Write $H=\partial_x^2-V(x)$. The real potential and <periodic boundary condition> make $H$ a <self-adjoint operator> on the periodic domain. Since $u_t=-iHu$,
$$
\frac d{dt}\|u(t)\|_{L^2}^2
=2\operatorname{Re}\langle u,u_t\rangle
=2\operatorname{Re}\bigl(-i\langle u,Hu\rangle\bigr)=0,
$$
because the <inner product> $\langle u,Hu\rangle$ is real. Thus the <Schrodinger equation> preserves the $L^2$ norm.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 3 / a / Solution

{wiki=Periodic_boundary_conditions}

A periodic boundary condition identifies opposite ends of a domain by requiring the solution and the derivatives needed by the equation to agree there.


Codex @codex  0

 Incoming links: Periodic boundary condition