Path integral formalism and Schrodinger's operator formalism giving different answers for $H=V(x)$ I'm calculating the propagator after a small time $dt$, that is $U(x-x',dt)$ assuming $H=V(x)$.
Schroedinger's operator formalism gives (since $H$ is diagonal in the $X$ basis):
$$U(x-x',dt)= e^{iV(x)dt} \delta (x-x').$$
Path integral gives (assuming $dt$ is so small that only straight line paths are possible, no in-between points between $x$ and $x'$):
$$U(x-x',dt)=Ae^{iV(x)dt},$$
because the action between $x$ and $x'$ is $V(x)dt$, assuming $dt$ is so small that there are no in-between points.
Why the discrepancy in the propagators obtained?
I know the Hamiltonian $H=V(x)$ is unphysical (since it doesn't have a kinetic term) but it should be fine mathematically.
 A: One should remember that the fundamental version of the path integral is the Hamiltonian phase space path integral. In the short-time limit, if we insert 1 complete set of momentum states, we calculate
$$\begin{align} \langle x| e^{-\frac{i}{\hbar}V \Delta t} | x^{\prime}\rangle~=~& \int_{\mathbb{R}} \! dp   \langle x|p\rangle e^{-\frac{i}{\hbar}V \Delta t}  \langle p| x^{\prime}\rangle \cr
~=~&\int_{\mathbb{R}} \!dp  \frac{e^{\frac{i}{\hbar}xp}}{\sqrt{2\pi\hbar}} e^{-\frac{i}{\hbar}V \Delta t}  \frac{e^{-\frac{i}{\hbar}x^{\prime}p}}{\sqrt{2\pi\hbar}}\cr
~=~&e^{-\frac{i}{\hbar}V \Delta t}\delta(x\!-\!x^{\prime}). \end{align}$$
Notice how the Dirac delta distribution emerges as expected.
A: The following argument is not rigorous but should give the right idea. The Feynman path integral is
\begin{equation}
U(x-x',dt) = \int_{x(0) = x'}^{x(dt) = x} \mathcal{D}x(t)\mathcal{D}p(t) \exp\{i\int_0^{dt} \textrm{d}t' [p(t') \dot{x}(t') - H(p(t'),x(t'))] \}.
\end{equation}
Note that the momentum integral has not been performed yet. In the absence of a kinetic term this gives a functional Dirac delta and we have
\begin{equation}
U(x-x',dt) = \int_{x(0) = x'}^{x(dt) = x} \mathcal{D}x(t)\delta[\dot{x}(t) ] \exp\{-i\int_0^{dt} \textrm{d}t'  V(x(t')) \}.
\end{equation}
This implies that the path integral receives contributions only from trajectories with vanishing velocity, i.e. from constant trajectories with $x = x'$. Hence,
\begin{equation} 
U(x-x',dt) =  \exp\{-i\int_0^{dt} \textrm{d}t'  V(x) \} \delta(x-x') =  \exp\{-i   V(x)dt \} \delta(x-x'). 
\end{equation}
