Checkpoint: Anneal to a Ground State

This checkpoint assembles everything in the module into one runnable algorithm: a complete, Trotterized single-qubit anneal that starts in the driver's ground state and finishes in the ground state of the problem Hamiltonian.

The problem

Take the problem Hamiltonian $H_P = -Z$ , whose ground state is $|0\rangle$ (energy $-1$ ). The driver is $H_B = -X$ , with ground state $|{+}\rangle$ . The linear schedule is

H(s) = (1-s)\,H_B + s\,H_P = -(1-s)\,X - s\,Z, \qquad s = t/T .

Run slowly enough and the system tracks the instantaneous ground state from $|{+}\rangle$ at $s = 0$ all the way to $|0\rangle$ at $s = 1$ . From the gap lesson, $\Delta_{\min} = \sqrt{2}$ , so a modest total time $T$ already makes the evolution comfortably adiabatic.

The recipe

Initialize. Prepare the driver ground state $|{+}\rangle = H|0\rangle$ .
Discretize. Split $T = 20$ into $N = 60$ steps of size $dt = T/N = \tfrac{1}{3}$ .
Trotterize each step. At step $k$ use the midpoint schedule value $s = (k + \tfrac12)/N$ and apply the first-order Trotter step $U_k = RX\bigl(-2\,dt\,(1-s)\bigr)\,RZ(-2\,dt\,s),$ i.e. the $RZ$ rotation first, then the $RX$ rotation.
Measure in the computational basis.

The midpoint rule $s = (k+\tfrac12)/N$ samples the schedule symmetrically across each interval, a standard improvement over using the left endpoint.

Why the answer concentrates on 0

Because $T = 20 \gg 1/\Delta_{\min}^2 = \tfrac12$ , the adiabatic theorem applies: the diabatic transition probability is tiny, so the system stays in the ground state. By $s = 1$ that ground state is $|0\rangle$ , and a computational-basis measurement returns outcome $0$ with probability close to 1. The small residual weight on outcome $1$ comes from finite $T$ and the $O(dt^2)$ Trotter error.

Try it

Implement the full anneal exactly as specified and measure. The grader checks the measurement distribution: outcome $0$ should dominate.

Run your code to see the quantum state.

After running, the Probabilities tab shows the bar for $0$ near 1 and the bar for $1$ near 0 — you have annealed into the ground state of $H_P = -Z$ .