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\begin{document}

\subsection{The free propagator}

In the case of a free particle, $\hat{H}_{0}=\hat{p}^{2}/(2m),$ the propagator
can be calculated by inserting a complete set of momentum eigenkets,
$\int\mathrm{d}k$ $\left|  k\right\rangle \left\langle k\right|  =1,$
\begin{align}
K(x,x_{0};t)\, &  =\left\langle x\right|  \exp\left(  -i\ \frac{\hat{H}_{0}%
t}{\hbar}\right)  \left|  x_{0}\right\rangle =\int\mathrm{d}k\,\,\left\langle
x\right|  \exp\left(  -i\ \hat{H}_{0}t/\hbar\right)  \left|  k\right\rangle
\,\left\langle k\right.  \left|  x_{0}\right\rangle \nonumber\\
&  =\int\mathrm{d}k\,\exp\left(  -i\ \frac{\hbar k^{2}t}{2m}\right)
\left\langle x\right.  \left|  k\right\rangle \,\left\langle k\right.  \left|
x_{0}\right\rangle \nonumber\\
&  =\frac{1}{2\pi}\int\mathrm{d}k\,\exp\left(  -i\ \frac{\hbar k^{2}t}%
{2m}+ik(x-x_{0})\right)  .
\end{align}
Here, we used the wavevector $k=p/\hbar$ and $\left\langle x\right.  \left|
k\right\rangle =(2\pi)^{-1/2}\exp\left(  ikx\right)  .$ Using $\int_{-\infty
}^{\infty}\exp\left(  -iax^{2}\pm ibx\right)  \,dx=$ $\sqrt{\pi/a}\exp\left[
ib^{2}/(4a)-i\pi/4\right]  $ for $a+i\varepsilon,b\in\mathbb{R}$ one finds
\begin{equation}
K(x,x_{0};t)\,=\sqrt{\frac{m}{2\pi i\hbar t}}\exp\left(  i\frac{\left(
x-x_{0}\right)  ^{2}m}{2\hbar t}\right)  \,.\label{propagatorfree}%
\end{equation}
It is worth noting that the phase in (\ref{propagatorfree}) is related to the
classical action $R(x,x_{0};t)$ of a free particle going in time $t$ from
$x_{0}$ to $x.$ Since the classical trajectory is $x^{\prime}(t^{\prime
})=x_{0}+(x-x_{0})\,t^{\prime}/t\,\,$the action reads%
\begin{equation}
R(x,x_{0};t)=\int_{0}^{t}\mathcal{L}\left(  x^{\prime}(t^{\prime}),\dot
{x}^{\prime}(t^{\prime})\right)  \,\mathrm{d}t^{\prime}=\int_{0}^{t}%
\frac{m}{2}\left[  \frac{\mathrm{d}}{\mathrm{d}t}x^{\prime}(t^{\prime
})\right]  ^{2}\,\mathrm{d}t^{\prime}=\frac{m}{2t}\left(  x-x_{0}\right)
^{2}.\label{freeaction}%
\end{equation}
We observe that one can write both the phase and the prefactor in
(\ref{propagatorfree}) in terms of the classical quantity (\ref{freeaction})
and $\hbar,$%
\begin{equation}
K(x,x_{0};t)\,=\frac{1}{\left(  2\pi i\hbar\right)  ^{1/2}}\left|
\frac{\partial^{2}R}{\partial x\partial x_{0}}\right|  ^{1/2}\exp\left(
i\frac{R(x,x_{0};t)}{\hbar}\right)  .\label{freepropagatoraction}%
\end{equation}
Note also that $\lim_{t\rightarrow0}K(x,x_{0};t)\,=\delta(x-x_{0}).$ Let us
now apply the 1d propagator to a gaussian wave packet..

\end{document}
