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 $|l, m\rangle\left|s, m_{s}\right\rangle$.