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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

TheQuantumMan
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  • 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? – OkThen Sep 02 '18 at 00:11
  • @OkThen Yes, that is a mistake. Let me edit it. Your input also gives a partial answer to what I am looking for ! Thanks! – TheQuantumMan Sep 02 '18 at 00:13
  • My question is also related to Question 1 in https://physics.stackexchange.com/questions/411116/identities-of-pauli-matrices-in-two-component-spinor-formalism?rq=1 – TheQuantumMan Sep 02 '18 at 00:19
  • 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. – OkThen Sep 02 '18 at 00:23
  • @OkThen I was using the appendix of Zee's QFT nutshell and after that, the following paper: https://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)! – TheQuantumMan Sep 02 '18 at 00:31
  • I was also using Shifman's Advanced Quantum Field theory, and in contrast to his other chapters that I've read, I found his approach to this subject to be awful; he's just defining and stating things. – TheQuantumMan Sep 02 '18 at 00:32

3 Answers3

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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}

Prahar
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  • Why is$(\sigma_{\mu\nu}){\alpha\beta}=(\sigma{\mu\nu})_{\beta\alpha}$? – TheQuantumMan Sep 02 '18 at 03:08
  • To prove this use the definition of $\sigma_{\mu\nu}$ in terms of $\sigma_\mu$ and ${\bar \sigma}\mu$ as well as the property $(\sigma\mu){\alpha{\dot\beta}} ({\bar \sigma}\nu)^{ {\dot \beta} \alpha} = - 2 \eta_{\mu\nu}$. – Prahar Sep 02 '18 at 03:10
  • 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 $(\sigma^{\mu\nu})^{\delta}{\ \ \gamma}\ \text{or}\ (\sigma^{\mu\nu})_\gamma^{\ \ \ \delta}$? – TheQuantumMan Sep 02 '18 at 03:10
  • They are equal due to the symmetry property I mentioned. – Prahar Sep 02 '18 at 03:12
  • Hello again, you said to use the identity for $(\sigma_\mu){\alpha\dot{\beta}}(\bar{\sigma}\nu)^{\dot{\beta}\alpha}$, but in the definition of $(\sigma_{\mu\nu})^{\ \ \beta}\alpha$, we have two free indices. So, we have terms like $(\sigma\mu){\alpha\dot{\beta}}(\bar{\sigma}\nu)^{\dot{\beta}\beta}$. – TheQuantumMan Sep 02 '18 at 18:57
  • Yes and you need to contract the spinor indices. But you need to think a little. I'm not going to give you the answer. Why would one need to contract spinor indices?? Remember, $(\sigma_{\mu\nu})_{\alpha\beta}$ is a $2\times2$ matrix for each $\mu$ and $\nu$. What do you need to prove to show that it is symmetric? – Prahar Sep 02 '18 at 19:02
  • So, let's see if I got this right: $(\sigma^{\mu\nu})\alpha^{\ \ \alpha}=\epsilon^{\alpha\beta}(\sigma^{\mu\nu}){\alpha\beta}$. So, if we show that $(\sigma^{\mu\nu})\alpha^{\ \ \alpha}=0$, it means that either $(\sigma^{\mu\nu}){\alpha\beta}=0$ (which is not except for the case where $\mu=0$, $\nu=0$) or $(\sigma^{\mu\nu}){\alpha\beta}=(\sigma^{\mu\nu}){\beta\alpha}$ . Is this correct? – TheQuantumMan Sep 02 '18 at 19:43
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    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. – Prahar Sep 02 '18 at 19:47
  • I used the identity you referenced and it worked out! I have one last question though, that's related to what I directly asked in my question. I asked if $\epsilon_{\beta\gamma}(\sigma^{\mu\nu})\alpha^{\ \ \beta}\epsilon^{\alpha\delta}=\epsilon{\gamma\beta}\epsilon^{\delta\alpha}(\sigma^{\mu\nu})\alpha^{\ \ \beta}=(\sigma^{\mu\nu})^{\delta}{\ \ \gamma}\ \text{or}\ (\sigma^{\mu\nu})_\gamma^{\ \ \ \delta}$. – TheQuantumMan Sep 02 '18 at 20:03
  • If I lower the index $\beta$, I get $\epsilon_{\beta\gamma}(\sigma^{\mu\nu})\alpha^{\ \ \beta}\epsilon^{\alpha\delta}=-\delta\gamma^\rho \epsilon^{\alpha\delta}(\sigma^{\mu\nu}){\alpha\rho}=(\sigma^{\mu\nu})^\delta{\ \ \gamma}$. But, if I use the fact that $(\sigma^{\mu\nu}){\alpha\rho}=(\sigma^{\mu\nu}){\rho\alpha}$ I get $=(\sigma^{\mu\nu})\gamma^{\ \ \delta}$. This seems to imply that $(\sigma^{\mu\nu})\alpha^{\ \ \beta}=(\sigma^{\mu\nu})^\beta_{\ \ \alpha}$, which is not the case since for every $\mu$ and $\nu$, $(\sigma^{\mu\nu})$ is either zero or proportional to a Pauli matrix – TheQuantumMan Sep 02 '18 at 20:05
  • You said that they are equal, but for example $\sigma^{0i}=-\frac{1}{2}\sigma^i$, and $\sigma^i$ is not equal to its transpose but to its Hermitian conjugate. – TheQuantumMan Sep 02 '18 at 20:14
  • Let us continue this discussion in chat. – TheQuantumMan Sep 02 '18 at 21:12
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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

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

InertialObserver
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  • 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 $b$. – TheQuantumMan Sep 02 '18 at 02:55
  • 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. – TheQuantumMan Sep 02 '18 at 02:56
  • We can move this to chat. – InertialObserver Sep 02 '18 at 02:58
  • Could we do it tomorrow? In my country it's 6am right now! – TheQuantumMan Sep 02 '18 at 02:59
  • Sure. I think that this answer will be extremely helpful https://physics.stackexchange.com/q/305520/56599 – InertialObserver Sep 02 '18 at 20:37
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\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)

vyali
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