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Each of the four operator terms in ˜ H (1) cond takes ρ (0) ss to another eigenstate of L $^{(0)}$; for example, b$_{p}$$_{α}$ ˜ s † p + $_{k$_{L}$}$ρ (0) ss has the eigenvalue − c ( p + k$_{L}$ ) / 2 + i ν$_{α}$ , the first term from | ˜ 1 〉〈 ˜ 0 | = ˜ s † p + $_{k$_{L}$}$| ˜ 0 〉〈 ˜ 0 | and the second term from the b$_{p}$$_{α}$ . We can hence apply $^{[}$L (0) ] − 1 by simply inserting the reciprocal eigenvalues. Using the short hand notation 

this yields 

<img src='content_image/429253.jpg'>

<img src='content_image/429252.jpg'>

This inverse is not unique: we could add in an arbitrary L (0) = 0 state. However, doing so would have no effect on the final result, as it would be turned into an L (0) ̸ = 0 state by the second L (1) then annihilated by P . 

We now apply the second L $^{(1)}$, and finally apply P , i.e. keep only terms which satisfy Eqs. (A.12) and (A.13). This gives 

<img src='content_image/429251.jpg'>

where Ξ is defined by Eq. (A.24) and the Hermitian conjugate comes from $^{[}$L (0) ] − 1 $^{(}$ρ (0) ss ˜ H (1) † cond $^{)}$= Ξ $^{†}$. Using Eq. (23), we find that the first term in this equation is given by 

<img src='content_image/429250.jpg'>

Analogously, remembering that P ( | ˜ 1 〉〈 ˜ 1 | ) = | ˜ 0 〉〈 ˜ 0 | , we find that 

<img src='content_image/429249.jpg'>

Moreover, we obtain from Eq. (27), 

<img src='content_image/429248.jpg'>

