Derive the form of an isometric extension of the erasure channel

Question

Show that an isometric extension of the erasure channel is $$U^N_{A\to BE} =\sqrt{1−\epsilon}\left(|0\rangle_B \langle 0|_A +|1\rangle_B \langle 1|_A \right)\otimes|e\rangle_E+ \sqrt{\epsilon}|e\rangle_B \langle0|_A \otimes |0\rangle_E + \sqrt{\epsilon}|e\rangle_ B \langle 1|_A \otimes |1\rangle_E$$ $$=\sqrt{1−\epsilon} \text{ I}_{A \to B}\otimes|e\rangle_E+\sqrt{\epsilon}\text{ I}_{A \to E}\otimes\otimes|e\rangle_B$$ where, the Erasure Channel implements the following: $\rho \to (1 − \epsilon) \rho + \epsilon|e\rangle\langle e|$.

I know that the Kraus operators for the quantum erasure channel are the following: $\left\{\sqrt{1−\epsilon}\left(|0\rangle_B \langle 0|_A +|1\rangle_B \langle 1|_A \right),\sqrt{\epsilon}|e\rangle_B \langle0|_A,\sqrt{\epsilon}|e\rangle_B \langle1|_A\right\}$. I also know that for a quantum channel $N_{A\to B}$ with the following Kraus representation: $$N_{A\to B}(\rho_A) = \sum_jN_j\rho_A N_j^{\dagger},$$ an isometric extension of the channel $N_{A\to B}$ is the following linear map: $$U^N_{A \to BE} =\sum_jN_j \otimes|j\rangle$$ .

Using this, the $N_1$, $N_2$ and $N_3$ in our case are $\sqrt{1−\epsilon}\left(|0\rangle_B \langle 0|_A +|1\rangle_B \langle 1|_A \right)$,$\sqrt{\epsilon}|e\rangle_B \langle0|_A$ and $\sqrt{\epsilon}|e\rangle_B \langle1|_A$ are respectively. I am just confused on, which $|j\rangle$ to choose. How do I know the appropriate orthonormal vectors in this case?

Thanks for the help!

Answer 1

The problem is essentially already solved. If you know that a channel $\Phi$ has Kraus operators $A_a$, then its corresponding isometry $V$ such that $\Phi(X)=\operatorname{Tr}_2[VXV^\dagger]$ is the matrix built putting the Kraus operators $A_a$ one above the other, as in $$V=\begin{pmatrix}A_1 \\ A_2 \\\vdots\\ A_n\end{pmatrix}.$$ Note that this being an isometry is equivalent to the condition $\sum_a A_a^\dagger A_a=I$.