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  Superalgebra and BRST supersymmetry 1: odd vs even

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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
asked Oct 25, 2020 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]

1 Answer

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  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.

--

$^1$ 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
answered Oct 25, 2020 by Qmechanic (3,110 points) [ no revision ]

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