Path integral in quantum mechanics with definite momentum states as the boundary states $\newcommand{\bra}[1]{\left\langle#1\right|}$
$\newcommand{\ket}[1]{\left|#1\right\rangle}$
Consider a quantum mechanical non-relativistic particle with a Hamiltonian $$\hat{H} = \frac{\hat{p}^2}{2m}+\hat{V}(x)$$ Let $\hat{U}(t_1,t_2)$ be the time evolution operator and $\ket{p_1}$ and $\ket{p_2}$ be two momentum eigenstates.
To compute the matrix element $\bra{p_2}\hat{U}\left(t_1,t_2\right)\ket{p_1}$, using the path integral method, can we just add up all the paths in $(x,t)$ space (with the weight factor $e^{iS(x(t))/\hbar}$) that start with slope $v_1 = \frac{p_1}{m}$ and end with slope $v_2 = \frac{p_2}{m}$. Initial and final position coordinates are kept unrestricted.

I tried calculating this for the case of free particle and after integrating over all the paths, everything in the exponential just cancelled out and gave me $\bra{p_2}\hat{U}\left(t_1,t_2\right)\ket{p_1} = e^0 = 1$ up to normalization factors. Now I may have made some mistakes in the calculations but I just wanted to confirm:
Would adding all the paths that start with slope $v_1 = \frac{p_1}{m}$ and end with slope $v_2 = \frac{p_2}{m}$ in $(x,t)$ space give us $\bra{p_2}\hat{U}\left(t_1,t_2\right)\ket{p_1}$? The initial and final positions would be kept free in the path integral.
Edit: I know that one of the correct methods to find $\bra{p_2}\hat{U}\left(t_1,t_2\right)\ket{p_1}$ is to express the momentum states into a linear combination of position states and then use the path integral for position states. This will lead to the Fourier transform with respect to the initial and final position coordinates.  I understand that the free particle the propagator written in momentum basis has to be proportional to a delta function $\delta(p_1 - p_2)$.
What I am asking is whether a very specific method of calculating $\bra{p_2}\hat{U}\left(t_1,t_2\right)\ket{p_1}$ (described above) works or not?
 A: *

*In the general case, use the Fourier transform/overlap between momentum and position states
$$\begin{align} 
\langle p_2,t_2 &| p_1,t_1 \rangle\cr
~=~&\int_{\mathbb{R}} \! dx_2\int_{\mathbb{R}} \! dx_1 \langle p_2,t_2 | x_2,t_2 \rangle \langle x_2,t_2 | x_1,t_1 \rangle \langle x_1,t_1 | p_1,t_1 \rangle \cr
~=~& \int_{\mathbb{R}}\! dx_2\int_{\mathbb{R}} \! dx_1 \frac{e^{\frac{i}{\hbar}(p_1x_1-p_2x_2)}}{2\pi\hbar} \langle x_2,t_2 | x_1,t_1 \rangle. 
\end{align} \tag{1} $$


*In the free case, it is
$$ \langle p_2,t_2 | p_1,t_1 \rangle~=~\delta(p_2-p_1)e^{-\frac{i}{\hbar}\frac{p^2_2}{2m}\Delta t}.\tag{2}$$


*In the general case, this can be written as a Hamiltonian phase space path integral
$$\begin{align}  \langle p_2,t_2 | p_1,t_1 \rangle ~=~&\int_{p(t_1)=p_1}^{p(t_2)=p_2} \!{\cal D}x {\cal D}p~  \exp\left\{\frac{i}{\hbar}S[x,p]\right\}, \cr
S[x,p]~=~&\int_{t_1}^{t_2}\!dt\left(-x\dot{p} -H(x,p,t)\right),
 \end{align}\tag{3}$$
where the Hamiltonian action $S[x,p]$ has been chosen to be compatible with the boundary conditions.
