quantum mechanics - Are propagators in QFT really Green functions?

In many textbooks the notions Green function and propagator are used interchangeably. But are they really the same thing?

This popular answer argues that a retarded propagator function $D_R(x,t,x',t')$ in quantum field theory is a Green function since it can be understood as the product of the kernel $K(x,t,x',t')$ and a Heaviside function $\theta(t-t')$: $$ D_R(x,t,x',t') = K(x,t,x',t') \Theta(t-t') . \tag{1} $$ The defining property of a Green function is that $$ DD_R(x,t,x',t') = \delta(t-t') \delta(x-x') , \tag 2 $$ where $D$ is the differential operator in question. Moreover, the defining condition of a kernel is $$ DK(x,t,x',t') = 0 .$$ Therefore, we can check whether or not Eq. 1 is correct: \begin{align} D D_R(x,t,x',t')&= D\Theta(t-t') K(x,t,x',t') \\ &= \Big( D\Theta(t-t')\Big) K(x,t,x',t') + \Theta(t-t') \Big( D K(x,t,x',t')\Big)\\ &= \delta(t-t') K(x,t,x',t'). \end{align} Thus we can see that this is equal to Eq. 2 only if $$K(x,t,x',t)=\delta(x-x').\tag{3}$$

In quantum field theory, however, the kernel is given by the Wightman function $$W(x,t,x^\prime,t') = \langle0| \varphi(x,t) \varphi(x^\prime,t') |0\rangle\,, $$ which is not equal to a delta distribution for $t=t'$. Instead, we have $$ \langle 0| \varphi(x,t) \varphi(x',t)|0\rangle=\frac{m}{4\pi^2 r} K_1(mr),$$ where $K_1$ is the modified Bessel function.

Thus it seems as if a QFT propagator, in general, is not necessarily a Green function. (It's still possible that some propagator (e.g. the Feynman propagator) is a Green function. However, so far I haven't found a source which clarifies which propagators are actually Green functions and which are not. There is the additional complication that there are different definitions e.g. for the retarded and advanced propagator which may explain some of the confusion.)

Answer

No, propagator is sometimes used to mean Green function, sometimes used to mean kernel, as explicitly written in the answer cited in the question itself.

EDIT. Also, the argumentation in the question has an issue: Eq. (1) from the cited answer is incorrect in general. By the obvious reason that when the operator $D$ is multiplied by 2, a kernel remains a kernel (whatever boundary conditions are) but a Green function cannot remain a Green function. A related issue is replacing $D\Theta(t-t')$ by $\delta(t-t')$. They need not be equal; this depends on a particular differential operator $D$.

But Eq.~(1) is correct e.g. for wave equation, see this answer for a detailed proof and further references.