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I am having some trouble deriving the transormation laws for the weyl spinors, equation (3.37) in the Peskin Schroesder book on quantum field theory.

Beginning with the relation $\psi\to(1-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu})\psi$ from (3.30) and the form of the transformation matrices in equations (3.26) and (3.27), I get

$1-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu} = 1-\frac{i}{2}\omega_{0\nu}S^{0\nu} + \frac{i}{2}\omega_{i\nu}S^{i\nu} = 1 - \frac{i}{2}\omega_{00}S^{00} + \frac{i}{2}\omega_{0i}S^{0i} + \frac{i}{2}\omega_{i0}S^{i0} - \frac{i}{2}\omega_{ij}S^{ij}$

$ = 1 - 0 + i\omega_{0i}S^{0i} - \frac{i}{2}\omega_{ij}S^{ij} = 1+i\omega_{0i}\frac{-i}{2}\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} - \frac{i}{2}\omega_{ij}\frac{1}{2}\epsilon^{ijk}\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix} $

The discussion at the end of section 3.1, leading to equations (3.20) and (3.21) then suggest the identification $\omega_{0i} = \beta_i$ and $\omega_{ij} = \epsilon_{ijk}\theta^k$. Plugging this in gives

$1 + \frac{1}{2}\beta_i\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} + \frac{1}{4}\epsilon_{ijl}\theta^l\epsilon^{ijk}\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix}$

Using the identitiy $\epsilon_{ijl}\epsilon^{ijk} = 2\delta_l^k$ gives

$1 + \frac{1}{2}\beta_i\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} + \frac{1}{2}\theta^k\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix}$

$ = \begin{pmatrix}1 + \frac{1}{2}\beta_i\sigma^i - \frac{1}{2}\theta^k\sigma^k & 0 \\ 0 & 1 - \frac{1}{2}\beta_i\sigma^i - \frac{1}{2}\theta^k\sigma^k \end{pmatrix}$

$ = \begin{pmatrix}1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} + \frac{1}{2}\vec{\beta}\cdot\vec{\sigma} & 0 \\ 0 & 1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} - \frac{1}{2}\vec{\beta}\cdot\vec{\sigma} \end{pmatrix}$

Making the identification $\psi = \begin{pmatrix}\psi_L \\ \psi_R\end{pmatrix}$, this then gives

$\psi_L\to(1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} + \frac{1}{2}\vec{\beta}\cdot\vec{\sigma})\psi_L$

$\psi_R\to(1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} - \frac{1}{2}\vec{\beta}\cdot\vec{\sigma})\psi_R$

for the Weyl transformations, which is the oposite order to how it appears in the book. Given that it is only a small difference, I initially thought it might just be a typo in the book, although I encountered similar sign errors later that lead me to think this calculation is wrong, although I can't seem to find where.

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This is almost correct, you have just made a mistake in the beginning in the index summation :

$$\begin{align}1-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu} &= 1-\frac{i}{2}\omega_{0\nu}S^{0\nu} \color{red}{-} \frac{i}{2}\omega_{i\nu}S^{i\nu}=1 \color{red}{-} \frac{1}{2}\beta_i\begin{pmatrix}\sigma^i & 0 \\ 0 & -\sigma^i\end{pmatrix} + \frac{1}{2}\theta^k\begin{pmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{pmatrix} \\&= \begin{pmatrix}1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} \color{red}{-} \frac{1}{2}\vec{\beta}\cdot\vec{\sigma} & 0 \\ 0 & 1 - \frac{1}{2}\vec{\theta}\cdot\vec{\sigma} \color{red}{+} \frac{1}{2}\vec{\beta}\cdot\vec{\sigma} \end{pmatrix} \end{align}$$

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  • $\begingroup$ Isn't $\omega_{\mu\nu}S^{\mu\nu}$ a Lorentz contraction, following the usual summation rule $v_\mu v^\mu = v_0v^0 - v_iv^i = (v^0)^2 - \vec{v}^2$? $\endgroup$
    – Flumpo
    Commented May 1, 2021 at 14:04
  • $\begingroup$ The first equality is wrong. The minus sign comes from the metric, not the summation. You have:$v_{\mu}v^{\mu}=g_{\mu\nu}v^{\mu}v^{\nu}=g_{00}(v^0)^2+g_{ii}(v^i)^2=(v^0)^2-(v^i)^2$ with the metric signature $(1,-1,-1,-1)$ $\endgroup$
    – Free_ion
    Commented May 1, 2021 at 14:59

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