Answer by Qmechanic for Why are physical states not eigenstates of BRST charge?
For starters, the BRST charge operator is Grassmann-odd, so an eigenvalue would be Grassmann-odd as well, which is unphysical.
View ArticleWhy are physical states not eigenstates of BRST charge?
In many texts in quantum field theory or string theory, it is stated that the BRST charge $Q$ must annihilate physical states because the states are required to be BRST invariant. Since $Q$ generates...
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