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.

• You can take the square root of a positive matrix pretty much in the same way that you take the square root of a positive number, via the spectral theorem. – lcv Mar 29 at 17:22

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.
• @Sito: I am having a hard time figuring out what your lecturer meant. My "usual" exressions have a direct physical meaning. My $\chi_\alpha$ is the particle spin in its rest fram, and the rest of the spinor solution is the result of it's boost from rest to momentm ${\bf k}$. Your lecturer seems to be abandoning physics for a kind of formal "simplicity" which is not actually simple. Notice that icv's comment does not apply as as $\sigma\cdot {\bf k}$ does not have positive eigenvalues, so it does not have a square root. – mike stone Mar 29 at 22:24
• @Sito. I don't have a copy of Peskin to hand. I'll look up what he writes when I get into my office, but again $\sigma\dot {\bf p}$ does not have all positive eigenvalues, so it does not have "positive roots" – mike stone Mar 30 at 19:20