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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 $\{|1\;1\rangle,|1\;0\rangle,|1\;-1\rangle,|0\;0\rangle\}$ ?

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$$\sigma_{tot}^2 = (\vec{\sigma}_1+\vec{\sigma}_2)^2 = \sigma_1^2 + 2\vec{\sigma_1} \cdot \vec{\sigma_2} + \sigma_2^2 $$ $$ \vec{\sigma_1} \cdot \vec{\sigma_2} = \frac{1}{2} (\sigma_{tot}^2 - \sigma_1^2 - \sigma_2^2 ) $$ 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

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