Confusion Regarding the Derivation of Graphene Dispersion Using Annihilation and Creation Operators

My understanding:

$h(k) = -tbegin{pmatrix} 0 & f(k) f^*(k) & 0end{pmatrix}$ is indeed the matrix form of the Hamiltonian in $textbf{k}$-representation. That is we define $|a,textbf{k}rangle = a^dagger_{textbf{k}}|0rangle$ and $|b,textbf{k}rangle = b^dagger_{textbf{k}}|0rangle$. Then it can be shown that the matrix elements $$langle a,textbf{k}|H|b,textbf{k}rangle=-tsum_vec{delta}e^{-ivec{k}cdotvec{delta}} quad text{and} quad langle b,textbf{k}|H|a,textbf{k}rangle=-tsum_vec{delta}e^{ivec{k}cdotvec{delta}}$$ 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}'|-tsum_{delta,textbf{k}}(e^{-itextbf{k}cdotdelta}a_{textbf{k}}^dagger b_{textbf{k}}+e^{itextbf{k}cdotdelta}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&=-tsum_{delta}e^{-itextbf{k}'cdotdelta}langle a,textbf{k}'| a_{textbf{k}'}^dagger b_{textbf{k}'}|b,textbf{k}'rangle &=-tsum_{delta}e^{-itextbf{k}'cdotdelta}langle 0|0rangle &=-tsum_{delta}e^{-itextbf{k}'cdotdelta} end{align} similarly for other matrix elements.