Angular momentum addition

Consider a system of two particles with angular momentum

\begin{align*} \ket \psi = \ket{j_1,m_1} \otimes \ket{j_2,m_2} = \ket{j_1,j_2,m_1,m_2}. \end{align*}

In the uncoupled basis, we describe this state by the 4 quantum numbers corresponding to the observables $\{ (J^{(1)})^2, (J^{(2)})^2, J_z^{(1)}, J_z^{(2)} \}$ . We’ve seen how we can construct the Bell basis for spin-1/2 particles in particular.

We will define a coupled basis that more naturally describes the system as a whole, for a general system of angular momentum particles. Define the total angular momentum $J = J^{(1)} + J^{(2)}$ .

$J$ is an angular momentum. The terms in each component $J_i = J_i^{(1)} \otimes 1 + 1 \otimes J_i^{(2)}$ commute, and if we consider the commutator the cross terms,

\begin{align*} [J_i,J_j] &= [J_i^{(1)},J_j^{(2)}] + [J_i^{(2)},J_j^{(2)}] \\ &= i\hbar \epsilon_{ijk} (J_k^{(1)} + J_k^{(2)}) \\ &= i\hbar \epsilon_{ijk} J_k. \end{align*}

We form a complete set of commuting observables $\{J^2, J_z, (J^{(1)})^2, (J^{(2)})^2 \}$ . Consider the eigenstates by first writing down all states in the original basis.

For any total $j \in \big[|j_1-j_2|,j_1+j_2\big]$ , we can form a state of total angular momentum $j$ by a superposition of the states in the uncoupled basis. In a product state, if one particle has angular momentum $+j_1$ and the other $-j_2$ , then intuitively the total angular momentum should be $j=j_1-j_2$ . We can’t achieve a total angular momentum smaller than $|j_1-j_2|$ .

We can factorize the angular momentum system

\begin{align*} j_1 \otimes j_2 = (j_1 + j_2) \oplus (j_1 + j_2 - 1) \oplus \cdots \oplus |j_1 - j_2|. \end{align*}

Where each factorized (reduced) term beahves as a single particle spin- $j$ system. With this in mind we can rearrange the states in this basis

Each row corresponds to a particular $m$ eigenvalue of $J_z$ . Each column corresponds to a $J^2$ eigenvalue $j$ . We can define the total lowering operator $J_- = J_-^{(1)} + J_-^{(2)}$ . It is easy to show that this operator lowers the total $m$ quantum number by one.

Within a row, all states correspond to a superposition of the uncoupled states spanning the $m_1 + m_2 = m$ subspace. It’s easy to write a set of states that span this space. Then, finding the superposition corresponding to a particular $j$ can be done analytically.

For $j=m$ , the total raising operator $J_+ \ket{j,m} = 0$ . Write the superposition in terms of unknown coefficients $\alpha,\beta,\ldots$ , and apply the raising operator, then set to zero. Solve for the coefficients and normalize.

These eigenstates are orthogonal, so if we have some $\ket{j,m}$ state, a state $\ket{j',m}$ must be orthogonal. Both states can be written in terms of the spanning $m$ states. Manually computing an orthogonal vector is another way to “move” across the row.