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 do I determine the form of $J_{\mu\nu}$ from Eq. (\ref{1}), and in addition, to determine the form of the spin component of $J_{\mu\nu}$?
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.
EDIT: MY ATTEMPT
If I invert Eq. (\ref{1}), I can see that $J^{\mu\nu}$ takes the form
$$J^{\mu\nu}=-\frac{1}{2}\Big( J_{\alpha\beta}\epsilon_{\dot{\alpha}\dot{\beta}}+ \epsilon_{\alpha\beta}\tilde{J}_{\dot{\alpha}\dot{\beta}} \Big) \sigma^{\mu\alpha\dot{\alpha}} \sigma^{\mu\beta\dot{\beta}}$$ where I observe that $J^{\mu\nu}$ does not have any free spinor indices. To see whether or not does $J^{\mu\nu}$ contain information about the spin of a given particle, I contract one of the Lorentz indices of $J^{\mu\nu}$ with the momentum four-vector (projecting out, in this way, the orbital part) by defining the Pauli-Lubanski operator, $$hp_{\rho}= \epsilon_{\rho\mu\nu\lambda}J^{\mu\nu}p^{\lambda}= \epsilon_{\rho\mu\nu\lambda}S^{\mu\nu}p^{\lambda}$$ where I have assumed that the total angular momentum can be split into the orbital angular momentum operator and another operator which is related to the spin of the particle this operator is going to act upon.
Now, I act with the operator $hp_{\rho}$ to a fermion wave-function, describing a particle with $p^{\mu}=(E,0,0,E)$ and helicity $h=\frac{1}{2}$, namely, $$u_+(\vec{p})=\sqrt{2E}\pmatrix{0\\0\\1\\0}= \sqrt{2E}\pmatrix{0\\\tilde{\lambda}^{\dot{\alpha}}}$$ Then, using $J_{\alpha\beta}u_+(\vec{p})=0$ and choosing, say to act on $u_+(\vec{p})$ with $\epsilon_{0123}J^{12}p^3$, we get $$\epsilon_{0123}J^{12}p^3 u_+(\vec{p})= \pmatrix{0\\ \frac{1}{2}\Big( \tilde{\lambda}_{\dot{1}} \frac{\partial}{\partial \tilde{\lambda}^{\dot{2}}}+ \tilde{\lambda}_{\dot{2}} \frac{\partial}{\partial \tilde{\lambda}^{\dot{1}}} \Big)p^3 \tilde{\lambda}^{\dot{\alpha}}}= \frac{1}{2}E\pmatrix{0\\ \tilde{\lambda}^{\dot{1}}}=hp_0u_+(\vec{p})$$ Does this tell us that $J^{\mu\nu}=L^{\mu\nu}+S^{\mu\nu}$, with $S^{\mu\nu}$ being the spin operator that contains information about the internal degrees of freedom of particles? Can someone elaborate on whether or not I am in the right direction?
