# Non-interacting causal Green function in localized-orbitals representation

Say I have a one-partilce hamiltonian $$\hat{h}=\sum_{\alpha} \epsilon_\alpha \hat{n}_\alpha+\sum_{\alpha \neq \beta, } t_{\alpha,\beta} \hat{c}^\dagger_{\alpha }\hat{c}_{\beta }$$ (I will ignore spin for simplicity, since it does not play a relevant part in my doubt), where the $\hat{c}_\mu,\hat{c}_\mu^\dagger$ are anihilation/destruction operators of electrons in states $\vert \chi_\mu \rangle .$

Quick version of my question: What is the causal Green function i frequency space for this problem?

Detailed question and problems: I want to write down explicitely what are the Green functions $g(\mu,t)= \frac{-i}{\hbar} \langle T [\hat{c}_\mu(t) \hat{c}_\mu^\dagger(t^\prime)]\rangle$ in frequency space, $g(\mu,\omega),$ and I want to do it by two methods: 1) Applying the direct definition and explicitely computing the Fourier transform and 2) By using that, if we call $h$ to the matrix of elements $\langle \chi_\nu \vert \hat{h} \vert \chi_\mu \rangle,$ it holds that $g(\mu,\omega)=(\omega \ \text{Id}-h)^{-1}(\mu,\mu)$ (this holds for one-electron hamiltonians and shows that, in that case, the notion of Green function in many-body theory coincides with the concept of Green function in pure mathematics).

1) First method: For simplicity, we assume $t^\prime =0$ and I will drop the hats from the operators, since there's no possible confussion We have: $\langle T [\hat{c}_\mu(t) \hat{c}_\mu^\dagger(0)]\rangle= \Theta(t)\langle c_\mu(t)c_\mu^\dagger\rangle-\Theta(-t) \langle c_\mu^\dagger c_\mu(t) \rangle.$ We insert in both terms the resolution of the identity, $\text{Id}=\sum_{m} \vert m \rangle\langle m \vert$, where $m$ is an index which runs over all eigenstates of the system. On the other hand, recall that $c_\mu(t)=e^{i h t}c_\mu e^{-i h t}$ (setting $\hbar=1$) and that $e^{-i h t} \vert m \rangle = e^{-i E_m t} \vert m \rangle.$ Keeping this in mind is easy to get to: $$\Theta(t)\sum_{m}e^{i(\omega_0-\omega_m)t}\langle 0 \vert c_\mu \vert m \rangle \langle m \vert c_\mu^\dagger \vert 0 \rangle-\Theta(-t)\sum_{m}e^{i(\omega_m-\omega_0)t}\langle m \vert c_\mu \vert 0 \rangle \langle 0 \vert c_\mu^\dagger \vert m \rangle.$$

Next, to Fourier transfrom this expression we notice that for the first term we need a convergent factor $i\eta$ (that is, we'll have pole sin the upper semi-plane), and for the second term we need to plug in a convergent factor $-i\eta$ (so the poles lie in the lower semi-plane).

After computing those Fourier transforms and replugging the prefactor $-i$, we get:

$$\sum_ {m} \dfrac{\langle 0 \vert c_\mu \vert m \rangle \langle m \vert c_\mu^\dagger \vert 0 \rangle}{\omega-(\omega_m-\omega_0)+i\eta}+\sum_{m}\dfrac{\langle m \vert c_\mu \vert 0 \rangle \langle 0 \vert c_\mu^\dagger \vert m \rangle}{\omega-(\omega_0-\omega_m)-i\eta}.$$