# 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. Apr 12 '19 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. Apr 12 '19 at 21:36
• If you have access to scribd, here. Apr 12 '19 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... Apr 12 '19 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 $$\vec{\sigma}_{\alpha\beta}\cdot\vec{\sigma}_{\gamma\delta}= \tfrac{3}{2} \delta_{\alpha\delta} \delta_{\gamma\beta} -\tfrac{1}{2}\vec{\sigma}_{\alpha\delta}\cdot\vec{\sigma}_{\gamma\beta} ~~~.$$