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3 votes
1 answer
84 views

Confused about the relation between BRST invariant states and 'group averaging'

According to (3.6) a Dirac state $\Psi(q)$ is related to a BRST invariant state $\Upsilon$ by $$ \Psi(q)=\int d\mu ~d\Pi~ d\bar\Pi~\Upsilon(q,-i\mu,\Pi,\bar\Pi). … \tag{*}$$ But how does it follow that $\Upsilon$ is actually BRST invariant? …
dennis's user avatar
  • 786
2 votes
2 answers
73 views

Does the following limit exist in the BRST formalism?

Consider the BRST operator $\Omega$ (which has ghost number $+1$) and the gauge fermion operator $\rho$ which has ghost number $-1$. Given an exact state $|\Phi\rangle$ (i.e. …
dennis's user avatar
  • 786
0 votes
Accepted

Does the following limit exist in the BRST formalism?

If we choose $\rho = {}^*\Omega$ (the co-BRST operator), then the answer is yes! … This is because $[\Omega,{}^*\Omega]_+$ is the BRST Laplacian which is a semi-positive definite (self-adjoint) operator. …
dennis's user avatar
  • 786
1 vote
1 answer
67 views

An explicit form for the co-BRST operator?

The BRST operator $\Omega$ has ghost number $+1$ and has an expansion in powers of $c$: \begin{equation} \Omega=\sum_{n\geq 0}\left(c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\right)_W~M^{ … W.r.t. the positive-definite inner product, however, (supposing it exists), the adjoint of $\Omega$ is known as the co-BRST operator $^*\Omega$ (which I suppose has ghost number $-1$). …
dennis's user avatar
  • 786
0 votes

An explicit form for the co-BRST operator?

From https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.64.2863 it appears that, w.r.t. the positive-definite inner product, $c^\alpha$ is the adjoint of $b_\alpha$. Therefore $^*\Omega$ is \be …
dennis's user avatar
  • 786
2 votes
2 answers
150 views

Square of BRST operator

The BRST operator $\Omega$ can be expanded in powers of the ghost fields $c^{\alpha}$ and their conjugates $b_{\alpha}$ (which satisfy $\{c^\alpha,b_\beta\}=\delta^{\alpha}_{\beta}$): $$ \Omega=c^{\alpha …
dennis's user avatar
  • 786
5 votes
1 answer
206 views

BRST structure functions in gravity?

In the classical Hamiltonian BRST formalism, there arise structure functions $\Omega^{\beta_1...\beta_n}_{\alpha_1... … These are such that the BRST charge $$ \Omega=\sum_{n\geq0}c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\Omega^{\beta_1...\beta_n}_{\alpha_1... …
dennis's user avatar
  • 786
1 vote
Accepted

BRST structure functions in gravity?

Take a general system and assume it is rank one. Then $\{\Omega,\Omega\}=0$ is equivalent to \begin{align} \frac{1}{2}\{\Omega_{\alpha_1},\Omega^{\beta_1}_{\alpha_2\alpha_3}\}+\Omega^{\beta}_{\alpha_1 …
dennis's user avatar
  • 786
0 votes
Accepted

Square of BRST operator

It appears that the (quantum) expression for $\Omega$ is NOT \begin{equation} \Omega=\sum_{n\geq 0}c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}~M^{\beta_1...\beta_n}_{\alpha_1...\alpha_{n+ …
dennis's user avatar
  • 786