In the calculation of the path integral, we of course have to sum over all possible paths between the initial and final positions. However, it can be useful to consider what contributions to the path integral look like for individual trajectories,

to get a clearer idea of how the action varies for paths away from the classical trajectory. In this problem, we will focus on evaluation of the action
\[\begin{equation*}S[x]=\int_{0}^{T} d t\left(\frac{m}{2} \dot{x}^{2}-U(x)\right) \tag{11.162}\end{equation*}\]
where \(x(t)\) is the trajectory that starts at \(x_{i}\) at \(t=0\) and ends at \(x_{f}\) at \(t=T\), and \(U(x)\) is the potential energy.
(a) Consider the system of a free particle which has trajectory \(x(t)\) given by
\[x(t)=\left\{\begin{array}{cc}x_{i}+\frac{2 t}{T}\left(x^{\prime}-x_{i}\right), & 0Plot the value of the action corresponding to this trajectory as a function of the intermediate position \(x^{\prime}\).
(b) Consider the system of a free particle which has trajectory \(x(t)\) given by
\[\begin{equation*}x(t)=x_{i}+\frac{t}{T}\left(x_{f}-x_{i}\right)+A \sin \left(\frac{\pi t}{T}\right) . \tag{11.164}\end{equation*}\]
Plot the value of the action corresponding to this trajectory as a function of the oscillation amplitude \(A\).
(c) Consider the system of a free particle which has trajectory \(x(t)\) given by
\[\begin{equation*}x(t)=x_{i}+\frac{t}{T}\left(x_{f}-x_{i}\right)+\left(x_{f}-x_{i}\right) \sin \left(\frac{n \pi t}{T}\right) . \tag{11.165}\end{equation*}\]
Plot the value of the action corresponding to this trajectory as a function of the frequency number \(n \in \mathbb{Z}\), the integers.
(d) Repeat parts (a)-(c) for the harmonic oscillator with angular frequency \(\omega\). What do you notice that is different between the plots of the corresponding actions of the free particle and those of the harmonic oscillator?