# About Vanishing of BRST commutator in path integral

In Witten's paper Topological Quantum Field Theory, about formula (3.2), the property $\langle\{Q,\mathcal{O}\}\rangle=0$ depends on the assertion that,

$$Z_{\varepsilon}(\mathcal{O})= \int \mathcal{D}X \exp(\varepsilon Q) \left[\exp\left(-\frac{\mathcal{L'}}{e^2}\right)\mathcal{O} \right]$$

is independent on $\varepsilon$. Where does the assertion come from?

Witten clearly writes the justification just on the line above the equation (3.2): the integral is independent because the integration measure is invariant under supersymmetry – the symmetry generated by $Q$.
Just to be sure, $Q$ is the infinitesimal generator which is why $\exp(\varepsilon Q)$ is a finite transformation generated by this generator: $\varepsilon$ is the argument ("Grassmann angle") of the transformation. And $\exp(\varepsilon Q) [{\mathcal A}]$ is the transformed operator ${\mathcal A}$ by this transformation and the integral is a device that produces a scalar out of a function of $X$.
The independence on $\varepsilon$ holds because the SUSY-transformed integral of the SUSY-transformed (operator-valued) function is the same thing as the original integral of the SUSY-transformed function: I could have erased the adjective "SUSY-transformed" in front of the integral because the integration is SUSY-invariant. Because it doesn't matter whether we transform the integrand by $\exp(\varepsilon Q)$, it's the same thing as saying that the integral is independent of $\epsilon$ because it has the same value as the value for $\varepsilon=0$.
• Thanks Lubos. My puzzle is just why the integral is $Q$ invariant? In my opinion, the operator $\mathcal{O}$ is not $Q$ invariant, while the other two parts $\mathcal{D}X$ and Lagrangian are $Q$ invariant. Why the thole integral is $Q$ invariant? Maybe I make some stupid mistakes, and sorry for that. – thone Mar 26 '13 at 10:59
• Dear Craig, the only statement here is the the integration, the procedure, is invariant: $\int Q(f) = \int f$ where $Q(f)$ is the finite-SUSY-transformation-transformed $f$. So the integral of a SUSY-transformed quantity/function gives you the same result. This is the analogous statement to the statement that $\int d^4x \,f(x)$ is invariant under translations $x\to x+\Delta x$, i.e. by $f(x)\to f(x+\Delta x)$, just translations in $x$ are replaced by SUSY which are some kind of translations in $\theta$. You don't need $f(x)$ to be point-wise invariant under translations for the integral to be. – Luboš Motl Mar 27 '13 at 7:18