very strong constraint on trial states. If we take two trial quantum states for $\rho_1$ and $\rho_2$, we cannot simply take their tensor product and use it as a trial state for $\rho_1+\rho_2$. The tensor product does not have the fermionic symmetry, and if we anti-symmetrize it the density is not equal to $\rho_1+\rho_2$ anymore.

For this reason, we will use an incomplete partition of unity with holes, in order to make sure that the local quantum states are not overlapping. Since we want to get the exact density, holes are however in principle not allowed. Instead of filling the holes with electrons, our idea is to rather average over all the possible rotations, translations and dilations of the partition of unity, which will make the holes disappear in average. All the arguments of this section apply the same to a tiling made of cubes, but for a better matching with the lower bound we will consider a tiling made of tetrahedra. Our lower bound relies on the Graf-Schenker inequality [@GraSch-95] which requires the use of tetrahedra.

Let us consider the unit cube $C_1=(-1/2,1/2)^3$, which is the union of 24 disjoint identical tetrahedra ${\mathbf{\Delta}}_1,...,{\mathbf{\Delta}}_{24}$, all of volume $1/24$. Since the 