# Spinor map and Lorentz transformations

1. Aug 3, 2013

### parton

At the moment I'm reading a little bit about the connection between the Lorentz group $SO^{+}(1,3)$ and its covering group $SL(2,\mathbb{C})$.

On wiki I found this:
https://en.wikipedia.org/wiki/Lorentz_group#Elliptic

But I think there is a sign error in the $P_{1}$ matrix. It should be:
$$Q_1 = \left[ \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) & 0 \\ 0 & -\sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right]$$

Could anyone confirm (or disporve) this?

Here my calculation:

I started with:
$$X = x^{\mu} \sigma_{\mu} = \left[ \begin{matrix} t+z & x-iy \\ x+iy & t-z \end{matrix}, \right]$$
where $x^{\mu} = (t, x, y, z)$ and $\sigma_{\mu} = (1_{2 \times 2}, \vec{\sigma})$
and
$$P_1 = \left[ \begin{matrix} \exp(i \theta/2) & 0 \\ 0 & \exp(-i \theta/2) \end{matrix} \right]$$

Computing $X' \equiv P_{1} X P_{1}^{\dagger}$ leads to:
$$X' = \left[ \begin{matrix} t+z & \exp(i \theta/2) (x-iy) \\ \exp(-i \theta/2) (x+iy) & t-z \end{matrix} \right].$$
and using the relation $(x')^{\mu} = \dfrac{1}{2} \mathrm{tr} \left( \underline{\sigma}^{\mu} X' \right)$, where $\underline{\sigma}^{\mu} = \sigma_{\mu}$ I found:
$$(x')^{1} = \dfrac{1}{2} \mathrm{tr} \left( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} X' \right) = x \cdot \cos \theta + y \cdot \sin \theta$$

If I check the other matrices like $P_{2}$ everything is ok.