Superalgebra and BRST supersymmetry 1: odd vs even

Question

It is said that most of theories, the even elements of the superalgebra correspond to bosons and odd elements to fermions (but this is not always true; for example, the BRST supersymmetry is the other way around)..

• Why BRST supersymmetry is the other way around: odd elements of the superalgebra correspond to bosons and even elements to fermions ?

This post imported from StackExchange Physics at 2020-12-12 20:04 (UTC), posted by SE-user annie marie heart

Answer 1

1. It is not completely clear what the quote from the Wikipedia page (October, 2020) mentioned by OP refers to.

   • Perhaps it refers to that e.g. the Yang-Mills Grassmann-even gauge parameter $\xi^a$ in the gauge transformation $\delta A_{\mu}=D_{\mu}\xi^a$ is related to the Grassmann-odd FP ghost $c^a$ in the BRST formulation.

   • Perhaps it refers to the shifted Grassmann-grading in the antibracket/Gerstenhaber algebra used in the Batalin-Vilkovisky (BV) formulation.

2. Bosons and fermions are by definition$^1$ physical particles that follow Bose-Einstein and Fermi-Dirac statistics, respectively, but e.g. Faddeev-Popov (FP) ghosts (and many auxiliary fields in the BRST formulation) are not physical fields, and therefore not bosons or fermions. Instead they first and foremost carry Grassmann-parity, although they (in the Hamiltonian formulation) often satisfy a CCR/CAR.

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$^1$ In supermathematics, which deals with supernumbers, it is common to refer to Grassmann-even and Grassmann-odd variables as bosons and fermions, respectively, but it is strictly speaking a misnomer.

This post imported from StackExchange Physics at 2020-12-12 20:04 (UTC), posted by SE-user Qmechanic