C042 derivation

Vectorized iteration

One cycle is the linear superoperator $S$ acting on $\mathrm{vec}(\rho)$ (column stacking), established in C040. Iterating $N$ cycles is therefore $$\mathrm{vec}(\rho_N) = S^N\,\mathrm{vec}(\rho_0), \qquad \rho_N = \mathrm{unvec}\!\left(S^N\,\mathrm{vec}(\rho_0)\right).$$

Spectral decomposition

$S$ is diagonalizable (generically) with eigenvalues $\lambda_k$, right eigenvectors $|r_k\rangle\!\rangle$ and dual left eigenvectors $\langle\!\langle l_k|$ normalized so $\langle!\langle l_j| r_k\rangle!\rangle =\delta_{jk}$: $$S = \sum_k \lambda_k\, |r_k\rangle\!\rangle\langle\!\langle l_k|, \qquad S^N = \sum_k \lambda_k^{N}\, |r_k\rangle\!\rangle\langle\!\langle l_k|.$$ For a trace-preserving, relaxing map there is a unique unit eigenvalue $\lambda_{ss}=1$ whose right eigenvector is the steady state $\mathrm{vec}(\rho_{ss})$; all other $|\lambda_k|<1$. Hence $$\rho_N = \rho_{ss} + \sum_{k:\,|\lambda_k|<1} \lambda_k^{N}\, c_k\, R_k, \qquad c_k = \langle\!\langle l_k|\,\mathrm{vec}(\rho_0)\rangle\!\rangle,$$ with $R_k=\mathrm{unvec}(|r_k\rangle\!\rangle)$.

Fidelity decay

The singlet fidelity is the linear functional $F(N)=\langle\Psi^-|\rho_N|\Psi^-\rangle$. With $\rho_{ss}=|\Psi^-\rangle \langle\Psi^-|$ (the parameter-independent fixed point, C008), $F_\infty=1$, so $$\epsilon(N)=1-F(N) = -\sum_{k:\,|\lambda_k|<1} \lambda_k^{N}\, c_k\, \langle\Psi^-|R_k|\Psi^-\rangle = \sum_{k} a_k\,\lambda_k^{N}.$$ Order the sub-unit eigenvalues by modulus; the largest is $\lambda_{\max}$. For $N\gg 1$ the term with the largest modulus dominates: $$\epsilon(N) \simeq C_0\,\lambda_{\max}^{N} = C_0\, e^{N\ln\lambda_{\max}} = C_0\, e^{-N/N_0},$$ $$\boxed{\,N_0 = -\frac{1}{\ln\lambda_{\max}}\,}, \qquad \text{convergence rate } 1/N_0 = -\ln\lambda_{\max}.$$ This is exactly the claimed form $F(N)\simeq 1 - C_0\lambda_{\max}^N =1-C_0 e^{-N/N_0}$. (If $\lambda_{\max}$ is real positive the decay is purely exponential; a complex pair would add an oscillation, but at the optimal/working parameters $\lambda_{\max}$ is real positive — verified numerically.)

The numeric check (run.py) confirms: a least-squares fit of $\ln\epsilon(N)$ vs $N$ over a large-$N$ window has slope $\ln\lambda_{\max}$ to high precision, and $\lambda_{\max}$ equals the second-largest-modulus eigenvalue of $S$.