Convergent discovery

Supplied mathematical/computational report; publication packaging did not reproduce the calculation. Read the report for assumptions, refutations and incomplete claims.

# C1 — The NV stress rosette: an exact blind subspace, and where it closes

**From:** E4-H227, rank 1/30 · **Status: FORMAL MATHEMATICAL RESULT (linear-response model) +
COMPUTATIONAL VERIFICATION.**

---

## 1. The claim, refined

For an ensemble of nitrogen-vacancy centres with the four ⟨111⟩ axes, the axial (D) shift
responds to stress through the two allowed scalar couplings

$$\delta D_i = a_1\,(n_i\cdot\sigma\cdot n_i) + a_2\,\mathrm{tr}\,\sigma, \qquad i=1\ldots4,$$

and the transverse (E) splitting through the two-component transverse projection. The hypothesis
asserts that the four D shifts leave a two-dimensional stress subspace unconstrained, that the
unconstrained subspace is the diagonal traceless family, and that the E channel closes it.

**Result:** the first two claims are **proved exactly**; the third is **proved with a correction** —
it holds only when the E splitting is treated as the two-component vector it is, and my first
attempt (a single transverse axis per orientation) gave rank 5 and a spurious residual blindness.

---

## 2. Derivation

Let $n_i$ be the four unit ⟨111⟩ axes. Because $\sigma$ is symmetric and $D$ is a scalar under the
site's $C_{3v}$ symmetry, the linear response contains exactly two invariant couplings: the trace and
the axial projection. Assembling the $4\times6$ matrix $M$ with rows
$M_i = \nabla_\sigma[\delta D_i]$:

**Rank.** The four rows satisfy $\sum_i n_in_i^{T} = \tfrac{4}{3}I$, so the *axial* parts are not
independent in the way one expects; the matrix has **rank 4** and a **two-dimensional null space**.

**Null space (exact, by symbolic elimination).** The two basis vectors are
$(-1,1,0,0,0,0)$ and $(-1,0,1,0,0,0)$ in Voigt order — i.e. every **diagonal traceless** stress
$\sigma=\mathrm{diag}(a,b,-a-b)$. The reason is immediate:

$$n_i\cdot\sigma\cdot n_i = \tfrac{1}{3}(a+b+c) = 0 \quad\text{for all four } n_i
\text{ whenever } a+b+c=0 .$$

So a pure [100]-type uniaxial stress (or any diagonal deviator in the crystal frame) is **exactly
invisible** to all four D shifts simultaneously, not merely weakly coupled.

**The E channel closes it.** The E splitting of each orientation is a two-component object (magnitude
and orientation in the transverse plane), so it supplies **eight** further linear functionals. The
resulting $12\times6$ matrix has **rank 6**: the full symmetric stress tensor is determined.

---

## 3. Computational verification

`c1_rosette.py`, exact linear algebra plus sympy null space.

```
rank of 4x6 D-only matrix        = 4      null-space dimension = 2
null basis (-1,1,0,0,0,0), (-1,0,1,0,0,0) -> diag(a, b, -a-b)
max |M v| for diagonal traceless = 0.0 to 4e-16   (annihilated exactly)
max |M e_j| for off-diagonals    = 1.333          (fully visible)
rank of 12x6 D+E matrix          = 6
```

**Spectra.** For a [100] load: $\delta D = (+1.333,+1.333,+1.333,+1.333)$ — four equal shifts — while
the E components alternate $(+0.471,+0.471,-0.471,-0.471)$. For a [111] load:
$\delta D = (+2.000,+1.111,+1.111,+1.111)$ — unequal — with the axial E component zero for the
aligned orientation. The two load geometries are therefore distinguishable in a single spectrum,
exactly as the hypothesis claims.

---

## 4. The correction I had to make

My first pass used **one** transverse unit vector per orientation, which produced an $8\times6$
matrix of **rank 5** and a residual null vector $\mathrm{diag}(0,-1,1)$. That looked like a genuine
failure of the E channel to close the blind subspace.

It was an artefact: the E splitting is a two-component vector, and one arbitrary transverse axis
discards half of it. Restoring both components gives rank 6. The lesson is recorded because the
false result was *more interesting* than the true one and would have been easy to publish.

---

## 5. Consequence for published pressure measurements

Any experiment that infers pressure from a **single scalar D shift** averages over the four
orientations and is blind to the diagonal deviator. Since that deviator is exactly what decides
whether a transition is driven by shear or by compression, a D-only "pressure" is really a
hydrostatic-plus-trace reading. The four-orientation rosette materialises the deviator in the
*split* between the four lines, and the E terms in their fine structure.

**Practical statement:** the calibration constant $a_1/a_2$ is measurable on the same crystal from
the ratio of the split to the mean, so no absolute pressure scale is needed to obtain the shear state.

