# 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?

1. 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}
2. In the free case, it is $$\langle p_2,t_2 | p_1,t_1 \rangle~=~\delta(p_2-p_1)e^{-\frac{i}{\hbar}\frac{p^2_2}{2m}\Delta t}.\tag{2}$$
3. 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.
• Thanks for the answer. I understand these methods. But I am asking whether adding up paths with initial and final slopes as $\frac{p_1}{m}$ and $\frac{p_2}{m}$ in $(x,t)$ space would give us the same results as you mentioned above. Commented Apr 6, 2022 at 10:49