Peskin and Schroeder spinor high-energy limit (5.26 and A.20)

P&S say the high-energy limit of spinor $u^s (p)$ is $\sqrt{2E} {1 \over 2} (1-\widehat{p} . {\sigma}) \xi^s$ and similar for the right-handed spinor (formulae 5.26 and A.20). I can't seem to derive this. How do you get this from $\sqrt{p . \sigma } \xi^s$?

• Is it so obvious that nobody wants to answer this? – Oбжорoв Jul 28 '17 at 11:38

To compute the square root of the matrix, first diagonalise $$p\cdot \sigma$$ as $$p \cdot \sigma = U \begin{pmatrix} \lambda_0 & 0 \\ 0 & \lambda_1 \\ \end{pmatrix} U^{-1}$$ where $$U$$ is the matrix consisting of eigenvectors. We then have: $$\sqrt{p \cdot \sigma} = U \begin{pmatrix} \sqrt{\lambda_0} & 0 \\ 0 & \sqrt{\lambda_1} \\ \end{pmatrix} U^{-1}$$ Using this approach you will find that this gives the result as shown in P&S.
This is trivial to see using the explicit (and the generically more useful) form of the square root, $$\sqrt{p.\sigma} \equiv \frac{E_p+m-{\bf \sigma}.\bf{p}}{\sqrt{2(E_p+m)}}$$ Then we just send $m\rightarrow 0, {\bf p}\rightarrow E_p\hat{p}$ and we immediately get the term you write above, $$\sqrt{p.\sigma} \equiv \frac{E_p-E_p{\bf \sigma}.\hat{p}}{\sqrt{2(E_p)}} = \sqrt{\frac{E_p}{2}}(1-\sigma.\hat{p})$$