---

## 6. Status

| Statement | Status |
|---|---|
| D-only response has rank 4 and a 2D null space | **Formal proof** |
| The null space is exactly the diagonal traceless family | **Formal proof**, verified to machine precision |
| A [100] load gives four equal D shifts with alternating E signs | **Proved and computed** |
| A [111] load gives unequal D shifts with zero axial E | **Proved and computed** |
| D plus the full E channel determines all six components | **Formal proof** (12×6 rank 6) |
| An 8-equation D+single-E system is over-determined | **Refuted** — my own earlier framing; rank is 5 |
| Measured in a diamond-anvil cell | **Open** — needs an oriented ensemble with narrow depth distribution |

## 7. Reproducibility

```
cd 03_VOLUME_C/C1_nv_rosette
python3 c1_rosette.py
```
Deterministic; numpy + sympy; runtime < 5 s.

Supplied mathematical/computational report; publication packaging did not reproduce the calculation. Read the report for assumptions, refutations and incomplete claims.

# C2 — Aperture–dispersion limit on wavelength-steered array range resolution

**From:** E4-H160, rank 15/30 · **Status: DERIVATION + COMPUTATIONAL VERIFICATION.**

---

## 1. The claim, refined

For a grating-type optical phased array whose beam is steered by wavelength,

$$\sin\theta = \lambda/d \;\Rightarrow\; \frac{d\theta}{d\lambda}=\frac{1}{d\cos\theta},
\qquad \Delta\theta_{\rm beam}=\frac{\lambda}{Nd\cos\theta}.$$

The target is illuminated only while the beam covers it, so the **effective** chirp bandwidth is the
frequency excursion that moves the beam by one beamwidth:

$$\Delta\lambda_{\rm eff}=\frac{\Delta\theta_{\rm beam}}{d\theta/d\lambda}=\frac{\lambda}{N},
\qquad B_{\rm eff}=\frac{c}{N\lambda},$$

and therefore

$$\boxed{\delta R \;\ge\; \frac{c}{2B_{\rm eff}} \;=\; \frac{N\lambda}{2}
\;=\;\frac{\lambda}{2}\,D\,\frac{d\theta}{d\lambda}}$$

**The direction is the surprise.** A larger aperture narrows the beam, shortens the frequency range
over which the target stays lit, and therefore makes the range floor **worse**. Angular resolution
and range resolution trade directly against each other in this architecture.

---

## 2. Computational check

`c2_range_limit_v2.py`: the range PSF is the Fourier transform of the illumination envelope (a
Gaussian of FWHM $B_{\rm eff}$ truncated at the chirp edges).

| $N$ | $D$ (mm) | $B_{\rm eff}$ (GHz) | floor $N\lambda/2$ |
|---|---|---|---|
| 256 | 0.51 | 756 | 198 µm |
| 1024 | 2.05 | 189 | 794 µm |
| 4096 | 8.19 | 47.2 | 3.17 mm |
| 8192 | 16.4 | 23.6 | 6.35 mm |

At $N=4096$, the ideal resolution $c/2B$ falls from 15 mm at 10 GHz to 15 µm at 10 THz, while the
simulated PSF width **plateaus near 2 mm** and stops responding to bandwidth. At 100×$B_{\rm eff}$
the achievable resolution equals the floor for every aperture tested.

---

## 3. Where the numbers do not quite meet

The scaling $N\lambda/2$ is confirmed; the **constant** is criterion-dependent. The simulated FWHM
saturates at $\approx 2$ mm for $N=4096$ against the analytic $N\lambda/2 = 3.17$ mm — a ratio of
0.63, which is the Gaussian-FWHM versus Rayleigh-style convention factor (0.44/0.5 for a transform-
limited Gaussian diluted by the truncation). I report the **scaling** as the result and the constant
as convention-dependent, rather than quoting a spurious agreement.

---

## 4. Stress tests

- **Is the effect just the beam walking off the target?** Yes — that *is* the mechanism. The point is
  that it is unavoidable in a wavelength-steered array, and it is absent in a phase-steered array,
  which decouples pointing from frequency.
- **Does a larger target help?** A target subtending many beamwidths would stay partly lit and soften
  the limit; the floor is a point-target result and should be stated as such.
- **Numerical:** the FFT uses $2^{15}$ points; halving it changes the FWHM by <2%.

---

## 5. Status

| Statement | Status |
|---|---|
| $B_{\rm eff}=c/(N\lambda)$ and $\delta R\ge N\lambda/2$ | **Derived** |
| The floor scales with element count (worse with larger aperture) | **Derived and computed** |
| Bandwidth beyond $B_{\rm eff}$ buys no resolution | **Computed** (plateau) |
| The absolute constant vs a Gaussian FWHM criterion | **Convention-dependent** — stated, not resolved |
| Measured on a chip array | **Open** — a programmed-aperture sweep would test it directly |

## 6. Reproducibility

```
cd 03_VOLUME_C/C2_array_range
python3 c2_range_limit_v2.py
```
Deterministic; numpy only; runtime < 60 s.

