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Therefore, we can write g(r) = 1 + 1 ρ ∑ k=even ∞ ∑ n=1 ⟨ψ 0|ˆR(k) n |ψ 0⟩es(k) n r =1 − λ 0 ρ ∑ k=even ∞ ∑ n=1 |⟨ψ 0|ψ k(s(k) n + β P)⟩|2 λ ′ k(s(k) n + β P) es(k) n r

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