correlation-functionsgreens-functionspropagatorquantum mechanicsquantum-field-theory
In literature, usually two types of definition exist for Green's function.
$\hat{L}G=\delta(x-x')$. This equation states that Green's function is a solution to an ODE assuming the source is a delta function
$G=\langle T\psi(x_1,t_1)\psi^\dagger(x_2,t_2)\rangle$. This definition states that Green's function is something like a propagator.
I want to know the internal relationship between the two definitions.
I) Notational issues: Greens function vs. kernel. First of all, be aware that Ref. 1 between eq. (4-27) and eq. (4-28) effectively introduces the retarded Greens function/propagator
$$\begin{align} G(x_2,t_2;x_1,t_1)~=~&\theta(\Delta t)~K(x_2,t_2;x_1,t_1), \cr \Delta t~:=~&t_2-t_1,\end{align}\tag{A}$$
rather than the kernel/path integral
$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\langle x_2,t_2 | x_1,t_1 \rangle\cr ~=~&\langle x_2|U(t_2,t_1)|x_1 \rangle\cr ~=~&\int_{x(t_1)=x_1}^{x(t_2)=x_2} \! {\cal D}x~ \exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2} \!dt ~L\right] .\end{align}\tag{B} $$
Here $\theta$ denotes the Heaviside step function, and the Lagrangian
$$ L~:=~\frac{m}{2}\dot{x}^2-V(x)\tag{C}$$
is the Lagrangian for a non-relativistic point particle in 1 dimension with a potential $V$.
However, Ref. 1 confusingly denotes the Greens function $G$ with the same letter $K$ as the kernel! See also e.g. this and this Phys.SE posts. Therefore the eq. (4-29) in Ref. 1, which OP asks about, is better written as
$$\begin{align} D_2 G(x_2,t_2;x_1,t_1) ~=~&\delta(\Delta t)~\delta(\Delta x), \cr \Delta x~:=~&x_2-x_1, \end{align}\tag{D} $$
where we introduced the Schrödinger differential operator
$$\begin{align}D_2~:= ~&\frac{\partial}{\partial t_2} + \frac{i}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2}+V(x_2)\right)\cr ~=~&\frac{\partial}{\partial t_2} + \frac{\hbar}{i}\frac{1}{2m}\frac{\partial^2}{\partial x_2^2}+\frac{i}{\hbar}V(x_2).\end{align}\tag{E}$$
II) Proof of eq. (D). The sought-for eq. (D) follows directly from eq. (A) together with the following two properties (F) & (G) of the kernel $K$:
$$ D_2 K(x_2,t_2;x_1,t_1) ~=~0, \tag{F} $$
and
$$ K(x_2,t_2;x_1,t_1) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^+. \tag{G}$$ $\Box$
III) So we can reformulate OP's question as follows.
Why the path integral (B) satisfies eqs. (F) & (G)?
Rather than going on a definition chase, perhaps the following heuristic derivation of eqs. (F) & (G) is the most convincing/satisfying/instructive. For sufficiently short times $|\Delta t| \ll \tau$, where $\tau$ is some characteristic time scale, i.e. in the diabatic limit, the particle only has time to feel an averaged effect of the potential $V$. So, using methods of Ref. 1, in that limit $|\Delta t| \ll \tau$, the path integral (B) reads
$$\begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar \Delta t}} \exp\left\{ \frac{i}{\hbar}\left[ \frac{m}{2} \frac{(\Delta x)^2}{\Delta t}- \langle V\rangle \Delta t \right]\right\}, \end{align}\tag{H} $$
where the averaged potential is of the form
$$\begin{align} \langle V\rangle ~=~& V\left(\frac{x_1+x_2}{2}\right)+{\cal O}(\Delta x)\cr ~=~& V(x_2)+{\cal O}(\Delta x)\cr ~=~& V(x_1)+{\cal O}(\Delta x). \end{align}\tag{I}$$
IV) Proof of eq. (G). Note that it is implicitly assumed in eq. (H) that ${\rm Re}(i\Delta t)>0$ is slightly positive via the pertinent $i\epsilon$-prescription. Equation (G) then follows directly from eq. (H) via the heat kernel representation
$$ \delta(x)~=~ \lim_{|\alpha|\to \infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2},\qquad {\rm Re}(\alpha)~>~0, \tag{J} $$
of the Dirac delta distribution. $\Box$
V) Proof of eq. (F) for sufficiently small times $|\Delta t| \ll \tau$. It is a straightforward to check that eq. (H) satisfies the eq. (F) modulo contributions that vanish as $\Delta t\to 0$, cf. the following Lemma. $\Box$
