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By Dave Morris Bacon

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2 Fixed basis OSR A tool which we will find useful later in our derivation of master equations is the fixed basis form of the OSR[42, 10]. 3) for expanding each of the operators Ai (t) in the OSR: Ai(t) = biα (t)Fα . 14) αβ where biα (t)b∗iβ (t). 15) i Eq. 14) is the fixed basis or chi representation of the OSR. Normalization requires that b∗iα biβ F†α Fβ = χβα F†α Fβ = I. 16) iαβ αβ Taking the trace of this equation we find that χαα = d. 17) 18 The χαβ (t) matrix is a positive hermitian matrix which specifies the OSR in a given basis.

Suppose that instead of the environment being in the initial state |+ +| it is in the state |0 0|. In this case, if we us the basis |+ , |− to calculate the OSR we find that 1 A1 (t) = +|E (cos(λt)I − i sin(λt)σ z ⊗ σ z ) |0 E = √ (cos(λt)I − i sin(λt)σ z ) , 2 A2 (t) = −|E (cos(λt)I − i sin(λt)σ z ⊗ σ z ) |0 E = A1 (t). 24) 19 If we instead use the basis |0 , |1 to calculate the OSR, we find that ˜ 1 (t) = A ˜ 2 (t) = A 0|E (cos(λt)I − i sin(λt)σ z ⊗ σ z ) |0 1|E (cos(λt)I − i sin(λt)σ z ⊗ σ z ) |0 E E = (cos(λt)I − i sin(λt)σ z ) , = 0.

67) β=0 for α = 0. If we require the new basis to maintain the trace inner product, then Tr F†α Fβ = Tr ∗ gαµ gβν G†µ Gν = µ,ν ∗ gαν gβν = δαβ . 68) ν ∗ ∗ Thinking about gαν as a matrix, this implies that gαν is a unitary matrix. The non-Hamiltonian generator of the SME, Eq. 3) is defined as L [ρ] = 1 aαβ [Fα ρ, F†β ] + [Fα , ρF†β ] . 69) The change of basis, Eq. 67), transforms this generator to L [ρ] = 1 ∗ aαβ gαν gβµ [Gν ρ, G†µ ] + [Gν , ρG†µ ] . 71) where a′µν = ∗ aαβ gαν gβµ . 72) 29 ∗ Since gαν can be any unitary matrix, and aαβ is a hermitian matrix, we can choose ∗ gαν such that this matrix diagonalizes a′µν .