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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 \begin{equation} \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). \end{equation} 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 \begin{equation} \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)} \end{equation} \begin{equation} \bar{G}^{R}(\mathbf{k},\omega) = \frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}+i\eta}, \end{equation} \begin{equation} \bar{G}^{A}(\mathbf{k},\omega) = \frac{1}{\omega - \hbar^{-1}\xi_{\mathbf{k}}-i\eta}, \end{equation} where we will take $\eta \to 0^{+}$ in the end. However, it is not obvious to me that \begin{equation} \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). \end{equation} 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 \begin{equation} \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} \end{equation} which is not (or at least that that obvious?) equal to \begin{equation} \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)}. \end{equation} 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!

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1 Answer 1

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You are completely correct that the identity of the two equations is not trivial. Keep in mind however, that the Fourier-transform of a real-time-Green-function is indeed a distribution. Interpreting the r.h.s. in the distributional sense and using the Sokhotski-Plemelj-theorem helps to resolve your issues.

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