# Product of Pauli spin matrices is diagonal in $|s\;m_s\rangle$ basis

Given the Pauli spin matrices $\;\sigma_1,\;\sigma_2\;$ for two identical spin $1/2$ fermions, how do I show that their product is diagonal in the $|s\;m_s\rangle$ basis $\{|1\;1\rangle,|1\;0\rangle,|1\;-1\rangle,|0\;0\rangle\}$ ?

$$\sigma_{tot}^2 = (\vec{\sigma}_1+\vec{\sigma}_2)^2 = \sigma_1^2 + 2\vec{\sigma_1} \cdot \vec{\sigma_2} + \sigma_2^2$$ $$\vec{\sigma_1} \cdot \vec{\sigma_2} = \frac{1}{2} (\sigma_{tot}^2 - \sigma_1^2 - \sigma_2^2 )$$ But as you said, both particles have spin $\frac{1}{2}$ so for any case, $$\sigma_1^2 = \sigma_2^2 = \hbar^2 \cdot \frac{1}{2} \cdot (\frac{1}{2}+1) = \frac{3}{4} \hbar^2$$ And we get that $$\vec{\sigma_1} \cdot \vec{\sigma_2} = \frac{1}{2} (\sigma_{tot}^2 - \frac{3}{4}\hbar^2- \frac{3}{4}\hbar^2) = \frac{1}{2} (\sigma_{tot}^2 - \frac{3}{2}\hbar^2)$$ Meaning $\sigma_1 \cdot \sigma_2$ is the same like $\sigma_{tot}$ (with constants) , so they have the same diagonal basis