Angular momentum matrices (Schiff section 27)

On page 203 3rd edition of Schiff we are given the angular momentum matrices ${J}$ for $j=1$.

I am curious as to how these relate to orbital angular momentum for $j = 1$. If we take the corresponding 3x3 matrices for spin given on page 198. Lets just use

$$S_x = i \hbar \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix}\,\,\,\mathrm{and}\,\,\,S_y = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{pmatrix}$$

With

$$J_x = \frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}\,\,\,\mathrm{and}\,\,\,J_y = \frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & -i & 1 \\ i & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix}$$

Using $\vec{J} = \vec{L} + \vec{S}$ we can solve for $L_x$ and $L_y$ by subtracting $J$ from $S$.

$$L_x = J_x-S_x = \begin{pmatrix} 0 & \frac{\hbar}{\sqrt{2}} & 0 \\ \frac{\hbar}{\sqrt{2}} & 0 & i\hbar + \frac{\hbar}{\sqrt{2}} \\ 0 & -i\hbar + \frac{\hbar}{\sqrt{2}} & 0 \\ \end{pmatrix}$$ $$L_y = J_y-S_y = \begin{pmatrix} 0 & -\frac{i\hbar}{\sqrt{2}} & -i \hbar \\ \frac{i \hbar}{\sqrt{2}} & 0 & -\frac{i\hbar}{2} \\ i \hbar & \frac{i \hbar}{\sqrt{2}} & 0 \\ \end{pmatrix}$$

$$L_z = J_z - S_z = \begin{pmatrix} \hbar & i \hbar& 0\\ -i \hbar & 0 & 0 \\ 0 & 0 & -\hbar \\ \end{pmatrix}$$ My question is that when I do this, they don't obey the standard commutation relations $[L_i,L_j] = i \hbar \epsilon_{ijk} L_k$. More curious is the 2x2 case for $j=1/2$ on page 203 again, you get null orbital angular momentum matrices. What went wrong here and what am I missing?

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Hi John M. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. –  Qmechanic Oct 25 '13 at 0:19

Your mistake is taking $\vec{J}=\vec{L}+\vec{S}$.

A more complete statement is $\vec{J} = \vec{L}\otimes\mathbb{1}_S + \mathbb{1}_L\otimes\vec{S}$. The operators $\vec{L}$ and $\vec{S}$ act on independent components of the Hilbert space of the particle. The complete Hilbert space is given as the direct product of space and spin components: $\mathcal{H}\simeq\mathcal{H}_L\otimes\mathcal{H}_S$.

In short, the computation that you describe has no meaning.

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