The three spin Pauli matrices are: $ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $
According to problem 2.2.3 (Mathematical methods for physicists, Arfken Weber & Harris, Seventh Edition) Complex numbers, a + ib, with a and b real, may be represented by (or are isomorphic with) 2 × 2 matrices as follows:
$ a + ib = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} $
This implies that: $ i = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, -i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $
And therefore according to the above proposition. $ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} = i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = i(-i) = 1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, $
By exchanging the rows in $\sigma_1$, we can also write it as the identity matrix. Therefore according to this we get $\sigma_1 = \sigma_2$ which to me is quite surprising. They have the property that $\sigma_1^2 = \sigma_2^2 = I$. One can also prove that $\sigma_1 \sigma_2 = i \sigma_3$. Since $\sigma_1 = \sigma_2 = I$, $\sigma_1 \sigma_2 = I$. The RHS becomes:
$ i\sigma_3 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $
The problem arises when I try to prove the identity $\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}I_2 $ in Problem 2.2.11(c). This clearly does not hold since $\sigma_2 \sigma_3 + \sigma_3 \sigma_2$ $= I \sigma_3 + \sigma_3 I \neq 0$
Where am I making a mistake here ?
