Parity superselection obstructs monogamy of mutual information in free fermions

Abstract

We prove that free fermions in the spin (tensor product) factorization violate monogamy of mutual information: $I_3^{\mathrm{spin}} > 0$ for adjacent strips of width $w \le 6$ at all Fermi momenta, and for all $w$ at $z = k_F w < z^* \approx 1.329$. Many-body computation at $w=6$ via the $G$-matrix formula maps the scaling-limit function $I_3^{\mathrm{spin}}(z)$: it has a minimum of $0.100$ at $z \approx 1.5$, numerically establishing the conjecture $I_3^{\mathrm{spin}} > 0$ for all $z$ and $w$. The proof rests on an exact identity: the fermionic and spin reduced density matrices of disjoint regions $A$, $D$ separated by $B$ differ by the parity insertion $(-1)^{N_B}$ in the partial trace. A Perron--Frobenius argument proves element-wise coherence damping; for free fermions, an independent Gaussian bound gives the entropy ordering $\Delta S_{AD} \ge 0$. Exact diagonalization confirms this for interacting fermions. DMRG on the $t$-$V$ chain shows that the factorization contribution exceeds the genuine interaction contribution to $I_3$ by a factor of 8, accounting for $\sim 80\%$ of the deviation in spin-basis numerics. Strong repulsion ($K \lesssim 0.7$) restores monogamy. Conversely, $Z_2$ parity superselection enforces $I_3 \le 0$ at all fillings (proved for $w \le 3$), with the ratio of parity entropy to quantum excess approaching $2\ln 2/(3\ln(4/3)) = 1.606$. Any use of $I_3$ as a diagnostic for holographic duality, quantum chaos, or Fermi surface topology must specify the operator algebra; without this, the sign of $I_3$ is ambiguous.