Schrödinger equation

Remarks

In mathematics one often sets $\hbar = 1$ and $2m = 1$, giving $i\partial_t u = -\Delta u + Vu$.

When $H$ is self-adjoint on $L^2(\R^n)$, Stone's theorem gives the solution $\psi(t) = e^{-itH/\hbar}\psi(0)$ for every initial value in $L^2(\R^n)$, and $e^{-itH/\hbar}$ is unitary, so $\norm{\psi(t)}_{L^2} = \norm{\psi(0)}_{L^2}$.

Example: with $V = 0$, $\hbar = 1$, $2m = 1$ and $k \in \R^n$, the plane wave $u(x, t) = e^{i(k \cdot x - \abs{k}^2 t)}$ is a solution. Indeed $i\partial_t u = i \cdot (-i\abs{k}^2) u = \abs{k}^2 u$ and $-\Delta u = \abs{k}^2 u$.

Non-example: the heat equation $\partial_t u = \Delta u$ differs only by the factor $i$, but it is not a Schrödinger equation. Its solutions decay in $L^2$ instead of keeping their norm: $u = e^{i k\cdot x - \abs{k}^2 t}$ solves it, and its modulus $e^{-\abs{k}^2 t}$ decreases.