Addition of angular momentum

For two subsystems, the total operator is $\mathbf{J}={\mathbf{J}}_1\otimes I+I\otimes {\mathbf{J}}_2$, and $J_{\pm }=J_{1\pm }+J_{2\pm }$. The coupled basis diagonalizes ${\mathbf{J}}^2$ and $J_z$.

For dimensionless angular momentum operators and a Hamiltonian $H=a\mathbf{I}\cdot \mathbf{J}$ at fixed $I,J$, addition of angular momentum gives $E_F=a[F(F+1)-I(I+1)-J(J+1)]/2$. Subtracting neighbouring energy levels proves the displayed rule. For fine structure substitute $\mathbf{L},\mathbf{S}$ and total $J$; for magnetic-dipole hyperfine structure use electronic $J$, nuclear spin $I$ and total $F$. Electric-quadrupole interactions and mixing with nearby energy levels can violate the rule.

Consecutive positive gaps obey ${\nu }_F=|A|F/h$ for an isolated hyperfine multiplet governed by the magnetic-dipole hyperfine interaction. Their common difference estimates $|A|/h$, and the first gap divided by that scale identifies the first higher F. The endpoints of the resulting F ladder constrain I and J. Gap magnitudes alone do not fix the sign of A or an absolute nuclear g-factor.

Let $d_M$ count states of fixed angular momentum projection $M$ in a finite rotational representation, and $n_J$ its spin-$J$ multiplicity. Every spin-$J$ irreducible representation contributes one state at each $M=-J,\dots ,J$. Thus $d_M={\sum }_{J\ge M}{n}_J$ for nonnegative $M$, and subtraction gives $n_J=d_J-d_{J+1}$. Counting occupation states first incorporates Bose-Einstein statistics before decomposing identical-boson angular states.

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.

For $N$ identical fermions restricted to one angular-momentum multiplet $j$, with all other one-particle labels fixed, the Pauli exclusion principle permits distinct angular momentum projections $j,j-1,\dots ,j-N+1$. Their sum is $Nj-N(N-1)/2$, provided $N\le 2j+1$. The resulting Slater determinant is killed by the total raising angular momentum ladder operator because raising any occupied state either exceeds $j$ or duplicates an occupied state. It is therefore a highest-weight vector of precisely this total angular momentum. The formula is invariant under replacing particles by holes in the full shell.

Coupling orbital angular momentum $L>0$ to spin one-half produces $J=L\pm 1/2$. The dimensions add as $(2L+2)+2L=2(2L+1)$. A change between coupled and product bases changes labels, not the dimension of the Hilbert space.

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.

An atomic addition of angular momentum scheme in which the individual orbital angular momentum vectors first combine into $\mathbf{L}$ and the individual spins into $\mathbf{S}$, then these combine into total angular momentum $\mathbf{J}$. When spin-orbit coupling is weak compared with the interactions between electrons described by electrostatics, $L$ and $S$ remain useful approximate quantum numbers. The Landé g-factor describes the resulting weak-field magnetic splitting.

For electrons in distinct radial orbitals, angular coupling permits all $L$ from $|l_1-l_2|$ to $l_1+l_2$ and both total spins $S=0,1$. A p-s pair therefore gives singlet and triplet P terms. A nonequivalent p-p pair gives singlet and triplet S, P and D terms. The equivalent-subshell exclusion restrictions do not remove these nonequivalent terms. Counting $(2S+1)(2L+1)$ gives 12 p-s and 36 p-p microstates, matching products of one-electron state counts.

Two equivalent $l=1$ electrons require an antisymmetric complete state. Orbital exchange has parity $(-1{)}^L$ and the spin singlet is antisymmetric, giving allowed $L+S$ even. The terms are ${}^3{P}_{0,1,2}$, ${}^1{D}_2$ and ${}^1{S}_0$, with $9+5+1=15$ states, equal to $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund ordering puts ${}^3{P}_0$ lowest for the less-than-half-filled configuration.

