I am trying to prove that :

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


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

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

  • $\begingroup$ Briefly looking at your conventions, I see that you always appear to raise and lower indices by contracting with the second index of epsilon. However, when you are manipulating $\sigma^{\mu\nu}$ you lower the $\beta$ index contracting it with the first entry. Why? $\endgroup$
    – OkThen
    Commented Sep 2, 2018 at 0:11
  • $\begingroup$ @OkThen Yes, that is a mistake. Let me edit it. Your input also gives a partial answer to what I am looking for ! Thanks! $\endgroup$ Commented Sep 2, 2018 at 0:13
  • $\begingroup$ My question is also related to Question 1 in physics.stackexchange.com/questions/411116/… $\endgroup$ Commented Sep 2, 2018 at 0:19
  • $\begingroup$ Spinor indices are somewhat awkward because a convention always needs to be known. It might be useful if you tell us what book or set of notes are you using. In any case, there is only one way to prove what you want and it is to use the definition of $\sigma^{\mu\nu}$ in terms of $\sigma^{\mu}$ and work out the symmetry properties. $\endgroup$
    – OkThen
    Commented Sep 2, 2018 at 0:23
  • $\begingroup$ @OkThen I was using the appendix of Zee's QFT nutshell and after that, the following paper: arxiv.org/pdf/0812.1594.pdf . I suppose I should do it with $\sigma^\mu$ as you say, but this doesn't solve this because you might have any other quantity with lots of spinor indices, and you want to know the convention for contracting its indices. And, yeah, spinor indices are awkward (at least for a beginner)! $\endgroup$ Commented Sep 2, 2018 at 0:31

3 Answers 3


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}

  • $\begingroup$ Why is$(\sigma_{\mu\nu})_{\alpha\beta}=(\sigma_{\mu\nu})_{\beta\alpha}$? $\endgroup$ Commented Sep 2, 2018 at 3:08
  • $\begingroup$ To prove this use the definition of $\sigma_{\mu\nu}$ in terms of $\sigma_\mu$ and ${\bar \sigma}_\mu$ as well as the property $(\sigma_\mu)_{\alpha{\dot\beta}} ({\bar \sigma}_\nu)^{ {\dot \beta} \alpha} = - 2 \eta_{\mu\nu}$. $\endgroup$
    – Prahar
    Commented Sep 2, 2018 at 3:10
  • $\begingroup$ Also, this does not solve the problem with raising/lowering the spinor indices. I mean, what is the answer to $\epsilon_{\beta\gamma}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}\epsilon^{\alpha\delta}=\epsilon_{\gamma\beta}\epsilon^{\delta\alpha}(\sigma^{\mu\nu})_\alpha^{\ \ \beta}$. Is it $(\sigma^{\mu\nu})^{\delta}_{\ \ \gamma}\ \text{or}\ (\sigma^{\mu\nu})_\gamma^{\ \ \ \delta}$? $\endgroup$ Commented Sep 2, 2018 at 3:10
  • $\begingroup$ They are equal due to the symmetry property I mentioned. $\endgroup$
    – Prahar
    Commented Sep 2, 2018 at 3:12
  • 1
    $\begingroup$ Yes, every matrix can be separated into its symmetric part and its antisymmetric part. For $2\times2$ matrices its antisymmetric part is proportional to $\epsilon_{\alpha\beta}$. If you can show that this part is zero, then the matrix must be symmetric. $\endgroup$
    – Prahar
    Commented Sep 2, 2018 at 19:47

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

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

  • $\begingroup$ Thanks for the answer, but I am still confused on how to raise/lower indices, because now we just have $\sigma^{\mu\nu}$, but we could just as well have anything else with many spinor indices. What I am trying to say is that, we could had something without the symmetry in the indices $\bar{a}$ and $b$. $\endgroup$ Commented Sep 2, 2018 at 2:55
  • $\begingroup$ And also, I can't really figure out how you concluded that we have the symmetry that you mention. It probably has to do with the fact that I'm really confused with this notation at the moment. $\endgroup$ Commented Sep 2, 2018 at 2:56
  • $\begingroup$ We can move this to chat. $\endgroup$ Commented Sep 2, 2018 at 2:58
  • $\begingroup$ Could we do it tomorrow? In my country it's 6am right now! $\endgroup$ Commented Sep 2, 2018 at 2:59
  • $\begingroup$ Sure. I think that this answer will be extremely helpful physics.stackexchange.com/q/305520/56599 $\endgroup$ Commented Sep 2, 2018 at 20:37

\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}


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


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