Understanding solutions of the Dirac equation 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.
 A: The more usual expressions are 
$$
u_\alpha = \frac{1}{\sqrt{2E(E+m)}} \left[\matrix{ (E+m)\chi_\alpha \cr  ({ \sigma}\cdot {\bf k} )\chi_\alpha}\right],
$$
$$
v_\alpha =  \frac{1}{\sqrt{2|E|(|E|+m)}} \left[\matrix{ - ({ \sigma}\cdot {\bf k} ) \chi_\alpha \cr  (|E|+m)\chi_\alpha}\right].
$$
Alternatively one can use the rapidity ${\bf k}= {\bf n}\cosh s $ and write
$$
u_\alpha= \frac{1}{\sqrt{2m(E+m)}}\left[\matrix{(E+m)\chi_\alpha \cr  ({ \sigma}\cdot {\bf k} )\chi_\alpha}\right]
=\left[\matrix{\phantom{({\bm \sigma}\cdot {\bf n} )} \cosh (s/2)\chi_\alpha \cr \sinh(s/2) ({ \sigma}\cdot {\bf n} )\chi_\alpha}\right].
$$
 The second way of writing $u_\alpha$ is designed to stress that the eigenstate   depends only on the geometry of the Lorentz boost, and not  on the rest mass $m$.
I had a look at P&S ch 3 and he explains what he means in eq 3.49, the second line of which coincides with my "alternately" expression. The proof is also given there as he starts from an obvious solution and boosts it. My comment about "not being positive" was mistaken because I thought that
$\sigma\cdot {\bf p}$
meant the operator with eigenvalue $±|{\bf p}|$
rather than the operator which is written the same, but has positive eigenvalues $E\pm |{\bf p}|$.
 In the second $\sigma$ is not the threee vector but
is meant to mean $(1,{\bf \sigma})$
You did say this in your question, but I missed it. Sorry for misunderstanding.
