Let be an arbitrary assignment of numerical outcomes to a 3-design reference device in quantum theory. Let be the corresponding operator. The resolution of the identity tells us that An unbiased 3-design has the remarkable property that A little algebra later where , and .
We can further simplify this formula if we assume . By the 2-design property, , so that in particular, from which we obtain relating the second-moments of, on the one hand, a von Neumann measurement of , and on the other, a reference device measurement of it—which, flipped around, is Thus if we consider the inequality it is saturated iff . Since is diagonalizable, this implies that . Now how can it be that a quantum mechanical observable has zero variance? Suppose and can be simultaneously diagonalized, that is, they commute . Then we can write , for some common set of projectors . The interpretation is that are the numerical outcomes of the von Neumann measurement, and are the probabilities for each outcome, given the preparation . The variance will be zero if and have disjoint support. In fact, then . Can this happen in any other way? Thus if is a pure state, may have at most rank-, whereas if is full rank, the only observable which gives equality is 0. Thus for example, when is rank-1, will saturate the lower bound.
Finally, we observe that since the first moment with respect to a von Neumann measurement and the first moment with respect to the reference device agree, saturation of the inequality implies that , or which simplifies to , or .