Sign choice for sigma-matrices I'm trying to figure out the consequences of the sign choice
$$
\sigma^\mu = (\mathbf{1},\vec\sigma)\qquad\text{vs.}\qquad \sigma^\mu = (-\mathbf{1},\vec\sigma) \,.
$$
This choice does not affect the Clifford algebra, in terms of gamma-matrices the change is $\gamma^0\to-\gamma^0$, and $\gamma_5\to-\gamma_5$, which suggests that one might need to re-define the notion of left-handed and right-handed, I haven't yet tried to trace the consequences of this in detail.
What I was wondering more about is what other things are affected by this choice, and what needs to be adjusted. In particular, what happens to the terms that are linear in $\sigma^\mu$, such as $\sigma^\mu\partial_\mu$, that appear in the kinetic term for fermions, as well as in the superalgebra. The transition from $(\partial_t + \vec\sigma\vec\nabla)$ to $(-\partial_t + \vec\sigma\vec\nabla)$ seems non-trivial, and I'm not quite sure what this means.
 A: If we write spinor indices explicitly $\sigma^\mu{}^{AB'} = (\mathbf{1},\vec\sigma)$
where $A,B=1,2,\quad A',B'=1',2'\;.$ 
The transformations of spinor indices read 
$$
\epsilon^{AB}\chi_B=\chi^A,\quad \xi^B=\xi^A \epsilon_{BA}=-\xi^A \epsilon_{AB}=-\epsilon_{AB}\xi^A \;,\\
\epsilon^{A'B'}\chi_{B'}=\chi^{A'},\quad \xi^{B'}=\xi^{A'} \epsilon_{B'A'}=-\xi^{A'} \epsilon_{A'B'}=-\epsilon_{A'B'}\xi^{A'} \;.
$$
Note that if spinor indices were written explicitly the position of the terms in spinorial equation does not impornant and we have introducing $$ \epsilon^{AB}=\epsilon_{AB}=\epsilon^{A'B'}=\epsilon_{A'B'}=\left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right)  \;.$$
Therefore if you want to find $\sigma^\mu{}_{AB}$, you will  try to lowering spinor indices by $\epsilon$ 
$$
\sigma^\mu{}_{AB'} =\sigma^\mu{}^{CD'}\epsilon_{CA}\epsilon_{D'B'}
$$
In order to use matrix manipulations we will rearrange indices this way
$$
\sigma^\mu{}_{AB'} =-\epsilon_{AC}\sigma^\mu{}^{CD'}\epsilon_{D'B'}\;.
$$
The term-by-term we have
\begin{eqnarray}
\sigma^0{}_{AB'} &=&-\left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right) \left( \begin{matrix} 1 & 0\\ 0 & 1\end{matrix}  \right) \left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right)=\left( \begin{matrix} 1 & 0\\ 0 & 1\end{matrix}  \right)=\sigma^0{}^{AB'} \;\\
\sigma^1{}_{AB'} &=&-\left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right) \left( \begin{matrix} 0 & 1\\ 1 & 0\end{matrix}  \right) \left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right)=\left( \begin{matrix} 0 & -1\\ -1 & 0\end{matrix}  \right) =-\sigma^1{}^{AB'}\;\\
\sigma^2{}_{AB'} &=&-\left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right) \left( \begin{matrix} 0 & -i\\ i & 0\end{matrix}  \right) \left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right)=\left( \begin{matrix} 0 & i\\ -i & 0\end{matrix}  \right)=-\sigma^2{}^{AB'} \;\\
\sigma^3{}_{AB'} &=&-\left( \begin{matrix} 0 & 1\\ -1 & 0\end{matrix}  \right) \left( \begin{matrix}1 & 0\\ 0 & -1\end{matrix}  \right) \left( \begin{matrix}0 & 1\\ -1 & 0 \end{matrix}  \right)=\left( \begin{matrix} -1 & 0\\ 0 & 1\end{matrix}  \right) =-\sigma^3{}^{AB'}\;.
\end{eqnarray}
So now we have  $\sigma^\mu{}_{AB'} = (\mathbf{1},-\vec\sigma)$. The thing usually called $\overline{\sigma}^\mu$ is indeed this quantity. In the full spinorial indices version we define it as
$$
\overline \sigma^\mu{}_{B'A} :=\sigma^\mu{}_{AB'} = (\mathbf{1},-\vec\sigma)=\overline{\sigma}^\mu
$$
We flip the primed and un-primed indices because it will make more sense when multiply with ${\sigma}^\mu{}^{AB'}$ to produce the structure equation (Don't know how to call exactly) of spinor algebra
$$
\sigma^\mu \overline \sigma^\nu + \sigma^\nu \overline \sigma^\mu = -2\;\eta_{\mu\nu}\otimes \mathbb 1_{2\times 2}\;,
$$
where $\eta_{\mu\nu}=diag(-1,1,1,1)$. May be enough for understand the point or for further thoughts.
