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.
I) OP is right, ideologically speaking. Ideologically, OP's first eq.
$$ \tag{1} \left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Turns out to be ultimately wrong!}) $$
is the statement that a particle that is initially localized at a spacetime event $(x_i,t_i)$ must with probability 100% be within $x$-space $\mathbb{R}$ at a final time $t_f$, as our QM model does not allow creation or annihilation of particles.
However, such notion of absolute probabilities of the Feynman kernel $K(x_f,t_f;x_i,t_i)$ cannot be maintained when ideology has to be converted into mathematical formulas. E.g. for the harmonic oscillator, one has
$$\tag{A} \left| \int_{\mathbb{R}}\!\mathrm{d}x_f ~ K(x_f,t_f;x_i,t_i)\right|
~=~\frac{1}{\sqrt{\cos\omega \Delta t}}, \qquad \Delta t ~:=~t_f-t_i,$$
which only becomes unity for $\omega \Delta t \to 0$. The problem can ultimately be traced to the fact that there is no normalizable uniform probability distribution on the real axis $\mathbb{R}$, i.e. the $x$-position space. In general, OP's first eq. (1) only holds for short times $\Delta t\ll \tau$, where $\tau$ is some characteristic time scale of the system.
II) Let us review how normalization appears in the Feynman path integral from first principles. The main tool to determine the Feynman propagator/kernel/amplitude $K(x_b,t_b;x_a,t_a)$ is the (semi)group property
$$\tag{B} K(x_f,t_f;x_i,t_i) ~=~ \int_{\mathbb{R}}\!\mathrm{d}x_m ~ K(x_f,t_f;x_m,t_m) K(x_m,t_m;x_i,t_i). $$
III) Equivalently, if we identify
$$\tag{C} K(x_f,t_f;x_i,t_i)~=~\langle x_f,t_f \mid x_i,t_i \rangle$$
with an overlap of instantaneous$^1$ position eigenstates in the Heisenberg picture, then eq. (B) follows from the (first of) the completeness relations
$$\tag{D} \int \!\mathrm{d}x ~|x,t \rangle \langle x,t |~=~{\bf 1},
\qquad \text{and} \qquad
\int \!\mathrm{d}p~ |p,t \rangle \langle p,t |~=~{\bf 1}.$$
These instantaneous position and momentum eigenstates have overlap$^2$
$$\tag{E} \langle p,t \mid x,t \rangle~=~\frac{1}{\sqrt{2\pi\hbar}}\exp\left[\frac{px}{i\hbar}\right].$$
IV) OP's first eq. (1) is equivalent to the statement that
$$\tag{F} \left| \langle p_f=0,t_f \mid x_i,t_i \rangle \right| ~\stackrel{?}{=}~\frac{1}{\sqrt{2\pi\hbar}},\qquad(\leftarrow\text{ Ultimately wrong!}) $$
due to the identification (C) and
$$\tag{G} \langle p_f,t_f \mid x_i,t_i \rangle
~\stackrel{(D)+(E)}{=}~\int_{\mathbb{R}}\!\frac{\mathrm{d}x_f}{\sqrt{2\pi\hbar}}\exp\left[\frac{p_fx_f}{i\hbar}\right] \langle x_f,t_f \mid x_i,t_i \rangle. $$
Eq. (F) is violated for e.g. the harmonic oscillator, where one has
$$\tag{H} \left| \langle p_f,t_f \mid x_i,t_i \rangle \right| ~=~\frac{1}{\sqrt{2\pi\hbar\cos\omega \Delta t}}. $$
V) For sufficiently short times $\Delta t\ll \tau$, one derives from the Hamiltonian formulation (without introducing arbitrary normalization/fudge factors!) that
$$\begin{align} \langle x_f,t_f \mid x_i,t_i\rangle
~\stackrel{(D)}{=}~&\int_{\mathbb{R}} \!\mathrm{d}p~
\langle x_f,t_f \mid p,\bar{t} \rangle
\langle p,\bar{t} \mid x_i,t_i\rangle \cr
~=~&\int_{\mathbb{R}} \!\mathrm{d}p~\langle x_f,\bar{t} \mid
\exp\left[-\frac{i\Delta t}{2\hbar}\hat{H}\right]\mid p,\bar{t} \rangle
\langle p,\bar{t} \mid \exp\left[-\frac{i\Delta t}{2\hbar}\hat{H}\right]\mid x_i,\bar{t}\rangle\cr
~\approx~&\int_{\mathbb{R}} \!\mathrm{d}p~
\langle x_f,\bar{t} \mid p,\bar{t} \rangle
\langle p,\bar{t} \mid x_i,\bar{t}\rangle
\exp\left[-\frac{i\Delta t}{\hbar} H(\bar{x},p) \right]\cr
~\stackrel{(E)}{=}~& \int_{\mathbb{R}} \!\frac{\mathrm{d}p}{2\pi\hbar}
