# Fierz identities and Weinberg operator

I've been told that

$$(\bar{L_i^c}\widetilde{\phi}^{\ *})(\widetilde{\phi}^\dagger L_k) = -\frac{1}{2}(\bar{\widetilde{L}}_i \vec{\sigma}L_k)(\widetilde{\phi}^\dagger \vec{\sigma} \phi) \tag1$$

by Fierz identity (see source below). Also, it's been applied that

$$\bar{L_i^c}\widetilde{\phi}^{\ *} = -\bar{\widetilde{L}}_i \phi.$$

Latin indices are flavour/family indices, $$\phi$$ is the Higgs boson, $$\widetilde{\phi} = i\sigma^2\phi^{\ *}$$ with $$\sigma^2$$ the second Pauli matrix, $$L_i$$ is the $$SU(2)_L$$ lepton doublet for $$i$$th family. $$c$$ upper-index represents the charge conjugation.

I've been looking for information about Fierz identities and all I've found is for 4-spinors while here we have eleements that under $$SU(2)_L$$ you can consider 2-spinors. So, my problem is that I don't know how to apply Fierz identities to get Eq. (1). Any help?

Source: page 5 from https://arxiv.org/abs/hep-ph/0210271

• These are not γ matrix Fierz identities, they are SU(2) Fierz identities of tensor products of Pauli matrices, as you may find, e.g., in the appendices of Okun's book, which is not a bad book to study well at all for this sort of thing. – Cosmas Zachos Apr 12 at 21:25
• Okun, Leptons & Quarks, ISBN: 9789814603140, appendix, 29.2.4. In my parochial experience, there is hardly a point in teaching particle physics to students who have not basically mastered this book. – Cosmas Zachos Apr 12 at 21:36
• If you have access to scribd, here. – Cosmas Zachos Apr 12 at 21:44
• Tasteful move. You could actually reproduce these Fierz identities from the completeness relation of Pauli matrices, but Okun makes it so easy and natural for you... – Cosmas Zachos Apr 12 at 22:03

I'm jotting down the Pauli matrix Fierz transposition ur-identity, the completion relation, $$\vec{\sigma}_{\alpha\beta}\cdot\vec{\sigma}_{\gamma\delta} = 2 \delta_{\alpha\delta} \delta_{\beta\gamma} - \delta_{\alpha\beta}\delta_{\gamma\delta}$$ for use in deriving any transposition relation you'd desire, symmetrizing and antisymmetrizing indices suitably.
You might rewrite this, with Okun's appendix 29.2.4, as $$\vec{\sigma}_{\alpha\beta}\cdot\vec{\sigma}_{\gamma\delta}= \tfrac{3}{2} \delta_{\alpha\delta} \delta_{\gamma\beta} -\tfrac{1}{2}\vec{\sigma}_{\alpha\delta}\cdot\vec{\sigma}_{\gamma\beta} ~~~.$$