# How to derive the retarded Green's function matrix for a quadratic Hamiltonian?

Start with the quadratic Hamiltonian for fermion: $$\hat{H}=\sum_{ij}H_{ij}\hat{c}_i^\dagger \hat{c}_j$$

and the definition of retarded Green's functon in time domain: $$G_{i,j}^r(t_1,t_2)=-i\theta(t_1-t_2)\langle[\hat{c}_i(t_1),\hat{c}^\dagger_j(t_2)]_{+}\rangle=\langle\langle\hat{c}_i(t_1)|\hat{c}^\dagger_j(t_2)\rangle\rangle.$$

How does one derive the following matrix Green's function in energy domain? $$G^r(E)=\dfrac{1}{E+i\eta-H}.$$ Here $$H$$ is the Hamiltonian matrix with element $$H_{ij}$$ in our Hamiltonian $$\hat{H}$$.

• I have correctified the typo in the definition of Green's function. – Jack Dec 26 '18 at 0:43

$$\mathbf{Note :}$$ Here equation of motion method is used to deduce the result OP is after.

With the OP's hamiltonian : $$\hat H = \sum_{i,j=1,1}^{N,N}\mathbb{H}_{ij}^{}\hat c_{i}^{\dagger} \hat c_{j}^{}$$ and the definition of OP's retarded Green's function as : $$\mathbb{G}_{ij}^{}(t_{1}^{},t_{2}^{})=-\frac{i}{\hbar}\Theta(t_{1}^{}-t_{2}^{})\langle \{ \hat c_{i}^{}(t_{1}^{}),\hat c_{j}^{\dagger} (t_{2}^{}\} \rangle$$ assumed here is the standard variant, presuming OP's question has a typo.

Here $$\hat O(t)= e_{}^{\frac{i}{\hbar}\hat H t} \hat O e_{}^{-\frac{i}{\hbar}\hat H t}.$$

It can be shown that $$\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})$$ (matrix of $$\mathbb{G}_{ij}^{}(t_{1}^{},t_{2}^{})$$'s) satisfies the following differential equation's :

$$\boxed{\left[i\hbar\frac{\partial}{\partial t_{1}^{}}\mathbb{I} -\mathbb{H}\right]\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})=\delta(t_{1}^{}-t_{2}^{})\mathbb{I}}$$ and $$\boxed{\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})\left[-i\hbar\frac{\partial}{\partial t_{2}^{}}\mathbb{I} -\mathbb{H}\right]=\delta(t_{1}^{}-t_{2}^{})\mathbb{I}}$$ here $$\mathbb{I}$$ is an identity matrix of dimension, $$N \times N$$.

The solution of above two equations should clearly be of the form (the infamous jump discontinuity) : $$\boxed{\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})=\Theta(t_{1}^{}-t_{2}^{})\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})+\Theta(t_{2}^{}-t_{1}^{})\mathbb{G}_{}^{<}(t_{1}^{},t_{2}^{})}$$ with (the magnitude of jump discontinuity) $$\boxed{\mathbb{G}_{}^{>}(t_{}^{},t_{}^{})-\mathbb{G}_{}^{<}(t_{}^{},t_{}^{}) = -\frac{i}{\hbar}\mathbb{I}}$$ which can be deduced by integrating either first differential equation with respect to $$t_1$$ between $$t_2+0^+$$ and $$t_2-0^+$$ and using the above ansatz proposed followed by the identification $$t_2=t$$ or integrating second differential equation with respect to $$t_2$$ between $$t_1+0^+$$ and $$t_1-0^+$$ and using the above ansatz proposed followed by the identification $$t_1=t$$.

Further from the definition of $$\mathbb{G}_{ij}^{}(t_{1}^{},t_{2}^{})$$ it can be seen that $$\mathbb{G}_{}^{<}(t_{1}^{},t_{2}^{})=\mathbb{O}$$. Hence $$\mathbb{G}_{}^{>}(t_{}^{},t_{}^{}) = -\frac{i}{\hbar}\mathbb{I}$$.

Using thus deduced results and the resultant ansatz i.e., $$\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})=\Theta(t_{1}^{}-t_{2}^{})\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})$$ the following equations results,

$$\left[i\hbar\frac{\partial}{\partial t_{1}^{}}\mathbb{I} -\mathbb{H}\right]\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})=\mathbb{O}$$ and $$\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})\left[-i\hbar\frac{\partial}{\partial t_{2}^{}}\mathbb{I} -\mathbb{H}\right]=\mathbb{O}$$ which can be easily integrated from $$t$$ to $$t_{1}^{}$$ and $$t$$ to $$t_{2}^{}$$ respectively to get $$\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})=e_{}^{-\frac{i}{h}\mathbb{H} (t_{1}^{}-t)}\mathbb{G}_{}^{>}(t_{}^{},t_{}^{})e_{}^{\frac{i}{h}\mathbb{H} (t_{2}^{}-t)}$$ which upon using $$\mathbb{G}_{}^{>}(t_{}^{},t_{}^{}) = -\frac{i}{\hbar}\mathbb{I}$$ gives : $$\mathbb{G}_{}^{>}(t_{1}^{},t_{2}^{})=-\frac{i}{\hbar}e_{}^{-\frac{i}{h}\mathbb{H} (t_{1}^{}-t_{2}^{})}$$ using which finally the following falls out : $$\boxed{\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})=-\frac{i}{\hbar}\Theta(t_{1}^{}-t_{2}^{})e_{}^{-\frac{i}{h}\mathbb{H} (t_{1}^{}-t_{2}^{})}}.$$ clearly $$\mathbb{G}_{}^{}(t_{1}^{},t_{2}^{})$$ is time translationally invariant, hence without loss off generality, the following can be considered instead : $$\mathbb{G}_{}^{}(t)=-\frac{i}{\hbar}\Theta(t)e_{}^{-\frac{i}{h}\mathbb{H} t}.$$ Fourier transforming this as $$\tilde{\mathbb{G}}_{}^{}(E)=\int_{-\infty}^{+\infty}dt\mathbb{G}_{}^{}(t)e_{}^{\frac{i}{\hbar}E t}$$ finally reveals the sought after result : $$\boxed{\tilde{\mathbb{G}}_{}^{}(E)=\left[E\mathbb{I}-\mathbb{H}+i 0_{}^{+}\mathbb{I}\right]_{}^{-1}}.$$