Embed the input qubit as the span of $|0\rangle,|1\rangle$ in the three-dimensional output. Because the input is pure, its only purification has a trivial reference and <entanglement fidelity> reduces to its survival probability:
$$
\begin{aligned}
F_e(|+\rangle\langle+|,\mathcal E_{1/2})
&=\langle+|\mathcal E_{1/2}(|+\rangle\langle+|)|+\rangle\\
&=\frac12+\frac12|\langle+|e\rangle|^2.
\end{aligned}
$$
Since $|e\rangle$ is orthogonal to the embedded qubit subspace,
$$
\boxed{F_e=\frac12}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 1 / a / iii / Solution

For a purification $|\Psi_\rho\rangle_{RA}$ of $\rho$, the <entanglement fidelity> is
$$
F_e(\rho,\Lambda)=
\langle\Psi_\rho|
(\operatorname{id}_R\otimes\Lambda)
(|\Psi_\rho\rangle\langle\Psi_\rho|)
|\Psi_\rho\rangle.
$$
Using a <Kraus representation> $\Lambda(X)=\sum_kA_kXA_k^\dagger$ gives
$$
F_e(\rho,\Lambda)
=\sum_k\left|
\langle\Psi_\rho|I\otimes A_k|\Psi_\rho\rangle
\right|^2.
$$
In a <Schmidt decomposition> of the purification,
$$
\langle\Psi_\rho|I\otimes A_k|\Psi_\rho\rangle
=\sum_j\lambda_j\langle j|A_k|j\rangle
=\operatorname{Tr}(A_k\rho).
$$
Therefore
$$
\boxed{F_e(\rho,\Lambda)
=\sum_k|\operatorname{Tr}(A_k\rho)|^2}.
$$

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 1 / a / ii / Solution

{title2=$F_e(\rho,\Lambda)$}
{wiki}

For a purification $|\Psi_\rho\rangle_{RA}$, the entanglement fidelity of a channel $\Lambda$ is
$$
F_e(\rho,\Lambda)=
\langle\Psi_\rho|(\operatorname{id}_R\otimes\Lambda)
(|\Psi_\rho\rangle\langle\Psi_\rho|)
|\Psi_\rho\rangle.
$$
It is independent of the chosen purification.


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