Checkpoint: Trotterize a Hamiltonian

This checkpoint asks you to assemble a complete first-order Trotter circuit for a concrete two-qubit Hamiltonian — combining every skill from this module into a single working program.

The target Hamiltonian

Consider the two-qubit Hamiltonian

H = \frac{\pi}{8}\, Z \otimes Z \;+\; \frac{\pi}{8}\, X \otimes I.

The first term couples the two spins through a $ZZ$ interaction; the second drives single-qubit rotations on qubit 0 alone. Both coefficients equal $\pi/8$ , which keeps the angles tidy.

Applying the first-order Trotter formula

Because $Z \otimes Z$ and $X \otimes I$ do not commute — you can verify that $(X \otimes I)(Z \otimes Z) \neq (Z \otimes Z)(X \otimes I)$ since $XZ \neq ZX$ — the exact evolution $e^{-iHt}$ cannot be split naively. The first-order Trotter (Lie–Trotter) product formula gives the approximation

e^{-iHt} \approx e^{-i H_1 t}\, e^{-i H_2 t}, \qquad H_1 = \tfrac{\pi}{8} ZZ, \quad H_2 = \tfrac{\pi}{8} (X \otimes I),

with an error of order $t^2 [H_1, H_2]$ . For $t = 1$ and the given coefficients this is a single Trotter step — enough to demonstrate the structure without needing multiple repetitions.

Compiling each factor to gates

Factor 1 — $e^{-i(\pi/8)\, ZZ}$ . The $ZZ$ eigenvalue of $|xy\rangle$ is $(-1)^{x \oplus y}$ , so a CNOT from qubit 0 to qubit 1 maps the parity into qubit 1, and a single $R_z$ on qubit 1 applies the phase:

e^{-i\theta\, ZZ} = \mathrm{CNOT}_{0 \to 1}\cdot\bigl(I \otimes R_z(2\theta)\bigr)\cdot\mathrm{CNOT}_{0 \to 1}, \qquad \theta = \tfrac{\pi}{8}.

In code: c.cx(0, 1); c.rz(2 * Math.PI/8, 1); c.cx(0, 1);

Factor 2 — $e^{-i(\pi/8)\, (X \otimes I)}$ . Only qubit 0 participates. Using $e^{-i\theta X} = R_x(2\theta)$ :

e^{-i(\pi/8)\, X \otimes I} = R_x\!\bigl(\tfrac{\pi}{4}\bigr) \otimes I.

In code: c.rx(2 * Math.PI/8, 0);

Try it

Implement both factors in the correct order. The grader checks the full unitary matrix of your two-qubit circuit, so an empty or partial circuit will not pass.

Run your code to see the quantum state.