Generally, in path integral formalism a propagator between two states with definite position is computed, something like, $$K(x_1,t_1;x_0,t_0)=\int_{x_0(t_0)}^{x_1(t_1)}\mathcal{D}x(t)\exp\left(\frac{i}{\hbar}S[x(t)]\right).$$ However, if we want a propagator between two states, where the initial state has a definite momentum and final state has definite position, that is, $$K(x_1,t_1;p_0,t_0),$$ then how does one goes on compute such a quantity? Are there books/notes discussing this? Can someone construct this for the free non-relativistic particle case?
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$$ K(x_1,t_1;p_0,t_0)~=~\frac{1}{\sqrt{2\pi\hbar}} \int_{\mathbb{R}} \! dx_0~ K(x_1,t_1;x_0,t_0)\exp\left(\frac{ip_0x_0}{\hbar}\right), $$ cf. e.g. this Phys.SE post.
answered May 14, 2021 at 2:32
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$\begingroup$ I see. But what I really wanted to know is that what the whole thing becomes? As there is a $\mathcal{D}x(t)$ integral inside $dx_0$ integral. Is there any way we can do the $dx_0$ integral before we do the $\mathcal{D}x(t)$ integration? $\endgroup$ Commented May 14, 2021 at 2:38