---
title: Addition of Angular Momenta and Clebsch–Gordan Coefficients
module: Angular Momentum
moduleNumber: 6
lessonNumber: 3
order: 603
summary: >
  Two angular momenta combine into a total whose allowed magnitudes run from the
  difference to the sum of the parts in integer steps. The change from the
  uncoupled product basis to the coupled total-angular-momentum basis is carried
  out with the lowering operator and orthogonality, and its matrix of overlaps is
  the table of Clebsch–Gordan coefficients. Two spin-halves split into a triplet
  and a singlet, the prototype for every composite spin.
topics: [Angular Momentum]
sources:
  - book: Sakurai & Napolitano
    ref: "§3.8 Addition of Angular Momenta"
  - book: Griffiths & Schroeter
    ref: "§4.4.3 Addition of Angular Momenta"
  - book: Cohen-Tannoudji
    ref: "Ch. X — Addition of Angular Momenta"
draft: false
---

A hydrogen electron carries orbital angular momentum $\vec L$ and spin $\vec S$;
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 $\vec J = \vec J_1 + \vec J_2$, because it is $\vec J$,
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](/quantum-mechanics/angular-momentum/the-angular-momentum-algebra)
of the previous lesson supplies every tool: each $\vec J_i$ obeys the standard
commutators, and so does the sum.

## Two bases for the combined system

Let $\vec J_1$ and $\vec J_2$ be two angular momenta acting on distinct degrees of
freedom, so every component of the first commutes with every component of the
second, $[J_{1i}, J_{2j}] = 0$. Each has its own multiplet structure with fixed
$j_1$ and $j_2$. The combined space is the tensor product, of dimension
$(2j_1 + 1)(2j_2 + 1)$, and there are two natural bases for it.

- **Uncoupled basis.** The simultaneous eigenstates of
  $\{J_1^2, J_{1z}, J_2^2, J_{2z}\}$,
  $$
  \ket{j_1, m_1}\ket{j_2, m_2},
  \qquad m_1 = -j_1, \dots, j_1,\;\; m_2 = -j_2, \dots, j_2,
  $$
  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 $\ket{j, m; j_1, j_2}$ (the fixed
  $j_1, j_2$ often suppressed), in which the _total_ has a definite magnitude and
  projection.

Define the total angular momentum $\vec J = \vec J_1 + \vec J_2$. It obeys the
algebra, since

$$
[J_i, J_j] = [J_{1i} + J_{2i},\, J_{1j} + J_{2j}]
  = i\hbar\varepsilon_{ijk}J_{1k} + i\hbar\varepsilon_{ijk}J_{2k}
  = i\hbar\varepsilon_{ijk}J_k,
$$

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
$J_{1z}$ or $J_{2z}$ individually:

$$
[J^2, J_{1z}] \ne 0,
\qquad [J^2, J_1^2] = 0,
\qquad [J_z, J_{1z}] = 0 .
$$

The last two identities are why $\{J^2, J_z, J_1^2, J_2^2\}$ is a valid commuting
set while $\{J^2, J_{1z}, J_{2z}\}$ is not. Choosing between the bases is choosing
which projections to sharpen: the individual ones, or the total.

> **Definition (Uncoupled and coupled bases).** The uncoupled basis diagonalizes
> the individual projections $J_{1z}, J_{2z}$; the coupled basis diagonalizes the
> total $J^2$ and $J_z$. Both span the same $(2j_1+1)(2j_2+1)$-dimensional space,
> and the unitary matrix relating them is the array of Clebsch–Gordan
> coefficients.

## The range of the total quantum number

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

$$
J_z\,\big(\ket{j_1,m_1}\ket{j_2,m_2}\big)
  = \hbar(m_1 + m_2)\,\ket{j_1,m_1}\ket{j_2,m_2},
\qquad m = m_1 + m_2 .
$$

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 $\ket{j_1,j_1}\ket{j_2,j_2}$. 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:

> **Theorem (Clebsch–Gordan series).** Coupling $j_1$ and $j_2$ produces total
> angular momenta
> $$
> j = |j_1 - j_2|,\; |j_1 - j_2| + 1,\; \dots,\; j_1 + j_2,
> $$
> each value occurring exactly once. Symbolically $j_1 \otimes j_2 = \bigoplus_{j}
> j$ over that range.

