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Let $p=|\alpha {|}^2$, so $|\beta {|}^2=1-p$. The projective measurement has outcome 0 with probability $p$ and leaves the state $|0⟩$; outcome 1 has probability $1-p$ and leaves $|1⟩$. Alice knows her actual result, so her conditional description is a pure state. Her fidelity of quantum states with the initial state is $p$ after outcome 0 or $1-p$ after outcome 1. Its outcome average is $p^2+(1-p{)}^2$.

Bob knows the preparation but not the outcome, so his density matrix is the nonselective postmeasurement stateIt is mixed for $0<p<1$ and pure only at the endpoints. The relative phase has disappeared because the classical outcome record is unavailable to him.

The undisclosed qubit measurement outcome with an unspecified preparation prior requires care: Charlene has no specified prior over preparations. Consequently the information given does not determine a unique posterior probability for the two outcomes. Under a symmetric ignorance or maximum-entropy prior, her best assignment isMore generally, if her prior mean of $|\alpha {|}^2$ is $q$, her assignment is ${\rho }_C=\mathrm{diag}(q,1-q)$ and its fidelity of quantum states with the actual initial state is $pq+(1-p)(1-q)$. Knowing that a measurement occurred guarantees diagonality in its basis, but does not by itself prove $q=1/2$. Different descriptions here reflect different available records, rather than different physical outcomes. ^[ref]

For any joint lower-register state, the probability is $P_0=(1+⟨S⟩)/2$, where $S$ is its exchange operator. The Bell-state eigenvalues from part a therefore giveFor the spin singlet state, controlled exchange adds a minus sign to the control-1 branch, so the final Hadamard gate puts the ancilla in $|1⟩$. For the three symmetric states the ancilla returns to $|0⟩$.

The product-state overlap formula cannot be applied to arbitrary entangled lower inputs. All four Bell states have identical one-qubit reduced states $I/2$, yet the spin singlet state result differs maximally. The network measures exchange symmetry of the joint state, rather than fidelity of quantum states inferred from its two marginals.

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.