Time ordering

Time ordering places operators with later time arguments to the left. Interchanging fermionic operators also contributes the sign of the required permutation.

The time-ordered exponential solves $i\hslash {\partial }_{t}U=H(t)U$ with initial condition $U(t_0,t_0)=I$. Its expansion is the Dyson series. For bounded continuous Hamiltonian operators the series converges, as in differentiation of the Dyson time-ordered exponential. If the generators commute at all times, time ordering is unnecessary and an ordinary operator exponential suffices.

For $\dot{U}=A(t)U$ and $U(0)=I$, the Magnus expansion writes the logarithm of the propagator as iterated integrals of commutators. Its first terms are ${\Omega }_1={\int }_0^{T}A(t)dt$ and ${\Omega }_2=\frac{1}{2}{\int }_0^{T}dt{\int }_0^{t}ds[A(t),A(s)]$. For a Quantum Hamiltonian, $A=-iH/\hslash$, so a noncommuting time-dependent control has corrections beyond the exponential of its integrated Quantum Hamiltonian. This formula is a local convergent expansion under suitable smallness conditions, or a formal expansion; it must not be assumed globally convergent for arbitrary pulse durations.

For $H(t)={\Omega }_1(t)A+{\Omega }_2(t)B$ with fixed noncommuting Hermitian operators, $[H(t),H(s)]=[{\Omega }_1(t){\Omega }_2(s)-{\Omega }_2(t){\Omega }_1(s)][A,B]$. A constant ratio of the envelopes is sufficient to remove time ordering. Independently changing that ratio generally produces commutator corrections in the Magnus expansion, and a product of successive pulses need not equal the exponential of their summed areas.

A time-ordered product sorts operator insertions from latest time on the left to earliest time on the right. Reordering odd fermionic operators contributes the sign of their permutation. For bosonic $A,B$, $T[A(t)B(t^{\prime })]=\theta (t-t^{\prime })A(t)B(t^{\prime })+\theta (t^{\prime }-t)B(t^{\prime })A(t)$. Vacuum expectations of free time-ordered products are evaluated by the Wick theorem.