Theorem: Exact occupation correspondence and its limits

For every message input, including external references:

1. the joint occupation channel is the replacement channel

   $$X⟼\mathrm{tr}(X){\sigma }_O;$$

2. its occupation-state and joint count-law comparison with (eq:Hbaseline) is exactly

   $$\boxed{d_{\mathrm{tr}}({\hat{\sigma }}_O,{\rho }_{\mathrm{H}})=d_{\mathrm{TV}}(w_1\otimes w_2,p_{\mathrm{H}})=t_{\mathbf{J}}:=1-{\prod _{i=1}}^2(1-{q_i}^{J_i+1});}$$

3. the corresponding replacement channels have the same half-diamond distance $t_{\mathbf{J}}$;

4. the finite occupation mean is

   $$\boxed{{\mathbb{E}}_{\mathrm{model}}{N}_{O_i}={\nu }_i(1-{q_i}^{J_i}),{\nu }_i-{\mathbb{E}}_{\mathrm{model}}{N}_{O_i}={\nu }_i{q}_i^{J_i}.}$$

Each bounded occupation effect changes in probability by at most $t_{\mathbf{J}}$. No all-observable or local-field correspondence is asserted.

Proof. The occupation emissions act on independent factors and have the exact output densities in (eq:numberemission). Tracing all payload systems therefore gives the stated replacement channel.

The finite law is the deterministic clipping of the product geometric law. In the common countable number basis, mass above the finite rectangle is removed and added to its boundary. The positive inside change and negative outside change occupy orthogonal number records. Their half trace norm is the outside probability, which is (eq:Htail). The geometric marginals are independent for the chosen baseline.

For replacement channels,

Complete trace-norm contraction of partial trace gives the upper bound $\Vert \sigma -\tau {\Vert }_1$; a normalized input attains it. Thus the half-diamond distance is the state distance.

Finally,

An effect expectation is Lipschitz with constant one in trace distance, proving the bounded-observable statement.