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Write $c=⟨e_0|e_1⟩=v{e}^{i\alpha }$. The Cauchy-Schwarz inequality gives $0\le v\le 1$ because the two environment states are normalized. Immediately after the interaction, the joint quantum state isApplying the last Hadamard gate givesSince the environment is unobserved, the Born rule uses the squared norm of the environment-valued coefficient of $|0⟩$:Thus the Hadamard interference with environment overlap formula isThe plus sign follows from the explicitly chosen phase gate and the order $⟨e_0|e_1⟩$ of the inner product. Equivalently, before the last Hadamard gate the partial trace over the environment givesThe complex environment overlap in qubit dephasing both reduces the quantum coherence by $v$ and adds a relative quantum phase $\alpha$. Orthogonal environment states have $v=0$ and give $P_0=1/2$ for every $\phi$. If $v=1$, the quantum interference fringe has full contrast and is only shifted; the qubit has not become mixed. For $v=0$, $\alpha$ is immaterial.

Moving the interaction to between the first Hadamard gate and the phase gate givesthe identical joint quantum state. Therefore the expression is unchanged. This is the commutation of computational-basis dephasing and phase gates; it uses the specified population-preserving interaction with the same environment states, rather than an arbitrary model of quantum decoherence.

Let $c=⟨R_0|R_1⟩$. The joint state after the environment interaction is $\alpha |0⟩|R_0⟩+\beta |1⟩|R_1⟩$. Taking its partial trace over the environment gives the exact reduced density operatorThe conjugate in the upper off-diagonal entry follows from tracing $|R_0⟩⟨R_1|$. Normalization of the environment states and the Cauchy-Schwarz inequality imply $|c|\le 1$.

For comparison, a phase-flip channel sends $\rho$ to $(1-p)\rho +pZ\rho Z$, multiplying its off-diagonal entries by the real number $1-2p$. Consequently the claimed representation is valid, for a general input, provided the environment overlap is real, withThis is a genuine probability between zero and one. The environment maps printed in the question do not themselves imply that $c$ is real. For example, choose $|R_1⟩=i|R_0⟩$, so $c=i$, and take $\alpha =\beta =1/\sqrt{2}$. The qubit becomes $(|0⟩+i|1⟩)/\sqrt{2}$, whose imaginary off-diagonal entries cannot come from any probabilistic mixture of $\rho$ and $Z\rho Z$. These environment maps are a valid isometry and can be extended to a unitary interaction, so this is an actual counterexample to the unrestricted premise.

The general complex environment overlap in qubit dephasing has an exact corrected description. Write $c=r{e}^{i\phi }$, $0\le r\le 1$, and let $D_{\phi }=\mathrm{diag}(1,e^{i\phi })$. ThenAt $r=0$, choose any phase. The dephasing channel includes both random phase errors and, when the overlap is complex, a coherent diagonal rotation. The subsequent probability formulas use the intended real-overlap phase-flip model.