The Trotter Product Formula

The time-evolution operator $e^{-iHt}$ for a sum of Hamiltonians $H = H_1 + H_2$ cannot generally be split into $e^{-iH_1 t} e^{-iH_2 t}$ , because matrices that do not commute satisfy $e^{A+B} \ne e^A e^B$ in general. The Lie–Trotter product formula makes this splitting an approximation that becomes exact in a controlled limit.

The first-order Trotter formula

For any two operators $A$ and $B$ ,

e^{A+B} = \lim_{r \to \infty} \left(e^{A/r} e^{B/r}\right)^r.

Truncating to a single step ( $r = 1$ ) gives the first-order Trotter approximation:

e^{-i(H_1 + H_2)t} \approx e^{-iH_1 t}\, e^{-iH_2 t}.

The error in the unitary can be bounded using the Baker–Campbell–Hausdorff expansion. To leading order,

e^{-iH_1 t} e^{-iH_2 t} = e^{-i(H_1+H_2)t}\, e^{-\tfrac{1}{2}[H_1,H_2]t^2 + O(t^3)},

so the operator error is $O(t^2 \|[H_1, H_2]\|)$ . When $H_1$ and $H_2$ commute the commutator $[H_1, H_2] = 0$ and the approximation is exact for any $t$ .

Mapping a Trotter step to gates

Consider the Hamiltonian

H = Z_0 Z_1 + X_0,

where $Z_0 Z_1$ is the two-qubit Pauli interaction from the previous lesson and $X_0$ acts only on qubit 0. The two terms do not commute — $[ZZ, X \otimes I] = 2i\, Y \otimes Z$ — so Trotterization introduces an approximation error proportional to $t^2$ .

A single first-order Trotter step at time $t$ decomposes as:

e^{-it(Z_0 Z_1 + X_0)} \approx e^{-it X_0}\; e^{-it Z_0 Z_1}.

Reading the product right-to-left (standard operator convention), this means the $ZZ$ factor is applied to the state first, followed by the $X_0$ factor — matching the left-to-right gate order in the circuit diagram below.

Each factor maps directly to known gate patterns:

Factor 1 (applied first to the state): $e^{-it Z_0 Z_1}$

As derived in the previous lesson, the $ZZ$ rotation is implemented with a CNOT sandwich:

e^{-it Z_0 Z_1} = \mathrm{CNOT}_{0 \to 1} \cdot \bigl(I \otimes R_z(2t)\bigr) \cdot \mathrm{CNOT}_{0 \to 1}.

Factor 2 (applied second to the state): $e^{-it X_0}$

Using the general Pauli rotation identity $e^{-i\theta P} = R_P(2\theta)$ ,

e^{-it X_0} = R_x(2t).

Chaining these two factors gives the full Trotter step as a four-gate circuit:

cx(0, 1)       // parity encode
rz(2t, 1)      // ZZ rotation
cx(0, 1)       // parity decode
rx(2t, 0)      // X_0 rotation

Generalising to many terms

The same recipe extends to any Hamiltonian written as a sum of Pauli strings:

H = \sum_{j} c_j P_j.

Each $e^{-i c_j t P_j}$ can be implemented with at most $O(n)$ CNOT gates (one CNOT ladder of length $n$ ), so a single Trotter step over an $n$ -qubit system with $k$ terms requires $O(kn)$ two-qubit gates in total. Larger $r$ (more repetitions) trades circuit depth for accuracy.

Try it

Implement one first-order Trotter step for $H = Z_0 Z_1 + X_0$ at time $t = \pi/4$ . The grader checks the full unitary matrix, so an empty circuit will not pass.

Run your code to see the quantum state.

After grading, try changing $t$ to a different value (such as $\pi/8$ or $\pi/2$ ) and observe how the unitary changes — but make sure your referenceSolution matches whatever $t$ you choose.