# How to treat spin orbit operator directly

## Main Question or Discussion Point

For an electron the spin operator $S_z$is represented by a $2×2$ matrix, with spin up and down as its bases. Consider the angular momentum operator $L_z$ with $l=1$ which is a $3×3$ matrix. How can we treat the $L_z S_z$ operator directly in matrix form?

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If with $L_z S_z$ you mean the total angular momentum $J_z$ than you have that $J=L+S$, so $j=l+s=\frac{3}{2}$. The eigenvalues of $J_z$ are $m_j=-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$ ($-j \le m_j \le j$). By choosing a basis where $J^2$ and $J_z$ are diagonal, your matrix operator of $J_z$ is:

$J_z = \hbar \begin{pmatrix} \frac{3}{2} & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & -\frac{3}{2} \end{pmatrix}$

P.S.: i'm studying this right now, so maybe i could probably make some mistakes

vanhees71
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In non-relativistic physics (I guess that's all about non-relativistic quantum theory) spin and orbital angular momentum commute, and total angular momentum is given by
$$\vec{J}=\vec{L}+\vec{S}.$$
Then according to representation theory of the rotation group, if the orbital angular angular momentum has definite $l$ and spin $s$, then the total angular momentum has $j \in \{|l-s|,|l-s|+1,\ldots,l+s \}$, i.e., in your case you have $j \in \{1/2,3/2 \}$.

Indeed the total-angular momentum space is $(2l+1)(2s+1)$, i.e., in your case $3 \cdot 2=6$ dimensional. The possible values for $j$ imply the same dimension, i.e., $2+4=6$.

For the eigenvalues of $j_z=l_z+s_z$. You should look this up in a good textbook. Look for Clebsch-Gordan coefficients.

Thanks, but my question is how to multiply the two matrixes directly without solving the problem this way.

Khashishi
The usual way to get $L S$ is to use the identities
$J^2 = (L + S)^2 = L^2 + S^2 + 2LS$
$S$ is a constant, so the result depends on $L^2$ and $J^2$. With $l=1$ fixed, you can write out the basis states of $j$ in terms of the basis states of $s$.