0
$\begingroup$

We can find a correspondence between the restricted Lorentz group and the group $SL(2,\mathbb{C})$ if to each coordenate $x^{\mu}$ we associate a $2\times 2$ hermitian matrix $X$ given by $$X = x^{\mu} \sigma_{\mu},$$ where $\sigma_{0} = 1$ (the identity matrix) and $\sigma_{i}$, $i = 1,2,3$, are the Pauli matrices. It's easy to see that $$det X = (x^{0})^{2} - (x^{1})^{2} - (x^{2})^{2} - (x^{3})^{2},$$ which is a invariant. Now, if I do a Lorentz transformation $x^{\mu} \rightarrow x'^{\mu} = \Lambda^{\mu}_{\ \ \ \nu}\ x^{\nu}$, the transformation in the matrix $X$ is $$X \rightarrow X' = A X A^{\dagger} = x'^{\mu} \sigma_{\mu}.$$ Since $det X = det X'$, $|det A | = 1$ and I can choose $det A = 1 \Rightarrow A \in SL(2,\mathbb{C})$. So, for each Lorentz transformation $\Lambda$ I can associate two elements $\pm A$ of the group $SL(2,\mathbb{C})$. My first question is: do the matrices $A(\Lambda)$ forms a spinorial representation of the Lorentz group? How can I show it?

The book of Wu-Ki Tung says that these matrices satisfy the relation $A \epsilon A^{T} = \epsilon$, where

$$\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$

is a kind of metric tensor for the matrices $A$. I can see that this equation is analogous to the equation $\Lambda^{T} g \Lambda = g$, where $g$ is the metric tensor of the Minkowsk space. But how can I deduce this relation? (mainly the form of the matrix $\epsilon$) the book does'nt explain it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.