In the section 4.1 of Quantum Computation by Adiabatic Evolution, Farhi et al proposes a quantum adiabatic algorithm to solve the $2$-SAT problem on a ring of spin chain. To compute the complexity of the algorithm, the authors computed the energy gap between the ground and first excited states of the adiabatic Hamiltonian.
After being expressed in momentum representation the Hamiltonian is:
$$\tilde{H}(s) = \sum_{p = \pm 1, \pm 3, \ldots, \pm \left(n-1\right)} A_p (s)$$ where $$ A_p (s) = 2 (1-s)[\beta^\dagger_p \beta_p + \beta^\dagger_{-p} \beta_{-p}] + s \left\{1 - \cos\frac{\pi p}{n} [\beta^\dagger_p \beta_p - \beta_{-p} \beta^\dagger_{-p}] + i \sin \frac{\pi p}{n}[\beta^\dagger_{-p} \beta^\dagger_{p} - \beta_{p} \beta_{-p}]\right\}. $$
Now, The authors compute the eigenvalues of $A_p (s)$.
The authors than express $A_p (s)$ in the basis of $\Omega_p$ and $\Sigma_p$ where $\Omega_p$ is annihilated by $\beta_{\pm p}$ and $\Sigma_p = \beta^\dagger_{-p} \beta^\dagger_{p} \Omega_p$.
$$ A_p (s) = \begin{bmatrix} s + s \cos \pi p /n & i s (\sin \pi p / n)\\ -i s (\sin \pi p / n) & 4 - 3 s - s \cos \pi p /n \end{bmatrix} $$
Here is my understanding of how $A_p (s) = \begin{bmatrix} s + s \cos \pi p /n & i s (\sin \pi p / n)\\ -i s (\sin \pi p / n) & 4 - 3 s - s \cos \pi p /n \end{bmatrix}$ has been constructed.
We can reexpress $A_p (s)$ as follows.
$$ A_p (s) = 2 (1-s)[\beta^\dagger_p \beta_p + \beta^\dagger_{-p} \beta_{-p}] + s \left\{1 - \cos\frac{\pi p}{n} [\beta^\dagger_p \beta_p - \beta_{-p} \beta^\dagger_{-p}] + i \sin \frac{\pi p}{n}[\beta^\dagger_{-p} \beta^\dagger_{p} - \beta_{p} \beta_{-p}]\right\}\\ = (2-2s)\beta^\dagger_p \beta_p + (2-2s)\beta^\dagger_{-p} \beta_{-p} + s -s \cos\frac{\pi p}{n} [\beta^\dagger_p \beta_p -s \beta_{-p} \beta^\dagger_{-p}] + i s \sin \frac{\pi p}{n}[\beta^\dagger_{-p} \beta^\dagger_{p} - s \beta_{p} \beta_{-p}]\\ = 2 \beta^\dagger_p \beta_p-2s \beta^\dagger_p \beta_p + 2\beta^\dagger_{-p} \beta_{-p}-2s \beta^\dagger_{-p} \beta_{-p} + s - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p + s\cos\frac{\pi p}{n} \beta_{-p} \beta^\dagger_{-p} + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}\\ = 2 \beta^\dagger_p \beta_p-2s \beta^\dagger_p \beta_p - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p + 2\beta^\dagger_{-p} \beta_{-p}-2s \beta^\dagger_{-p} \beta_{-p} + s\cos\frac{\pi p}{n} \beta_{-p} \beta^\dagger_{-p} + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}+ s\\ = 2 \beta^\dagger_p \beta_p-2s \beta^\dagger_p \beta_p - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p + 2\beta^\dagger_{-p} \beta_{-p}-2s \beta^\dagger_{-p} \beta_{-p} + s\cos\frac{\pi p}{n} \left(1-\beta^\dagger_{-p} \beta_{-p}\right) + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}+ s\\ = 2 \beta^\dagger_p \beta_p-2s \beta^\dagger_p \beta_p - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p + 2\beta^\dagger_{-p} \beta_{-p}-2s \beta^\dagger_{-p} \beta_{-p} -s\cos\frac{\pi p}{n}\beta^\dagger_{-p} \beta_{-p} + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}+ s + s\cos\frac{\pi p}{n} $$
Now, we can see how $A_p (s)$ acts on $|\Omega_p\rangle = |0\rangle$ and $|\Sigma_p\rangle = |1\rangle$.
