# General expression of time-ordered thermal average Green's function does not reproduce non-interacting limit (Fetter ch. 31 Eq. (31.24))

Hi I am going through Fetter's Quantum Theory of Many-Particle Systems Dover Edition.

In ch. 31 he computed the relation between $$\bar{G}(\mathbf{k},\omega)$$, $${\bar{G}}^{R}(\mathbf{k},\omega)$$ and $$\bar{G}^{A}(\mathbf{k},\omega)$$ where each of them are the real-time Fourier transformation of the corresponding real-time thermal averaged Green's function. The thermal average is done in the grand canonical ensemble.

It is proved that we can express $$\bar{G}(\mathbf{k},\omega)$$ for Fermi system as $$$$\bar{G}(\mathbf{k},\omega) = [1 + e^{-\beta \hbar \omega}]^{-1}{\bar{G}}^{R}(\mathbf{k},\omega)+[1+e^{\beta \hbar \omega}]^{-1} {\bar{G}}^{A}(\mathbf{k},\omega).$$$$ This is in Eq. (31.24) of Fetter and I totally agree with this expression as I can really derive it. However, things get weird when I try to apply this expression to non-interacting Fermi system.

Suppose we have a non-interacting Fermi system with grand canonical Hamiltonian $$K = \sum_{\mathbf{k}} \left( \epsilon_{\mathbf{k}} - \mu\right)c^{\dagger}_{\mathbf{k}}c_{\mathbf{k}} = \sum_{\mathbf{k}}\xi_{\mathbf{k}}c^{\dagger}_{\mathbf{k}}c_{\mathbf{k}}$$ where $$\epsilon_{\mathbf{k}}$$ is the kinetic energy and $$\mu$$ is the chemical potential, then it is also known that $$$$\bar{G}(\mathbf{k},\omega) = \frac{1}{1+e^{-\beta \xi_{\mathbf{k}}}}\cdot\frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}+i\eta} + \frac{1}{1+e^{\beta \xi_{\mathbf{k}}}}\cdot\frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}} - i\eta}, \text{ (Eq. (31.38) in Fetter)}$$$$ $$$$\bar{G}^{R}(\mathbf{k},\omega) = \frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}+i\eta},$$$$ $$$$\bar{G}^{A}(\mathbf{k},\omega) = \frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}-i\eta},$$$$ where we will take $$\eta \to 0^{+}$$ in the end. However, it is not obvious to me that $$$$\bar{G}(\mathbf{k},\omega) = [1 + e^{-\beta \hbar \omega}]^{-1}{\bar{G}}^{R}(\mathbf{k},\omega)+[1+e^{\beta \hbar \omega}]^{-1} {\bar{G}}^{A}(\mathbf{k},\omega).$$$$ can help us obtain the correct $$\bar{G}(\mathbf{k},\omega)$$. This is because if we just plug in the non-interacting $$\bar{G}^{A}(\mathbf{k},\omega)$$ and $$\bar{G}^{R}(\mathbf{k},\omega)$$ which I have shown above, we will obtain $$$$\bar{G}(\mathbf{k},\omega) = \frac{1}{1 + e^{-\beta \hbar \omega}}\cdot\frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}+i\eta}+\frac{1}{1+e^{\beta \hbar \omega}}\cdot \frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}-i\eta}$$$$ which is not (or at least that that obvious?) equal to $$$$\bar{G}(\mathbf{k},\omega) = \frac{1}{1+e^{-\beta \xi_{\mathbf{k}}}}\cdot\frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}+i\eta} + \frac{1}{1+e^{\beta \xi_{\mathbf{k}}}}\cdot\frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}} - i\eta} \text{ (Eq. (31.38) in Fetter)}.$$$$ In the first expression we still have a $$\omega$$ dependence in $$\frac{1}{1 + e^{-\beta \hbar \omega}}$$ and $$\frac{1}{1+e^{\beta \hbar \omega}}$$ but in the second expression we have instead $$\frac{1}{1+e^{-\beta \xi_{\mathbf{k}}}}$$ and $$\frac{1}{1+e^{\beta \xi_{\mathbf{k}}}}$$. Are there suggestion to reconcile this? Thanks!