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Since the operators L (0) at , p and L (0) phn , p α all act on differ- ent modes, they all commute with each other. We can hence choose a basis of simultaneous eigenstates of all these terms, which will also be eigenstates of the whole. Denoting the eigenstates of L (0) by ρ$_{λ}$ with eigenvalues λ such that L $^{(0)}$( ρ$_{λ}$ ) = λρ$_{λ}$ , these are of the form 

with 

The relevant atomic eigenstates ρ$_{λ}$$_{at}$$_{,}$$_{p}$ and their respec- tive eigenvalues λ$_{at}$$_{,}$$_{p}$ are 

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Here | ˜ n$_{p}$ 〉 = 1 √ n $_{!}$(˜ s † $_{p}$) $^{n}$| ˜ 0$_{p}$ 〉 is the state with n excitations in the mode p . The tildes indicate that these are in terms of the ˜ s$_{p}$ operators. In terms of the original s$_{p}$ , | ˜ 0$_{p}$ 〉 is a coherent state of amplitude √ N Ω / | c ( k$_{L}$ ) | for p = k$_{L}$ , and a vacuum state for all other wavevectors p . (0) 

Note that L does not correspond to a unitary evolu- tion, and its eigenstates are hence not orthogonal. In par- ticular, if P is the projection onto the zeroth order steady states L (0) = 0, P ( | ˜ 1$_{p}$ 〉〈 ˜ 1$_{p}$ | ) = P ( | ˜ 1$_{p}$ 〉〈 ˜ 1$_{p}$ | − | ˜ 0$_{p}$ 〉〈 ˜ 0$_{p}$ | + | ˜ 0$_{p}$ 〉〈 ˜ 0$_{p}$ | ) = | ˜ 0$_{p}$ 〉〈 ˜ 0$_{p}$ | , not zero. 

The phonon eigenstates are simply given by 

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Since the eigenvalue only depends on the difference in phonon energy between the right and left sides of ρ , and phonons of all wavevectors p have the same zeroth or- der energy ν$_{α}$ , these phonon eigenstates are highly de- generate. In particular, the zeroth order steady states ρ$_{ss}$ = ρ$_{λ}$$_{=0}$ are 

where the atomic part | ˜ 0 〉 = ⊗ p | ˜ 0$_{p}$ 〉 is the vacuum state of 

all the ˜ s$_{p}$ , while the phonon part ρ (0) phn can be any phonon state which is diagonal in the total phonon energy (i.e. it can be a mixture, but not a superposition, of energies), 

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We see below that this degeneracy in the steady state is broken by the higher order terms, giving a unique ρ (0) phn when terms up to second order in η$_{α}$ are considered. 

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The first order terms in η$_{α}$ in Eq. (A.1) are 

The unknown in this equation is ρ (1) ss , and solving for it gives 

Here L (1) is formed in the same way as L (0) (in Eq. (A.3)), but using the first order terms H (1) cond and R $^{(1)}$( ρ ), also given in Eqs. (23) and (27). The relevant terms, i.e. those which are non-zero when applied to ρ (0) ss , are ex- tracted in Eq. (A.21). Because L (0) has zero eigenvalues, the inverse $^{[}$L (0) ] − 1 is not defined on every density ma- trix. However, the L (0) = 0 states ρ (0) ss are diagonal in the total phonon energy (Eq. (A.13)). Every term in L (1) ei- ther creates or destroys exactly one phonon. These terms hence transform L (0) = 0 states to only L (0) ̸ = 0 states, on which the inverse $^{[}$L (0) ] − 1 is defined. This means all the degenerate zeroth order steady states ρ (0) ss give valid first order steady states. Hence, the first order terms do not break the degeneracy (non-uniqueness) of ρ (0) $_{phn}$. 

We therefore consider the second order terms in Eq. (A.1), which we will find do break the degeneracy to give a unique steady state solution. These terms are given by 

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Substituting ρ (1) ss from Eq. (A.15) and solving for ρ (2) ss , we find that 

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The L term comes from processes with a single second order phonon vertex (the first diagram in Fig. 5), while L (1) $^{[}$L (0) ] − 1 L (1) comes from processes with two first or- der vertices. Eq. (A.17) only has a solution if the inverse [ L $^{(0)}$] − 1 exists on the state it acts on, or equivalently if this state (in the large round brackets) has no component in the L (0) = 0 subspace. This condition can be written as 

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where P is the (non-orthogonal) projection operator which selects all the density matrices that are of the form of ρ (0) ss in Eq. (A.12). Otherwise, the operator [ L $^{(0)}$] − 1 cannot be applied to this expression. 

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## 2. First order 

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## 3. Second order 