For a composite quantum system, the total angular momentum operator is the sum of the constituent angular momenta, with each summand acting on its own tensor factor. Its simultaneous eigenstates with $J_z$ organize the state space into total-spin sectors.

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.

Coupling three distinguishable spin-one systems giveswhere the labels denote total spin and coefficients denote multiplicities. The dimensions are respectively $1$, $3\cdot 3$, $2\cdot 5$, and $7$, summing to $27$.

The completely symmetric subspace of three spin-one systems has dimensionand decomposes into one total-spin-$3$ multiplet and one total-spin-$1$ multiplet:The highest-weight product state generates the seven-dimensional spin-$3$ multiplet; its three-dimensional invariant complement is spin $1$.

The spin-one Cartesian basis $|x⟩,|y⟩,|z⟩$ transforms as a three-dimensional vector under rotations. In its two-site tensor square, the scalar, antisymmetric, and symmetric traceless subspaces are respectively the total-spin $0$, $1$, and $2$ sectors.

For a spin-one Hamiltonian $H=A{S}_z^2+B(S_x^2-S_y^2)$, Cartesian basis energies are ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and zero along the $x,y,z$ basis labels. The $S_x=\hslash$ state combines the latter two basis vectors equally. Their relative phase produces the displayed spin-reversal probability. Complete reversal occurs at odd multiples of $\pi /(|A+B|\hslash )$ if $A+B\ne 0$; when it vanishes those two components remain degenerate. Individual matrix signs depend on basis conventions, while the commutators and physical probability must remain consistent.

For two spins of magnitude $s$, ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=\frac{1}{2}({\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2)$. For two spin-one particles its eigenvalues in total-spin sectors $J=0,1,2$ are $-2,-1,1$ in units of ${\hslash }^2$.

For $s_1=s_2=3/2$, angular momentum addition gives one copy of $j=0,1,2,3$. The identity $2{\mathbf{S}}_1\cdot {\mathbf{S}}_2={\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2$ gives the four eigenvalues in the title, with state degeneracy $1,3,5,7$. These exhaust the $4\times 4$ spin space. If the particles are identical fermions, an appropriate spatial exchange parity accommodates each sector; only a specified spatial restriction can remove sectors.

The tensor product of spin-$j_1$ and spin-$j_2$ irreducible representations decomposes into one copy of each spinThe change-of-basis entries between uncoupled and coupled states are Clebsch--Gordan coefficients.

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.

With $J=j_1+j_2$, the maximal state is the product $|J,J⟩=|j_1,j_1⟩|j_2,j_2⟩$. Lowering it once givesThe orthogonal combination is the highest-weight state $|J-1,J-1⟩$.

Two equal spins $j$ have the unique total-spin-zero stateIt is invariant under joint rotations.

A singlet state is the one-dimensional rotation representation; this construction supplies the angular-momentum-zero case for two equal spins.

In the product basis $|m_1,m_2⟩$ for two spin-one particles, total magnetic quantum number zero allows only $|1,-1⟩$, $|0,0⟩$ and $|-1,1⟩$. Applying the total ladder operators to $\alpha |1,-1⟩+\beta |0,0⟩+\gamma |-1,1⟩$ givesA singlet state is annihilated by $S_{+}$ and $S_{-}$, so $\alpha =\gamma =-\beta$. Wavefunction normalization gives the displayed state, up to an overall phase. Here the ket on the left of the title formula has total-spin labels, while its right-hand kets have individual magnetic labels. The state is symmetric under exchanging the particles, although no exchange constraint is needed for distinguishable particles.

For two spin-one-half particles, the singlet isEvery component of the total spin annihilates it, and ${\mathbf{S}}^2$ has eigenvalue zero.

The symmetric two-particle statesform the spin-one triplet. On this subspace, ${\mathbf{S}}^2$ has eigenvalue $1(1+1){\hslash }^2=2{\hslash }^2$.