# How to build an antisymmetric selfdual tensor out of two 4-vectors?

In problem C of section 1.4 of Ramon's Field Theory: A Modern Primer, we are asked to build a field bilinear in $$\chi_L$$ and $$\psi_L$$, two left-handed weyl spinors, which transforms as the (1,0) representation of $$\text{SL}(2,\mathbb{C})$$. This representation is equivalent to the behavior of rank 2 tensors $$B_{\mu\nu}$$ which are antisymmetric and selfdual, i.e., $$B_{\mu\nu}=-B_{\mu\nu}\\B_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}B^{\rho\sigma}.$$ One can check that both $$i\psi_L^\dagger \sigma^\mu\psi_L$$ and $$i\chi_L^\dagger\sigma^\mu\chi_L$$ are 4-vectors. I think the correct way to proceed is to use them to build the field $$B_{\mu\nu}$$. My first approavh was to antisymmetrize and selfdualize $$\psi_L^\dagger \sigma^\mu\psi_L\chi_L^\dagger \sigma^\nu\chi_L$$. This of course fails to be bilinear in the fields. Does any body have any clue? Would something like $$\psi_L^\dagger\sigma^\mu\sigma^\nu\chi_L$$ work?

Since both fields transform as $$(\tfrac{1}{2},0)$$ we can treat them as $$SU(2)$$ spinors and ignore everything else. So the question is how do you multiply two $$j=\tfrac{1}{2}$$ spinors to get a $$j=1$$ vector? Easy enough, just use regular Pauli matrices: $$b_L^{\pm,0} = \psi_{L\alpha} \sigma^{\pm,3}_{\alpha\beta} \chi_{L\beta}$$ or $$\vec{b}_L = \psi_{L,\alpha} \vec{\sigma}_{\alpha\beta} \chi_{L,\beta}$$
To see that this is equivalent to an anti-symmetric Lorentz tensor that satisfies $$B_{\mu\nu} = \tilde{B}_{\mu\nu}$$ write $$B_{\mu\nu} = \begin{pmatrix}0 & -b_L^x & -b_L^y & -b_L^z \\ b_L^x & 0 & -b_L^z & b_L^y \\ b_L^y & b_L^z & 0 & -b_L^x \\ b_L^z & -b_L^y & b_L^x & 0\end{pmatrix}$$
• Are you sure $(1,0)$ is not the antisymmetric selfdual tensors? Your argument for showing it is not only takes into account antisymmetry. Selfduality further restricts dimension. Maxwell's field tensor tranforms under $(1,0)\oplus(0,1)$. I will need some time to understand your answer! Thanks! – Iván Mauricio Burbano Dec 28 '18 at 21:31
• How us it that $b_L^{\pm,0}$ is a spin 1 vector? I think I don't understand the notation properly. – Iván Mauricio Burbano Dec 28 '18 at 21:32
• Oh sorry, by self duality I mean that the condition $B_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}B^{\rho\sigma}$ is fulfilled. – Iván Mauricio Burbano Dec 28 '18 at 21:36