= Solution
Put $P_1=\sigma_1\otimes\sigma_2$ and $|\psi\rangle=|A\rangle^{\otimes2}$. After the first <Hadamard gate> and the two controlled Pauli gates, the joint state is
$$
\frac{|0\rangle|\psi\rangle+|1\rangle P_1|\psi\rangle}{\sqrt2}.
$$
The final Hadamard gate changes this to
$$
\frac12\left[
|0\rangle(I+P_1)|\psi\rangle
+|1\rangle(I-P_1)|\psi\rangle
\right].
$$
Conditioned on ancilla outcome $s\in\{0,1\}$, the normalized data state is therefore
$$
\boxed{
|\Psi_a\rangle=
\frac{[I+(-1)^sP_1]|A\rangle^{\otimes2}}
{\sqrt{2[1+(-1)^s\langle A|^{\otimes2}P_1|A\rangle^{\otimes2}]}}}.
$$
Write the input in the two <eigenspaces> of $P_1$ as
$$
|A\rangle^{\otimes2}=\alpha|p_+\rangle+\beta|p_-\rangle,
\qquad
P_1|p_\pm\rangle=\pm|p_\pm\rangle,
$$
where the displayed eigenstates are normalized. A direct PBC measurement gives
$$
|\Psi_b\rangle=
\begin{cases}
|p_+\rangle,&\text{outcome }+1,\\
|p_-\rangle,&\text{outcome }-1,
\end{cases}
$$
with probabilities $|\alpha|^2$ and $|\beta|^2$. Hence the ancilla circuit and the <Pauli measurement> have identical outcome distributions and conditional data states after identifying the Pauli outcome with $(-1)^s$.
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