 Solution

Let $H|n⟩=E_n|n⟩$, $\beta =1/T$, $Z={\sum }_n{e}^{-\beta E_n}$, and $A_{nm}=⟨n|A|m⟩$. Inserting two energy resolutions into the commutator and carrying out the one-sided Fourier integral gives the Lehmann representationIf the trace in the question is intentionally unnormalized, the same formula holds without $1/Z$.

For $H=\epsilon a^{†}a$,Thereforeand the Green function requested literally for $A=a^{†}$ isPhysically, a number-conserving oscillator has no anomalous response connecting two creation operators. The nonzero normal retarded propagator pairs annihilation with creation: $G_{a{a}^{†}}^R(\omega )=1/(\omega -\epsilon +i{0}^{+})$, whose pole is the one-quantum excitation at energy $\epsilon$.