I am studying the continuum limit as described in section II B of this paper. The tight binding Hamiltonian for graphene is given by
$$ H = -t \sum_{\langle i,j \rangle, \sigma}\left(a^\dagger_{\sigma,i}b_{\sigma , j} + \text{h.c.}\right)$$
as show in Eq. (4). It states that in order to derive a theory that is valid close to the Dirac points, we approximate the fields as
$$ a_n \approx e^{-i K \cdot R_n} a_{1,n} +e^{-iK' \cdot R_n} a_{2,n}$$
and similarly for $b_n$, as shown in Eq. (17). Upon substituting this into the Hamiltonian $H$, we arrive at
$$ H \approx -iv_F\int dx dy \left( \psi_1^\dagger \sigma \cdot \nabla \psi_1 + \psi^\dagger_2 \sigma^* \cdot \nabla \psi_2\right)$$
where $ \sigma = (\sigma^x , \sigma^y)$ and $\psi_i^\dagger = (a_i^\dagger,b_i^\dagger)$, as shown in Eq. (18).
My questions
The answer to this question states that as the lattice spacing $\delta \rightarrow 0$, only the features close to the Dirac points remain at finite energy. However, this was not given any more detail. How do I show this mathematically? What is our definition of "close to the Dirac points"?
Upon substituting the approximations for $a_n,b_n$ into the Hamiltonian $H$, I find cross terms such as $a^\dagger_1 b_2 $, however the continuum limit quoted above does not have terms like this, i.e., no coupling between $1$ and $2$. I suppose a physical reason for this is that a low-energy process should not evolve the system near one Dirac point to the other, however is there a precise mathematical reason why this is the case?