I've recently been trying to compute the Green's function for a non-interacting system of fermions. Since this is a site for mathematicians, for context, let me provide the following definition:



> 
> **Definition:** A **noninteracting system of fermions** is a quantum dynamical system
>  along with the following data:
> 
> 
> * A single-particle Hilbert space $\mathfrak h$, for which the full Hilbert space of the dynamical system is the exterior algebra
>  $\Lambda(\mathfrak h):=\oplus\_{n\geq 0}\,\Lambda^n(\mathfrak h)$.
> * A non-interacting Hamiltonian $H$, which, for some basis $f\_1,\cdots f\_{\dim \mathfrak h}$ of the single particle Hilbert space reads as
>  $H=\sum\_{ij}A\_{ij}c^\dagger(f\_i)c(f\_j).$
> 
> 
> 


I've been puzzled by how physicists go about computing the Green's function of a non-interacting Hamiltonian $H$. To see what I mean, here is a theorem:



> 
> **Theorem (Classification of Non-interacting Systems):** For $(\Lambda(\mathfrak h), A\_{ij})$ a general non-interacting system of
>  fermions, the time-evolution of any $k$-particle state factors in the
>  following manner: 
>  \begin{align\*} e^{itH}(g\_1\wedge \cdots \wedge g\_k)=e^{it\mathcal H}g\_1\wedge \cdots \wedge e^{it\mathcal H}g\_k
> \end{align\*} 
>  Where $\mathcal H= \sum\_{ij}A\_{ij}\,\left|f\_i\right>\left<f^j\right|$ is a single-particle Hamiltonian, and the raised index indicates dualization. In other words, a "non-interacting system of identical fermions" always factors into a set of identical, non-interacting, single-particle systems, where the single-particle dynamics has the replacements \begin{align\*} c^\*(f\_i)\,c(f\_j)\mapsto \left|f\_i\right>\left<f\_j\right|.\\ \end{align\*}
> 
> 
> 


Enough background. When physicists say, "The Green's function of this non-interacting Hamiltonian $H$", I would think that they mean 
$$G:=\frac{1}{\frac{i}{\hbar}H-\partial\_t},~~~~~~ H=\sum\_{ij}A\_{ij}c^\dagger(f\_i)c(f\_j). $$
However, they really mean the Green's function of the associated single-particle Hamiltonian:
$$G':=\frac{1}{\frac{i}{\hbar}\mathcal H-\partial\_t},~~~~~~\mathcal H= \sum\_{ij}A\_{ij}\,\left|f\_i\right>\left<f^j\right|.$$
However, this does not generalize straightforwardly to interacting systems, and therefore, I am actually curious: ***is there a nice formula for $G$ in terms of $G'$?*** Naively, using the direct-sum decomposition $\Lambda(\mathfrak h)=\oplus\_{k\geq 0}\,\Lambda^k(\mathfrak h)$, we get
$$\frac{1}{\frac{i}{\hbar}H-\partial\_t}=\bigoplus\_{k\geq 0} \frac{1}{\frac{i}{\hbar}H\_k-\partial\_t}$$
So this reduces to computing the Green's function of the $k$-particle Hamiltonian: $(\frac{i}{\hbar}H\_k-\partial\_t)^{-1}$. However, this is as far as I can get on my own.



Is there a true many-body green's function for interacting systems?
You ask for a relation between the Green's function of the single-particle Hamiltonian and the Green's function of the many-particle Hamiltonian, in the case of non-interacting particles (fermions or bosons). Let me try to explain that the *retarded* Green's functions are identical.


• For the *single-particle* Hamiltonian $H(x)=-\partial\_x^2+V(x)$ the retarded Green's function is defined by
$$i\partial\_t G(x,x';t,t')=H(x)G(x,x';t,t'),\;\;t>t',$$
with the condition that $G(x,x';t,t')\equiv 0$ for $t<t'$ and 
$$\lim\_{t\downarrow t'}G(x,x';t,t')=-i\delta(x'-x).$$


• The retarded *many-particle* Green's function is defined as the ground state expectation value $\langle\cdots\rangle$ of the (anti-)commutator of field operators $\hat\psi(x,t)$,
$${\cal G}(x,x';t,t')=-i\langle\hat\psi(x,t)\hat\psi^\dagger(x',t')\pm\hat\psi^\dagger(x',t')\hat\psi(x,t)\rangle\theta(t-t').$$
The function $\theta(t)$ is the unit step function, the $+$ sign is for fermions and the $-$ sign for bosons. The field operator satisfies the operator equation
$$i\partial\_t\hat\psi(x,t)=[\hat\psi(x,t),\hat{\cal H}],$$
with $\hat{\cal H}$ the many-particle Hamiltonian operator and $[\cdot,\cdot]$ the commutator. Note also the equal-time (anti-commutation) relation
$$\hat\psi(x,t)\hat\psi^\dagger(x',t)\pm\hat\psi^\dagger(x',t)\hat\psi(x,t)=\delta(x-x').$$


• Now let us compare the two functions $G(x,x';t,t')$ and ${\cal G}(x,x';t,t')$. Both vanish for $t<t'$ and both satisfy the delta-function limit when $t\downarrow t'$. But in general the function ${\cal G}$ is not the Green's function of any differential equation, unlike $G$.


However, for *non-interacting* particles we have
$$[\hat\psi(x,t),\hat{\cal H}]=(-\partial\_x^2+V(x))\hat\psi(x,t)=H(x)\hat\psi(x,t),$$
hence ${\cal G}$ satisfies the same differential equation
$$i\partial\_t {\cal G}(x,x';t,t')=H(x){\cal G}(x,x';t,t'),$$
as $G$ and we conclude that they are the very same function.




---


I hope this clarifies what physicists mean when they speak of the "many-particle Green's function". The name suggests otherwise, but in general this function is not the Green's function of any differential equation. (Its equation of motion is nonlinear when the particles interact.) The reason why physicists call ${\cal G}$ a Green's function is because it reduces to the Green's function $G$ of the Schrödinger equation in the absence of interactions.


All of this is for the retarded Green's function. The time-ordered Green's function or thermal Green's function are different quantities, that do not reduce to the single-particle Green's function even in the absence of interactions.


