \]

Unit vector in the same \(m=\tfrac12\) plane orthogonal to \(\big|\tfrac32,\tfrac12\big\rangle\). Condon–Shortley fixes the phase by demanding \(\langle j_1\,j_1;\,j_2,\,m{-}j_1\,|\,j\,j\rangle>0\), i.e. the coefficient of the highest-\(m_\ell\) product state (\(m_\ell{=}1\)) is real positive at the top of the doublet; propagating the sign gives the coefficients shown, with \(\big(\sqrt{1/3}\big)^2+\big(\sqrt{2/3}\big)^2=1\). C

\[ 1\otimes\tfrac12=D^{(3/2)}\oplus D^{(1/2)}\ \Longleftrightarrow\ {}^2P_{3/2}\ (4\text{ states})\ \oplus\ {}^2P_{1/2}\ (2\text{ states}) \]

Reading. The general result \(\big|\ell+\tfrac12,m\big\rangle=\sqrt{\tfrac{\ell+m+1/2}{2\ell+1}}\,|m-\tfrac12\rangle|\!\uparrow\rangle+\sqrt{\tfrac{\ell-m+1/2}{2\ell+1}}\,|m+\tfrac12\rangle|\!\downarrow\rangle\) reproduces the numbers above at \(\ell=1,m=\tfrac12\). Units check. Coefficients dimensionless; spin–orbit energy \(\propto\langle\hat{\vec{L}}\cdot\hat{\vec{S}}\rangle=\tfrac{\hbar^2}{2}[j(j+1)-\ell(\ell+1)-s(s+1)]\) carries the \(\hbar^2\), giving \(+\tfrac{\hbar^2}{2}\) for \(j=\tfrac32\) and \(-\hbar^2\) for \(j=\tfrac12\).

Problems

State all allowed total \(j\) for \(j_1=2,\ j_2=1\), and verify the dimension count.
Solution
Triangle rule: \(j=|2-1|,\dots,2+1=1,2,3\). Dimensions \((2\cdot3+1)?\) no: \((2j_1+1)(2j_2+1)=(5)(3)=15\). Check \(\sum(2j+1)=(3)+(5)+(7)=15.\ \checkmark\) So \(2\otimes1=D^{(1)}\oplus D^{(2)}\oplus D^{(3)}\).
For \(j_1=j_2=1\), how many total states have \(m=0\), and how many total-\(j\) multiplets contain an \(m=0\) state?
Solution
\(1\otimes1=0\oplus1\oplus2\) (dim \(1+3+5=9=3^2\)). Product states with \(m_1+m_2=0\): \((1,-1),(0,0),(-1,1)\Rightarrow N(0)=3\). Every \(j=0,1,2\) contains an \(m=0\) state, so \(3\) multiplets — consistent with \(N(0)=3\).
Compute the CG coefficient \(\langle 1\,0\,;\,1\,0\,|\,0\,0\rangle\) for coupling two \(j=1\) momenta to a total singlet.
Solution
Build \(|0,0\rangle\propto\sum_{m}(-1)^{1-m}|1,m\rangle|1,-m\rangle\) (antisymmetric scalar). Normalized: \(|0,0\rangle=\tfrac{1}{\sqrt3}\big(|1,1\rangle|1,-1\rangle-|1,0\rangle|1,0\rangle+|1,-1\rangle|1,1\rangle\big)\). Hence \(\langle 1\,0\,;1\,0|0\,0\rangle=-\tfrac{1}{\sqrt3}\). (Check normalization: \(3\times(1/\sqrt3)^2=1.\ \checkmark\))
A p-electron (\(\ell=1\)) has spin \(\tfrac12\). Using the spin–orbit shift \(\Delta E=\tfrac{\zeta\hbar^2}{2}[j(j+1)-\ell(\ell+1)-s(s+1)]\) with \(\zeta>0\) a constant of dimension energy/\(\hbar^2\), find the splitting between \(^2P_{3/2}\) and \(^2P_{1/2}\).
Solution
For \(j=\tfrac32\): \(\tfrac{\zeta\hbar^2}{2}[\tfrac{15}4-2-\tfrac34]=\tfrac{\zeta\hbar^2}{2}(1)=+\tfrac{\zeta\hbar^2}{2}\). For \(j=\tfrac12\): \(\tfrac{\zeta\hbar^2}{2}[\tfrac34-2-\tfrac34]=\tfrac{\zeta\hbar^2}{2}(-2)=-\zeta\hbar^2\). Splitting \(\Delta=E_{3/2}-E_{1/2}=\tfrac{\zeta\hbar^2}{2}+\zeta\hbar^2=\tfrac{3}{2}\zeta\hbar^2\). The Landé interval rule (\(\Delta\propto j_{\text{upper}}=\tfrac32\)) is reproduced.
Two identical spin-1 bosons occupy the same spatial orbital. Which total-spin states \(S\) are allowed by Bose symmetry?
Solution
\(1\otimes1=S{=}0\oplus S{=}1\oplus S{=}2\). Exchange symmetry of coupled two-particle states of equal \(j\) is \((-1)^{2j-S}=(-1)^{2-S}\): symmetric for \(S=2\) (\((-1)^0=+\)) and \(S=0\) (\((-1)^2=+\)), antisymmetric for \(S=1\) (\((-1)^1=-\)). A symmetric spatial part (same orbital) requires a symmetric spin part for bosons, so only \(S=0\) and \(S=2\) are allowed; \(S=1\) is forbidden.