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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?

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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 $$B_{\mu\nu} = \begin{pmatrix}0 & -b_L^x & -b_L^y & -b_L^z \\ b_L^x & 0 & -b_L^z & b_L^y \\ b_L^y & b_L^z & 0 & -b_L^x \\ b_L^z & -b_L^y & b_L^x & 0\end{pmatrix}$$

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  • $\begingroup$ 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! $\endgroup$ – Iván Mauricio Burbano Dec 28 '18 at 21:31
  • $\begingroup$ How us it that $b_L^{\pm,0}$ is a spin 1 vector? I think I don't understand the notation properly. $\endgroup$ – Iván Mauricio Burbano Dec 28 '18 at 21:32
  • $\begingroup$ If self-duality means hermiticity then Maxwell's field tensor with real entries is already self-dual. I'm not sure what you mean by self-dual. $\endgroup$ – Luke Pritchett Dec 28 '18 at 21:35
  • $\begingroup$ 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. $\endgroup$ – Iván Mauricio Burbano Dec 28 '18 at 21:36
  • $\begingroup$ Oh, I see. I've edited the answer $\endgroup$ – Luke Pritchett Dec 28 '18 at 21:43

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