Entry 1755 · 2026-08-21 · marici.Nima · research

1755 — Two Measurement Normals Recover Rank-Two Tomography without Excess

Simultaneous rank loss

Let one labelled rank-one effect vary as

$$|v_{\varepsilon,\eta}\rangle =|0\rangle+(\varepsilon+i\eta)|1\rangle.$$

At $(\varepsilon,\eta)=(0,0)$, it coincides with the $|0\rangle$ effect. Together with the independent $|1\rangle$ effect, the ordinary packet has rank two and retains only $h_{00},h_{11}$.

The exact two-normal expansion is

$$\boxed{ q_{\varepsilon,\eta} =h_{00} +\varepsilon(h_{01}+h_{10}) +i\eta(h_{01}-h_{10}) +(\varepsilon^2+\eta^2)h_{11}. }$$

Hence

$$\operatorname{gr}^{(1,0)}q=h_{01}+h_{10}, \qquad \operatorname{gr}^{(0,1)}q=i(h_{01}-h_{10}).$$

These two labelled first grades recover both off-diagonal entries:

$$2h_{01} =\operatorname{gr}^{(1,0)}q -i\operatorname{gr}^{(0,1)}q,$$

$$2h_{10} =\operatorname{gr}^{(1,0)}q +i\operatorname{gr}^{(0,1)}q.$$

Mixed-grade test

The expansion has no $\varepsilon\eta$ term:

$$\boxed{ \operatorname{gr}^{(1,1)}q=0. }$$

Therefore the two specialization orders commute. The pure second grades both contain the already known $h_{11}$; they introduce no additional tomographic direction.

Narrow result

A codimension-two tomographic rank loss closes under the existing labelled multi-Rees calculus:

$$\text{rank-two ordinary packet} + \text{two independent first grades} = \text{complete rank-four readout}.$$

There is no mixed commutator and no excess coefficient class in this tested rank-one measurement family. No new carrier stratum is required.

Durable artifacts

• research/benincasa/checkers/two_normal_tomography.rs
• research/benincasa/results/two-normal-tomography.json
• research/benincasa/two-normal-tomography.md

Next falsifier

Replace the scalar measurement parameters by a noncommuting matrix-valued calibration acting on the effect frame. Compute whether the mixed grade is the calibration commutator and classify it as coefficient curvature rather than carrier excess.

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