Let us consider the barred sigma matrices defined as $$(\overline{\sigma}^\mu)^{\dot{\alpha}\alpha}:=\epsilon^{\alpha \beta} \epsilon^{\dot{\alpha} \dot{\beta}} (\sigma^\mu)_{\beta \dot{\beta}}. \tag{1}\label{eq:1}$$ I want to manipulate this expression to write this as a product of matrices. I manipulated the above expression as $$\epsilon^{\alpha \beta} \epsilon^{\dot{\alpha} \dot{\beta}} (\sigma^\mu)_{\beta \dot{\beta}}= \epsilon^{\dot{\alpha} \dot{\beta}}.(\sigma^\mu_{\dot{\beta} \beta})^\mathrm{T}.(-\epsilon^{\beta \alpha})=\epsilon^{\dot{\alpha} \dot{\beta}}. (\sigma^\mu_{\dot{\beta} \beta})^\mathrm{T}.(-\epsilon^{\alpha \beta})^T.$$ So, I thought $$\overline{\sigma}^\mu=\dot{\epsilon}. \sigma^T .(-\epsilon^T)$$ where both $\epsilon$ and $\dot{\epsilon}$ look like $$\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \end{pmatrix}.$$ But it doesn't work for $\mu=0$ for which I am supposed to get $\overline{\sigma}^0=\sigma^0=\mathbb{I}_2$ but I am getting a $-\mathbb{I}$. What is going wrong in this i.e. how to obtain the matrix form of \eqref{eq:1}?
1 Answer
You must be careful translating the index notation to matrices, since not all properties of $\sigma^\mu$ and $\bar \sigma^\mu$ are independent of the representation. For example, calculating $\sigma^0 \cdot \sigma^0 = \mathbb I$ isn't a legal thing in the index notation. Also, transposing is a subtle buisness.
Mathematically, this means that the linear maps represented by the matrices cannot be composed, because the domains and targets don't match.
I guess, the best thing is to write out the Einstein sum convention. E.g.,
$$\bar \sigma^{0}{}^{\dot 1 1} = \sum_{\beta \dot \beta}\epsilon^{1\beta} \epsilon^{\dot 1 \dot\beta} \sigma^0_{\beta \dot \beta} = \epsilon^{12} \epsilon^{\dot 1 \dot2} \sigma^0_{2 \dot 2} = 1$$.
To avoid confusion, let's distinguish $\epsilon^{\alpha \beta}$ and $\dot \epsilon^{\dot \alpha \dot\beta}$ The correct expression is given $$ \bar\sigma^\mu =( \epsilon \sigma^\mu \dot \epsilon^T)^T$$
- the transposition of $\dot \epsilon$ is needed, to ensure that always the second index of the $\epsilon$'s is contracted.
- The overall transposition is needed, to bring the indices of the result in the right order. You may check that the definition of matrix multiplication yields the correct index contractions.
In particular, we have $$\bar \sigma^0 = (\epsilon \sigma^0 \dot \epsilon^T)^T= (\epsilon \mathbb I \epsilon^T)^T = (-\epsilon\epsilon)^T = \mathbb I^T = \mathbb I$$
