Question

In Chapter, Problem , you are asked to prove some identities among the Pauli spin matrices (called A, B, C, in that problem). Call the Pauli spin matrices ${\mathit{\sigma }}_{\mathbf{1}}\mathbf{,}{\mathit{\sigma }}_{\mathbf{2}}\mathbf{,}{\mathit{\sigma }}_{\mathbf{3}}$; then show that the identities can be written in the following summation forms:

$$\begin{gathered} {\mathit{\sigma }}_{\mathbf{k}}{\mathit{\sigma }}_{\mathbf{m}}\mathbf{=}\mathit{i}{\mathit{\epsilon }}_{\mathbf{k}\mathbf{m}\mathbf{n}}{\mathit{\sigma }}_{\mathbf{n}}\mathbf{+}{\mathit{\delta }}_{\mathbf{k}\mathbf{m}} \\ {\mathit{\sigma }}_{\mathbf{k}}{\mathit{\sigma }}_{\mathbf{m}}{\mathit{\epsilon }}_{\mathbf{k}\mathbf{m}\mathbf{n}}\mathbf{=}\mathbf{2}\mathit{i}{\mathit{\sigma }}_{\mathbf{n}} \end{gathered}$$

Short answer

The statement has been proven.

Step-by-step solution

Given Information

The Pauli spin matrices are given below.

Definition of a cartesian tensor

The first rank tensor is just a vector. A tensor of the second rank has nine components (in three dimensions) in every rectangular coordinate system.

Find the value

The Pauli spin matrices are given below.

$$\begin{gathered} {\sigma }_1=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ {\sigma }_2=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right] \\ {\sigma }_3=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \end{gathered}$$

Identities are mentioned below.

$\begin{array}{rcl}\left\{{\sigma }_i,{\sigma }_j\right\}& =& 2{\delta }_{ij}\\ \left[{\sigma }_i,{\sigma }_j\right]& =& 2i{\epsilon }_{ijk}{\sigma }_k\end{array}$

Add the identities, equation becomes as follows.

$\begin{array}{rcl}\left\{{\sigma }_i,{\sigma }_i\right\}+\left[{\sigma }_i,{\sigma }_i\right]& =& 2i{\epsilon }_{ijk}+2{\delta }_{ij}\\ 2{\sigma }_i{\sigma }_j& =& 2i{\epsilon }_{ijk}+2{\delta }_{ij}\\ {\sigma }_i{\sigma }_j& =& i{\epsilon }_{ijk}+{\delta }_{ij}\end{array}$

Multiply by $\begin{array}{rcl}& & {\epsilon }_{ijk}\end{array}$ the equation becomes as follows.

$\begin{array}{rcl}{\sigma }_i{\sigma }_j{\epsilon }_{ijl}& =& i{\epsilon }_{ijk}{\epsilon }_{ijl}{\sigma }_k+{\delta }_{ij}{\epsilon }_{ijl}\\ {\epsilon }_{ijk}{\epsilon }_{ijl}& =& 2{\delta }_{kl}\\ {\delta }_{ij}{\epsilon }_{ijl}& =& 0\\ {\sigma }_i{\sigma }_j{\epsilon }_{ijl}& =& 2i{\sigma }_i\\ & & \end{array}$

Hence, the statement has been proven.