This is a somewhat broad question, because there are a number of different Green's functions in quantum physics. Perhaps the simplest one is the resolvent Green's function for a single-particle system. Its definition is $$G(\omega^{\pm})=\lim_{\delta \rightarrow 0^+}\left[ \omega \pm i\delta - H \right]^{-1}\equiv \frac{1}{\omega\pm i\delta - H},$$ where $H$ is the Hamiltonian. Ignoring the $\delta$, and replacing $\omega \rightarrow E$, you can see how this relates to the time-independent Schrödinger equation: $$H\left|\psi \right>=E\left|\psi \right> \longrightarrow \left[ E - H\right]\left|\psi \right> = 0.$$ Basically, if you apply the operator $G$ to a solution to the Schrödinger equation, $\left|\psi_n \right>$ with energy level $E_n$, you get a pole (the "denominator" vanishes) at $\omega \pm i\delta = E_n$. That is, $$G_{nn}(\omega^{\pm})\equiv\left<\psi_n \right|G(\omega^{\pm}) \left|\psi_n \right>=\frac{1}{\omega \pm i\delta - E_n},$$ has a pole. Then it is clear that the poles of (Tr = trace) $$\mathrm{Tr}\{G(\omega^{\pm})\}\equiv\sum_{n}^{\mathrm{all\,states}}G_{nn}(\omega^{\pm}),$$ give you the full spectrum. In fact, you can show that the quantity (Im = imaginary part) $$\rho(\omega)=-\frac{1}{\pi}\mathrm{Im}\{\mathrm{Tr}\{G(\omega^+)\}\},$$ gives you the density of states of the Hamiltonian $H$.
The resolvent Greens function is valid for a single-particle system, but the concept caries over well to many-body physics, and thus to quantum field theory. However, the definition of the Green's function is not as transparent in those cases. The Green's function is given by the probability amplitude that a particle will be added to the vacuum state at a time $t$ and state $n$, and that after time evolution it will be removed at time $t'$ and state $n'$: $$\mathcal{G}(t',n';t,n)=i\left<\Omega \right|\psi_{n'}(t')\psi_n^\dagger (t) \left|\Omega \right>,$$
where the factor of $i$ appears just as a convention, and $\left|\Omega \right>$ represents the vacuum state. Once again, the definition is very transparent, because this quantity is the answer to the question: if in the quantum vacuum, which is (to quote Lawrence Krauss' favorite phrase) "a boiling bubbling brew of particles popping in and out of existence," a particle were to pop into existence at time-state $(t,n)$, what is the probability (amplitude) that it would propagate to time-state $(t',n')$? There are a number of different versions of this type of Green's function, each useful for a different thing. One thing all of these have in common is that handling them in this form is complicated, and it is much easier to work with their Fourier transforms $$\mathcal{G}_{n'n}(\omega^{\pm})=\int\mathrm{d}(t'-t)\,\mathrm{e}^{i\omega (t'-t) \pm \delta}\mathcal{G}(t',n';t,n).$$ As you can see, a complex shift $\delta$ is introduced here, which has to do with the convergence of the integral. This is precisely the origin of it in the resolvent Green's function. In fact, if you have a system where the particles don't interact with each other, and instead each particle simply follows the Hamiltonian $H$ with energies $\{E_n\}$, then you can show that $$\mathcal{G}_{n'n}(\omega^{\pm})=\frac{\delta_{n',n}}{\omega\pm i\delta - E_n},$$ and you get a full analogy with the single-particle Green's function, and the answer to the question above is "if the particle is created at an eigen-energy of the Hamiltonian, and propagates to be annihilated at the same eigen-energy, you get a peak in probability (a pole); if it is not, you get zero."
However, things get more interesting when you do have interactions, and your Hamiltonian becomes $H + U$. For that case, it can be shown that $$\mathcal{G}_{n'n}(\omega^\pm) = \frac{1}{\omega - E_n - \Sigma_{n'n}(\omega^\pm)},$$ where $\Sigma_{n'n}(\omega^\pm)$ is called the self energy, and the equation above is called the Dyson equation. In many cases this will have a simple form when you calculate specifically $\mathcal{G}_{nn}$: $$\mathcal{G}_{nn}(\omega) = \frac{1}{\omega - (E_n + \Lambda(\omega) ) - i \Gamma_{n}(\omega) }.$$ This has a very simple meaning, when you compare it to the non-interacting case: first, if you ignore the $\Gamma$, you have that because of the interactions the pole has moved from $E_n$ to $E_n + \Lambda$. Then, if you consider $\Gamma$, you will find that the pole is not infinitely "narrow," but instead has a width $\approx \Gamma$. In fact, this expression is nothing but a type of Lorentz distribution. Basically, because of interactions, your energy level is not longer infinitely lived, but now decays into the vacuum in a time scale given by $\hbar/\Gamma$. $\Gamma$ is sometimes called the scattering rate, or decay rate. But because there is still an analogy with the resolvent Green's function, one says that this state, which has a finite lifetime, is an energy state for a new particle---a quasi-particle---that emerges from the interacting system.
