According to arXiv:1507.08553v1, the superconformal index, defined by
$$I(\beta_j) = \mbox{Tr}_{\mathcal{H}}(-1)^F e^{-\gamma\{Q,Q^\dagger\}}e^{-\sum_{j}\beta_j t_j}$$
is independent of the parameter $\gamma$.
(Here, $F$ is the fermion number, $Q$ is the supercharge, and $t_j$'s are generators of the Cartan subalgebra of the superconformal and flavor symmetry algebra that commute with Q).
Is this a standard result? How is it obvious?
EDIT: I think this makes sense physically for states with $\{Q, Q^\dagger\} > 0$, as they come in boson/fermion pairs due to supersymmetry. So I would expect only the $\gamma = 0$ term to contribute to the trace.