Quantum master equation in the Batalin-Vilkovisky formalism I am reading the Section 15.9 of Weinberg's book "The Quantum Theory of Fields, vol. 2". Under a shift $\delta\Psi[\chi]$ in $\Psi[\chi]$, we have
$$
    \begin{split}
     \delta Z&=i\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big)
      \left(\frac{\delta_RS[\chi,\chi^{\ddagger}]}{\delta\chi_n^{\ddagger}}
      \right)_{\chi^{\ddagger}=\delta\Psi/\delta\chi}\left(
      \frac{\delta(\delta\Psi[\chi])}{\delta\chi^n}\right) \\
     &=i\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big)\left\{\frac{\delta_L}
      {\delta\chi^n}\left(\frac{\delta_RS}{\delta\chi_n^{\ddagger}}
      \delta\Psi\right)-\frac{\delta_R}{\delta\chi_n^{\ddagger}}
      \frac{\delta_LS}{\delta\chi^n}\delta\Psi\right\}_{\chi^{\ddagger}
      =\delta\Psi/\delta\chi} \\
     &=\int[d\chi]\,\exp\big(iI_{\Psi}[\chi]\big)\left\{
      \frac{\delta_RS[\chi,\chi^{\ddagger}]}{\delta\chi_n^{\ddagger}}
      \frac{\delta_LI_{\Psi}[\chi]}{\delta\chi^n}-i\Delta 
      S[\chi,\chi^{\ddagger}]\right\}_{\chi^{\ddagger}=\delta\Psi/
      \delta\chi}\delta\Psi[\chi]  
    \end{split} 
$$
The last line is exactly the same to Eq. (15.9.33). Referring to the definition of antibracket 
$$
  (F,G)=\frac{\delta_RF}{\delta\chi^n}\frac{\delta_LG}{\delta\chi_n^{\ddagger}}-
  \frac{\delta_RF}{\delta\chi_n^{\ddagger}}\frac{\delta_LG}{\delta\chi^n}
$$
we can see that the quantum master equation reads
$$
 -(S,S)-2i\Delta S=0 
$$
which has an extra minus sign. I are not sure whether this is a typo or not.
Could someone help me to check this derivation? 
Besides, I am also confused by $\delta_L$ and $\delta_R$. Any clarifications 
will be appreciated.
Many thanks in advance!
 A: I) Let us first clarify the left and right derivatives. Left derivatives is explained between eq. (15.8.9) and (15.8.10) in Ref. 1. A left derivative means a derivative that acts from the left. E.g. if $F= \chi G$, where $G$ does not depend on $\chi$, then  $\frac{\delta_LF}{\delta\chi}=G$. Similarly, a right derivative acts from the right. E.g. if $F= G\chi$, then $\frac{\delta_RF}{\delta\chi}=G$. One may then work out that left and right derivatives are equal up to a sign factor:
$$\tag{A} \frac{\delta_LF}{\delta\chi}~=~(-1)^{(|F|+1)|\chi|}\frac{\delta_RF}{\delta\chi}.$$
Here $|F|$ denotes the Grassmann parity of $F$. Note in particular that the left and right derivative of the gauge fermion $\Psi[\chi]$ are the same:
$$\tag{B} \frac{\delta_L\Psi}{\delta\chi}~=~\frac{\delta_R\Psi}{\delta\chi}, \qquad |\Psi|~=~1.
$$
II) Now let us consider the Batalin-Vilkovisky formalism. We start with the full quantum master action $S[\chi,\chi^{\ddagger}]$, which depends on fields $\chi^n$ and antifields $\chi^{\ddagger}_n$.
The odd Laplacian is originally defined in eq. (16b) of Ref. 2 as
$$\tag{16b} \Delta_{BV}~:=~ \frac{\delta_R}{\delta\chi^n}\frac{\delta_L}{\delta\chi^{\ddagger}_n}. $$ 
Ref. 1 defines (wrongly) the odd Laplacian as
$$\tag{15.9.34} \Delta_{SW}~:=~ \frac{\delta_R}{\delta\chi^{\ddagger}_n}\frac{\delta_L}{\delta\chi^n}. $$ 
One may show that the two definitions (16b) and (15.9.34) are related as
$$\tag{C} \Delta_{BV}F~=~(-1)^{|F|+1}\Delta_{SW}F. $$
In particular the two definitions (16b) and (15.9.34) differ by a sign 
$$\tag{D} \Delta_{BV} S~=~-\Delta_{SW} S, \qquad |S|~=~0, $$
when applied to the action $S$, which is Grassmann-even $|S|=0$.
III) The Quantum Master Equation (QME) reads in Ref. 2
$$\tag{16a} \frac{1}{2}(S,S)~=~i\hbar\Delta_{BV} S,$$
while the QME in Ref. 1 reads
$$\tag{15.9.35} \frac{1}{2}(S,S)~=~i\hbar\Delta_{SW} S.$$
So OP is right. Eqs. (15.9.34) and (15.9.35) are mutually inconsistent. There is a wrong sign in Ref. 1 in either eq. (15.9.34) or eq. (15.9.35).
References:


*

*S. Weinberg, The Quantum Theory of Fields, Vol. 2, 1996.

*I.A. Batalin and G.A. Vilkovisky, Gauge Algebra and Quantization, Phys. Lett. B 102 (1981) 27–31. 
