Confusion Regarding the Derivation of Graphene Dispersion Using Annihilation and Creation Operators I am going through a text which derives the energy bands in graphene (https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/3/67057/files/2018/09/graphene_tight-binding_model-1ny95f1.pdf) and am stuck on a step.
Given $H = -t\sum_{\vec{k}, \vec{\delta}}e^{i\vec{k}\cdot\vec{\delta}}a^\dagger(k)b(k)$ + h.c. the author then writes:
$H = -t\sum_\vec{k}\psi^\dagger(k)h(k)\psi(k) $
Where $\psi(k)=\begin{pmatrix}a(k)\\b(k) \end{pmatrix}$
and $h(k) = \begin{pmatrix} 0 & f(k) \\ f^*(k) & 0\end{pmatrix}$
with $f(k) = \sum_\vec{\delta}e^{i\vec{k}\cdot\vec{\delta}}$
The author then proceeds to find the eigenvalues of $h(k)$, based on the idea that they are the eigenvalues of $H$ as well. My question is why is diagonalizing $h(k)$ sufficient when searching for the eigenvalues of $H$?
 A: My understanding:
$h(k) = -t\begin{pmatrix} 0 & f(k) \\ f^*(k) & 0\end{pmatrix}$ is indeed the matrix form of the Hamiltonian in $\textbf{k}$-representation. That is we define $|a,\textbf{k}\rangle = a^\dagger_{\textbf{k}}|0\rangle$ and $|b,\textbf{k}\rangle = b^\dagger_{\textbf{k}}|0\rangle$. Then it can be shown that the matrix elements $$\langle a,\textbf{k}|H|b,\textbf{k}\rangle=-t\sum_\vec{\delta}e^{-i\vec{k}\cdot\vec{\delta}} \quad \text{and} \quad \langle b,\textbf{k}|H|a,\textbf{k}\rangle=-t\sum_\vec{\delta}e^{i\vec{k}\cdot\vec{\delta}}$$
where $H$ is given by Eq(5) in the note. It can also be shown that the matrix elements $\langle a,\textbf{k}|H|a,\textbf{k}\rangle=\langle b,\textbf{k}|H|b,\textbf{k}\rangle=0$
For example, to compute $\langle a,\textbf{k}'|H|b,\textbf{k}'\rangle$
\begin{align}
\langle a,\textbf{k}'|-t\sum_{\delta,\textbf{k}}(e^{-i\textbf{k}\cdot\delta}a_{\textbf{k}}^\dagger b_{\textbf{k}}+e^{i\textbf{k}\cdot\delta}b_{\textbf{k}}^\dagger a_{\textbf{k}})|b,\textbf{k}'\rangle
\end{align}
Compute first term(second term is $0$ as $a_{\textbf{k}}|b,\textbf{k}'\rangle =0$ ), note that $a_{\textbf{k}}|a,\textbf{k}\rangle =|0 \rangle$ and $\langle a,\textbf{k}|a_{\textbf{k}}^\dagger =\langle 0|$. Also only if $\textbf{k}=\textbf{k}'$ we have a non-zero term.
So we have
\begin{align}
\langle a,\textbf{k}'|H|b,\textbf{k}'\rangle&=-t\sum_{\delta}e^{-i\textbf{k}'\cdot\delta}\langle a,\textbf{k}'| a_{\textbf{k}'}^\dagger b_{\textbf{k}'}|b,\textbf{k}'\rangle \\
&=-t\sum_{\delta}e^{-i\textbf{k}'\cdot\delta}\langle 0|0\rangle \\
&=-t\sum_{\delta}e^{-i\textbf{k}'\cdot\delta}
\end{align}
similarly for other matrix elements.
