Understanding Example 36.7 in the Blundell's Quantum field theory I am reading Blundell's Quantum field thoery for the gifted amateur, p.332, Example 36.7 and stuck at understanding some calculation.
In the example, he expresses
$$ \Sigma_{s=1}^{2}u^{s}(p)\bar{u}^{s}(p)=\Sigma_s \begin{pmatrix} \sqrt{p\cdot \sigma}\xi^{s} \\ \sqrt{p\cdot \bar{\sigma}\xi^{s}} \end{pmatrix} \begin{pmatrix} \xi^{s\dagger}\sqrt{p \cdot\bar{\sigma}} & \xi^{s\dagger}\sqrt{p\cdot \sigma} \\  \end{pmatrix}, $$
where $u^{s}(p)$ is dirac spinor(c.f. his book (36.43) , and $\sigma := (I, \boldsymbol{\sigma}), \bar{\sigma}:= (I, -\boldsymbol{\sigma})$, where $\boldsymbol{\sigma}:= (\sigma^{1}, \sigma^{2}, \sigma^{3})$ is the pauli matrices. And maybe $\xi^{1}:= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\xi^{2}:=\begin{pmatrix} 0  \\ 1  \end{pmatrix}$ (his book p.331) (It seems that it is too long to write down all definitions of the notations in the above equality own by own. So please understanding some omittion-please refer to his book).
Q.1.) My first question is, in the above equation, why $\bar{u}^{s}(p)= \begin{pmatrix} \xi^{s\dagger}\sqrt{p \cdot\bar{\sigma}} & \xi^{s\dagger}\sqrt{p\cdot \sigma} \\  \end{pmatrix}$?
What is the definition of $\bar{u}^{s}(p)$ ?
Q.2) My second question is, in the derivation of (36.51) in the example, he calculate
$ \sqrt{p\cdot \sigma} \sqrt{p \cdot \bar{\sigma}} = m$. Why is this true?
Can anyone help?
 A: *

*The adjoint spinor is defined as $\bar{u}^s(p) = u^{s\dagger}(p)\gamma^0$, with $\bar{u}^s(p)u^t(p) = 2m\delta^{st}$.


*The spinor matrix $p\cdot\sigma = p_\mu \sigma^\mu$ 'squares' to give the mass, since
$$
p\cdot\sigma p\cdot\bar{\sigma} = p^\mu p^\nu \sigma_\mu\bar{\sigma}_\nu = \frac12 p^\mu p^\nu\{\sigma_\mu,\bar{\sigma}_\nu\} = p^2,
$$
where we used the clifford algebra $\{\sigma_\mu,\bar{\sigma}_\nu\} = 2\eta_{\mu\nu}$.
To see why this is the case, we note that $p\cdot\sigma$ is in fact a bispinor, where $\sigma$ is known as an Infeld–Van der Waerden symbol, given by
$$
\sigma^\mu_{\alpha\dot{\beta}} = \{1,\sigma^i\}= \left\{\begin{pmatrix}
 1 & 0 \\
 0 & 1
\end{pmatrix},\begin{pmatrix}
 0 & 1 \\
 1 & 0
\end{pmatrix},\begin{pmatrix}
 0 & -i \\
 i & 0
\end{pmatrix},\begin{pmatrix}
 1 & 0 \\
 0 & -1
\end{pmatrix}\right\}
$$
and
$$
\bar{\sigma}^{\mu~\alpha\dot{\beta}} = \{1,-\sigma^i\}
$$
The bispinor is then given by
$$
p_\mu\sigma^\mu_{\alpha\dot{\beta}} = \begin{pmatrix}
 p_0+p_3 & p_1-ip_2 \\
 p_1+ip_2 & p_0-p_3
\end{pmatrix}.
$$
You will find that $\frac{1}{2}p_{\alpha\dot{\beta}}p^{\alpha\dot{\beta}} = p_0^2 - p_1^2 - p_2^2 - p_3^2$.
The notation $\sqrt{p\cdot\sigma}$ really means that you need to take the square root of this matrix, which is not nice notation IMO.One example is to take the particle to have momentum $p_\mu = (E,0,0,|p|)$, such that
$$
\sqrt{p\cdot\sigma}\xi =(0,\sqrt{E-|p|}). 
$$
A better solution is to use spinor helicity variables, see e.g. Tales of 1001 gluons section 3.3.2.
