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Results tagged with brst
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user 242254
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BRST Symmetry and Single Particle States
According to Weinberg, the BRST charge operator obeys the commutation relation $[Q, \alpha_{\mu}^{\dagger}(\vec{k})]=k_{\mu}c^{\dagger}(\vec{k})$, where $c^{\dagger}(\vec{k})$ is the creation operator … Then, acting on the above-mentioned state with the BRST charge operator yields
$$Q|r,\vec{k};\psi\rangle=-\epsilon^{\mu}_r(\vec{k})k_{\mu}
c^{\dagger}(\vec{k})|\psi\rangle=
-k\cdot\epsilon_r(\vec{k})|g …
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