The dimension count confirms completeness. With $j_1 \ge j_2$ the multiplets run
from $j_1 - j_2$ to $j_1 + j_2$, a total of $2j_2 + 1$ values, and the dimensions
sum by the arithmetic series to

$$
\sum_{j = j_1 - j_2}^{j_1 + j_2} (2j + 1)
  = (2j_2 + 1)\cdot\frac{[2(j_1 - j_2)+1] + [2(j_1 + j_2)+1]}{2}
  = (2j_2 + 1)(2j_1 + 1),
$$

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

$$
% caption: Coupling j1 = 1 with j2 = 1 gives three total multiplets of lengths
% five, three, and one, whose rungs total the nine product states.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% j=2 ladder: 5 rungs
\foreach \y in {0,1,2,3,4}
  \draw[acc, very thick] (-0.6,\y*0.75) -- (0.6,\y*0.75);
\node[anchor=north, font=\scriptsize] at (0,-0.35) {j = 2};
\node[anchor=south, font=\scriptsize, black!70] at (0,3.15) {5 states};
% j=1 ladder: 3 rungs
\foreach \y in {0,1,2}
  \draw[acc, very thick] (2.4,\y*0.75+0.75) -- (3.6,\y*0.75+0.75);
\node[anchor=north, font=\scriptsize] at (3,-0.35) {j = 1};
\node[anchor=south, font=\scriptsize, black!70] at (3,2.4) {3 states};
% j=0 ladder: 1 rung
\draw[acc, very thick] (5.4,1.5) -- (6.6,1.5);
\node[anchor=north, font=\scriptsize] at (6,-0.35) {j = 0};
\node[anchor=south, font=\scriptsize, black!70] at (6,1.65) {1 state};
% plus signs
\node[font=\large] at (1.5,1.5) {$+$};
\node[font=\large] at (4.5,1.5) {$+$};
\node[anchor=west, font=\scriptsize, black!70] at (7.0,1.5) {5 + 3 + 1 = 9};
\end{tikzpicture}
$$

$$
% caption: Uncoupled states on the (m1, m2) lattice for j1 = j2 = 1. Each
% anti-diagonal collects the states of one total projection m = m1 + m2; the nine
% states regroup into total multiplets j = 2, 1, 0.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[->, black] (-2.0,0) -- (2.0,0) node[anchor=west, black] {$m_{1}$};
\draw[->, black] (0,-2.0) -- (0,2.0) node[anchor=south, black] {$m_{2}$};
% anti-diagonal lines of constant m = m1+m2
\foreach \s in {-2,-1,0,1,2} {
  \draw[black] (\s-1.4,1.4) -- (\s+1.4,-1.4);
}
% lattice dots for m1,m2 in {-1,0,1}
\foreach \a in {-1,0,1} {
  \foreach \b in {-1,0,1} {
    \fill[acc] (\a,\b) circle (2.8pt);
  }
}
% tick labels
\foreach \a/\lab in {-1/{-1}, 1/{+1}}
  \node[anchor=north, font=\scriptsize, black!70] at (\a,-0.12) {\lab};
\foreach \b/\lab in {-1/{-1}, 1/{+1}}
  \node[anchor=east, font=\scriptsize, black!70] at (-0.1,\b) {\lab};
% annotate the top corner (m=2, single state)
\node[acc, anchor=south west, font=\scriptsize] at (1.05,1.05) {m = +2};
\end{tikzpicture}
$$

## Constructing the coupled states

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

$$
\ket{j_1 + j_2,\; j_1 + j_2} = \ket{j_1, j_1}\ket{j_2, j_2}.
$$

Apply the total lowering operator $J_- = J_{1-} + J_{2-}$ to descend. The left
side uses the [ladder formula](/quantum-mechanics/angular-momentum/the-angular-momentum-algebra)
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.

$$
% caption: The coupled basis is built column by column: lower within a multiplet
% with the total J-minus operator, then orthogonalize in the shared m-subspace to
% seed the next lower multiplet.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0,
  st/.style={draw, minimum width=20mm, minimum height=8mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
% top state
\node[st, draw=acc, text=acc, thick] (top) at (0,3) {top state\\(highest m)};
% lower down first multiplet
\node[st] (a1) at (0,1.5) {next m, same j};
\node[st] (a2) at (0,0) {lower again};
\draw[->, black, thick] (top) -- (a1) node[midway, right, font=\scriptsize] {lower};
\draw[->, black, thick] (a1) -- (a2) node[midway, right, font=\scriptsize] {lower};
% orthogonal seed of next multiplet
\node[st, draw=acc, text=acc] (b1) at (4.2,1.5) {orthogonal\\combination};
\draw[->, acc, thick, dashed] (a1) -- (b1) node[midway, above, font=\scriptsize] {orthogonalize};
\node[anchor=west, font=\scriptsize, black!70] at (5.6,1.5) {seeds next j};
\node[anchor=west, font=\scriptsize, acc] at (0.9,3) {seeds top j};
\end{tikzpicture}
$$

