Annealing Schedules

The schedule is the recipe for turning the driver off and the problem Hamiltonian on. It controls both whether the evolution stays adiabatic and how the spectral gap behaves along the way. This lesson studies the simplest non-trivial case — a single qubit — where the instantaneous ground state can be written down exactly.

The interpolating Hamiltonian

The most common schedule is the linear interpolation

More general schedules replace the two coefficients with smooth functions $A(s)$ and $B(s)$ with $A(0) = 1, A(1) = 0$ and $B(0) = 0, B(1) = 1$:

Good schedules slow down — keep $\tfrac{ds}{dt}$ small — near the minimum gap, where the system is most fragile, and can move quickly where the gap is wide.

A single-qubit example

Take $H_B = -X$ and $H_P = -Z$ with the linear schedule:

Writing $a = 1 - s$ and $b = s$, this is $H(s) = -(a X + b Z)$. The operator $aX + bZ$ has eigenvalues $\pm\sqrt{a^2 + b^2}$, so the gap is

which is smallest at $s = \tfrac12$, where $\Delta = \sqrt{2}$. The ground state of $H(s)$ is the $+1$ eigenstate of $aX + bZ$ — a spin pointing along the Bloch-sphere direction

Preparing it with one rotation

A $Y$-rotation of $|0\rangle$ produces a state whose Bloch vector lies in the $xz$-plane:

Matching $(\sin\theta, 0, \cos\theta)$ to the direction $(1-s, 0, s)$ gives

At the midpoint $s = \tfrac12$ this is $\theta = \operatorname{atan2}(\tfrac12, \tfrac12) = \tfrac{\pi}{4}$, so the ground state is $\cos\tfrac{\pi}{8}\,|0\rangle + \sin\tfrac{\pi}{8}\,|1\rangle$.

Try it

Prepare the instantaneous ground state of $H(s) = -(1-s)X - sZ$ at $s = 0.5$ using a single $RY$ rotation. The grader checks the statevector.

After running, the statevector panel shows amplitudes $\cos(\pi/8) \approx 0.924$ on $|0\rangle$ and $\sin(\pi/8) \approx 0.383$ on $|1\rangle$.

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