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.

• 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 in the absence of interactions.

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.

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.

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.

• 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 in the absence of interactions.