Hence, the main problem becomes: for given E and βQ , can we find an appropriate ρ$_{B}$ and U$_{AB}$ such that Eqs. (4) and (11) are satisfied? (ii) If Q is a possible HTO for an operation E , then for any positive number a , the operator ˜ Q = Q + a 1$_{A}$ is also a possible HTO for the same operation. In other words, it is always possible to dump an additional heat a whose amount is independent of the state of the device. To show this, suppose that U$_{AB}$ on bath B is a realization of E that yields the HTO Q . Let C be a two-level system with Hamiltonian H$_{C}$ = Δ σ$_{z}$/ 2 where σ$_{z}$ denotes the associated Pauli spin matrix. Now, take BC as the new bath and apply the isometry ˜ U$_{ABC}$ = U$_{AB}$ ⊗ ( σ$_{x}$ )$_{C}$ to the composite system ABC. Since the transformation that C is subjected to is independent of how AB is changed, it is obvious that ˜ U$_{ABC}$ realizes the same operation E . But, the heat dumped to the bath BC is now where Δ E$_{C}$ = Δ tanh(Δ / 2 k$_{B}$T ). If the value of Δ is chosen such that Δ E$_{C}$ = a , then we get Δ E$_{BC}$ = tr ρ$_{A}$ ˜ Q . This shows that ˜ Q is also a possible HTO for the same operation. (iii) The set of two-tuples ( E , Q ) formed from quantum operations E and allowed HTOs Q is a convex set. In other words, if Q$_{i}$ is a possible HTO for the operation E$_{i}$ for i = 1 , 2 , . . . , n , then for any λ$_{i}$ ≥ 0 with ∑ $_{i}$λ$_{i}$ = 1, the operator Q = ∑ $_{i}$λ$_{i}$ Q$_{i}$ is a possible HTO for the operation E = ∑ $_{i}$λ$_{i}$ E$_{i}$ . Before showing this, first note that realizations for each E$_{i}$ can be assumed to involve the same bath B. This can always be done by taking the bath sufficiently large. As a result, without loss of generality, suppose that U ( i ) AB is an isometry on AB that realizes E$_{i}$ and yields the associated HTO Q$_{i}$ . Let C be an n -level quantum system and let {| 1 〉 , · · · , | n 〉} be an orthonormal basis in the Hilbert space of C. Let the initial state of C be ρ$_{C}$ = ∑ $_{i}$λ$_{i}$ | i 〉 〈 i | . 9 Δ E$_{BC}$ = tr( ρ$_{A}$Q ) + Δ E$_{C}$ , (12) βQ = − tr$_{B}$ [ ( 1$_{A}$ ⊗ ρ$_{B}$ ) U † $_{AB}$( 1$_{A}$ ⊗ ln ρ$_{B}$ ) U$_{AB}$ ] + 1$_{A}$ tr$_{B}$( ρ$_{B}$ ln ρ$_{B}$ ) . (11) (i) First, note that any restriction that the HTOs obey can be expressed entirely in terms of E and the ratio Q/k$_{B}$T = βQ . To see this, note that both E and βQ can be expressed exclusively in terms of U$_{AB}$ and ρ$_{B}$ , with no explicit T (or H$_{B}$ ) dependence, i.e., Eq. (4) and