Kitaev Chain Spectrum (Unpaired Majorana Fermions in quantum wires) How does one arrive at the spectrum equation(13):
$$\epsilon (q)=\pm \sqrt{(2w \cos q +\mu)^2+4\cdot \mid {\Delta} \mid^2 \sin ^{2} q}$$
from the initial Hamiltonian.
Also, shouldn't (12) in the paper be a diagonal matrix as per the canonical Hamiltonian in (11) in terms of b' and b''?
The paper can be found here.
EDIT: The related Hamiltonian is: 
$H_1 = \sum_{j} [-w(a_j^\dagger a_{j+1} + a_{j+1} ^\dagger a_j)-\mu(a_j^\dagger a_j - \frac{1}{2}) + \Delta a_j a_{j+1} + \Delta ^{*} a_{j+1}^\dagger a_j^\dagger]$
It is from the famous paper by Kitaev on unpaired Majorana Modes at the ends of a chain on the surface of a p-wave superconductor.
The answer below outlines the general method of finding the spectrum for any general hamiltonian which was what I was looking for.
 A: The energy spectra of Equation (13) is the bulk energy spectra as a function of the momentum $q$. It describes the energy levels of the chain where the first and the last sites 1 and $N$ are connected in a ring. The spectra can be obtained in three steps: 
1) first, Fourier transform Eq. 4, so that you will obtain a Hamiltonian in momentum space, whose terms are of the operators $a_q$ and $a^\dagger_q$, i.e., which annihilate/create a particle with momentum $q$. (the old operators $a_i$ and $a^\dagger_i$ annihilate/create particles at lattice sites $i$)
2) you may notice that you will get terms like $a^\dagger_q a^\dagger_{-q}$ and $a_q a_{-q}$. You can write this Hamiltonian in the Bogoliubov-de Gennes (BdG) form as a 2$\times$2 matrix, corresponding to particle and hole states.
3) At this point, you can diagonalize your BdG Hamiltonian in momentum space and obtain the spectra.
It is not so difficult to obtain the spectra so I didn't write all the steps but I just outlined the general way to do it. If you have further question please comment me.
Regarding your 2nd question about Eq. 12, please notice that this matrix correspond to the Hamiltonian in Majorana representation, which is not a Hermitian matrix. 