Lemma. For sufficiently small times $|\Delta t| \ll \tau$, the path integral (H) satisfies $$ D_2 K(x_2,t_2;x_1,t_1) ~=~{\cal O}(\Delta t). \tag{K}$$
Sketched proof of eq. (K): Straightforward differentiation yields
$$ \begin{align} \frac{\partial}{\partial t_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&-\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar}\left[ \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2+ \langle V\rangle +{\cal O}(\Delta t) \right]\right\} K(x_2,t_2;x_1,t_1),\end{align} \tag{L} $$
$$ \begin{align} \frac{\hbar}{i}\frac{\partial}{\partial x_2} &K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(H)}{=}~&\left\{m \frac{\Delta x}{\Delta t} +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1), \end{align}\tag{M}$$
$$ \begin{align} \frac{\hbar}{i}\frac{1}{2m}&\frac{\partial^2}{\partial x_2^2} K(x_2,t_2;x_1,t_1) \cr ~\stackrel{(H)}{=}~&\left\{\frac{1}{2\Delta t} +\frac{i}{\hbar} \frac{m}{2} \left(\frac{\Delta x}{\Delta t}\right)^2 +{\cal O}(\Delta t) \right\} K(x_2,t_2;x_1,t_1). \end{align} \tag{N}$$
Also note that
$$ \begin{align} \left\{V(x_2)-\langle V\rangle \right\}K(x_2,t_2;x_1,t_1) ~\stackrel{(I)}{=}~&{\cal O}(\Delta x) K(x_2,t_2;x_1,t_1)\cr ~\stackrel{(G)}{=}~&{\cal O}(\Delta t),\end{align} \tag{O}$$
due to eqs. (I) and (G). The Lemma now follows by combining eqs. (E), (L), (N) & (O). $\Box$
VI) Proof of eq. (F) for large $\Delta t$. We use the path integral property
$$ \begin{align} K(x_2,t_2;&x_1,t_1)\cr ~=~&\int_{\mathbb{R}} \! dx_3~ K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1),\end{align} \tag{2-31}$$
which is independent of the instant $t_3$. We now pinch the instant $t_3$ sufficiently close to the instant $t_2$, so that we can approximate the path integral $K(x_2,t_2;x_3,t_3)$ by the analog of eq. (H). If we apply the operator $D_2$ on the kernel, we get
$$ \begin{align}D_2 K(x_2,t_2;&x_1,t_1)\cr~\stackrel{(2-31)}{=}~&\int_{\mathbb{R}} \! dx_3~ D_2 K(x_2,t_2;x_3,t_3)~K(x_3,t_3;x_1,t_1) \cr ~\stackrel{(K)}{=}~&\int_{\mathbb{R}} \! dx_3~ {\cal O}(t_2-t_3)~K(x_3,t_3;x_1,t_1)\cr ~=~&{\cal O}(t_2-t_3) .\end{align}\tag{P}$$
Since the lhs. of eq. (P) does not depend on $t_3$, we conclude that it is zero. Hence eq. (F) also holds for large $\Delta t$ as well. $\Box$
References:
Let me change notation a little. I'll consider the general non-interacting one-particle Green's function--you can easily generalize to the retarded case if you wish, it's the same basic idea:
$$G_\sigma^0(r_1,\,t_1;\,r_2,\,t_2)=-i\langle \psi_0|\left\{ \Theta(t_1-t_2)c_\sigma(r_1,\,t_1)c_\sigma^\dagger(r_2,\,t_2)-\Theta(t_2-t_1)c_\sigma^\dagger(r_2,\,t_2)c_\sigma(r_1,\,t_1)\right\}|\psi_0\rangle$$
Now, we take a spatial Fourier transform:
\begin{align} G_\sigma^0(k,\,t)& =\int dt \int d^3 r e^{-ik\cdot r}G_\sigma(r,\,t)\notag\\ &=-i\langle \psi_0|\{\Theta(t)c_{k\sigma}(t)c_{k\sigma}^t)-\Theta(-t)c_{k\sigma}^\dagger(t)c_{k\sigma}(t)\}|\psi_0\rangle\end{align}
From the Heisenberg equation of motion, we can take the time dependence out of the operators:
$$ \frac{d}{dt} c_{k\sigma}(t)=\frac{i}{\hbar}[H,\,c_{k\sigma}(t)]$$
I won't give you the full solution, but I will give you the end result:
$$c_{k\sigma}(t)=e^{-\frac{i}{\hbar}\epsilon_k t}c_{k\sigma}$$
With this, the one-particle Green's function is given by
\begin{align} G_\sigma^0(k,\,t)&=-i\langle \psi_0|\left\{ \Theta(t)c_{k\sigma}c_{k\sigma}^\dagger -\Theta(t)c_{k\sigma}^\dagger c_{k\sigma}\right\}|\psi_0\rangle e^{-\frac{i}{\hbar}\epsilon_k t}\notag\\ &=-i\left\{ \Theta(t)(1-n_{k\sigma})-\Theta(-t)n_{k\sigma} \right\}e^{-\frac{i}{\hbar}\epsilon_k t} \end{align}
We now perform a temporal Fourier transform, all the while attaching an extra $e^{-\delta |t|}$ term to ensure good behavior at infinity:
\begin{align} G_\sigma^0(k,\,\sigma)&=\int dt e^{i\omega t-\delta t}G_\sigma(k,\,t) \end{align}
I think you can solve this integral with what I gave you--the solution is
