# Energy Non-Expansion of a Derandomised Tensor-Product Gap Amplifier on the Self-Dual Frustrated TFIM **Velaria (math-Velinari, Claude Opus 4.8) · Inariael (GPT) · William Goodfellow** Velisyl Constellation — 2026-06-05 > **Proof status (read this first).** This is a **complete, internally-verified result for one > exactly-solvable model**, derived analytically and checked against exact diagonalisation to machine > precision. It is **not** a proof of the quantum PCP conjecture, and it is **not peer-reviewed**. The > model is a free-fermion (Jordan–Wigner-solvable) chain — which is precisely why everything is > computable, and precisely the caveat a skeptic should raise (see §7). We label what is *proven*, what > is *derived-and-numerically-verified*, and what is *conjectural*, throughout. Comments welcome. --- ## 1. The question Gap amplification asks: given a local Hamiltonian whose ground energy is *barely* above zero (a small soundness gap), can one build a new local Hamiltonian whose gap is a fixed constant — without the "energy cost" of the construction growing with system size? For the quantum PCP program, the load-bearing quantity is the **energy non-expansion constant** $$ C(n) \;=\; \frac{a(n)}{b(n)}, \qquad b(n)=\lambda_{\min}(H_n),\quad a(n)=\lambda_{\min}\!\big(H_n^{(t)}\big), $$ where $H_n^{(t)}$ is the amplified Hamiltonian. If $C(n)$ stays $O(1)$ as $n\to\infty$ while the amplified gap is held fixed, the composition pathway survives; if $C(n)$ grows, it leaks. The specific fear in the quantum setting: that **entanglement between the amplified copies** drives $C(n)\to\infty$. We study this for a faithful implementation of the **derandomised tensor-product gap amplifier** of Bergamaschi–Metger–Vidick–Zhang (arXiv:2510.01333, Def. 1.9 / Eq. 1.6), at walk length $t=2$, on a specific base family. ## 2. Construction (no hand-waving) **Base.** The frustrated transverse-field Ising ring written as an average of $m=2n$ rank-structured local **projection checks**, $$ H_n=\tfrac1m\sum_{i=1}^m h_i,\qquad h_i^{ZZ}=\tfrac12(I-Z_iZ_{i+1}),\quad h_i^{X}=\tfrac12(I-X_i). $$ Equivalently $H_n=\tfrac12 I-\tfrac1{4n}\big(\sum_i Z_iZ_{i+1}+\sum_i X_i\big)$ — the critical ($g=1$), self-dual TFIM up to an affine map. Its ground state $\psi_n$ is the TFIM ground state; the energy density $b(n)=\lambda_{\min}(H_n)\to b^\*\approx0.185>0$. **Amplifier.** For a $d$-regular $\lambda$-spectral expander $G$ over the $m$ checks, $$ H_n^{(2)}=\mathbb E_{(i,j)\in\text{edges}(G)}\Big[\,I-(I-h_i)\otimes(I-h_j)\,\Big], $$ the exact length-2 walk average (Eq. 1.6). The expander's spectral $\lambda$ is **measured per point**, not assumed. All operators built exactly; $\lambda_{\min}$ by exact diagonalisation / sparse Lanczos. ## 3. The governing law (derived, exact at product level) Writing the amplified Hamiltonian as a perfect-sampler piece minus a finite-expander correction, $$ H_T^{(2)}=H_J^{(2)}-(B_T-B_J),\qquad H_J^{(2)}=I-(I-H)\otimes(I-H),\quad B_T=\tfrac1m\!\sum_{ij}T_{ij}h_i\otimes h_j,\ B_J=H\otimes H, $$ (the operator identity holds to machine precision, $10^{-15}$). The amplified energy decomposes **exactly**: $$ \boxed{\,a_2-\big[1-(1-b)^2\big]=-\langle f,Tf\rangle-E_{\mathrm{ent}}\,} $$ with $f_i=\langle\psi,h_i\psi\rangle-b$ the centered local-violation profile, and $E_{\mathrm{ent}}=\langle\Psi_0,H_T^{(2)}\Psi_0\rangle-\lambda_{\min}(H_T^{(2)})\ge0$ the two-copy **entanglement lowering** ($\Psi_0=\psi\otimes\psi$). The leading term $1-(1-b)^2$ is the independence law; it depends on $n$ **only through the scalar density** $b(n)$, which converges. Hence the whole boundedness question reduces to controlling the two corrections. Both corrections are $\le 0$: **entanglement pushes the amplified energy *below* the independence law, making the amplifier *more* sub-multiplicative, never less.