\exp\left[\frac{i}{\hbar}\left(p\Delta x -\left(\frac{p^2}{2m} + V(\bar{x})\right)\Delta t\right) \right]\cr
~=~& \sqrt{\frac{A}{\pi}}
\exp\left[-A(\Delta x)^2-\frac{i}{\hbar}V(\bar{x})\Delta t\right],
\qquad A~:=~\frac{m}{2 i\hbar} \frac{1}{\Delta t},\cr ~=~&\sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}}
\exp\left[ \frac{i}{\hbar}\left(\frac{m}{2}\frac{(\Delta x)^2}{\Delta t}-V(\bar{x})\Delta t\right)\right],\end{align}\tag{I}$$
where
$$\tag{J} \Delta t~ :=~t_f-t_i, \quad \bar{t}~ :=~ \frac{t_f+t_i}{2}, \quad \Delta x~ :=~x_f-x_i, \quad \bar{x}~ :=~ \frac{x_f+x_i}{2} .$$
The oscillatory Gaussian integral (I) over momentum $p$ was performed by introducing the pertinent $\Delta t\to\Delta t-i\epsilon$ prescription. Eq. (I) implies that
$$\tag{K} K(x_f,t_f;x_i,t_i) ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad \Delta t \to 0^{+}, $$
which in turn implies OP's first eq. (1) in the short time limit $\Delta t \to 0^{+}$. More generally, Eq. (I) implies OP's first eq. (1) for $\Delta t\ll \tau$.
VI) Note that the short time probability
$$\tag{L} P(x_f,t_f;x_i,t_i)~=~|K(x_f,t_f;x_i,t_i)|^2~\stackrel{(I)}{\approx}~\frac{m}{2\pi \hbar} \frac{1}{\Delta t} , \qquad \Delta t\ll \tau, $$
is independent of initial and final positions, $x_i$ and $x_f$, respectively. For fixed initial position $x_i$, the formula (L) can be interpreted as a uniform and unnormalizable probability distribution in the final position $x_f\in\mathbb{R}$. This reflects the fact that the instantaneous eigenstate $|x_i,t_i \rangle$ is not normalizable in the first place, and ultimately dooms the notion of absolute probabilities.
VII) For finite times $\Delta t$ not small, the interaction term $V$ becomes important. In the general case, the functional determinant typically needs to be regularized by introducing a cut-off and counterterms. But regularization is not the (only) source of violation of OP's first eq. (1), or equivalently, eq. (F). Rather it is a generic feature that the $px$ matrix elements of an unitary evolution operator
$$\tag{M} \frac{\langle p,t \mid
\exp\left[-\frac{i\Delta t}{\hbar}\hat{H}\right]
\mid x,t\rangle}{\langle p,t \mid x,t\rangle} $$
is not just a phase factor away from the short time approximation $\Delta t\ll \tau$.
VIII) Example: Consider the Hermitian Hamiltonian
$$\tag{N} \hat{H}~:= \frac{\omega}{2}(\hat{p}\hat{x}+\hat{x}\hat{p})
~=~ \omega(\hat{p}\hat{x}+\frac{i\hbar}{2}). $$
Then
$$ \begin{align}\frac{\langle p,t \mid
\exp\left[-\frac{i\Delta t}{\hbar}\hat{H}\right]
\mid x,t\rangle}{\langle p,t \mid x,t\rangle}
~=~&1 - \omega\Delta t\left(\frac{1}{2}-i\frac{px}{\hbar} \right)\cr
&+\frac{(\omega\Delta t)^2}{2}\left(\frac{1}{4}-2i\frac{px}{\hbar} - \left(\frac{px}{\hbar} \right)^2\right)
+{\cal O}\left((\omega\Delta t)^3\right),\end{align}\tag{O} $$
which is not a phase factor if $\omega\Delta t\neq 0$. To see this more clearly, take for simplicity $px=0$.
References:
R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965.
J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 2.5.
--
$^1$ Instantaneous eigenstates are often introduced in textbooks of quantum mechanics to derive the path-integral formalism from the operator formalism in the simplest cases, see e.g. Ref. 2. Note that the instantaneous eigenstates $\mid x,t \rangle $ and $\mid p,t \rangle $ are time-independent states (as they should be in the Heisenberg picture).
$^2$ Here we assume that possible additional phase factors in the $px$ overlap (E) have been removed via appropriate redefinitions, cf. this Phys.SE answer.
Best Answer
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.