I am reading Evidence for a new Soft Graviton theorem, by Cachazo and Strominger. At some point, they express the relation $$J_{\mu\nu}\sigma^{\mu}_{\alpha\dot{\alpha}}\sigma^{\mu}_{\beta\dot{\beta}} =J_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}} +\epsilon_{\alpha\beta}\tilde{J}_{\dot{\alpha}\dot{\beta}} \tag{1}\label{1}$$ where $$J_{\alpha\beta}=\frac{i}{2}\Big(\lambda_{\alpha} \frac{\partial}{\partial \lambda^{\beta}}+ \lambda_{\beta} \frac{\partial}{\partial \lambda^{\alpha}}\Big),\ \tilde{J}_{\dot{\alpha}\dot{\beta}}=\frac{i}{2}\Big(\tilde{\lambda}_{\dot{\alpha}} \frac{\partial}{\partial \tilde{\lambda}^{\dot{\beta}}}+ \tilde{\lambda}_{\dot{\beta}} \frac{\partial}{\partial \tilde{\lambda}^{\dot{\alpha}}}\Big)$$ with the normalization $\sigma_{\mu\alpha\dot{\alpha}}\sigma_{\nu}^{\alpha\dot{\alpha}}=2\eta_{\mu\nu}$ being used.

I have read how to connect the total angular momentum $J_{\mu\nu}$ with $J_{\alpha\beta}$ and $\tilde{J}_{\dot{\alpha}\dot{\beta}}$, i.e. to prove Eq. (\ref{1}), in the PSE post Derivation of conformal generators in spinor helicity formalism. My question is how does $J_{\mu\nu}$ know about the spin of the particle carrying total angular momentum $J_{\mu\nu}$?

I am asking this because in the PSE post I cite, the starting point in proving Eq. (\ref{1}) is $$J_{\alpha\beta}=\frac{1}{2}(\sigma^{\mu\nu})_{\alpha\beta}M^{\mu\nu}$$ where $$\sigma^{\mu\nu}_{\alpha\beta}=\frac{i}{4}(\sigma^{\mu}\bar{\sigma}^{\nu}- \sigma^{\nu}\bar{\sigma}^{\mu})_{\alpha\beta}$$ and $$M^{\mu\nu}=p^{\mu}\frac{\partial}{p_{\nu}}- p^{\nu}\frac{\partial}{\partial p_{\mu}}$$ $M^{\mu\nu}$ doesn't know about the spin of the particle, this is the reason I am asking. It would be a lot more intuitive if instead of $M^{\mu\nu}$ we had $M^{\mu\nu}+\Sigma^{\mu\nu}$, where $\Sigma^{\mu\nu}$ is somehow connected to the spin degrees of freedom of the particle.

Also, what exactly is $J^{\mu\nu}$ in Eq. (\ref{1})?