 Solution

Normalize the second qubit as $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. Since a Pauli X gate changes $|-⟩$ to $-|-⟩$, the Boolean quantum oracle acts asThis quantum phase kickback leaves the second qubit uncorrelated with the first. Put $b=f(0)\oplus f(1)$: $b=0$ for a constant function and $b=1$ for a balanced Boolean function. Apart from the irrelevant global phase $(-1{)}^{f(0)}$, the first qubit and environment reachThe population-preserving interaction commutes with the quantum oracle on this answer-register input, so its placement before or after the oracle, between the two Hadamard gates, gives the same result. Substitute $\phi =\pi b$ into b:The printed Deutsch algorithm uses the rule zero means constant function; one means balanced Boolean function. With that rule, the Deutsch algorithm with pure dephasing succeeds for either class withFor real nonnegative environment overlap this is $(1+v)/2$, ranging from certainty at $v=1$ to random guessing at $v=0$. For general complex overlap, a coherent shift can invert the printed answer: for example $v=1,\alpha =\pi$ makes the printed rule always wrong. If the overlap is known but the printed quantum measurement is retained, choose the better labeling of its outcomes; this givesThere is a further improvement if a known $\alpha$ can be compensated by the phase gate $P_{-\alpha }$ before the final Hadamard gate. It removes the coherent shift without reversing any actual quantum decoherence, and yieldsThis is optimal when only the first qubit is accessible, with equal prior probabilities for the two classes. Before the final Hadamard gate, their reduced density matrices areThe difference ${\rho }_0-{\rho }_1$ has eigenvalues $\pm v$, giving trace distance $v$. The Holevo–Helstrom theorem therefore bounds the success probability by $(1+v)/2$, exactly attained by the compensated procedure. A known phase shift and loss of coherence have different effects: at $v=0$ the two density matrices coincide and no measurement on the available qubit can reveal the answer.