Quantum mechanics

Remarks

Example: take $\mathcal{H} = \C^2$ and $H = \mathrm{diag}(E_0, E_1)$ with real $E_0, E_1$. Then $H$ is self-adjoint, and $U(t) = \mathrm{diag}(e^{-iE_0 t/\hbar}, e^{-iE_1 t/\hbar})$ is unitary, so $(\C^2, H)$ is a quantum-mechanical system, a two-level system.

Non-example: on $\C^2$, the matrix $H = iI$ is not self-adjoint. Then $e^{-itH/\hbar} = e^{t/\hbar} I$ is not unitary, and it does not keep the trace of a density operator equal to $1$.

When the state is pure and given by a unit vector $\psi$, clause 3 becomes $\psi(t) = U(t)\psi$, the solution of the abstract Schrödinger equation $i\hbar\, \psi'(t) = H\psi(t)$ for $\psi \in D(H)$.