Insert a complete set of <momentum eigenstates> into the <quantum-mechanical propagator>:
$$
\begin{aligned}
K(x_f,T;x_i,0)
&=\langle x_f|e^{-i\widehat p^2T/(2m\hbar)}|x_i\rangle\\
&=\int_{-\infty}^{\infty}\frac{dp}{2\pi\hbar}
\exp\!\left[\frac{i}{\hbar}p(x_f-x_i)-\frac{iT}{2m\hbar}p^2\right].
\end{aligned}
$$
Completing the square and evaluating the resulting <Gaussian integral>, with the usual <i-epsilon prescription>, gives the <free-particle propagator>
$$
\boxed{K(x_f,T;x_i,0)=
\sqrt{\frac{m}{2\pi i\hbar T}}
\exp\!\left[\frac{im(x_f-x_i)^2}{2\hbar T}\right]}.
$$
The square-root branch is fixed by requiring $K(x_f,T;x_i,0)\to\delta(x_f-x_i)$ as $T\downarrow0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 1 / a / i / Solution

For a free particle of mass $m$ on the line,
$$
K(x_f,T;x_i,0)=\sqrt{\frac{m}{2\pi i\hbar T}}
\exp\!\left(\frac{im(x_f-x_i)^2}{2\hbar T}\right).
$$


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