Isometric Extension of an Erasure Channel

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!

• Does it matter which $|\,j\,\rangle$ you choose? Why? Nov 28 '18 at 1:46
• @PeterShor I can always take $E$ to be spanned by $|0\rangle$, $|1\rangle$ and $|e\rangle$.
– tattwamasi amrutam
Nov 28 '18 at 18:19