Let's start with a simple example where $H_i$ and $H_f$ commute because they are both diagonal:
$H_i=
\begin{pmatrix}1 & 0\\
0 & -1
\end{pmatrix}
$
$H_p=
\begin{pmatrix}-1 & 0\\
0 & -0.1
\end{pmatrix}
$
The eigenvector with lowest eigenvalue (i.e. the ground state) of $H_i$ is $|1\rangle $ so we start in this state.
The ground state of $H_f$ is $|0\rangle$ so this is what we're looking for.
Remember the minimum runtime for the AQC to give the correct answer to within an error $\epsilon$:
$\tau\ge \max_t\left(\frac{||H_i - H_f||^2}{\epsilon E_{\rm{gap}}(t)^3}\right)$.
This is given and explained in Eq. 2 of Tanburn et al. (2015).
- Let's say we want $\epsilon = 0.1$.
- Notice that $||H_i - H_f||^2 = 0.1 $ according Eq. 4 of the same paper.
- Notice that $\frac{||H_i - H_f||^2}{\epsilon}=1$ (I've chosen $\epsilon$ so that this would happen, but it doesn't matter).
- We now have $\tau \ge \max_t\left(\frac{1}{E_{\rm{gap}}(t)^3}\right)$
So what is the minimum gap between ground and first excited state (which gives the $\max_t$) ?
When $t=20\tau/29$, the Hamiltonian is:
$H=\frac{9}{29}H_i + \frac{20}{29}H_p$
$H=\frac{9}{29}\begin{pmatrix}1 & 0\\
0 & -1
\end{pmatrix} + \frac{20}{29}\begin{pmatrix}-1 & 0\\
0 & -0.1
\end{pmatrix}$
$
H=\begin{pmatrix}\frac{9}{29} & 0\\
0 & -\frac{9}{29}
\end{pmatrix}+\begin{pmatrix}-\frac{20}{29} & 0\\
0 & -\frac{2}{29}
\end{pmatrix}
$
$
H=\begin{pmatrix}\frac{-11}{29} & 0\\
0 & -\frac{11}{29}
\end{pmatrix}
$
So when $t=\frac{20}{29}\tau$, we have $E_{\rm{gap}}=0$ and the lower bound on $\tau$ is essentially $\infty$.
So the adiabatic theorem still applies, but when it states that the Hamiltonian needs to change "slowly enough", it turns out it needs to change "infinitely slowly", which means you will not likely ever get the answer using AQC.