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:
- R.P. Feynman & A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.
Hello handsome poster (why thank you kind stranger). The answer to your question it turns out is in Rovelli's "Quantum Gravity", at least insofar as the free scalar field is concerned. This is done in the following way. As you may recall (from Feynman and Hibbs), through various arguments about doing the path integral as a perturbation on the classical path, the path integral of a Gaussian lagrangian for a point particle
$$L = a(t) \dot x^2 + b(t) \dot x x + c(t) x^2 + d(t) \dot x + e(t) x + f(t)$$
will be
$$K(x_a, t_a; x_b, t_b) = e^{\frac{i}{\hbar} S_{cl} [x_b, x_a]} \int_0^0 \exp\{ \frac{i}{\hbar} \int_{t_a}^{t_b}[a(t) \dot y^2 + b(t) \dot y y + c(t) y^2]dt\}\mathcal Dy(t)$$
Where the second term will be some function only depending on the beginning and end time, as it does not depend on the position.
Through a rather tedious calculation, you can work out that the classical action for the harmonic oscillator is
$$S_{cl} = \frac{m\omega}{2\sin(\omega T)} ((x_a^2 + x_b^2) \cos (\omega T) - 2 x_a x_b)$$
with the remaining term of the kernel
$$\int_0^0 \exp\{ \frac{i}{\hbar} \int_{t_a}^{t_b} \frac{m}{2}[\dot y^2 - \omega y^2]dt\}\mathcal Dy(t)$$
As the function $y(t)$ goes from $0$ to $0$ in the interval $T = t_b - t_a$, it can be written as the Fourier series
$$y(t) = \sum_n a_n \sin(\frac{n\pi t}{T})$$
The action can thus be transformed into
$$\int_{t_a}^{t_b} \frac{m}{2}[\dot y^2 - \omega y^2]dt = \frac{mT}{4} \sum_n [(\frac{n\pi}{T})^2 - \omega^2] a_n^2$$
With a path integral being a simple infinite product of the individual gaussians for each $a_n$, which becomes (cf Feynman Hibbs for more details)
$$F(T) = \sqrt{\frac{m\omega}{2 \pi i \hbar \sin(\omega T)}}$$
Now the free scalar field can be decomposed as an infinite number of harmonic oscillators via some Fourier transform
\begin{eqnarray}
S &=& \int_{t_a}^{t_b} dt \int d^3x \frac{1}{2}[\partial_\mu\varphi \partial^\mu \varphi - m^2 \varphi^2]\\
&=& \int \frac{d^3k}{(2\pi)^3} \{\int_{t_a}^{t_b} dt \frac{1}{2}[\dot \varphi(k)^2 -(\vec k^2 + m^2)\vert \varphi(k) \vert^2]\}
\end{eqnarray}
(with some use of Parseval's theorem) which is just the action of the harmonic oscillator. Then, via some generous physicist magic, we can write
$$\exp[\frac{i}{\hbar}\int \frac{d^3k}{(2\pi)^3} S_{SHO}[\varphi(\vec k)]] \approx \prod_k \exp[\frac{i}{\hbar (2\pi)^3} S_{SHO}[\varphi(\vec k)]]$$
For every mode, the transition amplitude will be
$$\exp[\frac{i}{\hbar(2\pi)^3} \frac{\omega}{2\sin(\omega T)} ((\varphi_a^2(k) + \varphi_b^2(k)) \cos (\omega T) - 2 \bar \varphi_a(k) \varphi_b(k)) \sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}$$
Which will give us, once properly multiplied back,
$$\exp[\frac{i}{\hbar} \int \frac{d^3 k}{(2\pi)^3} \frac{\omega}{2\sin(\omega T)} ((\varphi_a^2(k) + \varphi_b^2(k)) \cos (\omega T) - 2 \bar \varphi_a(k) \varphi_b(k)) \prod_k [\sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}]$$
$\prod_k [\sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}]$ corresponds to a normalization factor, which Rovelli writes as being
$$\mathcal N \approx \prod_k [\sqrt{\frac{m\omega}\hbar}] \exp[-\frac{V}{2} \int \frac{d^3 k}{(2\pi)^3} \ln [\sin (\omega T)]]$$
which is apparently formally divergent (I assume due to the volume term $V$ - Rovelli does not seem to mention what it is tho), but I suppose this is similar to the Hamiltonian of the Hilbert space method also being divergent.
Best Answer
The point-particle path integral can allegedly be defined unambiguously in continuous time, but I'm not familiar with the high-brow details of that definition, so I'll do something barbaric: I'll discretize time. And to avoid issues with non-normalizable states, I'll use... normalizable states. Then the quantity of interest is $$ \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \la\tilde\psi,t_N|\psi,t_1\ra \equiv \int \prod_t dq_t\ e^{iS[q]} \tilde\psi^*(q_N)\psi(q_1). \tag{1} $$ This is well-defined as long as (a) $\psi$ and $\tilde\psi$ are normalizable, and (b) doing the integral over $q_1$ gives an expression of the same form but with a new "initial" state that is a normalizable function of $q_2$, and so on.
What happens if we add a total-derivative term to the lagrangian, so that the modified action $S'$ picks up terms that depend on $q_1$ and $q_N$? It modifies the initial and final states. It's the same model as before, with the same action as before, but with different initial and final states.
Now let's increase the number of spacetime dimensions from $1$ to $D$, still discretized. If we imposed periodic boundary conditions in the spatial dimensions, then nothing new would happen. The answer would be the same as above. But what if we don't impose periodic boundary conditions in the spatial dimensions? In that case, adding a total-derivative term to the lagrangian must change the boundary conditions on the spatial dimensions, in addition to changing the initial/final states. In other words, it changes the boundary conditions on all of the spacetime dimensions. I'm using the phrase "boundary conditions" in a broad sense: not just Neumann or Dirichlet, but also including generalizations like integrating over a function of the boundary-variables, just like we normally do to specify the initial/final states.
Just for fun, let's generalize even further. What if we cut a hole inside spacetime? Now we need to specify boundary conditions on the boundary of the hole, and adding a total-derivative term to the lagrangian will change those boundary conditions just like it changed the "outside" boundary conditions. How should we interpret this? If I'm not mistaken, cutting a hole and specifying its boundary conditions is equivalent to inserting a local operator into the path integral (I mean localized within the hole, not necessarily at a point), so that now we're calculating $\la\tilde\psi,t_N|O|\psi,t_1\ra$ for some nontrivial operator $O$. Which operator? That's what the boundary conditions specify. I infer that adding a total-derivative term to the lagrangian changes the operator that we inserted, in addition to changing the initial/final states and the boundary conditions at the far edges of space. Those two things (changing operators and changing states) look different in the canonical formulation, but from this perspective, they look like the same kind of thing, just in different parts of spacetime.
Disclaimer: I'm not an expert in this subject, and what little intuition I do have is based on local lattice QFT, which might not be sufficient to capture everything that can happen in local continuum QFT (cf the not-strictly-local overlap lattice Dirac operator).