Question

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{ - {\bf{1}}}\end{aligned}} \right)\),\(B = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}\end{aligned}} \right)\).These are Pauli spin matrices used in the study of electron spin in quantum mechanics. Show that \({A^{\bf{2}}} = I\), \({B^{\bf{2}}} = I\), and \(AB = - BA\). Matrices such that \(AB = - BA\) are said to anticommute.

Short answer

Hence, \({A^2} = I,{B^2} = I,\) and \(AB = - BA\) are proved.

Step-by-step solution

Compute \({A^{\bf{2}}}\)

\(\begin{aligned}{c}{A^2} = AA\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\\{A^2} = I\end{aligned}\)

Compute \({B^{\bf{2}}}\)

\(\begin{aligned}{c}{B^2} = BB\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}1&0\\0&1\end{aligned}} \right)\\{B^2} = I\end{aligned}\)

Check \(AB =  - BA\)

\(\begin{aligned}{c}AB = \left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\{ - 1}&0\end{aligned}} \right)\end{aligned}\)

\(\begin{aligned}{c} - BA = - \left( {\begin{aligned}{*{20}{c}}0&1\\1&0\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}1&0\\0&{ - 1}\end{aligned}} \right)\\ = - \left( {\begin{aligned}{*{20}{c}}0&{ - 1}\\1&0\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}0&1\\{ - 1}&0\end{aligned}} \right)\end{aligned}\)

Thus \( - BA = AB\).