Addition of Angular Momenta and Clebsch–Gordan Coefficients

A hydrogen electron carries orbital angular momentum $L$ and spin $S$ ; two electrons in an atom each carry spin; a proton and electron each carry spin in the hyperfine problem. In every case the physically relevant quantity is the total angular momentum $J = J_{1} + J_{2}$ , because it is $J$ , not the parts, that a rotationally invariant interaction conserves. Passing from the product states of the individual angular momenta to the eigenstates of the total is the addition problem, and its solution is a fixed table of numbers, the Clebsch–Gordan coefficients, that recurs everywhere from atomic spectra to particle physics.

The algebraic machinery of the previous lesson supplies every tool: each $J_{i}$ obeys the standard commutators, and so does the sum.

Two bases for the combined system

Let $J_{1}$ and $J_{2}$ be two angular momenta acting on distinct degrees of freedom, so every component of the first commutes with every component of the second, $[J_{1 i}, J_{2 j}] = 0$ . Each has its own multiplet structure with fixed $j_{1}$ and $j_{2}$ . The combined space is the tensor product, of dimension $(2 j_{1} + 1) (2 j_{2} + 1)$ , and there are two natural bases for it.

Uncoupled basis. The simultaneous eigenstates of ${J_{1}^{2}, J_{1 z}, J_{2}^{2}, J_{2 z}}$ , $∣ j_{1}, m_{1} ⟩ ∣ j_{2}, m_{2} ⟩, m_{1} = - j_{1}, \dots, j_{1}, m_{2} = - j_{2}, \dots, j_{2},$ in which each part has a definite projection. This is the basis you write down first.
Coupled basis. The simultaneous eigenstates of ${J^{2}, J_{z}, J_{1}^{2}, J_{2}^{2}}$ , written $∣ j, m; j_{1}, j_{2} ⟩$ (the fixed $j_{1}, j_{2}$ often suppressed), in which the total has a definite magnitude and projection.

Define the total angular momentum $J = J_{1} + J_{2}$ . It obeys the algebra, since

[J_{i}, J_{j}] = [J_{1 i} + J_{2 i}, J_{1 j} + J_{2 j}] = i ℏ ε_{ij k} J_{1 k} + i ℏ ε_{ij k} J_{2 k} = i ℏ ε_{ij k} J_{k},

the cross terms vanishing because the two spaces commute. The coupled labels are consistent because $J^{2}$ and $J_{z}$ commute with $J_{1}^{2}$ and $J_{2}^{2}$ , but not with $J_{1 z}$ or $J_{2 z}$ individually:

[J^{2}, J_{1 z}] \neq = 0, [J^{2}, J_{1}^{2}] = 0, [J_{z}, J_{1 z}] = 0.

The last two identities are why ${J^{2}, J_{z}, J_{1}^{2}, J_{2}^{2}}$ is a valid commuting set while ${J^{2}, J_{1 z}, J_{2 z}}$ is not. Choosing between the bases is choosing which projections to sharpen: the individual ones, or the total.

The range of the total quantum number

$J_{z} = J_{1 z} + J_{2 z}$ is diagonal in the uncoupled basis, so every product state already has a definite total projection:

J_{z} (∣ j_{1}, m_{1} ⟩ ∣ j_{2}, m_{2} ⟩) = ℏ (m_{1} + m_{2}) ∣ j_{1}, m_{1} ⟩ ∣ j_{2}, m_{2} ⟩, m = m_{1} + m_{2} .

Counting how many product states share each value of $m$ fixes which total $j$ -multiplets are present. The maximum projection is $m = j_{1} + j_{2}$ , reached by the single state $∣ j_{1}, j_{1} ⟩ ∣ j_{2}, j_{2} ⟩$ . A state with the highest projection in the whole space must be the top rung of a multiplet with $j = j_{1} + j_{2}$ , and no larger $j$ can occur. Stepping down, each value of $m$ is hit by one additional product state until $m = ∣ j_{1} - j_{2} ∣$ , and each new appearance starts one more multiplet. The result:

The dimension count confirms completeness. With $j_{1} \geq j_{2}$ the multiplets run from $j_{1} - j_{2}$ to $j_{1} + j_{2}$ , a total of $2 j_{2} + 1$ values, and the dimensions sum by the arithmetic series to

j = j_{1} - j_{2} \sum j_{1} + j_{2} (2 j + 1) = (2 j_{2} + 1) \cdot \frac{[ 2 ( j _{1} - j _{2} ) + 1 ] + [ 2 ( j _{1} + j _{2} ) + 1 ]}{2} = (2 j_{2} + 1) (2 j_{1} + 1),

exactly the dimension of the product space. No states are lost or double-counted.

