set of operators { M † $_{i}$M$_{j}$ } n i,j =1 is linearly independent[22]. Therefore, part (b) of the theorem above gives a complete characterization of the HTOs of all extreme quantum operations. Proof: For (a), the use of the definition of the HTO given in Eq. (9) directly leads to Eq. (38) where To show (b), first write the isometry condition U † $_{AB}$U$_{AB}$ = 1$_{AB}$ as In other words, each L$_{i}$ is an isometry on the bath that maps H$_{B}$ onto mutually perpendicular subspaces of H$_{B}$ . The last set of relations is actually equivalent to the statement that U$_{AB}$ is an isometry. Consider now a hypothetical n -level system R. Let {| i 〉} n i =1 be an orthonormal basis for the Hilbert space of R. The following operator on the composite system RB, is then an isometry. It can also be readily verified that W$_{RB}$ is an isometry if and only if U$_{AB}$ is also an isometry. The most important point in here is that W$_{RB}$ is a realization of a Landauer erasure on R. Furthermore, the HTO associated with this erasure is given by For n ≥ 2, the application of Theorem 3 proves the claim. The result for the special case of n = 1 also follows trivially from the discussion in this subsection. Part (c) follows from the converse of the argument used above for part (b). Suppose that q satisfies (39). In that case, Q ′ R given by Eq. (45) is an allowed HTO for a Landauer erasure. Hence one can construct the associated realization; let W$_{RB}$ be the isometry for this realization. We use Eq. (43) to define L$_{i}$ and then use Eq. (34) to define U$_{AB}$ . It is 20 Q ′ R = tr$_{B}$ [ ( 1$_{R}$ ⊗ ρ$_{B}$ ) W † $_{RB}$( 1$_{R}$ ⊗ H$_{B}$ ) W$_{RB}$ ] − E$_{B}$ 1$_{R}$ . (44) = ∑ i,j q$_{ij}$ | i 〉 〈 j | . (45) W$_{RB}$ = n ∑ i =1 ( | 1 〉 〈 i | )$_{R}$ ⊗ ( L$_{i}$ )$_{B}$ (43) L † $_{i}$L$_{j}$ = δ$_{ij}$ 1$_{B}$ . (42) The linear independence of { M † $_{i}$M$_{j}$ } n i,j =1 then leads to the following relations n ∑ i,j =1 ( M † $_{i}$M$_{j}$ )$_{A}$ ⊗ ( L † $_{i}$L$_{j}$ )$_{B}$ = 1$_{A}$ ⊗ 1$_{B}$ . (41) q$_{ij}$ = tr ( ρ$_{B}$L † $_{i}$H$_{B}$ L$_{j}$ ) − E$_{B}$δ$_{ij}$ . (40)