The Mixing Hamiltonian

QAOA alternates two unitaries: a cost unitary that encodes the objective, and a mixer unitary that drives the system away from a fixed point so it can explore the full solution space. Without the mixer, the optimizer would stay trapped in whatever initial state it started in.

The standard mixer Hamiltonian

Farhi, Goldstone, and Gutmann (2014) defined the mixing Hamiltonian as the sum of Pauli- $X$ operators over all $n$ qubits:

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

Because $X$ acting on qubit $i$ commutes with $X$ acting on any other qubit $j \neq i$ , the time evolution factorizes into independent single-qubit rotations:

$U_B(\beta) = e^{-i\beta B} = \prod_{i=0}^{n-1} e^{-i\beta X_i} = \bigotimes_{i=0}^{n-1} R_X(2\beta),$

where $R_X(\theta) = e^{-i\theta X/2}$ is the standard $X$ -axis rotation. The factor of 2 enters because the convention is $R_X(\theta) = e^{-i\theta X/2}$ , so matching $e^{-i\beta X}$ requires the argument $\theta = 2\beta$ .

Why X is the right choice

The Pauli- $X$ operator generates rotations that mix $|0\rangle$ and $|1\rangle$ without preferred direction. An $R_X(\pi)$ gate is just the $X$ gate — a full bit-flip. At intermediate angles it creates genuine superpositions. This means the mixer, parametrized by $\beta$ , can continuously tune how aggressively the quantum state tunnels between different bit-string configurations.

More formally, the mixer must not commute with the cost Hamiltonian $C$ , otherwise $[e^{-i\gamma C}, e^{-i\beta B}] = 0$ and the two unitaries could be absorbed into one, removing any quantum advantage. The standard $X$ -mixer satisfies this whenever the cost function is diagonal in the $Z$ basis.

The circuit for two qubits

For $n = 2$ qubits and mixing angle $\beta$ , the mixer layer is simply

$U_B(\beta) = R_X(2\beta) \otimes R_X(2\beta).$

No two-qubit gates are needed: the $X_i$ terms in $B$ each act on a single qubit, so the whole mixer decomposes into parallel single-qubit rotations.

Try it

Build the 2-qubit mixer unitary $U_B(\beta)$ for $\beta = \pi/4$ . The grader checks the full unitary matrix, so returning an empty circuit will fail.

Run your code to see the quantum state.

At $\beta = \pi/4$ each qubit undergoes $R_X(\pi/2)$ , a 90-degree rotation about the $x$ -axis. The resulting unitary is

\otimes \frac{1}{\sqrt{2}}\begin{pmatrix}1 & -i \\ -i & 1\end{pmatrix}.$$ In a QAOA circuit you would choose $\beta$ as a free variational parameter and tune it classically; here we fix $\beta = \pi/4$ simply to make the expected unitary well-defined for the grader.