Green function dependence on temperature Consider the retarded Green function for fermions
$$
G_{ij}^r(t)=-\frac{i}{\hbar} \theta(t)\langle[c_i(t),c^\dagger_j(0)]_{+}\rangle
$$
They can be understood as the $(i,j)$ entry of the matrix $G^r(t)$. The operators evolve in time as they are in the Heisenberg picture. The operator average means
$$
\langle A \rangle = \mathrm{Tr} \rho A
$$
where the density matrix has temperature dependence.
$$
\rho = \frac{1}{Z} e^{-\beta H}.
$$
In this answer, it is shown that the Fourier transform of $G^r(t)$ is
$$
G^r(E) = (E + i \eta - H)^{-1}.
$$
The LHS depends on temperature but the RHS does not?
 A: In the Heisenberg picture, the operators evolve according to the Heisenberg equation
$$i\hbar \frac{d}{dt} A = [A, H]_-$$
This governs the time evolution of the annihilation operator
\begin{align*}
i\hbar \frac{d}{dt} c_j(t) &= [c_j(t), H]_- \\
    &= \sum_{mn} H_{mn} [c_j(t), c_m^\dagger(t) c_n(t)]_- \\
    &= \sum_{mn} H_{mn} (c_j(t) c_m^\dagger(t) c_n(t) - c_m^\dagger(t) c_n(t) c_j(t)) \\
    &= \sum_{mn} H_{mn} ( (-c_m^\dagger(t) c_j(t) + \delta_{mj}) c_n(t) + c_m^\dagger(t) c_j(t) c_n(t)) \\
    &= \sum_{mn} H_{mn} \delta_{mj} c_n(t)  \\
    &= \sum_{n} H_{jn} c_n(t)
\end{align*}
To considering all $j$ at once, define the notation
\begin{align*}
\vec{c}(t) = \begin{pmatrix}c_1(t) \\ c_2(t) \\ \vdots \\ c_N(t) \end{pmatrix}
\end{align*}
Then this is a system of coupled differential equations, which can be written in a Schrodinger-equation-like form.
\begin{align*}
i\hbar \frac{d}{dt} \vec{c}(t) = H \vec{c}(t)
\end{align*}
The solution can be obtained by the integrating factor method, and is
\begin{align*}
\vec{c}(t) = e^{-\frac{i}{\hbar} H t} \vec{c}(0) \equiv U(t) \vec{c}(0)
\end{align*}
Or individually,
\begin{align*}
c_i(t) = \sum_{q} U_{iq}(t) c_q(0)
\end{align*}
where $U_{lq}(t)$ are the entries of the unitary matrix $U(t)$. By taking the hermitian conjugate, we get, similarly,
\begin{align*}
    c^\dagger_j(t) = \sum_{p} U^*_{jp}(t) c^\dagger_p
\end{align*}
We are now in a position to evaluate the electron Green function. The main thing is the anti-commutator at different times.
\begin{align*}
[c_i(t), c_j(0)]_+ &= \sum_{q} U_{iq}(t) [c_q(0), c_j(0)]_+ \\
&= \sum_{q} U_{iq}(t) \delta_{qj} \\
&= U_{ij}(t)
\end{align*}
The crucial fact is that the above anti-commutator is always a multiple of the identity operator. Thus,
\begin{align*}
G^r_{ij}(t) &= - \frac{i}{\hbar} \theta(t) \langle [c_i(t), c_j(0)^\dagger]_+ \rangle \\
&= - \frac{i}{\hbar} \theta(t) \mathrm{Tr}(\rho U_{ij}(t)) \\
&= - \frac{i}{\hbar} \theta(t) U_{ij}(t) 
\end{align*}
has no dependence on temperature.