Coupling j1 = 1 with j2 = 1 gives three total multiplets of lengths five, three, and one, whose rungs total the nine product states.

Uncoupled states on the (m1, m2) lattice for j1 = j2 = 1. Each anti-diagonal collects the states of one total projection m = m1 + m2; the nine states regroup into total multiplets j = 2, 1, 0.

Constructing the coupled states

The lowering operator builds each multiplet explicitly. Start from the unique top state, which is a bare product:

∣ j_{1} + j_{2}, j_{1} + j_{2} ⟩ = ∣ j_{1}, j_{1} ⟩ ∣ j_{2}, j_{2} ⟩ .

Apply the total lowering operator $J_{-} = J_{1 -} + J_{2 -}$ to descend. The left side uses the ladder formula for the total, the right side distributes over the two factors, and matching the results expresses each coupled state as a combination of products. Once a whole $j = j_{1} + j_{2}$ column is built, the top state of the next multiplet, $j = j_{1} + j_{2} - 1$ , is the unique combination in the $m = j_{1} + j_{2} - 1$ subspace orthogonal to the one already found. Repeating alternately, lowering within a multiplet and orthogonalizing to start the next, generates the entire coupled basis.

The coupled basis is built column by column: lower within a multiplet with the total J-minus operator, then orthogonalize in the shared m-subspace to seed the next lower multiplet.

Two spin-halves: the singlet and triplet

Take $j_{1} = j_{2} = \frac{1}{2}$ . The series gives $\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$ : a spin- $1$ triplet and a spin- $0$ singlet, four states in all. Write $↑, ↓$ for the $m = \pm \frac{1}{2}$ single-spin states. The triplet top is $∣ 1, 1 ⟩ =↑↑$ . Lowering with $J_{-} = J_{1 -} + J_{2 -}$ ,

J_{-} ∣ 1, 1 ⟩ = ℏ 2 ∣ 1, 0 ⟩, J_{-} (↑↑) = ℏ ↓↑ + ℏ ↑↓,

so equating gives the symmetric combination

∣ 1, 0 ⟩ = \frac{1}{2} (↑↓ + ↓↑) .

Lowering once more yields $∣ 1, - 1 ⟩ =↓↓$ . The singlet is the remaining $m = 0$ state, orthogonal to $∣ 1, 0 ⟩$ :

∣ 0, 0 ⟩ = \frac{1}{2} (↑↓ - ↓↑) .

The exchange symmetry is not incidental: the three triplet states are symmetric under swapping the two spins and the singlet is antisymmetric, a division that drives the exchange splitting of helium and the spin structure of hydrogen's hyperfine levels. The value of $J_{1} \cdot J_{2}$ separates them cleanly. From $J^{2} = J_{1}^{2} + J_{2}^{2} + 2 J_{1} \cdot J_{2}$ ,

J_{1} \cdot J_{2} = \frac{1}{2} (J^{2} - J_{1}^{2} - J_{2}^{2}) = \frac{ℏ ^{2}}{2} [j (j + 1) - \frac{3}{4} - \frac{3}{4}] = {+ \frac{1}{4} ℏ^{2}, - \frac{3}{4} ℏ^{2}, j = 1 (triplet), j = 0 (singlet) .

Two spin-halves span a four-dimensional space that splits into a symmetric total-spin-1 triplet and an antisymmetric total-spin-0 singlet.

