Why is $( \alpha_i r_i) (\alpha_j r_j ) = \frac{1}{2} \{ \alpha_i , \alpha_j\}r_i r_j$? Where $\alpha_i= \left(
\begin{matrix}
0 & \sigma_i \\
\sigma_i & 0
\end{matrix}
\right)$.
To me it should just be $( \alpha_i r_i) (\alpha_j r_j ) = \alpha_i \alpha_j r_i r_j$, but it is not. Why the difference?
 A: Without loss of generality,
$$
\alpha_i \alpha_j = \frac{1}{2} \left( [\alpha_i, \alpha_j] + \{\alpha_i, \alpha_j\}\right) 
$$
Note the first term is antisymmetric under interchange of $i$ and $j$, and the second term is symmetric.
Second, note that $r_i r_j$ is symmetric under interchange of $i$ and $j$. Therefore, $[\alpha_i, \alpha_j] r_i r_j = 0$.
This follows since the trace of a product of an antisymmetric matrix and a symmetric matrix is zero. If $A=-A^T$, and $S=S^T$, then ${\rm tr}(AS) = {\rm tr} ((AS)^T) = {\rm tr}(S^T A^T) = {\rm tr}(A^T S^T) = -{\rm tr}(A S)$, and therefore ${\rm tr}(AS)=0$. In this chain of equations I've used ${\rm tr}(AB)={\rm tr}(BA)$, ${\rm tr}(A)={\rm tr}(A^T)$, $(AB)^T=B^TA^T$, and ${\rm tr}(-A) = - {\rm tr}(A)$.
Combining the above facts, we conclude that
$$
\alpha_i \alpha_j r^i r^j = \frac{1}{2} [\alpha_i, \alpha_j] r^i r^j + \frac{1}{2}\left\{\alpha_i, \alpha_j\right\} r^i r^j = \frac{1}{2}\left\{\alpha_i, \alpha_j\right\} r^i r^j
$$
A: Since $r_ir_j = r_jr_i$ we have
$$
(\alpha_i r_i) (\alpha_jr_j) = \alpha_i\alpha_j r_ir_j\\
= \frac 12 \alpha_i\alpha_j  r_ir_j + \frac 12 \alpha_i\alpha_j  r_jr_i\\
= \frac 12 \alpha_i\alpha_j r_ir_j +\frac 12 \alpha_j\alpha_i r_ir_j\quad \text{relabeling $i\leftrightarrow j$ in second term} \\
= \frac 12 \{\alpha_i\alpha_j\}  r_ir_j
$$
