Trotter–Suzuki Hamiltonian simulation

Simulating $e^{-iHt}$ for a Hamiltonian $H=\sum_j A_j$ with non-commuting terms $A_j$ cannot generally be done by exponentiating each term separately, since $e^{-i(A+B)t}\neq e^{-iAt}e^{-iBt}$ unless $[A,B]=0$. The first-order Trotter formula approximates it anyway by splitting the evolution into $r$ short steps:

$$e^{-iHt} \approx \left(\prod_j e^{-iA_j t/r}\right)^{r},$$

with error $O(t^2/r)$ per Lloyd's original analysis (1996); tighter, commutator-dependent bounds (Childs, Su, Tran, Wiebe, Zhu, 2019) show the error scales with $\sum_{i<j}\|[A_i,A_j]\|$, vanishing exactly when the terms commute. Suzuki's symmetric (even-order) product formulas reduce the error to $O(t^{p+1}/r^p)$ for order $p$, at the cost of more exponentials evaluated per step.

Exactly solvable small instance. Take the single-qubit Hamiltonian $H=X+Z=\sqrt2\,\hat n\cdot\vec\sigma$ with $\hat n=(1,0,1)/\sqrt2$, and note $[X,Z]=-2iY\neq0$, so a Trotter error genuinely exists here. Exact evolution from $|0\rangle$ gives

$$e^{-iHt}|0\rangle = \left[\cos(\sqrt2 t) - i\frac{\sin(\sqrt2 t)}{\sqrt2}\right]|0\rangle - i\frac{\sin(\sqrt2 t)}{\sqrt2}|1\rangle,$$

so $P(|0\rangle) = 1-\tfrac12\sin^2(\sqrt2 t)$ and $P(|1\rangle)=\tfrac12\sin^2(\sqrt2 t)$. Choosing $t=\pi/(2\sqrt2)$ makes $\sqrt2 t=\pi/2$, giving $P(|0\rangle)=P(|1\rangle)=0.5$ exactly — a concrete analytic target that the first-order Trotter circuit (alternating $R_X$ and $R_Z$ layers) converges to as the number of steps $r$ grows, with discrepancy shrinking as $O(t^2/r)$.

Where it is used. Trotterization is the default way to run digital Hamiltonian simulation on gate-based hardware, and underlies quantum chemistry, condensed-matter, and QITE/VarQITE pipelines that need $e^{-iHt}$ or $e^{-H\tau}$ as a subroutine.