Clebsch–Gordan coefficients

Each coupled state expands in the uncoupled basis:

∣ j, m ⟩ = m_{1} + m_{2} = m \sum ⟨ j_{1}, m_{1}; j_{2}, m_{2} ∣ j, m ⟩ ∣ j_{1}, m_{1} ⟩ ∣ j_{2}, m_{2} ⟩ .

The overlaps $⟨ j_{1}, m_{1}; j_{2}, m_{2} ∣ j, m ⟩$ are the Clebsch–Gordan coefficients. They vanish unless two selection rules hold, both already derived:

Projection additivity. The coefficient is zero unless $m = m_{1} + m_{2}$ , because $J_{z} = J_{1 z} + J_{2 z}$ .
Triangle rule. The coefficient is zero unless $∣ j_{1} - j_{2} ∣ \leq j \leq j_{1} + j_{2}$ , the Clebsch–Gordan series.

With the Condon–Shortley convention (all coefficients real, and $⟨ j_{1}, j_{1}; j_{2}, j - j_{1} ∣ j, j ⟩ > 0$ ) the coefficients are unique. Because the change of basis is unitary and real, it is orthogonal, giving two orthogonality relations:

m_{1}, m_{2} \sum ⟨ j_{1} m_{1}; j_{2} m_{2} ∣ j m ⟩ ⟨ j_{1} m_{1}; j_{2} m_{2} ∣ j^{'} m^{'} ⟩ = δ_{j j^{'}} δ_{m m^{'}},

j, m \sum ⟨ j_{1} m_{1}; j_{2} m_{2} ∣ j m ⟩ ⟨ j_{1} m_{1}^{'}; j_{2} m_{2}^{'} ∣ j m ⟩ = δ_{m_{1} m_{1}^{'}} δ_{m_{2} m_{2}^{'}} .

The inverse expansion writes each product state as a sum over coupled states with the same coefficients. Reading a published table, the rows are labelled by $(m_{1}, m_{2})$ and the columns by $(j, m)$ ; a printed entry is the coefficient with the convention that a square-root sign is understood over each number, keeping the sign. The two spin- $\frac{1}{2}$ block reads:

$(m_{1}, m_{2})$	$∣ 1, 1 ⟩$	$∣ 1, 0 ⟩$	$∣ 0, 0 ⟩$	$∣ 1, - 1 ⟩$
$(+ \frac{1}{2}, + \frac{1}{2})$	$1$	$0$	$0$	$0$
$(+ \frac{1}{2}, - \frac{1}{2})$	$0$	$\frac{1}{2}$	$\frac{1}{2}$	$0$
$(- \frac{1}{2}, + \frac{1}{2})$	$0$	$\frac{1}{2}$	$- \frac{1}{2}$	$0$
$(- \frac{1}{2}, - \frac{1}{2})$	$0$	$0$	$0$	$1$

Each column is a coupled state read off down the rows, and each row is a product state read across the columns; the singlet column carries the lone relative minus sign that makes it antisymmetric.

Coupling orbital and spin angular momentum

The same construction applies to $J = L + S$ for a spin- $\frac{1}{2}$ particle, producing the $∣ n, ℓ, j, m_{j} ⟩$ basis in which the spin–orbit interaction is diagonal. With $s = \frac{1}{2}$ the series gives only two totals, $j = ℓ + \frac{1}{2}$ and $j = ℓ - \frac{1}{2}$ (for $ℓ \geq 1$ ).

The lesson's arithmetic, done once, is reused wherever two rotational degrees of freedom combine.¹²³

Sakurai & Napolitano, Modern Quantum Mechanics, 3rd ed. (Cambridge, 2021), §3.8 — addition of angular momenta, the Clebsch–Gordan series from projection counting, and the recursion relations fixing the coefficients. Publisher: https://doi.org/10.1017/9781108587280 ↩
Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed. (Cambridge, 2018), §4.4.3 — combining spins with the lowering operator, the singlet/triplet decomposition, and reading Clebsch–Gordan tables. Publisher: https://doi.org/10.1017/9781316995433 ↩
Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II (Wiley, 1977), Ch. X — the general theory of adding two angular momenta and the properties of the Clebsch–Gordan coefficients. Publisher: https://doi.org/10.1002/9783527617586 ↩