When one defines the path integral propagator, there is the need to normalize the propagator (since it would give you a probability density). There are two formulas which are used.
1) Original (v1+v2): The first formula (which I can intuitively agree with) says that:
$$\tag{1} \int_{Dx_b}dx_b\left|K(x_bt_b|x_at_a)\right|^2=1$$
for all values of $x_a$ on fixed values of $t_a, t_b$ and where $Dx_b$ means the domain of $x_b$.
1') Update (v3+v4): I changed my mind (to get more into agreement with the Born-rules). The first formula (which I can intuitively agree with) says that:
$$\tag{1'} \left|\int_{Dx_b}dx_bK(x_bt_b|x_at_a)\right|^2=1$$
for all values of $x_a$ on fixed values of $t_a, t_b$ and where $Dx_b$ means the domain of $x_b$.
2) The second formula (which is actually also very intuïtive) says that:
$$\tag{2}\lim\limits_{t_b\rightarrow t_a}K(x_bt_b|x_a,t_a) = \delta(x_b-x_a).$$
Now these are usually treated as equivalent, but I can't directly see how this can be the case. Isn't the second formula less restrictive ?
Best Answer
I) Conceptually, OP's original eq. (1)
$$\int_{\mathbb{R}}\! \mathrm{d}x_f~ \left| K(x_f,t_f;x_i,t_i) \right|^2 ~\stackrel{?}{=}~1 \qquad(\leftarrow\text{Wrong!})\tag{1} $$
clashes (as OP independently realized) with the fundamental principle of the Feynman path integral that the amplitude
$$K( x_f ,t_f ; x_i ,t_i )~=~\sum_{\rm hist.}\ldots$$
is a sum of histories, while the probability
$$P( x_f ,t_f ; x_i ,t_i )~=~|K( x_f ,t_f ; x_i ,t_i )|^2~\neq~\sum_{\rm hist.}\ldots $$
is not a sum of histories.
Concretely, the failure of eq. (1) may also be seen as follows. If we assume that$^1$
$$ K( x_i ,t_i ; x_f ,t_f ) ~=~ \overline{K( x_f ,t_f ; x_i ,t_i ) }, \tag{A}$$
and the (semi)group property of Feynman propagators/kernels
$$ K(x_f,t_f;x_i,t_i) ~=~ \int_{\mathbb{R}}\!\mathrm{d}x_m ~ K(x_f,t_f;x_m,t_m) K(x_m,t_m;x_i,t_i),\tag{B}$$
then the lhs. of OP's original first eq. (1) with $(x_i,t_i)=(x_f,t_f)$ is not equal to $1$, but instead becomes infinite
$$\begin{align} K(x_f,t_f;x_i,t_i) ~=~&\delta(x_f-x_i)~=~\delta(0)~=~\infty, \cr x_i~=~&x_f,\qquad t_i~=~t_f, \end{align}\tag{C}$$
because of OP's second formula (2).
II) The infinite normalization result (C) can be intuitively understood as follows. Recall that the paths in the path integral satisfy Dirichlet boundary condition $x(t_i)=x_i$ and $x(t_f)=x_f$. In other words, the particle is localized in $x$-position space at initial and final times. On the other hand, a particle localized in $x$-position space corresponds to a delta function wave function $\Psi(x)=\delta(x-x_0)$, which is not normalizable, cf. e.g this and this Phys.SE posts.
III) Conceptually, OP's first eq. (1')
$$\left| \int_{\mathbb{R}}\! \mathrm{d}x_f~K(x_f,t_f;x_i,t_i) \right| ~\stackrel{?}{=}~1 \quad(\leftarrow\text{Turns out to be ultimately wrong!}) \tag{1'} $$
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 a concept has to be converted into mathematical formulas, as discussed in detail in this Phys.SE post. 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.
IV) Example. Finally, let us consider the example of a non-relativistic free particle in 1D. The Feynman propagator then reads
$$\begin{align} K( x_f ,t_f ; x_i ,t_i ) ~=~& \sqrt{\frac{A}{\pi}} e^{-A(\Delta x)^2}\cr ~=~& \sqrt{\frac{m}{2\pi i\hbar} \frac{1}{\Delta t}} \exp\left[ \frac{im}{2\hbar}\frac{(\Delta x)^2}{\Delta t}\right],\cr A~:=~&\frac{m}{2 i\hbar} \frac{1}{\Delta t} , \cr \Delta x~:=~&x_f-x_i, \cr \Delta t~:=~&t_f-t_i ~\neq ~0. \end{align}\tag{D}$$
[It is an instructive exercise to show that formula (D) satisfies eqs. (A-C) and OP's second formula (2).] The Gaussian integral over $x_f$ is one
$$ \int_{\mathbb{R}}\!\mathrm{d}x_f ~ K(x_f,t_f;x_i,t_i)~=~1,\tag{E} $$
which shows that OP's first eq. (1') actually holds for a free particle. The integrand
$$ \begin{align} |K(x_f,t_f;x_i,t_i)|^2~=~& \frac{|A|}{\pi}~=~ \frac{m}{2\pi \hbar}\frac{1}{|\Delta t|}, \cr \Delta t ~\neq ~&0,\end{align}\tag{F}$$
on the lhs. of OP's original first eq. (1) is independent of the midpoint $x_m$. Hence the integral over $x_m$ (i.e. lhs. of OP's first eq. (1)) becomes infinite
$$\begin{align} \int_{\mathbb{R}}\!\mathrm{d}x_f ~ |K(x_f,t_f;x_i,t_i)|^2~=~& \frac{m}{2\pi \hbar}\frac{1}{|\Delta t|} \int_{\mathbb{R}}\!\mathrm{d}x_f ~=~\infty, \cr \Delta t ~\neq ~&0,\end{align}\tag{G} $$
in agreement with what we found in eq. (C) in section I.
References:
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$^1$ Note that Ref. 1 defines $K(x_f,t_f;x_i,t_i)=0$ if $t_i>t_f$, see Ref. 1 between eq. (4-27) and eq. (4-28). Here we assume property (A) instead.