How to derive the analytical expression for the retarded Green's function with quadratic Hamiltonian?

Question

For two operators, $A(t)$ and $B(t)$ the retarded Green’s function is defined as \begin{equation} G^R(t,t') \equiv \langle \langle A(t)|B(t) \rangle \rangle^R = -i\theta(t-t')\langle \{A(t),B(t')\} \rangle \end{equation}

then one can show that \begin{align*} i\dfrac{\partial}{\partial t} G^R(t,t') & = \delta(t-t')\langle \{A(t),B(t')\} \rangle - i \theta(t-t')\langle \{i\dfrac{\partial A(t)}{\partial t},B(t')\} \rangle \\ & = \delta(t-t')\langle \{A(t),B(t')\} \rangle - i \theta(t-t')\langle \{[A(t),H(t)],B(t')\} \rangle \\ & = \delta(t-t')\langle \{A(t),B(t')\} \rangle + \langle \langle [A(t),H(t)]|B(t') \rangle \rangle^R \end{align*} If the Hamiltonian $H$ in Schrodinger picture is independent of time, then the correlation functions depend on $(t-t')$, not on $t$ and $t'$ separately. We can go to the Fourier space,the EOM becomes \begin{equation} \boxed{\omega \langle \langle A|B \rangle\rangle^R = \langle \{A,B\} \rangle + \langle \langle [A,H]|B \rangle\rangle^R}. \end{equation}

Starting with this formula, I want to derive the analytical expression for retarded Green's function with the following Hamiltonian $H$ (fermionic system): \begin{equation} H = \sum_k x a_k^\dagger a_k + \sum_{m \neq n} y (a^\dagger_m a_n + a^\dagger_n a_m) \end{equation}

This following is my solution: \begin{equation} \boxed{A=a_s,B=a^\dagger_t} \Rightarrow \omega \langle \langle a_s|a^\dagger_t \rangle \rangle^R = \langle \{a_s,a^\dagger_t\} \rangle + \langle \langle [a_s,H]|a^\dagger_t \rangle \rangle^R \end{equation}

\begin{align*} [a_s,H] & =\left[a_s,\sum_k x a_k^\dagger a_k + \sum_{m \neq n} y (a^\dagger_m a_n + a^\dagger_n a_m) \right ]\\ & = \sum_k x \{a_s,a_k^\dagger\}a_k + \sum_{m \neq n} y \{a_s,a_m^\dagger\}a_n +\sum_{m \neq n} y \{a_s,a_n^\dagger\}a_m \\ & = x a_s+ \sum_{n} y a_n \qquad (s=m \neq n) \end{align*}

\begin{equation} \omega G_{st}^R = \delta_{st} + x G_{st}^R + \sum_n y \langle \langle a_n|a^\dagger_t \rangle \rangle^R \Rightarrow G_{st}^R = \dfrac{\delta_{st}+y\sum_n G^R_{nt}}{\omega-x} \end{equation}

But this result is the final solution? Or how can I further simplify my results?

Answer 1

$G_{st}$ is a matrix in the space of indices $s,t$, and the final equation (before $\Rightarrow$ symbol) is a matrix equation: $$ \omega G_{st}^R = \delta_{st} + x G_{st}^R + \sum_n y G^R_{nt}, $$ which can be written as $$ \left(\omega - x\right) G_{st}^R - \sum_n y G^R_{nt}= \delta_{st},\\ \sum_n\left[\left(\omega - x\right)\delta_{nt} - y\right] G^R_{nt}= \delta_{st},\\ \sum_n\left([G^R]^{-1}\right)_{s,n}G_{nt}^R=\delta_{st} $$ or in matrix form: $$ [G^R]^{-1}G_{nt}^R=1 $$ We thus essentially need to invert the matrix in the left-hannd-side: $$ \left([G^R]^{-1}\right)_{s,n}=\left(\omega - x\right)\delta_{nt} - y $$ E.g., in the case of two levels, we get: $$ [G^R]^{-1} = \begin{bmatrix} \omega - x&y\\y& \omega-x \end{bmatrix}, $$ so that $$ G^R=\frac{1}{(\omega - x)^2-y^2} \begin{bmatrix} \omega - x&-y\\-y& \omega-x \end{bmatrix}, $$

Answer 2

$\textbf{Quick hint:}$ Using this answer. You just have to identify a specific form of $\mathbb{H}=\left(x-y\right)\mathbb{I}_{N \times N}^{}+ y \mathbb{J}_{N \times N}^{}$, with $\mathbb{I}_{N \times N}^{}$ and $\mathbb{J}_{N \times N}^{}$ being identity matrix and matrix of ones respectively. $\mathbb{G}_{}^{R}(E)=\left[E\mathbb{I}-\mathbb{H}+i 0_{}^{+}\mathbb{I}\right]_{}^{-1}$ can be found analytically using Sherman-Morrison formula.