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Let $\sigma_x$, $\sigma_y$, $\sigma_z$ be the standard Pauli matrices.

Prove, if $\alpha \cdot \sigma = \alpha_x \sigma_x + \alpha_y \sigma_y + \alpha_z \sigma_z$, that $\alpha \cdot (\sigma \beta) \cdot \sigma = \alpha \cdot \beta + i \alpha \times \beta \cdot \sigma$.

What does $\times$ represent in this question, cross product or multiplication ?. And also what about $\alpha \cdot \sigma\beta$ , is it $\alpha_x \sigma_x \beta_x + \alpha_y \sigma_y \beta_y+ \alpha_z \sigma_z \beta_z$. If yes, how can we do the second dot operation $\cdot \sigma$ ?

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The formula is wrong as written in the question (in fact it does not make a lot of sense). The correct expression is $(\vec{\alpha} \cdot \vec{\sigma}) (\vec{\beta} \cdot \vec{\sigma}) = (\vec{\alpha}\cdot \vec{\beta}) I + i (\vec{\alpha} \times\vec{\beta})\cdot \vec{\sigma};$ $\times$ indicates the cross product and $I$ the $2\times2$ identity matrix.

The formula follows from the relation $\sigma_k \sigma_l = \delta_{kl} I + i \sum_m \epsilon_{klm} \sigma_m$ which the Pauli matrices fulfill ($\epsilon_{klm}$ is the total antisymmetric tensor).

You then have $ \begin{align}(\vec{\alpha} \cdot \vec{\sigma}) (\vec{\beta} \cdot \vec{\sigma}) &= \sum_{kl} \alpha_k \sigma_k \beta_l \sigma_l =\sum_{kl} \alpha_k \beta_l \left(\delta_{kl} I + i \sum_m \epsilon_{klm} \sigma_m\right) = (\sum_k \alpha_k \beta_k) I + i \sum_{klm}\epsilon_{klm} \alpha_k \beta_l \sigma_m \\&=(\vec{\alpha}\cdot \vec{\beta}) I + i (\vec{\alpha} \times\vec{\beta})\cdot \vec{\sigma}. \end{align}$