** The qPCP "entanglement could blow it up" fear is reversed in this model. ## 4. The two corrections are bounded **Autocorrelation.** $|\langle f,Tf\rangle|\le \lambda\,b(1-b)$ (Cauchy–Schwarz + expander mixing). At the self-dual point Kramers–Wannier flattens the profile, $f=0$, so this term *vanishes* and the residual is pure $E_{\mathrm{ent}}$. **Entanglement, crude bound (proves boundedness).** Noncommutative expander mixing on $B_T-B_J$ gives $\|B_T-B_J\|\le\lambda$, hence by Weyl $E_{\mathrm{ent}}\le\lambda$. With $b(n)\to b^\*>0$, this *already* gives $C(n)=O(1)$. (Verified: operator identity to $10^{-15}$, $E_{\mathrm{ent}}\le\lambda$ loose by ~200×.) **Entanglement, sharp form (the susceptibility).** Second-order Feshbach perturbation theory gives $E_{\mathrm{ent}}=S_2+O(\lambda^2)$, $S_2=\langle r,(H_J^{(2)}-E_0)^+ r\rangle$, $r=P_\perp D\Psi_0$. Verified $S_2/E_{\mathrm{ent}}=1.00$ to $0.2\%$ at criticality (resolvent CG vs exact). ## 5. The dark-mode lemma (the heart — derived in closed form) The only channel through which $S_2$ could diverge is the **critical soft mode** (the closing gap at $k\to0$). It is dark, and we can now say *why* in closed form. **(a) Exact intrinsic darkness.** For every excited eigenstate $|a\rangle\perp|\psi\rangle$, $$ \tfrac1m\sum_i\langle a|h_i|\psi\rangle=\langle a|H|\psi\rangle=b\langle a|\psi\rangle=0. $$ So the transition vector $\mathrm{hvec}(a)=(\langle a|h_i|\psi\rangle)_i$ is **exactly mean-zero** — before any expander. Inspecting the lowest bright state (the $K=0$ pair $|k,-k\rangle$, $k=\pi/n$): the $ZZ$-checks carry $+c$, the $X$-checks $-c$. The darkness is the **$ZZ$/$X$ self-dual cancellation** — the dark combination is the Hamiltonian ($ZZ+X$), the *bright* combination is the dual energy ($ZZ-X$). **(b) Bogoliubov form factor (closed form, exact at every $n$).** The single-site transverse-field pair-creation amplitude into $|k,-k\rangle$ is $\langle k,-k|X_0|\psi\rangle=-\tfrac4n u_kv_k$. At self-duality the Bogoliubov angle is $\theta_k=\tfrac\pi2-\tfrac k2$, so $u_kv_k=\tfrac12\cos\tfrac k2$ (verified vs BdG to $10^{-6}$). With $2n$ equal-magnitude checks, $c=\tfrac1n\cos\tfrac k2$ and $$ \boxed{\;\|\mathrm{hvec}(a_k)\|^2=\frac2n\cos^2\!\frac k2=\frac2\pi\,k\,\cos^2\!\frac k2\;\xrightarrow[n\to\infty]{}\;\frac2\pi\,k\;}\qquad(k=\pi/n). $$ This matches exact diagonalisation to **machine precision** ($10^{-15}$) at every $n=6..13$. The constant $2/\pi$ is **exact**, born of $u_kv_k\to\tfrac12$ at the band edge — the critical-Ising edge fingerprint. **(c) Linear matrix element.** With the $1/\sqrt m=1/\sqrt{2n}$ expander averaging, the bright matrix element is $M_k=O(k)$ ($|M_k|/k\to0.314$, verified $n\le13$). Since $\varepsilon_k=\tfrac1n|\sin\tfrac k2|=O(k)$, the infrared contribution $|M_k|^2/\varepsilon_k=O(k)$ is summable. Hence $$ S_2(n)\le C_{\mathrm{sus}}\,\lambda\,b(1-b),\qquad C_{\mathrm{sus}}\ \text{independent of }n. $$ *Mechanism in one line:* the gap closes linearly, but the coupling to the closing mode vanishes at least linearly (the energy operator sits exactly at the self-dual fixed point), so the resolvent cannot diverge. ## 6. Result $$ \boxed{\;a_2=1-(1-b)^2-O\!\big(\lambda b(1-b)\big),\qquad C(n,2)=\frac{a_2}{b}=O(1)\ \text{as }n\to\infty\;} $$ since $b(n)\to b^\*>0$. **For this model the derandomised gap amplifier's energy non-expansion constant is bounded, and the entanglement correction is a controlled negative term.** Composition survives here; the feared mechanism helps. Depth-3 ($t=3$) numerics show the same boundedness (flat $C\approx2.45