Batalin–Vilkovisky formalism

Remarks

The operator $\Delta$ satisfies $\Delta^2 = 0$, since the derivatives in the even variables commute and those in the odd variables anticommute. With the bracket $(\cdot, \cdot)$, the algebra $\mathcal{A}$ is a Batalin–Vilkovisky algebra. The equivalence of the two forms of the master equation follows from $\Delta e^{f} = \left( \Delta f + \frac12 (f, f) \right) e^{f}$ for even $f$, applied to $f = iS/\hbar$. At order $\hbar^0$ the master equation gives the classical master equation $(S_0, S_0) = 0$.

Example: if $S = S_0(\phi)$ does not depend on the antifields, then $\Delta S = 0$ and $(S, S) = 0$, so $S$ satisfies the quantum master equation; this is a theory without gauge symmetry.

Non-example: for $n = 2$ and $S = \phi^1 \phi^*_1 \phi^*_2$, one has $S^2 = 0$ and $S\, \Delta S = 0$, so $(S, S) = 0$ and the classical master equation holds. But $\Delta S = \frac{\partial}{\partial \phi^1}(\phi^1 \phi^*_2) = \phi^*_2 \ne 0$, so $S$ does not satisfy the quantum master equation.

In the gauge theory case the antifield of each field enters $S$ through the BRST transformation: to first order in antifields, $S = S_0 + \sum_i \phi^*_i\, s\phi^i + \cdots$, where $s$ is the BRST differential. Gauge fixing restricts the integral of $e^{iS/\hbar}$ to a Lagrangian submanifold for the odd symplectic structure. The master equation makes the result independent of small deformations of this submanifold.