7.3: Total Angular Momentum

In general, a particle may have both spin and orbital angular momentum. Since L and S have the same dimensions, we can ask what is the total angular momentum J of the particle. We write this as

\[\mathbf{J}=\mathbf{L}+\mathbf{S} \equiv \mathbf{L} \otimes \mathbb{I}+\mathbb{I} \otimes \mathbf{S},\tag{7.46}\]

which emphasizes that orbital and spin angular momentum are described in distinct Hilbert spaces.

Since \(\left[L_{i}, S_{j}\right]=0\), we have

\[\begin{aligned}
\left[J_{i}, J_{j}\right] &=\left[L_{i}+S_{i}, L_{j}+S_{j}\right]=\left[L_{i}, L_{j}\right]+\left[S_{i}, S_{j}\right] \\
&=i \hbar \epsilon_{i j k} L_{k}+i \hbar \epsilon_{i j k} S_{k}=i \hbar \epsilon_{i j k}\left(L_{k}+S_{k}\right) \\
&=i \hbar \epsilon_{i j k} J_{k}
\end{aligned}\tag{7.47}\]

In other words, J obeys the same algebra as L and S, and we can immediately carry over the structure of the eigenvalues and eigenvectors from L and S.

In addition, L and S must be added as vectors. However, only one of the components of the total angular momentum can be sharp (i.e., having a definite value). Recall that \(l\) and \(s\) are magnitudes of the orbital and spin angular momentum, respectively. We can determine the extremal values of J, denoted by \(\pm j\), by adding and subtracting the spin from the orbital angular momentum, as shown in Figure 3:

\[|l-s| \leq j \leq l+s.\tag{7.48}\]

For example, when \(l=1\) and \(s=\frac{1}{2}\), the possible values of \(j\) are \(j=\frac{1}{2}\) and \(j=\frac{3}{2}\).

The commuting operators for J are, first of all, \(\mathbf{J}^{2}\) and \(J_{z}\) as we expect from the algebra, but also the operators \(\mathbf{L}^{2}\) and \(\mathbf{S}^{2}\). You may think that \(S_{z}\) and \(L_{z}\) also commute with these operators, but that it not the case:

\[\left[\mathbf{J}^{2}, L_{z}\right]=\left[(\mathbf{L}+\mathbf{S})^{2}, L_{z}\right]=\left[\mathbf{L}^{2}+2 \mathbf{L} \cdot \mathbf{S}+\mathbf{S}^{2}, L_{z}\right]=2\left[\mathbf{L}, L_{z}\right] \cdot \mathbf{S} \neq 0\tag{7.49}\]

We can construct a full basis for total angular momentum in terms of \(\mathbf{J}^{2}\) and \(J_{z}\), as before:

\[\mathbf{J}^{2}\left|j, m_{j}\right\rangle=\hbar^{2} j(j+1)\left|j, m_{j}\right\rangle \text { and } J_{z}\left|j, m_{j}\right\rangle=m_{j} \hbar\left|j, m_{j}\right\rangle.\tag{7.50}\]

Alternatively, we can construct spin and orbital angular momentum eigenstates directly as a tensor product of the eigenstates

\[\mathbf{L}^{2}|l, m\rangle\left|s, m_{s}\right\rangle=\hbar^{2} l(l+1)|l, m\rangle\left|s, m_{s}\right\rangle \quad \text { and } \quad L_{z}|l, m\rangle\left|s, m_{s}\right\rangle=m \hbar|l, m\rangle\left|s, m_{s}\right\rangle,\tag{7.51}\]

and

\[\mathbf{S}^{2}|l, m\rangle\left|s, m_{s}\right\rangle=\hbar^{2} s(s+1)|l, m\rangle\left|s, m_{s}\right\rangle \quad \text { and } \quad S_{z}|l, m\rangle\left|s, m_{s}\right\rangle=m_{s} \hbar|l, m\rangle\left|s, m_{s}\right\rangle.\tag{7.52}\]

Since the \(L_{z}\) and \(S_{z}\) do not commute with \(\mathbf{J}^{2}\), the states \(\left|j, m_{j}\right\rangle\) are not the same as the states \(|l, m\rangle\left|s, m_{s}\right\rangle\).