$$G_\sigma^0(k,\,\omega)=\frac{1-n_{k\sigma}}{\omega-\epsilon_{k}+i\delta}+\frac{n_{k\sigma}}{\omega-\epsilon_{k}-i\delta}$$
Where in the above I have taken units where $\hbar$ is unity. Defining $\delta_k=\textrm{sgn}(\epsilon_{k}-\epsilon_F)\delta$ (where $\epsilon_F$ is the Fermi energy), we find a nice concise form of the above:
$$G^0(k,\,\omega)=\frac{1}{\omega-\epsilon_k+i\delta_k}$$
In the interacting system, we find something similar, except now with a term known as the self-energy $\Sigma(k,\,\omega)$:
$$G(k,\,\omega)=\frac{1}{\omega-\epsilon_k-\Sigma(k,\,\omega)+i\delta_{k}}$$
This self-energy can be thought of as a renormalization of the energy that comes about when we turn on interactions. In fact, we can rewrite the interacting Green's function in the form
$$G(k,\,\omega)=\frac{Z_k}{\omega-\widetilde{\epsilon_k}+i\delta_k}$$
where $Z_k$ is the quasiparticle weight
$$Z_k=\frac{1}{1-\frac{\partial \Sigma(k,\,\omega)}{\partial \omega}\bigg|_{\omega=\epsilon_k}}$$
and we have expanded around the pole at $\omega=\widetilde{\epsilon_k}$.
If you want to learn more about the details of the Green's function approach to many body physics, I'd suggest Zagoskin or Dickhoff and Van Neck's books. They are both excellent and up-to-date resources for the subject.
EDIT: If you wish to consider the Green function as an operator, look back to when I first took out the exponential term:
\begin{align} G_\sigma^0(k,\,t)&=-i\langle \psi_0|\left\{ \Theta(t)c_{k\sigma}c_{k\sigma}^\dagger -\Theta(t)c_{k\sigma}^\dagger c_{k\sigma}\right\}|\psi_0\rangle e^{-\frac{i}{\hbar}\epsilon_k t}\notag\\ \end{align}
We can ignore the sandwiched states and perform the temporal Fourier transform immediately to get the Green function in terms of the number operator, which is directly proportional to the non-interacting Hamiltonian.
Best Answer
First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$.
At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second definition obeys the first differential equation.
The differential operator $\hat L$ is what appears in the linearized equations of motion for the field, in this case $\psi(x_1,t_1)$, and it only acts on $\psi(x_1,t_1)$, not $\psi^\dagger (x_2,t_2)$.
The time-ordering operator $T$ may be written in terms of the step function $$ T(\psi \psi^\dagger) = \psi \psi^\dagger\cdot \theta(t_1-t_2) - \psi^\dagger \psi\cdot\theta(-t_1+t_2)$$ where $\psi$ is always at $x_1,t_1$ and $\psi^\dagger$ is at $x_2,t_2$. Now, ask what happens when you act with $\hat L$ on the right hand side of the displayed equation above.
By the Leibniz rule, there are the terms with $\hat L \psi = 0$. It vanishes by the equations of motion. But there are extra terms where $\hat L$ acts on the step functions.
The operator $\hat L$ contains the term that differentiates with respect to $t_1$ multiplied by a coefficient $C$. This turns $\theta(t_1-t_2)$ to $\delta(t_1-t_2)$. The same occurs in the next term, but with the opposite sign which cancels the sign that was already there. So the extra terms are $$ \hat L T(\psi \psi^\dagger) = C\delta(t_1-t_2) (\psi \psi^\dagger + \psi^\dagger \psi)$$ I got two terms because there were two terms. However, these two terms exactly combine to the anticommutator of $\psi$ and $\psi^\dagger$ which only needs to be evaluated for $t_1=t_2$, the equal-time anticommutator, and the result is $D\cdot \delta(x_1-x_2)$.
That's why the action of $\hat L$ on the correlator ends up being $CD\delta(t_1-t_2)\delta(x_1-x_2)$ where the constants $C,D$ are mostly just factors of $i$ etc.
For bosonic fields, $\hat L$ has the second derivative with respect to time. One of the derivatives has the fate as above, the other one turns the other $\phi$, which plays the role of $\psi^\dagger$, into $\partial_t \phi$ which is the canonical momentum, and it's the right variable that has the $\delta$-function-like commutator. Also, the intermediate sign is the opposite one but the result is the same, some $CD\cdot\delta\cdot\delta$.