Is the real part of Green's function directly observable? In many-body physics, the imaginary part of a Green's function corresponds to the signal intensity of some scattering experiments.
Does the real part of a Green function directly correspond to any experimental observable?
 A: Green's function itself is an abstract mathematical object, and per se is not directly observable. However, the observable quantities can be expressed in terms of Green's functions, thus giving us the idea of what the Green's function is like.
The most well-known case is the imaginary part of full Green's function, which corresponds to the density of states. E.g., if the Green's function of a level, broadened by interactions, is: 
$$ G(\omega) = \frac{1}{\omega -\epsilon_0 +\frac{i\Gamma}{2}},$$
(I took for simplicity a retarded Green's function, and take $\hbar =1$.)
then the associated density of states is
$$\rho(\omega) = -\frac{1}{\pi}\Im\left[G(\omega)\right] = \frac{\Gamma}{2\pi}\frac{1}{(\omega - \epsilon_0)^2+\frac{\Gamma^2}{4}}.$$
Considering a slightly more general case, the Green's function is given by
$$ G(\omega) = \frac{1}{\omega -\epsilon_0 +\Sigma(\omega)},$$
where $\Sigma(\omega)$ is itself a combination of Green's function. In a simple case of scattering from an entity with Green's function $g(\omega)$, we will have 
$$\Sigma(\omega) = |V|^2g(\omega),$$
which means that
$$\rho(\omega) = -\frac{1}{\pi}\Im\left[G(\omega)\right] = \frac{\Im[g(\omega)]}{\pi}\frac{1}{(\omega - \epsilon_0 - |V|^2\Re[g(\omega)])^2+\Im[g(\omega)]^2}.$$
If $\Im[g(\omega)]$ has a simple form (e.g., $\Im[g(\omega)]=\Gamma/2$) we can study $\Re[g(\omega)]$. Such study is in fact routinely done in all kinds of interference experiments, e.g., in the mesoscopic Aharonov-Bohm interferometers.  