### Two spin-halves: the singlet and triplet

Take $j_1 = j_2 = \tfrac12$. The series gives $\tfrac12 \otimes \tfrac12 = 1 \oplus 0$:
a spin-$1$ **triplet** and a spin-$0$ **singlet**, four states in all. Write
$\uparrow, \downarrow$ for the $m = \pm\tfrac12$ single-spin states. The triplet
top is $\ket{1,1} = \uparrow\uparrow$. Lowering with $J_- = J_{1-} + J_{2-}$,

$$
J_-\,\ket{1,1} = \hbar\sqrt{2}\,\ket{1,0},
\qquad
J_-\,(\uparrow\uparrow) = \hbar\,\downarrow\uparrow + \hbar\,\uparrow\downarrow,
$$

so equating gives the symmetric combination

$$
\ket{1,0} = \frac{1}{\sqrt 2}\big(\uparrow\downarrow + \downarrow\uparrow\big).
$$

Lowering once more yields $\ket{1,-1} = \downarrow\downarrow$. The singlet is the
remaining $m = 0$ state, orthogonal to $\ket{1,0}$:

$$
\ket{0,0} = \frac{1}{\sqrt 2}\big(\uparrow\downarrow - \downarrow\uparrow\big).
$$

> **Definition (Singlet and triplet).** Two spin-$\tfrac12$ particles combine into
> a symmetric triplet of total spin $1$,
> $$
> \ket{1,1} = \uparrow\uparrow,
> \quad
> \ket{1,0} = \tfrac{1}{\sqrt 2}(\uparrow\downarrow + \downarrow\uparrow),
> \quad
> \ket{1,-1} = \downarrow\downarrow,
> $$
> and an antisymmetric singlet of total spin $0$,
> $\ket{0,0} = \tfrac{1}{\sqrt 2}(\uparrow\downarrow - \downarrow\uparrow)$.

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](/quantum-mechanics/identical-particles/the-pauli-principle-atoms-and-the-periodic-table)
of helium and the [spin structure](/quantum-mechanics/approximation-methods/fine-structure-and-the-real-hydrogen-atom)
of hydrogen's hyperfine levels. The value of $\vec J_1 \cdot \vec J_2$ separates
them cleanly. From
$J^2 = J_1^2 + J_2^2 + 2\,\vec J_1 \cdot \vec J_2$,

$$
\vec J_1 \cdot \vec J_2
  = \frac{1}{2}\big(J^2 - J_1^2 - J_2^2\big)
  = \frac{\hbar^2}{2}\Big[j(j+1) - \tfrac34 - \tfrac34\Big]
  = \begin{cases} +\tfrac14\hbar^2, & j = 1 \;(\text{triplet}), \\[2pt]
  -\tfrac34\hbar^2, & j = 0 \;(\text{singlet}). \end{cases}
$$

$$
% caption: Two spin-halves span a four-dimensional space that splits into a
% symmetric total-spin-1 triplet and an antisymmetric total-spin-0 singlet.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0,
  st/.style={draw, minimum width=22mm, minimum height=8mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
% product space
\node[st, thick] (p) at (0,1) {combined space\\(4 states)};
% triplet block
\node[st, draw=acc, text=acc, thick] (t) at (5,2.1) {triplet (j = 1)};
\node[anchor=west, font=\scriptsize, black!70] at (7.1,2.1) {3 states, symmetric};
% singlet block
\node[st, draw=black, text=black, thick] (s) at (5,-0.1) {singlet (j = 0)};
\node[anchor=west, font=\scriptsize, black!70] at (7.1,-0.1) {1 state, antisymmetric};
\draw[->, acc, thick] (p) -- (t);
\draw[->, black, thick, dashed] (p) -- (s);
\end{tikzpicture}
$$

## Clebsch–Gordan coefficients

Each coupled state expands in the uncoupled basis:

$$
\ket{j, m} = \sum_{m_1 + m_2 = m}
  \braket{j_1, m_1; j_2, m_2 | j, m}\;\ket{j_1, m_1}\ket{j_2, m_2}.
$$

The overlaps $\braket{j_1, m_1; j_2, m_2 | j, m}$ 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 $J_z = J_{1z} + J_{2z}$.
- **Triangle rule.** The coefficient is zero unless
  $|j_1 - j_2| \le j \le j_1 + j_2$, the Clebsch–Gordan series.

