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The Haag-Lopuszanski-Sohnius (HLS) theorem yields a preference for the super-Poincare algebra. When the assumptions of the HLS theorem are not fulfilled, other non-trivial extensions of the spacetime Poincare algebra is possible, cf. e.g. this Phys.SE post.

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This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...

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Short answers Apply the Young calculus (per ACuriousMind's suggestion in the comments). For finding the multiplicity of the trivial representation in a tensor product of representations of $SU(n)$, note that each irreducible representation $D$ of $SU(n)$ has a unique conjugate irreducible representation $\bar D$ such that the Young calculus allows ...

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The spin group $Spin(3,1)\cong SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post and links therein.

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There is a long and formal way, and also an easy and dirty way. I will tell you the easy option. The algebra tells you that $[\delta_Q (\epsilon_1), \delta_Q (\epsilon_2)] = \delta_{P}(\xi^\mu_3)$ where $\epsilon$ is your SUSY parameter and $\xi^\mu_3 = \bar\epsilon_1 \gamma^\mu \epsilon_2$ is your translation parameter. Now, the only Lorentz vector that ...

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The simplification follows from the theorem which states that if such operator is conserved in Heisenberg sense, $$\frac{d\hat{Q}}{dt} = \frac{\partial \hat{Q}}{\partial t} - \frac{i}{\hbar}[\hat{Q}, \hat{H}] = 0,$$ than it commutes with S-operator: $$[\hat{Q}, \hat{S}] = 0$$ So that these two operators can be diagonalized simulatenously: in ...

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