Commuting observable group

The relation is not the one this record used to state. Its form was $G_k = \{P_j : [P_i,P_j] = 0\}$ — ordinary commutativity — and the cited paper shows that is the wrong condition. Two Pauli words that commute need not be simultaneously measurable by single-qubit projective measurements, and the paper's own counterexample is $[\hat{x}_1\hat{x}_2, \hat{y}_1\hat{y}_2] = 0$ while the two are not qubit-wise commuting. Their common eigenstates are entangled superpositions rather than product states, so no set of single-qubit measurements resolves both. The relation that does the work is the qubit-wise commutator of Eq. (4), which vanishes only when every single-qubit factor of $P_I$ commutes with its counterpart in $P_J$. Qubit-wise commuting implies commuting; the converse fails.

The property worth knowing about this object is that it is not a partition. Qubit-wise commutativity is reflexive and symmetric but not transitive — the paper's example is $[\hat{x}_1,\hat{y}_2]_{qw} = 0$ and $[\hat{y}_2,\hat{z}_1]_{qw} = 0$ while $[\hat{x}_1,\hat{z}_1]_{qw} \neq 0$. So it is not an equivalence relation, there are no equivalence classes, and a Hamiltonian has no unique grouping of its terms.

That is why the object is a group in the loose sense and never a partition, and it is the reason the optimization problem is a minimum clique cover over a graph rather than a sort into buckets. Represent each Pauli word as a vertex and join qubit-wise commuting pairs by an edge; a set of mutually measurable terms is then a clique, and the best grouping is a minimum cover by cliques — which the paper notes is NP-hard, hence the heuristics.

Why this record exists beside the method record, which cites the same paper. It was written to test whether it should. The answer is that the two carry different halves and neither is the other's summary: the algebraic facts above — which relation, why it is not transitive, why no unique grouping exists — are properties of the object and hold whatever procedure you use. What belongs to vqe-measurement-grouping instead is the procedure and its measured result: which heuristics were benchmarked, and that grouping reduced the operator count roughly threefold against the total number of Hamiltonian terms. An object record that repeated the threefold figure would be restating the method from the object's side, and this one does not.