= Time-evolution operator
{title2=$U(t)$}
The time-evolution operator maps an initial <quantum state> to its state at a later time. For a time-independent <Hamiltonian operator> $H$, it is the <unitary operator> $U(t)=e^{-iHt/\hbar}$.
More generally, write $|\psi(t)\rangle=U(t,t_0)|\psi(t_0)\rangle$. The <time-evolution operator> obeys
$$
U(t_0,t_0)=I,\qquad U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0),\qquad
U(t,t_0)^{-1}=U(t,t_0)^\dagger=U(t_0,t).
$$
For a constant <self-adjoint> <Hamiltonian operator>, $U(t,t_0)=\exp[-iH(t-t_0)/\hbar]$. With zero as the initial time, this gives the group law $U(t+s)=U(t)U(s)$. Successive evolution is reversible; a measurement followed by conditioning on an outcome is a separate operation described by <projective measurement>.
For a time-dependent <Hamiltonian operator>, the defining equation is $i\hbar\partial_tU(t,t_0)=H(t)U(t,t_0)$. In finite <dimension>, a continuous <Hermitian> $H(t)$ still gives <unitary time evolution>, since
$$
\frac{d}{dt}(U^\dagger U)
=\frac{i}{\hbar}U^\dagger H(t)U-\frac{i}{\hbar}U^\dagger H(t)U=0.
$$
The solution is a <time-ordered exponential>, with later factors acting on the left. One may replace it by $\exp[-i\int_{t_0}^tH(s)\,ds/\hbar]$ when $[H(s),H(s')]=0$ at all times; independently changing pulse axes generally requires <time ordering>.<Quantum Physics II (8.05), Fall 2013/Quantum Dynamics>[Quantum Physics II (8.05), Fall 2013; Zwiebach, Quantum Dynamics, section 2.3, printed pp. 16–17, equations (2.46)–(2.52) and Case 3]{ref}
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