Time-Dependent Hamiltonians

So far every Hamiltonian in this module was constant in time. Many real systems are not: a driven qubit, an adiabatic sweep, an annealing schedule, or a laser pulse all have a Hamiltonian $H(t)$ that changes as the experiment runs. This lesson shows how to simulate that.

Why time-dependence is harder

For a constant $H$ , evolution is simply $U(t) = e^{-iHt}$ . When $H = H(t)$ varies, the propagator is the time-ordered exponential

U(t) = \mathcal{T}\exp\!\left(-i\int_0^t H(t')\,dt'\right),

where $\mathcal{T}$ orders later times to the left. You cannot simply write $U(t) = \exp(-i\int_0^t H\,dt')$ in general, because $H(t_1)$ and $H(t_2)$ need not commute at different times — the order of application matters. The time-ordering symbol $\mathcal{T}$ is precisely what keeps track of that order.

Slicing time

The fix mirrors Trotterization in time rather than in operator structure. Chop $[0, t]$ into $r$ thin slices of width $\Delta t = t/r$ . Over a slice short enough that $H$ barely changes, treat it as constant and use the frozen propagator

U_k \approx \exp\!\bigl(-i\, H(t_k)\,\Delta t\bigr),

then compose the slices in time order, latest last:

U(t) \approx U_{r-1}\,U_{r-2}\cdots U_1\,U_0.

Each slice can itself be Trotterized if $H(t_k)$ has multiple non-commuting terms. Evaluating $H$ at the slice midpoint $t_k = (k + \tfrac12)\Delta t$ (rather than an endpoint) is a midpoint rule that gives second-order accuracy in $\Delta t$ for free — a cheap and worthwhile upgrade.

A driven single qubit

Take the time-dependent Hamiltonian $H(t) = g(t)\,X$ with a linear ramp $g(t) = t$ , evolving from $0$ to $T = 1$ . Each frozen slice is a single-qubit $X$ -rotation:

\exp\!\bigl(-i\,g\,X\,\Delta t\bigr) = R_X\bigl(2\,g\,\Delta t\bigr),

using $R_X(\theta) = e^{-i\theta X/2}$ . With two slices ( $\Delta t = 0.5$ ) and midpoint evaluation, slice 0 freezes $g = 0.25$ (angle $0.25$ ) and slice 1 freezes $g = 0.75$ (angle $0.75$ ). Because all the slice propagators commute here (they are all $X$ -rotations), they compose to a single $R_X(1.0)$ — which equals the exact answer $\int_0^1 t\,dt = \tfrac12$ doubled into the rotation angle. This commuting case is a sanity check; the slicing method is essential precisely when the slices do not commute.

Accuracy

Each frozen-slice approximation has error $O(\Delta t^2)$ (with the midpoint rule), so the total error over $r$ slices is $O(\Delta t^2 \cdot r) = O(t\,\Delta t) = O(t^2/r)$ — the same $1/r$ convergence as first-order Trotter. Finer slices track a faster-changing $H(t)$ more faithfully, at the cost of more gates.

Try it

Simulate $H(t) = g(t)\,X$ with the ramp $g(t) = t$ from $0$ to $T = 1$ , using 2 slices and midpoint evaluation. Apply one $R_X(2\,g\,\Delta t)$ per slice. The grader checks the final state vector.

Run your code to see the quantum state.