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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? …
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. …
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. …
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 $-1$). …
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 …
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 $\{c^\alpha,b_\beta\}=\delta^{\alpha}_{\beta}$):
$$
\Omega=c^{\alpha …
5
votes
1
answer
206
views
BRST structure functions in gravity?
In the classical Hamiltonian BRST formalism, there arise structure functions $\Omega^{\beta_1...\beta_n}_{\alpha_1... … These are such that the BRST charge
$$
\Omega=\sum_{n\geq0}c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\Omega^{\beta_1...\beta_n}_{\alpha_1... …
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 …
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+ …