## Anticommutativity of Pauli Spin Matrices

The Paul spin matrices operate on vectors representing the state of an electron so that the electron spin can be found.
The matrices are
$\sigma_x = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \: \sigma_y = \left( \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right), \: \sigma_z =\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$

where
$;^2 =-1$
.
To prove anticommutativity we must show
$AB = -BA$
for each pair of matrices.
$\sigma_x \sigma_y =\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) = \left( \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right)$

$\sigma_y \sigma_ =\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{cc} -i & 0 \\ 0 & i \end{array} \right)=- \sigma_y \sigma_x$

$\sigma_x \sigma_z -=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

$\sigma_z \sigma_x -=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)= - \sigma_x \sigma_z$

$\sigma_y \sigma_z -=\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) = \left( \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right)$

$\sigma_z \sigma_y -=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right) = \left( \begin{array}{cc} 0 & -i \\ -i & 0 \end{array} \right)= - \sigma_y \sigma_z$

Hence the set of Pauli Spin matrices are anticommutative.