With the Condon–Shortley convention (all coefficients real, and
$\braket{j_1, j_1; j_2, j - j_1 | j, j} > 0$) the coefficients are unique. Because
the change of basis is unitary and real, it is orthogonal, giving two orthogonality
relations:

$$
\sum_{m_1, m_2}
  \braket{j_1 m_1; j_2 m_2 | j\,m}\braket{j_1 m_1; j_2 m_2 | j'\,m'}
  = \delta_{jj'}\,\delta_{mm'},
$$
$$
\sum_{j, m}
  \braket{j_1 m_1; j_2 m_2 | j\,m}\braket{j_1 m_1'; j_2 m_2' | j\,m}
  = \delta_{m_1 m_1'}\,\delta_{m_2 m_2'} .
$$

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-$\tfrac12$ block reads:

| $(m_1, m_2)$ | $\ket{1,1}$ | $\ket{1,0}$ | $\ket{0,0}$ | $\ket{1,-1}$ |
| --- | --- | --- | --- | --- |
| $(+\tfrac12, +\tfrac12)$ | $1$ | $0$ | $0$ | $0$ |
| $(+\tfrac12, -\tfrac12)$ | $0$ | $\tfrac{1}{\sqrt 2}$ | $\tfrac{1}{\sqrt 2}$ | $0$ |
| $(-\tfrac12, +\tfrac12)$ | $0$ | $\tfrac{1}{\sqrt 2}$ | $-\tfrac{1}{\sqrt 2}$ | $0$ |
| $(-\tfrac12, -\tfrac12)$ | $0$ | $0$ | $0$ | $1$ |

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.

> **Worked example.** Read $\ket{0,0}$ from the table. The nonzero entries in its
> column sit in rows $(+\tfrac12, -\tfrac12)$ and $(-\tfrac12, +\tfrac12)$, with
> values $\tfrac{1}{\sqrt 2}$ and $-\tfrac{1}{\sqrt 2}$:
> $$
> \ket{0,0} = \frac{1}{\sqrt 2}\,\uparrow\downarrow - \frac{1}{\sqrt 2}\,\downarrow\uparrow,
> $$
> matching the singlet constructed by orthogonalization above. Reading the same
> table across the $(+\tfrac12, -\tfrac12)$ row instead gives the inverse relation
> $\uparrow\downarrow = \tfrac{1}{\sqrt 2}\ket{1,0} + \tfrac{1}{\sqrt 2}\ket{0,0}$:
> a product state is generally a superposition of different total spins.

### Coupling orbital and spin angular momentum

The same construction applies to $\vec J = \vec L + \vec S$ for a spin-$\tfrac12$
particle, producing the $\ket{n, \ell, j, m_j}$ basis in which the
[spin–orbit interaction](/quantum-mechanics/approximation-methods/fine-structure-and-the-real-hydrogen-atom)
is diagonal. With $s = \tfrac12$ the series gives only two totals,
$j = \ell + \tfrac12$ and $j = \ell - \tfrac12$ (for $\ell \ge 1$).

> **Worked example.** Couple $\ell = 1$ and $s = \tfrac12$, so
> $j \in \{\tfrac32, \tfrac12\}$. The stretched state is the bare product
> $\ket{\tfrac32, \tfrac32} = \ket{\ell=1, m_\ell=1}\!\uparrow$. Apply
> $J_- = L_- + S_-$:
> $$
> J_-\ket{\tfrac32,\tfrac32} = \hbar\sqrt{\tfrac32\cdot\tfrac52 - \tfrac32\cdot\tfrac12}\,\ket{\tfrac32,\tfrac12}
>   = \hbar\sqrt{3}\,\ket{\tfrac32,\tfrac12},
> $$
> $$
> (L_- + S_-)\,\ket{1,1}\!\uparrow
>   = \hbar\sqrt{2}\,\ket{1,0}\!\uparrow + \hbar\,\ket{1,1}\!\downarrow .
> $$
> Matching gives the coupled state as a Clebsch–Gordan superposition,
> $$
> \ket{\tfrac32,\tfrac12} = \sqrt{\tfrac23}\,\ket{1,0}\!\uparrow + \sqrt{\tfrac13}\,\ket{1,1}\!\downarrow .
> $$
> The other total-$j$ state at $m_j = \tfrac12$ is the orthogonal combination
> $\ket{\tfrac12,\tfrac12} = \sqrt{\tfrac13}\,\ket{1,0}\!\uparrow - \sqrt{\tfrac23}\,\ket{1,1}\!\downarrow$,
> reproducing the standard $\ell \pm \tfrac12$ coefficients
> $\sqrt{(\ell \pm m_j + \tfrac12)/(2\ell+1)}$ without a table.

The lesson's arithmetic, done once, is reused wherever two rotational degrees of
freedom combine.[^sakurai-add][^gs-add][^ct-add]

[^sakurai-add]: 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
[^gs-add]: 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
[^ct-add]: 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