The Pauli spin matrices in quantum mechanics are

$\mathbf{A}\mathbf{=}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\mathbf{,}\mathbf{}\mathbf{B}\mathbf{=}\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\mathbf{,}\mathbf{}\mathbf{C}\mathbf{=}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$

(You will probably find these called${\mathbf{\sigma }}_{\mathbf{x}\mathbf{,}}{\mathbf{\sigma }}_{\mathbf{y}\mathbf{,}}{\mathbf{\sigma }}_{\mathbf{z}}$in your quantum mechanics texts.) Show that${\mathbf{A}}^{\mathbf{2}}\mathbf{=}{\mathbf{B}}^{\mathbf{2}}\mathbf{=}{\mathbf{C}}^{\mathbf{2}}\mathbf{=}\mathbf{a}$unit matrix. Also show that any two of these matrices anticommute, that is, AB=-BA, etc. Show that the commutator of Aand B, that is AB=-BAis 2i C, and similarly for other pairs in cyclic order.

The unit matrix of dimension in linear algebra is the nxn square matrix with ones on the main diagonal and zeros everywhere else. When determining the inverse of a matrix, the unit matrix is utilized in proofs.

Two matrices A and B are anticommuting when the product AB equals the negation of the product BA.

The given matrices are

$\mathrm{A}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),\mathrm{B}=\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right),\mathrm{C}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$

The product ${\mathrm{A}}^2,{\mathrm{B}}^2$, and ${\mathrm{C}}^2$need to be a unit matrix and it needs to be proved that the product of any two of the given matrices anticommute.

Find the product ${\mathrm{A}}^2$.

$$\begin{gathered} {\mathrm{A}}^2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}\left(0\times 0\right)+\left(1\times 1\right)& \left(0\times 1\right)\left(1\times 0\right)\\ \left(1\times 0\right)+\left(0\times 1\right)& \left(1\times 1\right)+\left(0\times 0\right)\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

Find the product ${\mathrm{B}}^2$.

$$\begin{gathered} {\mathrm{B}}^2=\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{C}}^2=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ \mathrm{It}\mathrm{is}\mathrm{proved}\mathrm{that}{\mathrm{A}}^2={\mathrm{B}}^2={\mathrm{C}}^2=\mathrm{a}\mathrm{unit}\mathrm{matrix}. \\ \mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{AB}. \\ \mathrm{AB}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right) \\ =\left(\begin{array}{cc}\mathrm{i}& 0\\ 0& -\mathrm{i}\end{array}\right) \\ =\mathrm{i}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\mathrm{iC} \end{gathered}$$

Find the product -BA .

$$\begin{gathered} -\mathrm{BA}=-\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =-\left(\begin{array}{cc}-\mathrm{i}& 0\\ 0& \mathrm{i}\end{array}\right) \\ =-\left(-\mathrm{i}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\mathrm{iC} \end{gathered}$$

It is proved that AB=-BA.

The commutator of A and B is AB-BA .

$$\begin{gathered} \mathrm{AB}-\mathrm{BA}=\mathrm{iC}+\mathrm{iC} \\ =2\mathrm{iC} \end{gathered}$$

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{AC}. \\ \mathrm{AC}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right) \\ =-\mathrm{iB} \\ \mathrm{Find}\mathrm{the}\mathrm{product}-\mathrm{CA}. \\ -\mathrm{CA}=-\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =-\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right) \\ =-\mathrm{iB} \\ \mathrm{It}\mathrm{is}\mathrm{proved}\mathrm{that}\mathrm{AC}=-\mathrm{CA}. \\ \mathrm{The}\mathrm{commutator}\mathrm{of}\mathrm{A}\mathrm{and}\mathrm{C}\mathrm{is}\mathrm{AC}-\mathrm{CA}. \\ \mathrm{AC}-\mathrm{CA}=-\mathrm{iB}-\mathrm{iB} \\ =-2\mathrm{iB} \end{gathered}$$

$$\begin{gathered} \mathrm{Find}\mathrm{the}\mathrm{product}\mathrm{BC}. \\ \mathrm{BC}=\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ =\left(\begin{array}{cc}0& \mathrm{i}\\ \mathrm{i}& 0\end{array}\right) \\ =\mathrm{iA} \\ \mathrm{Find}\mathrm{the}\mathrm{product}-\mathrm{CB}. \\ -\mathrm{CB}=-\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}0& -\mathrm{i}\\ \mathrm{i}& 0\end{array}\right) \\ =-\left(\begin{array}{cc}0& -\mathrm{i}\\ -\mathrm{i}& 0\end{array}\right) \\ =-\mathrm{iA} \\ \mathrm{It}\mathrm{is}\mathrm{proved}\mathrm{that}\mathrm{BC}=-\mathrm{CB}. \\ \mathrm{The}\mathrm{commutator}\mathrm{of}\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{is}\mathrm{BC}-\mathrm{CB}. \\ \mathrm{BC}-\mathrm{CB}=\mathrm{iA}+\mathrm{iA} \\ =2\mathrm{iA} \\ \mathrm{Hence},\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{any}\mathrm{two}\mathrm{of}\mathrm{these}\mathrm{matrices}\mathrm{anticommute}\mathrm{and}{\mathrm{A}}^2={\mathrm{B}}^2={\mathrm{C}}^2=\mathrm{a}\mathrm{unit}\mathrm{matrix}. \end{gathered}$$