Why are time-ordered Greens functions equal to retarded Greens functions at zero temperature? When I calculate a photon polarization diagram:

I get the same answer:

*

*If I calculate it in equilibrium (retarded Greens functions) with finite chemical potential, in the limit of zero temperature, or


*If I calculate it as a scattering amplitude (time-ordered Greens functions) in a particle bath $|\psi\rangle = \prod_{|\vec{p}|<p_F}c^\dagger_{\vec{p}}|0\rangle$.
The equality follows from a weird cancellation from the Fermi-Dirac distribution at zero temperature. Is there a theorem or some argument that this should happen in general (i.e. for more general correlation functions)?
 A: This is generally not the case. The reason is that, in addition to zero temperature, you are averaging over the vacuum state with zero particles, which is killed by annihilation operators.
Let us take time ordered Green's function at zero temperature:
$$
G^t(t,t')=\langle 0|T[c(t)c^\dagger(t')]|0\rangle=\\
\langle 0|c(t)c^\dagger(t')|0\rangle\theta(t-t') \pm \langle 0|c^\dagger(t')c(t)|0\rangle\theta(t'-t)
$$
and the retarded Green's function
$$
G^r(t,t')=\langle 0|[c(t)c^\dagger(t')]_\pm|0\rangle\theta(t-t')=\\
\langle 0|c(t)c^\dagger(t')|0\rangle\theta(t-t') \pm \langle 0|c^\dagger(t')c(t)|0\rangle\theta(t-t')
$$
If
$$ c(t)|0\rangle=0$$ the second terms in both equations are zero and the two functions are identical.
However, e.g., if the ground state is a Fermi sea filled up to its chemical potential the result is different.
A: assuming you work in perturbation theory, at zero temperature the (unperturbed) state is the vacuum, meaning that $\psi_k(t)|0\rangle = 0$, so the only part of the retarded GF $$g^r(k, t) = \theta(t)\langle 0 | \psi_k^{\dagger}(0)\psi_k(t) + \psi_k(t) \psi_k^{\dagger}(0) | 0 \rangle$$ that survives will give the same result as the time-ordered one $$g(k, t) = \theta(t)\langle 0 | \psi_k(t)\psi_k^{\dagger}(0)|0\rangle + \theta(-t)\langle 0 | \psi^{\dagger}(0)\psi(t)|0\rangle$$
