= Solution
Let $H|n\rangle=E_n|n\rangle$, $\beta=1/T$, $Z=\sum_ne^{-\beta E_n}$, and $A_{nm}=\langle n|A|m\rangle$. Inserting two energy resolutions into the commutator and carrying out the one-sided Fourier integral gives the <Lehmann representation>
$$
\boxed{
G_{AA}^R(\omega)=\frac1Z\sum_{n,m}
\frac{(e^{-\beta E_n}-e^{-\beta E_m})A_{nm}A_{mn}}
{\omega+E_n-E_m+i0^+}.}
$$
If the trace in the question is intentionally unnormalized, the same formula holds without $1/Z$.
For $H=\varepsilon a^\dagger a$,
$$
a^\dagger(t)=e^{i\varepsilon t}a^\dagger.
$$
Therefore
$$
[a^\dagger(t),a^\dagger(0)]=0
$$
and the Green function requested literally for $A=a^\dagger$ is
$$
\boxed{G_{a^\dagger a^\dagger}^R(\omega)=0.}
$$
Physically, a number-conserving oscillator has no anomalous response connecting two creation operators. The nonzero normal retarded propagator pairs annihilation with creation: $G_{aa^\dagger}^R(\omega)=1/(\omega-\varepsilon+i0^+)$, whose pole is the one-quantum excitation at energy $\varepsilon$.
Back to article page
