\[\epsilon= \begin{cases}1&\textrm{fermions}\\-1&\textrm{bosons}\end{cases}\]
Here we introduce the following definitions:
\[f_F(E):=\pqty{e^{\beta E}+\epsilon}^{-1}\] \[S(\omega):=1-2 f_F(\omega-\mu)\]
Alternative notation used in the lecture notes: \[\ev{n}(\omega):=f_F(\omega-\mu)\] Here we write \(\ev{n}\) to disambiguate the function \(\ev{n}(\omega)\) from the particle number operator \(n_p=c^\dagger_p c_p\).
If we assume equilibrium, then the following relations hold:
\[\ev{n_p}=f_F(\epsilon-\mu)\] \[\ev{\bar n_p}=1-\epsilon \ev{n_p}\]
For \(T\rightarrow 0\): \[S(\omega)=-\mathrm{sgn}(\omega-\mu)=\mathrm{sgn}(\mu-\omega)\]
\[\lim_{\delta\rightarrow 0^+}\frac{1}{\omega-\epsilon_p\pm i\delta}= \mathcal{P}\frac{1}{\omega-\epsilon_p}\mp i\pi\delta(\omega-\epsilon_p)\] where \(\mathcal{P}\) denotes the Cauchy principal value. For a singularity at the finite number \(a < b < c\), it is1 \[\mathcal{P}\int_a^b f(x)\dd{x}=\lim_{\epsilon\rightarrow 0^+}\bqty{ \int_a^{b-\epsilon}f(x)\dd{x}+\int_{b+\epsilon}^c f(x)\dd{x} }\]
Hamiltonian: \[H_0=\sum_p\epsilon_p c^\dagger_p c_p\]
Creation- and annihilation operators: \[c_p^\dagger(t)=e^{i\epsilon_p t}c_p^\dagger(0),\; c_p(t)=e^{-i\epsilon_p t}c_p(0)\]
Retarded GF: \[G_r(p,\omega)=\frac{1}{\omega-\epsilon_p+i\delta}\]
Advanced GF: \[G_a(p,\omega)=\frac{1}{\omega-\epsilon_p-i\delta}\] \[G_a(p,\omega)=G_r(p,\omega)^*\]
Greater GF: \[G_>(p,\omega)=2\pi i\delta(\omega-\epsilon_p)\ev{\bar n_p}\]
Lesser GF: \[G_<(p,\omega)=2\pi i \epsilon \delta(\omega-\epsilon_p)\ev{n_p}\]
Keldysh GF: \[G_K(p,\omega)=-2\pi i \delta(\omega-\epsilon_p) (1-2\epsilon\ev{n_p})\]
\[G_K(p,\omega)=\bqty{G_r(p,\omega)-G_a(p,\omega)}S(\omega)\]
Let \(\underline A,\underline B\) be matrices in Keldysh space
\[\underline{A}=\left(\begin{array}{c|c} A^r & A^K\\ \hline 0 & A^a \end{array}\right),\, \underline{B}=\left(\begin{array}{c|c} B^r & B^K\\ \hline 0 & B^a \end{array}\right)\] where the components are matrices themselves. Then we have the following relations:
Product: \[\underline{A}\cdot\underline{B}=\left(\begin{array}{c|c} A^r B^r & A^r B^K + A^K B^a\\ \hline 0 & A^a B^a \end{array}\right)\]
Inverse: If \(B=A^{-1}\) then \[B^r = (A^r)^{-1},\;B^a = (A^a)^{-1}\] \[B^K = -(A^r)^{-1}A^K (A^a)^{-1}\] This implies: \[(B^{-1})^K = -A^r(A^{-1})^K A^a\]
\[\underline G=\underline G_0+\underline G_0*\underline \Sigma*\underline G\]
\[\underline G=\underline G_0+\underline G_0\cdot\underline \Sigma\cdot\underline G\]
\[\begin{equation} \label{eq:dyson2} \underline G^{-1}=\underline G_0^{-1}-\underline \Sigma \end{equation}\]
If the only perturbation is a single particle potential \[V=\Theta(t-t_0)\sum_p \mathcal{V}_p c_p^\dagger c_p\,,\] the self energy is2 \[\underline{\Sigma}_p=\mathcal{V}_p\cdot 1\!\!1_{2\times 2} =\begin{pmatrix}\mathcal{V}_p&0\\0&\mathcal{V}_p\end{pmatrix}\] which is diagonal in \(p,p'\) \[\underline\Sigma_{pp'}=\delta_{pp'}\underline\Sigma_p\]