Batalin and Vilkovisky define $^1$ an operation they call antibracket which is

$$(F,H) = \Big(\frac{\partial_r F}{\partial \Phi^A}\Big) \Big(\frac{\partial_l H}{\partial \Phi^* _A} \Big) - \Big(\frac{\partial_r F}{\partial \Phi^* _A}\Big) \Big(\frac{\partial_l H}{\partial \Phi^A} \Big) \tag{0} $$

where index $r$ and $l$ indicate respectively right and left derivative with respect to fields $\Phi$ and antifield $\Phi^*$.

This operation has some similar properties with super-Lie commutator such as Grassmann Parity

$$ \epsilon[(F,H)] = \epsilon(F) + \epsilon(H) + 1\tag{1} $$

and symmetry

$$ (F,H) = -(-1)^{(\epsilon_F + 1)(\epsilon_H+1)}(H,F)\tag{2} .$$

My question is about these properties. I am working on the way to prove them. I tried to use that $$FH = (-1)^{\epsilon_F\epsilon_H}HF\tag{3}$$ in second property but I didn't get to the right answer.


$^1$ I. A. Batalin and G. A. Vilkovisky, Phys. Lett. B102 (1981) 27.

  1. Fun fact: Eqs. (1) & (2) can be intuitively understood via the Koszul sign rule/convention as that the comma in the antibracket (0) is Grassmann-odd/Fermionic $|,|=1$. E.g. the parity for permuting $F \leftrightarrow ","\leftrightarrow H$ in eq. (2) is then after some Boolean algebra$^1$ $$|F||H| + |F||,| +|H||,| ~=~|F||H| + |F| +|H| ~=~ (|F|+1)(|H|+1)+1. \tag{A}$$ Of course eq. (A) is only a pseudo-explanation. Let's now turn to an actual proof.

  2. Sketched proof of eq. (2): Use the following rules: $$ |\Phi^{\ast}_A|~=~|\Phi^A|+1~:=~|A|+1, \tag{B}$$ and $$ \frac{\partial^LF}{\partial z}~=~(-1)^{(|F|+1)|z|}\frac{\partial^RF}{\partial z}. \tag{C}$$ So e.g. the parity for permuting the first term $$ \frac{\partial^RF}{\partial \Phi^A}\frac{\partial^LH}{\partial \Phi^{\ast}_A} \leftrightarrow \frac{\partial^RH}{\partial \Phi^{\ast}_A}\frac{\partial^LF}{\partial \Phi^A}\tag{D}$$ in the antibracket (0) is $$ \underbrace{\overbrace{(|F|+|A|)(|H|+|A|+1)}^{\text{rule } (3)} }_{=~(|F|+|H|)|A|+|F|(|H|+1)} +\underbrace{\overbrace{(|F|+1)|A|}^{\text{rule } (C)} +\overbrace{(|H|+1)(|A|+1)}^{\text{rule } (C)} }_{=~(|F|+|H|)|A|+(|H|+1)} ~=~(|F|+1)(|H|+1).\tag{E}$$ Note that the rhs. (E) does not depend on $|A|$ (as it shouldn't!), and that the rhs. (E) is opposite the sought-for result (A). This explains the relative sign between the 2 terms in the antibracket (0). $\Box$


$^1$ In this answer we use $$|\cdot|~\equiv~ \epsilon(\cdot)~\in~ \mathbb{Z}/2\mathbb{Z}~\cong~\{0,1\} \tag{F}$$ to denote Grassmann-parity.

  • $\begingroup$ Thank you for all the help, it was really usefull. $\endgroup$ – Fernandes Feb 11 '19 at 12:16

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