Clebsch–Gordan Coefficients

The change-of-basis coefficients

The coupled states $|j, m\rangle$ and the uncoupled states $|j_1, m_1; j_2, m_2\rangle$ are two bases for the same space. Each coupled state expands as

|j, m\rangle = \sum_{m_1, m_2} \langle j_1, m_1; j_2, m_2 \,|\, j, m\rangle\; |j_1, m_1; j_2, m_2\rangle .

The overlaps $\langle j_1, m_1; j_2, m_2 | j, m\rangle$ are the Clebsch–Gordan (CG) coefficients. They are the entries of the unitary matrix that rotates between the uncoupled and coupled bases.

Two rules that pin them down

The CG coefficients obey two restrictions that do most of the work:

$z$ -projection conservation. Because $J_z = J_{1z} + J_{2z}$ , a coefficient is nonzero only when $m = m_1 + m_2$ . This is why the sum above collapses to terms with matching projection.
Triangle / range rule. A coefficient is nonzero only for $|j_1 - j_2| \le j \le j_1 + j_2$ .

By the standard Condon–Shortley convention the coefficients are real, and the one with the largest $m_1$ for the stretched state $j = j_1 + j_2$ is chosen positive. With these conventions the CG coefficients are fixed numbers, tabulated once and for all.

Building them with ladder operators

The systematic construction starts from the stretched state of maximal projection,

|j_1 + j_2,\; j_1 + j_2\rangle = |j_1, j_1; j_2, j_2\rangle ,

which is unambiguous because only one uncoupled state has the maximal $m = j_1 + j_2$ . Repeatedly applying the total lowering operator $J_- = J_{1-} + J_{2-}$ , with the rule

J_- |j, m\rangle = \hbar\sqrt{j(j+1) - m(m-1)}\;|j, m-1\rangle ,

generates every state in the $j = j_1 + j_2$ family. The next-lower- $j$ family is then fixed by demanding orthogonality to the states already built, and the procedure repeats down to $j = |j_1 - j_2|$ .

Worked example: two spin-½

Take $j_1 = j_2 = \tfrac12$ . The stretched state is $|1,1\rangle = |\!\uparrow\uparrow\rangle$ . Lowering it,

J_-|1,1\rangle = \hbar\big(|\!\downarrow\uparrow\rangle + |\!\uparrow\downarrow\rangle\big), \qquad J_-|1,1\rangle = \hbar\sqrt{1\cdot2 - 1\cdot0}\;|1,0\rangle = \hbar\sqrt{2}\,|1,0\rangle .

Equating the two expressions,

|1,0\rangle = \tfrac{1}{\sqrt2}\big(|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle\big), \qquad\Rightarrow\qquad \langle\uparrow\downarrow|1,0\rangle = \tfrac{1}{\sqrt2}.

So the CG coefficient connecting $|\!\uparrow\downarrow\rangle$ to $|1,0\rangle$ is $1/\sqrt2 \approx 0.7071$ . The singlet $|0,0\rangle$ is then the orthogonal combination, with coefficient $-1/\sqrt2$ on $|\!\downarrow\uparrow\rangle$ .

Try it

This is a numerical exercise — your code should return a number. Reconstruct $|1,0\rangle$ from the lowering operator and return the Clebsch–Gordan coefficient $\langle\uparrow\downarrow|1,0\rangle$ .

Run your code to see the quantum state.