On the proof $\eta \sigma^{\mu\nu} \chi=-\chi \sigma^{\mu\nu} \eta$ (problem with spinor indices)

Question

I am trying to prove that :

$$\eta \sigma^{\mu\nu} \chi=-\chi \sigma^{\mu\nu} \eta$$

or

$$\eta^\alpha (\sigma^{\mu\nu})_\alpha^{\ \ \beta} \chi_\beta=-\chi^\alpha (\sigma^{\mu\nu})_\alpha^{\ \ \beta} \eta_\beta$$.

Here, $\mu,\ \nu$ are spacetime indices and $\alpha,\ \beta$ are spinor indices that are contracted with $\epsilon_{\alpha\beta}$ and $\epsilon^{\alpha\beta}$.
I am stuck at a particular point. I start out as follows:

$$ \begin{align} \eta^\alpha (\sigma^{\mu\nu})_\alpha^{\ \ \beta} \chi_\beta & =- \chi_\beta(\sigma^{\mu\nu})_\alpha^{\ \ \beta} \eta^\alpha \\ & =-(\epsilon_{\beta\gamma}\chi^\gamma)(\sigma^{\mu\nu})_\alpha^{\ \ \beta}(\epsilon^{\alpha\delta}\eta_\delta) \\ & =-\chi^\gamma\left[\epsilon_{\beta\gamma}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}\epsilon^{\alpha\delta}\right]\eta_\delta \end{align} $$

So, the quantity in the brackets should be equal to $(\sigma^{\mu\nu})_\gamma^{\ \ \ \delta}$ in order to complete the proof.

Now, here's the point where I where I can't figure out what to do: for the quantity in the brackets, I contact indices with $\epsilon$ as follows:

$$\epsilon_{\beta\gamma}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}\epsilon^{\alpha\delta}=\epsilon_{\gamma\beta}\epsilon^{\delta\alpha}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}=(\sigma^{\mu\nu})^{\delta}_{\ \ \gamma}\ \text{or}\ (\sigma^{\mu\nu})_\gamma^{\ \ \ \delta}$$

So, I guess that I do not know the convention well enough so that I can know which one is the correct (the latter of course, but I don't know why). Any help is appreciated.

EDIT: This is also related to Question 1 of: Identities of Pauli matrices in two-component spinor formalism

Answer 1

We note that $\psi^\alpha \chi_\alpha = - \psi_\alpha \chi^\alpha$. Thus, \begin{align} \eta \sigma_{\mu\nu} \chi &= \eta^\alpha ( \sigma_{\mu\nu})_\alpha{}^\beta \chi_\beta \\ &= - \eta^\alpha ( \sigma_{\mu\nu})_{\alpha\beta} \chi^\beta \\ &= - \eta^\alpha ( \sigma_{\mu\nu})_{\beta\alpha} \chi^\beta \qquad \qquad (\sigma_{\mu\nu})_{\alpha\beta} = (\sigma_{\mu\nu})_{\beta\alpha} \\ &= \chi^\beta ( \sigma_{\mu\nu})_{\beta\alpha} \eta^\alpha \qquad \qquad ~~~~~~\eta^\alpha \chi^\beta = -\chi^\beta \eta^\alpha \\ &= - \chi^\beta ( \sigma_{\mu\nu})_\beta{}^{\alpha} \eta_\alpha \\ &= - \chi \sigma_{\mu\nu} \eta \end{align}

Answer 2

In group theory, the raised and lower indices are somewhat of a red herring. They are used to distinguish between the $N$ and $\bar{N}$ reps of $SU(N)$. They are truly independent representations, except for the case of the fundamental, but it doesn't mess things up. If you're ever confused just use some other kind of index to indicate the other representation. For instance, we can write

$$\eta \sigma^{\mu\nu} \chi = \eta_{\bar{a}}\sigma^{\mu\nu}_{\bar{a}b}\chi_b =-\chi_b\sigma^{\mu\nu}_{\bar{a}b}\eta_{\bar{a}}$$

and we don't have to worry about raising or lowering anything like that. Now we just observe that $\sigma^{\mu\nu}$ is symmetric in its group indices. Hence

$$ -\chi_b\sigma^{\mu\nu}_{\bar{a}b}\eta_{\bar{a}} = -\chi_b\sigma^{\mu\nu}_{b\bar{a}}\eta_{\bar{a}}= - \chi\sigma^{\mu\nu}\eta.$$

Answer 3

\begin{align} \epsilon_{\beta\gamma}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}\epsilon^{\alpha\delta}=(\sigma^{\mu\nu})_\alpha^{\ \ \beta}\epsilon_{\beta\gamma}\,\epsilon^{\alpha\delta}&=(\sigma^{\mu\nu})_\gamma^{\ \ \beta}\epsilon_{\beta\alpha}\,\epsilon^{\alpha\delta} \quad \text{(see identity (B.7) in ref. [1])}\\ &=(\sigma^{\mu\nu})_\gamma^{\ \ \beta}\delta^\delta_\beta=(\sigma^{\mu\nu})_\gamma^{\ \ \delta} \end{align}

References:

[1] J. Wess and J. Bagger, Supersymmetry and supergravity, Princeton University Press, Princeton, NJ, U.S.A. (1992)