# Relation between Green's function and its Hamiltonian

I wanted to know how the relation between the definition of the retarded Green's function $$G^{R}(1,2)=\theta(t_1-t_2)\langle\{\Psi_{H}(1)\Psi^{\dagger}_{H}(1')\}\rangle$$ and its use in some places as the operator: $$G^{R}(w)=\frac{1}{H-w+i\delta}$$ comes to pass, where $H$ is non-interacting.

The closest I have come to this is the relationship between the causal Green's function and its Hamiltonian:

$$G(1,1')^{-1}=(-\frac{\partial}{\partial t_1}-H)\delta(1,1').$$

I have tried taking the derivative inside the delta function but the result doesn't add up.

I understand that in Quantum mechanics, the "one-body retarded green's function" has the same operator expression that I have given earlier. But I have seen this being used in some places in an essentially many-body context.

Let me change notation a little. I'll consider the general non-interacting one-particle Green's function--you can easily generalize to the retarded case if you wish, it's the same basic idea:

$$G_\sigma^0(r_1,\,t_1;\,r_2,\,t_2)=-i\langle \psi_0|\left\{ \Theta(t_1-t_2)c_\sigma(r_1,\,t_1)c_\sigma^\dagger(r_2,\,t_2)-\Theta(t_2-t_1)c_\sigma^\dagger(r_2,\,t_2)c_\sigma(r_1,\,t_1)\right\}|\psi_0\rangle$$

Now, we take a spatial Fourier transform:

\begin{align} G_\sigma^0(k,\,t)& =\int dt \int d^3 r e^{-ik\cdot r}G_\sigma(r,\,t)\notag\\ &=-i\langle \psi_0|\{\Theta(t)c_{k\sigma}(t)c_{k\sigma}^t)-\Theta(-t)c_{k\sigma}^\dagger(t)c_{k\sigma}(t)\}|\psi_0\rangle\end{align}

From the Heisenberg equation of motion, we can take the time dependence out of the operators:

$$\frac{d}{dt} c_{k\sigma}(t)=\frac{i}{\hbar}[H,\,c_{k\sigma}(t)]$$

I won't give you the full solution, but I will give you the end result:

$$c_{k\sigma}(t)=e^{-\frac{i}{\hbar}\epsilon_k t}c_{k\sigma}$$

With this, the one-particle Green's function is given by

\begin{align} G_\sigma^0(k,\,t)&=-i\langle \psi_0|\left\{ \Theta(t)c_{k\sigma}c_{k\sigma}^\dagger -\Theta(t)c_{k\sigma}^\dagger c_{k\sigma}\right\}|\psi_0\rangle e^{-\frac{i}{\hbar}\epsilon_k t}\notag\\ &=-i\left\{ \Theta(t)(1-n_{k\sigma})-\Theta(-t)n_{k\sigma} \right\}e^{-\frac{i}{\hbar}\epsilon_k t} \end{align}

We now perform a temporal Fourier transform, all the while attaching an extra $e^{-\delta |t|}$ term to ensure good behavior at infinity:

\begin{align} G_\sigma^0(k,\,\sigma)&=\int dt e^{i\omega t-\delta t}G_\sigma(k,\,t) \end{align}

I think you can solve this integral with what I gave you--the solution is

$$G_\sigma^0(k,\,\omega)=\frac{1-n_{k\sigma}}{\omega-\epsilon_{k}+i\delta}+\frac{n_{k\sigma}}{\omega-\epsilon_{k}-i\delta}$$

Where in the above I have taken units where $\hbar$ is unity. Defining $\delta_k=\textrm{sgn}(\epsilon_{k}-\epsilon_F)\delta$ (where $\epsilon_F$ is the Fermi energy), we find a nice concise form of the above:

$$G^0(k,\,\omega)=\frac{1}{\omega-\epsilon_k+i\delta_k}$$

In the interacting system, we find something similar, except now with a term known as the self-energy $\Sigma(k,\,\omega)$:

$$G(k,\,\omega)=\frac{1}{\omega-\epsilon_k-\Sigma(k,\,\omega)+i\delta_{k}}$$

This self-energy can be thought of as a renormalization of the energy that comes about when we turn on interactions. In fact, we can rewrite the interacting Green's function in the form

$$G(k,\,\omega)=\frac{Z_k}{\omega-\widetilde{\epsilon_k}+i\delta_k}$$

where $Z_k$ is the quasiparticle weight

$$Z_k=\frac{1}{1-\frac{\partial \Sigma(k,\,\omega)}{\partial \omega}\bigg|_{\omega=\epsilon_k}}$$

and we have expanded around the pole at $\omega=\widetilde{\epsilon_k}$.

If you want to learn more about the details of the Green's function approach to many body physics, I'd suggest Zagoskin or Dickhoff and Van Neck's books. They are both excellent and up-to-date resources for the subject.

EDIT: If you wish to consider the Green function as an operator, look back to when I first took out the exponential term:

\begin{align} G_\sigma^0(k,\,t)&=-i\langle \psi_0|\left\{ \Theta(t)c_{k\sigma}c_{k\sigma}^\dagger -\Theta(t)c_{k\sigma}^\dagger c_{k\sigma}\right\}|\psi_0\rangle e^{-\frac{i}{\hbar}\epsilon_k t}\notag\\ \end{align}

We can ignore the sandwiched states and perform the temporal Fourier transform immediately to get the Green function in terms of the number operator, which is directly proportional to the non-interacting Hamiltonian.

• You might want to add how G(k,w) can be represent as an operator. – Arnab Barman Ray Jun 23 '17 at 17:21
• Look back to where I first pulled out the exponential term. All you have to do is not apply the number operator to the state $|\psi_0\rangle$, perform the temporal Fourier transform, and you have the Green function in the form of an operator. – Joshuah Heath Jun 23 '17 at 17:25
• I mean you should add it to your answer for the sake of completeness since its the main thing that the question asks. Your answer is valid only for translationally invariant problems, which is okay since I have seen the operator being used only in such cases. Also, if I understand it correctly, G(k,w) is just the matrix representation of the operator G(w) in the eigen-basis of the Hamiltonian. Thanks a lot! – Arnab Barman Ray Jun 23 '17 at 17:33