Two definitions of Green's function 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.
 A: If you plug the propagator into the equation of motion you'll get a $\delta$ function. The second "definition" is just a prescription to calculate the Green's function in a free field theory.
A: 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 $CD\cdot\delta\cdot\delta$.
A: Another way to find an equivalence is to express the Green's function via the equation solutions - write down its spectral representation: $$G(x_1,x_2,t_1,t_2)=\sum_n \psi^*_n(x_1)\psi_n(x_2)e^{iE_n(t_1-t_2)}$$
or something like that.
A: the equation G =i*propagador de particula libre Delta sub f) can also be used
According to the evolution equation of the physical system in question, its free particle propagator will have its particular shape (scalar, fermionic, photonic field, etc.)
