In the literature one encounters a lot of different conventions for how left-handed spinor transforms (rotation angle $\phi$, rapidity $\beta$), among them
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,
$M_L = M_{(\frac{1}{2}, 0)} = e^{+i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$
Comparing various sources can be a pain, so I tried to establish a definite convention (as a reference for me), to compare against various sources, but I failed so far to do it consistently.
Specifically I tried to understand the conventions used in the following source:
Dreiner, H. K., Haber, H. E., & Martin, S. P. (2010). Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry. Physics Reports, 494(1-2), 1–196. doi:10.1016/j.physrep.2010.05.002 (http://arxiv.org/abs/0812.1594)
Is use the version with the $(-1, +1, +1, +1)$ convention that can be obtained from https://www.niu.edu/spmartin/spinors/ (whereas the arxiv version uses the (+1, -1, -1, -1) convention).
They state (page 10) that the transformation of a left-handed spinor ought to be:
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$
I then tried to get the corresponding lorentz-transformation by looking at the transformation of the hermitean matrix $X = \left( \begin{array}{cc} x^0+x^3 & x^1-ix^2\\ x^1+ix^2 & x^0-x^3 \end{array} \right)$
$A: X \rightarrow X' = A X A^\dagger$
To extract the lorentz-transformatin I use
${\Lambda^\mu}_\nu = -\frac{1}{2} \operatorname{tr}[\sigma^\mu A \bar{\sigma}_\nu A^\dagger]$
with $\sigma^\mu = (1, \vec{\sigma})$ and $\bar{\sigma}_\nu = (-1, -\vec{\sigma})$
This is not covered in their paper, but when I do it for
$R = e^{-i\phi\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-i\frac{1}{2}\phi} & 0 \\ 0 & e^{i\frac{1}{2}\phi} \end{array}\right)$ (rotation around the z-axis)
I get the lorentz transformation
${R^\mu}_\nu = \left( \begin{array}{cc} 1& 0& 0& 0\\ 0& \cos\phi& -\sin\phi & 0 \\ 0& \sin\phi& \cos\phi& 0 \\ 0& 0& 0& 1 \end{array} \right)$
Similar for
$B = e^{-\beta\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-\frac{1}{2}\beta} & 0 \\ 0 & e^{\frac{1}{2}\beta} \end{array}\right)$ (boost in z-direction)
I get
${B^\mu}_\nu = \left( \begin{array}{cc} \cosh\beta& 0& 0& -\sinh\beta\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\sinh\beta& 0& 0& \cosh\beta \end{array} \right)$
These correspond to active rotations (counterclockwise) around the z-axis, but passive boost in the z-direction.
But this is in constrast to their statement (page 10) that corresponding infinitesimal transformations are
$\vec{x} \rightarrow \vec{x} + (\vec{\phi} \times \vec{x})$
$\vec{x} \rightarrow \vec{x} + \vec{\beta}t$
which both are active rotations and boosts.
So either:
- something is wrong in my calculation from $SL(2, \mathbb{C})$ to the Lorentz Group (or may be also there is a hidden convention I got wrong ?)
- or the cited paper got it wrong (unlikely, since its already in its 5th incarnation, with a list of errata [https://www.niu.edu/spmartin/spinors/])
- or I am quite generally confused about how exactely spinors should transform, and specifically how the signs are determined.
