The <Euler-Lagrange equation> is $m\ddot x=0$. The unique path with the prescribed endpoints is therefore
$$
x_{\rm cl}(t)=x_i+\frac{x_f-x_i}{T}t.
$$
It is a minimum of the Euclidean action and a stationary point of the real-time action. Its classical action is
$$
S_{\rm cl}=\int_0^T\frac m2\dot x_{\rm cl}^{,2},dt
=\frac{m(x_f-x_i)^2}{2T}.
$$
The <principle of stationary action> consequently fixes the position-dependent phase of the <semiclassical propagator> as
$$
K(x_f,T;x_i,0)=C(T)e^{iS_{\rm cl}/\hbar}.
$$
Because the action is quadratic, the stationary-phase evaluation of the <path integral> is exact. Composition of propagators, or the <Van Vleck determinant>, gives $C(T)=\sqrt{m/(2\pi i\hbar T)}$, reproducing part i.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-304/1/a/ii/solution