# Measuring Green's Function

I am confused about Green's functions. I understand that $G^R(r,r',t,t')$ tells you the amplitude of a particle that was inserted into the system at position $r$ at time $t$ has to be at some position $r'$ at some later time $t'$. But what about $G^>(r,r',t-t')$ (assuming local in time) or $G^>(r,r',\omega)$? How would you measure $G^>(r,r',\omega)$?

The notations are not universal, so I do not know what you mean by $G^{<,>,R}$ but clearly a Green's function CAN NOT be measured. A Green's function is just a mathematical artefact, it exists in the form of two points correlators, like

$$G\left(x_{1},x_{2}\right)\sim\left\langle a\left(x_{1}\right)a^{\dagger}\left(x_{2}\right)\right\rangle$$

where the average $\left\langle \cdots\right\rangle$ is usually done with respect to a statistical average over a ground state, and $a$ is an annihilation operator (bosonic or fermionic), and $\sim$ means that I did not put the possible prefactor (usually $\pm i/\hbar$, $\pm i$, $\pm 1/\hbar$ or $\pm 1$ depending on the context and the chosen dimensionally) and any kind of time-ordering (eventually time-anti-ordering) operators or $\Theta$-like function which define the domain of validity of the Green's function in time.

Green's function also exists in the form of higher number of points, like for instance

$$G^{\left(4\right)}\left(x_{1},x_{2},x_{3},x_{4}\right)\sim\left\langle a\left(x_{1}\right)a\left(x_{2}\right)a^{\dagger}\left(x_{3}\right)a^{\dagger}\left(x_{4}\right)\right\rangle$$

where one sees the ordering of the operators might be of importance.

All this to say that ultimately, a Green's function has no intrinsic meaning at all. You can always say it measures the correlation between two points ... but it has no intrinsic meaning.

Once you define a Green's function, you still have to define all observables in term of this Green's function. In the same way a wave-function has no meaning at all. Once you got the Green's function and the set of rules for calculating observables, you can start doing calculations !

In short : a Green's function itself is NOT an observable, it is a mathematical tool.

Green's function method applied to many-body problem is reviewed quickly in

If you really want to use Green's functions for many-body applications, I suggest you to not open textbook on quantum field theory, since you will see many many terms related to relativistic problems only. At least not at the beginning. The subtle differences between the statistical field theory and the quantum field theory (that's the terminology for many-body vs. relativistic problems) might be quite confusing for a beginner. See e.g. this answer of mine for a few more details about the difference between the two.

Note finally that the imaginary part of a two-points Green's function is related to the decay time of a particle, that their poles are usually eigenvalues (up to prefactors) of Hamiltonians (in simple situations, for instance without interaction), and that many many authors forget to precise which convention they use for their Green's function. They sometimes do not even give the definition of the observables... So it is sometimes frustrating for beginners to try learning about this topic.

• Specifically, the Green's function I am interested in is: $\left<\Psi(r,t)\Psi^{\dagger}(r',t')\right>$, where $\Psi(r,t)$ is a the field operator that annihilates a particle at position r at time t. – Heracles Jan 27 '17 at 18:46