$$ A_p |0\rangle = 2 \beta^\dagger_p \beta_p |0\rangle - 2s \beta^\dagger_p \beta_p |0\rangle - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p |0\rangle + 2\beta^\dagger_{-p} \beta_{-p} |0\rangle -2s \beta^\dagger_{-p} \beta_{-p} |0\rangle -s\cos\frac{\pi p}{n}\beta^\dagger_{-p} \beta_{-p} |0\rangle + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} |0\rangle - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}|0\rangle+ s |0\rangle + s\cos\frac{\pi p}{n} |0\rangle\\ = 0 - 0 - 0 + 0 -0 -0 + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} |0\rangle - 0+ s |0\rangle + s\cos\frac{\pi p}{n} |0\rangle\\ = \left(si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} + s + s\cos\frac{\pi p}{n} \right)|0\rangle\\ = si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p}|0\rangle + s|0\rangle + s\cos\frac{\pi p}{n} |0\rangle\\ = si \sin \frac{\pi p}{n}|1\rangle + \left(s + s\cos\frac{\pi p}{n} \right)|0\rangle $$
$$ A_p |1\rangle = 2 \beta^\dagger_p \beta_p |1\rangle - 2s \beta^\dagger_p \beta_p |1\rangle - s\cos\frac{\pi p}{n}\beta^\dagger_p \beta_p |1\rangle + 2\beta^\dagger_{-p} \beta_{-p} |1\rangle -2s \beta^\dagger_{-p} \beta_{-p} |1\rangle -s\cos\frac{\pi p}{n}\beta^\dagger_{-p} \beta_{-p} |1\rangle + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} |1\rangle - si \sin \frac{\pi p}{n} \beta_{p} \beta_{-p}|1\rangle+ s |1\rangle + s\cos\frac{\pi p}{n} |1\rangle\\ = 2 |1\rangle - 2s |1\rangle - s\cos\frac{\pi p}{n} |1\rangle + 2|1\rangle -2s |1\rangle -s\cos\frac{\pi p}{n} |1\rangle + si \sin \frac{\pi p}{n}\beta^\dagger_{-p} \beta^\dagger_{p} |1\rangle - 0+ s |1\rangle + s\cos\frac{\pi p}{n} |1\rangle\\ = \left(2 - 2s - s\cos\frac{\pi p}{n} + 2 -2s -s\cos\frac{\pi p}{n} + s + s\cos\frac{\pi p}{n} \right)|1\rangle + si \sin \frac{\pi p}{n} \beta^\dagger_{-p} \beta^\dagger_{p} |1\rangle\\ = \left(4 - 3s - s\cos\frac{\pi p}{n} \right)|1\rangle + si \sin \frac{\pi p}{n} \beta^\dagger_{-p} \beta^\dagger_{p} |1\rangle\\ = \left(4 - 3s - s\cos\frac{\pi p}{n} \right)|1\rangle + si \sin \frac{\pi p}{n} \cdot 0\\ = \left(4 - 3s - s\cos\frac{\pi p}{n} \right)|1\rangle $$
So, $A_p (s)$ takes $|0\rangle$ to $|0\rangle$ with eigenvalue $\left(s + s\cos\frac{\pi p}{n} \right)$ and to $|1\rangle$ with eigenvalue $si \sin \frac{\pi p}{n}$.
And, $A_p (s)$ takes $|1\rangle$ to $|1\rangle$ with eigenvalue $\left(4 - 3s - s\cos\frac{\pi p}{n} \right)$ and not to $|0\rangle$.
So, $A_p (s)$ should be as follows.
$$ A_p (s) = \begin{bmatrix} s + s \cos \pi p /n & i s (\sin \pi p / n)\\ 0 & 4 - 3 s - s \cos \pi p /n \end{bmatrix} $$.
But in the paper it is:
$$ A_p (s) = \begin{bmatrix} s + s \cos \pi p /n & i s (\sin \pi p / n)\\ -i s (\sin \pi p / n) & 4 - 3 s - s \cos \pi p /n \end{bmatrix} $$
What am I missing?
It looks like you replaced $\beta_p\beta_{-p}|1\rangle$ by $0$ in your calculation of $A_p|1\rangle$. I think that's were you lost the missing term. More generally speaking, I'd suggest to try to make more use of symmetry and modularisation – your calculations seem unnecessarily lengthy and could probably be exhibited a lot more succinctly if you just first applied the various operator pairs to $|0\rangle$ and $|1\rangle$ instead of writing everything out many times in long lines – this sort of thing inevitably leads to mistakes sooner or later.