Square of BRST operator

Question

The BRST operator $\Omega$ can be expanded in powers of the ghost fields $c^{\alpha}$ and their conjugates $b_{\alpha}$ (which satisfy $\{c^\alpha,b_\beta\}=\delta^{\alpha}_{\beta}$): $$ \Omega=c^{\alpha_1}M_{\alpha_1}+c^{\alpha_1}c^{\alpha_2}b_{\beta_1}M^{\beta_1}_{\alpha_1\alpha_2}+c^{\alpha_1}c^{\alpha_2}c^{\alpha_3}b_{\beta_1}b_{\beta_2}M^{\beta_1\beta_2}_{\alpha_1\alpha_2\alpha_3}+O(c^4) $$ Notice that this is normal-ordered because the $c^\alpha$ are considered creation operators and the $b_\alpha$ are annihilation operators.

When we square $\Omega$ (i.e. $\Omega^2$) and normal-order it, we expect to get contributions with 'no contractions', '1 contraction', '2 contractions' and higher. However, in all the texts on this subject (see (4.3.9) of https://inspirehep.net/literature/221897 or (2.57) of https://arxiv.org/abs/hep-th/0201124) the authors discard contributions with 2 or more contractions. Why?

Answer 1

The BRST transformation rules for the ghost fields follow from requiring that $Q^2 = 0$. This implies some constraints on the coefficients $M_{\alpha_1}$, $M_{\beta_1 \alpha_1 \alpha_2}$ and $M_{\beta_1 \beta_2 \alpha_1 \alpha_2 \alpha_3}$. For example, if we consider the term $c_{\alpha_1} c_{\alpha_2} M_{\beta_1 \alpha_1 \alpha_2} b_{\beta_1}$ in $Q$, then its square will contain a term proportional to $c_{\alpha_1} c_{\alpha_2} c_{\beta_1} c_{\gamma_1} M_{\beta_2 \alpha_1 \alpha_2} M_{\gamma_2 \beta_1 \gamma_1} b_{\beta_2} b_{\gamma_2}$. This term should vanish because $Q^2 = 0$, so we must have $M_{\beta_2 \alpha_1 \alpha_2} M_{\gamma_2 \beta_1 \gamma_1} = -M_{\gamma_2 \alpha_1 \alpha_2} M_{\beta_2 \beta_1 \gamma_1}$. This is a Jacobi identity for the coefficients $M$.

Comparing your expression for $Q^2$ with equation (2.57), I think the reason why the first term in (*) is not present is because it is canceled by another term in $Q^2$. Specifically, if we look at the term $c_{\alpha_1} c_{\alpha_1} M_{\alpha_1} M_{\alpha_2}$ in $Q^2$, we see that it is symmetric under $\alpha_1 \leftrightarrow \alpha_2$, so it can be written as $\frac{1}{2} c_{\alpha_1} c_{\alpha_2} (M_{\alpha_1} M_{\alpha_2} + M_{\alpha_2} M_{\alpha_1})$. But we also have another term in $Q^2$ that comes from expanding $(c_{\alpha_3} M_{\beta _3})^2$, which is $-\frac{1}{2} c_{\alpha _3} c_{\beta _3} [M_{\beta _3}, M_{\alpha _3}]$. If we rename $\alpha _3 \rightarrow \alpha _1$ and $\beta _3 \rightarrow \alpha _2$, we get $-\frac{1}{2} c_{\alpha _1} c_{\alpha _2} [M_{\alpha _2}, M_{\alpha _1}]$, which cancels the previous term. Therefore, $Q^2$ does not contain any quadratic terms in $c$.

Answer 2

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+1}} \end{equation} but rather \begin{equation} \Omega=\sum_{n\geq 0}\left(c^{\alpha_1}...c^{\alpha_{n+1}}b_{\beta_1}...b_{\beta_n}\right)_W~M^{\beta_1...\beta_n}_{\alpha_1...\alpha_{n+1}} \tag{1} \end{equation} where $(\cdot)_W$ denotes Weyl ordering.

Computing $\{\Omega,\Omega\}=2\Omega^2$ with (1) gives \begin{align} \{\Omega,\Omega\}=&\sum_{n\geq0}\left(c^{\alpha_1}...c^{\alpha_{n+2}}b_{\beta_1}...b_{\beta_{n}}\right)_W \\ \times&\sum_{p=0}^{n}(-)^{np}\left([M^{\beta_1...\beta_{p}}_{\alpha_1...\alpha_{p+1}},M^{\beta_{p+1}...\beta_n}_{\alpha_{p+2}...\alpha_{n+2}}]+(p+1)(n-p+1)\{M^{\beta\beta_{1}...\beta_{p}}_{\alpha_{1}...\alpha_{p+2}},M^{\beta_{p+1}...\beta_{n}}_{\beta\alpha_{p+3}...\alpha_{n+2}}\}\right) \tag{2} \end{align} (This involves no more than 1 contraction between the indices, as expected.) To prove this use $$ (Q_1...Q_{k})_W=\frac{1}{2}(Q_1(Q_2...Q_k)_W+(-)^{k-1}(Q_2...Q_k)_WQ_1) \tag{3} $$ where $Q_i$ can stand for either the ghosts or their conjugates. You may find (3) in Gavazzi '89 (stated slightly wrongly):

The condition $\{\Omega,\Omega\}=0$ implies that (2), antisymmetrized among lower and upper indices, equals zero for all $n\geq 0$. Let $\stackrel{(m)}{M}\equiv M^{\beta_1...\beta_m}_{\alpha_1...\alpha_{m+1}}$. Assume $(-i)^m\stackrel{(m)}{M}$ is Hermitian for all $m<n+1$. Taking the Hermitian conjugate of (2) shows that $(-)^{n+1}\stackrel{(n+1)}{M}{}^\dagger$ satisfies the same equation as $\stackrel{(n+1)}{M}$. Since (2) is linear in $\stackrel{(n+1)}{M}$, we have that $\frac{1}{2}(\stackrel{(n+1)}{M}+(-)^{n+1}\stackrel{(n+1)}{M}{}^\dagger)$ satisfies the same equation as $\stackrel{(n+1)}{M}$. Therefore w.l.o.g. we can assert that $(-i)^{n+1}\stackrel{(n+1)}{M}$ is Hermitian. Assuming $\stackrel{(0)}{M}$ is Hermitian, we have, by the preceeding argument, that $(-i)^n\stackrel{(n)}{M}$ is Hermitian for all $n\geq0$.