# 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

## 2 Answers

A little late to the party, but hopefully this will also be of use to someone else.

In P&S a short remark can also be found on the calculation of the square root of the matrix on page 46, around equation 3.50, which I assume hints at starting with diagonalising the matrix using the diagonal matrix of eigenvalues and the matrix consisting of eigenvectors, and its inverse.

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, $$\sqrt{p.\sigma} \equiv \frac{E_p-E_p{\bf \sigma}.\hat{p}}{\sqrt{2(E_p)}} = \sqrt{\frac{E_p}{2}}(1-\sigma.\hat{p})$$

• Where does your first equation come from? – Oбжорoв Jul 31 '17 at 16:46
• It is the explicit form of the square root (see e.g., equation 2.106 in arxiv.org/abs/0812.1594) – JeffDror Jul 31 '17 at 19:42
• We seem to have a different concept of what an answer to a question is, or to what is trivial. Referring to an equation in an article, where the equation is not explained, is not an explanation. That is a strange way to do physics if you ask me. – Oбжорoв Aug 11 '17 at 12:02
• I am down voting your answer because of this. – Oбжорoв Aug 11 '17 at 12:02