I was trying to reproduce example 3.3 of Quantum Computation by Adiabatic Evolution by Edward Farhi et. al. This is an adiabatic algorithm to solve an instance of three qubits 2-SAT problem.
I think I have created the initial Hamiltonian, $H_B$ correctly. It is just an $8 \times 8$ Hadamard matrix.
$$H_B = \left( \begin{array}{cccccccc} \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} \\ \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} \\ \end{array} \right)$$
According to the example, the unique satisfying assignment is $011$. The problem Hamiltonian is a sum of three sub-Hamiltonians each correspond to a clause.
$$H_P = I_8 - H^{12}_{imply} - H^{13}_{disagree}-H^{23}_{agree}$$
Here are my results for sub-Hamiltonians.
$H^{12}_{imply}$ clause can be satisfied with any of $00$, $01$ or $11$. So,
$$H^{12}_{imply} = \left(\left(\frac{1}{\sqrt{6}}\right)\left(|000\rangle+|001\rangle+|010\rangle+|011\rangle+|110\rangle+|111\rangle\right)\right) \left(\left(\left(\frac{1}{\sqrt{6}}\right)\left(|000\rangle+|001\rangle+|010\rangle+|011\rangle+|110\rangle+|111\rangle\right)\right)\right)^{\dagger} $$
The ground state of this Hamiltonian satisfies the assignment constraint.
$H^{13}_{disagree}$ clause can be satisfied with any of $01$ or $10$. So,
$$H^{13}_{disagree} = \left(\left(\frac{1}{2}\right)\left(|001\rangle+|011\rangle+|100\rangle+|110\rangle\right)\right) \left(\left(\left(\frac{1}{2}\right)\left(|001\rangle+|011\rangle+|100\rangle+|110\rangle\right)\right)\right)^{\dagger}$$
The ground state of this Hamiltonian satisfies the assignment constraint.
$H^{23}_{agree}$ clause can be satisfied with any of $00$ or $11$. So,
$$H^{23}_{agree} = \left(\left(\frac{1}{2}\right)\left(|001\rangle+|011\rangle+|100\rangle+|110\rangle\right)\right) \left(\left(\left(\frac{1}{2}\right)\left(|001\rangle+|011\rangle+|100\rangle+|110\rangle\right)\right)\right)^{\dagger}$$
The ground state of this Hamiltonian satisfies the assignment constraint.
So, $$H_P = \left( \begin{array}{cccccccc} \frac{17}{6} & -\frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & 0 & 0 & -\frac{1}{6} & -\frac{1}{6} \\ -\frac{1}{6} & \frac{7}{3} & -\frac{1}{6} & -\frac{2}{3} & -\frac{1}{2} & 0 & -\frac{2}{3} & -\frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{17}{6} & -\frac{1}{6} & 0 & 0 & -\frac{1}{6} & -\frac{1}{6} \\ -\frac{1}{6} & -\frac{2}{3} & -\frac{1}{6} & \frac{7}{3} & -\frac{1}{2} & 0 & -\frac{2}{3} & -\frac{1}{6} \\ 0 & -\frac{1}{2} & 0 & -\frac{1}{2} & \frac{5}{2} & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 \\ -\frac{1}{6} & -\frac{2}{3} & -\frac{1}{6} & -\frac{2}{3} & -\frac{1}{2} & 0 & \frac{7}{3} & -\frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & 0 & 0 & -\frac{1}{6} & \frac{17}{6} \\ \end{array} \right)$$
This Hamiltonian, which represents the instance of the problem is expected to be satisfied only with the assignment 011. So, the ground state should be $|011\rangle$.
But in reality, the ground eigenvalue is greater than $0$ and the ground state is $\left(1,4,1,4,3,0,4,1\right)^{\dagger}$.
Here, $I_n$ is the $n \times n$ identity matrix.
What was I doing wrong? My Mathematica code is available here.
