Theorem: Conditional mixing with dephasing

If $J>0$ and ${\gamma }_{\mathrm{d}}>0$, then

For the fixed model there are $M,g>0$ such that

Consequently when ${\gamma }_{\mathrm{r}}=0,{\gamma }_{\mathrm{d}}>0$, all the entanglement diagnostics above tend to zero, and the full limiting state is the separable state

Proof. On all complex matrices, the conditional generator $\mathcal{A}(M)=-\mathrm{i}[h,M]-2{\gamma }_{\mathrm{d}}(M-\Delta M)$ obeys

If $\mathcal{A}(M)=\lambda M$, $M\ne 0$, and $\mathrm{Re}\lambda =0$, then $M$ is diagonal. The diagonal equation gives $\lambda M_{ii}=0$, hence $\lambda =0$. The neighboring off-diagonal equations give $J(M_{i+1,i+1}-M_{ii})=0$. Connectedness implies $M=c{I}_N$.

The invariant trace-zero subspace therefore has its finite spectrum strictly in the left half-plane. Jordan decomposition supplies an exponential bound, with $g$ chosen smaller than the spectral decay rate to absorb possible polynomial factors. Apply this bound to $S_0-I_N/N$. The limiting matrix has no off-diagonal entries; the entanglement formulas vanish and the displayed mixture is separable.