Single-qubit interference with conditional environment states ($P_0=\frac{1}{2}[1+v\cos (\varphi +\alpha )]$)

Prepare a qubit in $|+⟩$, apply a phase gate $P(\varphi )=\mathrm{diag}(1,e^{i\varphi })$, and correlate its computational basis states with normalized environment states $|e_0⟩,|e_1⟩$. Write $⟨e_0|e_1⟩=v{e}^{i\alpha }$. After a final Hadamard gate, the environment amplitude accompanying output zero is $(|e_0⟩+e^{i\varphi }|e_1⟩)/2$. Its squared norm is $P_0=[1+v\cos (\varphi +\alpha )]/2$. Thus $v$ is the interference visibility, while $\alpha$ shifts the fringe. This quantum decoherence map commutes with the phase gate, so placing it immediately before that gate gives the same probabilities.