# Help with understanding Green's Functions

I. The Green's Function Method

The Green's function is immensely useful as a tool in Solid State Physics. Using a Green's function, one can compute all relevant data from a physical system. For example, the Green's function for the time-independent Schrodinger equation (TISE), $$G(E):=\frac{1}{H-E}$$ yields both the density of states, $$-\frac{1}{\pi}\lim_{\epsilon\to 0^+}\text{Im}\,\text{Tr}\,G(E+i\epsilon)=\sum_n\delta(E-E_n)~~~~~~~~~~~~~~~$$ that is, the eigenvalues, and, letting $\{\psi_n\}$ denote the associated eigenstates, the Green's function also yields the projected density of states, $$-\frac{1}{\pi}\lim_{\epsilon\to 0^+}\text{Im}\,(f_0,G(E+i\epsilon)f_0)=\sum_n|(f_0,\psi_n)|^2\delta(E-E_n)$$ which is equivalent to the eigenstate data. Moreover, Green's functions allow us to efficiently formulate effective field theory, perturbation theory, and the renormalization group in the Hamiltonian picture, which is indispensable.

II. Many-Body Green's Functions: Where Everything Falls Apart:

However, this is all misleading: when physicists mention, "the Green's function of a non-interacting Hamiltonian $H$", which is, explicitly, $$(\psi_0, T\{\Psi^\dagger(x,t)\Psi(x',t')\}\psi_0),$$ Where $\psi_0$ is the groundstate of $H$, a literalist would think that they mean the Green's function for the many-body, time-dependent Schrodinger equation (TDSE): $$\frac{1}{\frac{i}{\hbar}H-\partial_t},~~~~~~ H=\sum_{ij}A_{ij}c^\dagger_ic_j.$$ However, close calculation shows that this is instead the Green's function of the associated single-particle Hamiltonian: $$(\psi_0, T\{\Psi^\dagger(x,t)\Psi(x',t')\}\psi_0)=\frac{1}{\frac{i}{\hbar}\mathcal H-i\partial_t},~~~~~~\mathcal H= \sum_{ij}A_{ij}\,\left|f_i\right>\left<f^j\right|.$$ However, this does not generalize straightforwardly to interacting systems. In particular, for an interacting system, there is no such single-particle hamiltonian! So the above virtues of the Green's function method no longer hold. We do not have the density of states, and we don't have the projected density of states.

So here's my question: what use is this method if it only characterizes non-interacting systems, which we already know how to solve? (Also, this gives a very boring renormalization group flow).

• You haven't computed anything with the first formulas, all you have done is to write down formal relationships that may, or may not be well defined in mathematical terms. They certainly do not represent a general solution theory for, not even for single particle QM (which, as far as I am aware, does not exist). – CuriousOne Feb 22 '16 at 8:33
• fair enough, but ignoring my mistakes there is still a valid concern here, namely that a Green's function seems to be being used for the wrong equation (i.e. not the TDSE) – David Roberts Feb 22 '16 at 8:37
• So you are complaining that life is tougher than you thought even for the single particle system and essentially unsolvable for general multi-particle systems? Welcome to physical reality. – CuriousOne Feb 22 '16 at 8:42
• Instead of ad hominem, how about you actually help me (and others) out by providing the actual Green's function, or explaining why the false Green's function actually still has physical relevance in interacting systems. – David Roberts Feb 22 '16 at 8:43
• Non-interacting green's functions are quickies to compute - when did I say those were hard? Interacting ones are ill-defined. Now that's a serious issue. – David Roberts Feb 22 '16 at 8:48

In mathematical physics the resolvent of the Hamiltonian $H$, the operator associated with the Green's function, is quite important in studying various aspects of $H$. See, for instance, the various volumes of "Methods of modern mathematical physics " by Reed and Simon. Consider the time evolution \begin{equation*} \psi (t)=\exp [-iHt]\psi (0) \end{equation*} Its complex Laplace transform is \begin{equation*} \hat{\psi}(z)=\int_{0}^{\infty }dt\exp [izt]\exp [-iHt]\psi (0),\;Im z>0 \end{equation*} Performing the integration \begin{equation*} \hat{\psi}(z)=i[z-H]^{-1}\psi (0) \end{equation*} Note that here the "energy " is non-real so the $E+i\varepsilon$ occurs naturally. In the simplest case the Green's function is \begin{equation*} G(x,y,z)=<x|[z-H]^{-1}|y> \end{equation*} and we can define \begin{equation*} G(x,y,E)=\lim_{\varepsilon \downarrow 0}G(x,y,E+i\varepsilon ) \end{equation*} if it exists which is usually not the case. The advantage of introducing $[z-H]^{-1}$ instead of $\exp [-iHt]$ is that it is much easier to study its properties and the perturbation theory of the spectrum of $H$.