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?


1 Answer 1


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 $\varepsilon=0$.

  • $\begingroup$ 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. $\endgroup$
    – thone
    Mar 26, 2013 at 10:59
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 27, 2013 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.