# Pauli matrices and the complex number matrix representation

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 ?

• Real 2x2 matrices. – NickD Oct 16 '17 at 20:46
• Yes where a and b are real – 256ABC Oct 16 '17 at 20:47
• I would look at the minimal polynomial $f_j(x)= x^2+b_j x+c_j$ of each matrix to obtain $\mathbb{R}[\sigma_j] \cong \mathbb{R}[x]/(f_j(x))$. The ring $\mathbb{R}[\sigma_1,\sigma_2,\sigma_3]$ is a non-commutative subring of $M_2(\mathbb{C})$ (the ring of $2\times 2$ complex matrices) – reuns Oct 16 '17 at 20:51
• Exchanging rows in $\sigma_1$ gives a completely different matrix. – Aretino Oct 16 '17 at 21:15
• The point is that the Pauli matrices are NOT of this form. – NickD Oct 16 '17 at 21:17

The mapping $a +ib \Rightarrow \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ where $a,b \in \mathbb{R}$ does not map any complex number to a Pauli matrix. The Pauli matrices are not of the right form. So saying that $1$ maps to $\sigma_2$ (or even worse, that it is equal to it) is meaningless.
• But you can bring $\sigma_1$ to this form by just interchanging the rows right ? – 256ABC Oct 16 '17 at 21:44