Problem and Driver Hamiltonians

Every adiabatic or annealing algorithm is built from two Hamiltonians: a driver that is easy to start from, and a problem Hamiltonian whose ground state is the answer. This lesson defines both and shows how to prepare the driver's ground state in the simulator.

The driver (transverse-field) Hamiltonian

The standard driver on $n$ qubits applies a transverse field to every qubit:

H_B = -\sum_{i=0}^{n-1} X_i .

Because $H_B$ is a sum of commuting single-qubit terms, its ground state factorizes. Each $-X_i$ is minimized by the $+1$ eigenstate of $X$ , which is

|{+}\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}} = H|0\rangle .

So the ground state of $H_B$ is the uniform superposition

|{+}\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{x=0}^{2^n - 1} |x\rangle ,

prepared by one Hadamard per qubit. Its energy is $E_0 = -n$ , since each term contributes $-1$ .

The problem Hamiltonian

The problem Hamiltonian $H_P$ is diagonal in the computational basis: every basis state $|x\rangle$ is already an eigenstate, with eigenvalue equal to a classical cost function $C(x)$ :

H_P\,|x\rangle = C(x)\,|x\rangle .

The ground state of $H_P$ is therefore the bit string $x^\star$ that minimizes $C$ . A typical construction uses $Z$ operators, for example a single-qubit field $H_P = -Z_0$ whose ground state is $|0\rangle$ (eigenvalue $-1$ ), or two-body couplings $Z_i Z_j$ for interacting spins. Because $H_P$ is diagonal, preparing its ground state is exactly the hard combinatorial problem you want to solve — which is why you start from the easy driver instead and interpolate.

Try it

Prepare the ground state of the 2-qubit driver $H_B = -(X_0 + X_1)$ . The grader checks the full statevector, so each of the four amplitudes should equal $1/2$ .

Run your code to see the quantum state.

After running, the statevector panel should show $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ each with amplitude $0.5$ — the uniform superposition $|{+}{+}\rangle$ .