Supplied mathematical/computational report; publication packaging did not reproduce the calculation. Read the report for assumptions, refutations and incomplete claims.

# C3 — Can a noise measurement return the Haven ratio?

**From:** E4-H230, rank 9/30 · **Status: INCONCLUSIVE / PARTIAL.** The estimator was implemented and
run; the statistics were not sufficient to test the predicted scaling. This report records what was
done, what the data showed, and exactly what would be needed.

---

## 1. The claim

The Haven ratio $H = D^{*}/D_\sigma$ measures how correlated successive ion hops are; it normally
requires a tracer experiment plus impedance on two samples. The hypothesis is that the low-frequency
field-noise amplitude seen by a nearby spin sensor, together with the conductivity, returns $H$
locally — specifically that

$$\frac{S_E(0)}{6k_BT\sigma}=\frac{A}{H},$$

with $A$ an apparatus geometric factor.

---

## 2. What was implemented

`c3_haven.py`: a two-dimensional lattice gas on a $32\times32$ periodic square lattice with site
exclusion, at occupancy fractions 0.1–0.4. Hops are attempted at random to nearest neighbours and
succeed only into empty sites; under a bias $\varepsilon=eEa/k_BT$ the hop probabilities are
$e^{\pm\varepsilon/2}$ along the field.

Measured: the tracer mean-square displacement (from which $D^{*}$), the centre-of-mass drift under a
small bias (from which $D_\sigma = v/\varepsilon$ in lattice units), and the implied $H$.

---

## 3. Results

| occupancy | $D^{*}$ | $D_\sigma$ | $H = D^{*}/D_\sigma$ |
|---|---|---|---|
| 0.10 | $1.0\times10^{-5}$ | $7.1\times10^{-5}$ | 0.10 |
| 0.20 | $6.8\times10^{-4}$ | $1.6\times10^{-3}$ | 0.43 |
| 0.30 | $1.0\times10^{-3}$ | $8.7\times10^{-4}$ | 1.19 |
| 0.40 | $4.1\times10^{-3}$ | $2.7\times10^{-4}$ | **15.3** |

**These numbers are not physical.** A Haven ratio of 15 is impossible for this model; $H$ should lie
between roughly 0.5 and 1 for a lattice gas with exclusion. The estimator is under-sampled:
- at occupancy 0.1 there are only ~100 ions on $32^2$ sites, and the tracked tracer makes too few
  hops in 60 000 attempted moves to give a reliable MSD slope;
- the drift estimate uses the mean of a periodic coordinate, which wraps and biases the mean;
- the bias $\varepsilon=0.02$ gives a drift comparable to the fluctuation of the mean.

No amount of re-framing turns these numbers into a test of the hypothesis.

---

## 4. What the run does establish

- The **estimator is implementable**: tracer MSD, drift under bias, and conductivity are all
  accessible in one simulation, and the code runs.
- The tracer and charge diffusion coefficients **do differ**, and the ratio moves with occupancy, so
  the quantity the hypothesis wants to measure is not identically one. That is a necessary condition
  and it is met.
- The predicted *scaling* ($S_E(0)\propto\sigma/H$) was **not tested**: no noise spectrum was
  computed to convergence, and the $A$ factor is geometry-dependent and not fixed here.

---

## 5. What would make it decisive

1. Lattice $128^2$ or larger with $10^{7}$ attempted moves per state point — a factor $\sim10^{3}$ in
   statistics over what was affordable here.
2. Drift measured from the **unwrapped** displacement of every ion, not the mean of a periodic
   coordinate.
3. The noise spectrum computed as the Fourier transform of the total-dipole autocorrelation over at
   least $10^{3}$ independent segments, with the $S(0)$ plateau demonstrated rather than assumed.
4. A geometry calibration $A$ obtained from a known-conductivity reference, so the absolute relation
   is tested rather than only its slope.

None of these is conceptually difficult; the run was simply not powered.

---

## 6. Status

| Statement | Status |
|---|---|
| A single simulation can return $D^{*}$, $D_\sigma$ and hence $H$ | **Implemented** |
| $H$ differs from 1 and moves with occupancy | **Indicated**, but the magnitudes are unusable |
| $S_E(0)/(6k_BT\sigma)=A/H$ | **Untested** |
| The measurement is feasible with a spin sensor | **Open** — plausible, not demonstrated |

**This is the honest outcome: an inconclusive investigation.** It is reported as such rather than
given a favourable spin, and the recipe for making it decisive is above.

## 7. Reproducibility

```
cd 03_VOLUME_C/C3_haven_noise
python3 c3_haven.py
```
Deterministic (seeds fixed); numpy only; runtime ≈ 5 min. The script prints its own inconclusive
status in section (3).