I am trying to understand one of the anti-commutation relations of SUSY algebra. The lecture notes "Supersymmetry and Extra Dimensions" (PDF) taken by Flip Tanedo, says on p.29, due to the index structure of the generators inside the brackets, the most general form of the anticommutator is $$\{Q_\alpha,Q^\beta\}=k(\sigma^{\mu\nu})_\alpha{}^{\beta}\ M_{\mu\nu}\tag{3.25}$$ where $k$ is some constant.

I want to know how the right-hand side is coming, I can't understand how the right-hand side is having such form.


1 Answer 1

  1. The anticommutator on the LHS of eq. (3.25) belongs to the left-handed representation $(1,0)$ (of the double cover) of the restricted Lorentz group in 3+1D, so the (graded) Lie algebra generators on the RHS must also belong to the same representation. This implies that the RHS must be a left-handed linear combination of the angular momentum generators $M_{\mu\nu}$. Recall that the angular momentum generators $M_{\mu\nu}$ belongs to the reducible representation $(1,0)\oplus(0,1)$ [as complex vector spaces], cf. e.g. this Phys.SE post.

  2. Whatever Tanedo means with the coefficient on the RHS of eq. (3.25), the main point is that they have to be zero, since the LHS commutes with $P_{\lambda}$ while the RHS doesn't always.

  • $\begingroup$ Thank you very much, I understand the LHS is (1,0) representation, so it should be the same for the RHS as well. I want to know how $M^{\mu\nu}$ is having (1,0) representation. $\endgroup$
    – Alex
    Jun 12, 2023 at 10:12
  • 1
    $\begingroup$ That is the topic of the linked post. $\endgroup$
    – Qmechanic
    Jun 12, 2023 at 16:30
  • $\begingroup$ I am a beginner to the subject and I am having difficulties grasping the abstractness of all the representations, so if you could explain how exactly angular momentum generators are (1,0) representations, I would be grateful. I thought $M_{\mu\nu}$ was an antisymmetric tensor and was having $(1,0)\oplus(0,1)$ represntation. $\endgroup$
    – Alex
    Jun 13, 2023 at 8:12
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 13, 2023 at 10:00
  • $\begingroup$ Thank you @Qmechanic $\endgroup$
    – Alex
    Jun 13, 2023 at 16:33

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