In one of the lectures that I'm currently taking we encountered the Dirac equation. The general solution was given as $$\psi ( x ) = \sum _ { s } \int \frac { d ^ { 3 } \bf { p } } { ( 2 \pi ) ^ { 2 } 2 \omega _ { p } } \left[ a _ { s } ( p ) u ^ { s } ( p ) e ^ { - i p \cdot x } + b _ { s } ^ { * } ( p ) v ^ { s } ( p ) e ^ { + i p \cdot x } \right],$$ where $$u^{s}(p)=\begin{pmatrix}{\sqrt{\sigma \cdot p} \xi^{s}} \\ {\sqrt{\overline{\sigma} \cdot p} \xi^{s}}\end{pmatrix} \quad\text{and}\quad v ^ { s } ( p ) = \begin{pmatrix} { \sqrt { \sigma \cdot p } \xi ^ { s } } \\ { - \sqrt { \bar { \sigma } \cdot p } \xi ^ { s } } \end{pmatrix}.$$ Note that we defined $\sigma^\mu \equiv (1,\vec{\sigma})$ and $\bar\sigma^\mu \equiv (1,-\vec\sigma)$ and $s\in\{+,-\}$ for $$\xi^+ \equiv \begin{pmatrix}1\\0\end{pmatrix},~\xi^-\equiv\begin{pmatrix}0\\1\end{pmatrix}.$$
My problem is now that I'm a bit confused on how to evalute the expression $\sqrt{p\cdot\sigma}\xi^s$. If I understood correctly we have $p\cdot \sigma = p_\mu\sigma^\mu$ which makes this expression a matrix. But how am I supposed to take the square-root now? So the questions boils down to explaining how one can evalute the expression $\sqrt{\sigma \cdot p}\xi^s$.
some notes: There was actually no proof given why $u^s(p)$ or $v^s(p)$ should solve the Dirac equation, only a statement that one could prove it using the identity $$(\sigma\cdot p)(\bar\sigma\cdot p)=p^2=m^2.$$ We were using the Wely-representation of the $\gamma$-matrices, if this should be relevant.