Why can $\hbar q_x$ and $\hbar q_y$ be replaced by $\hat{p}_x=-i\hbar\frac{\partial}{\partial x}$ and $\hat{p}_y=-i\hbar\frac{\partial}{\partial y}$?

Question

It is written in the book The Physics of Graphene (Page 10 and 17) that when the intervalley scattering is neglected, we can make the following substitution in the Hamiltonian of the graphene when using the tight-binding model: $$\hbar q_x\to\hat{p}_x=-i\hbar\frac{\partial}{\partial x}$$ $$\hbar q_y\to\hat{p}_y=-i\hbar\frac{\partial}{\partial y} $$ Then the effective Hamiltonian goes from $$\hat{H}_{\mathit{K},\mathit{K'}}(\vec{q})=\hbar v \begin{pmatrix} 0 & q_x \mp iq_y\\ q_x \pm iq_y & 0 \end{pmatrix}$$ to $$\hat{H}_{\mathit{K}}=-i\hbar v\vec{\sigma}\nabla$$ $$\hat{H}_{\mathit{K'}}=\hat{H}_{\mathit{K}}^{\text{T}}$$ where $\vec{q}=\vec{k}-\vec{\mathit{K}}$ denotes the relative momentum in the vicinity of the Dirac point $\mathit{K}$; $v=\frac{3a|t|}{2\hbar}$ is the electron velocity at the conical points; $\vec{\sigma} = (\sigma_x, \sigma_y)$ are the pauli matrices and T denotes a transposed matrix.

It hints that it corresponds to the effective mass approximation, or $\vec{k}\cdot\vec{p}$ perturbation theory, however, I still do not get the idea why this can be done.

Answer 1

The expansion of Hamiltonian near the degeneracy point of graphene is actually the equivalent of the effective mass approximation near the band bottom in more traditional materials. In this case one focuses on the long-wavelength component of the Bloch functions (i.e., on the short wave vectors), $$\psi(\mathbf{r})=e^{i\mathbf{k}\mathbf{r}}u_\mathbf{k}( \mathbf{r}),$$ omitting the factor $u_\mathbf{k}( \mathbf{r})$, which varies quickly on the scale of a unit cell (more precisely, we assume that it is constant). One says sometimes that the effective Hamiltonians obtained by such an expansion are Hamiltonians for the envelope function of a wave packet, composed of Bloch waves.

Once $u_\mathbf{k}( \mathbf{r})$ is assumed to be a constant, the relation between a wave function in position and momentum representations becomes a simple Fourier transform, as it is for free particles: $$ \psi(\mathbf{r})=\int d\mathbf{r}e^{i\mathbf{k}\mathbf{r}}\phi(\mathbf{k}),$$ and hence $$ -i\hbar\nabla\psi(\mathbf{r})=\int d\mathbf{r}e^{i\mathbf{k}\mathbf{r}}\hbar\mathbf{k}\phi(\mathbf{k}),$$ which is, of course, a basic property of the Fourier transform.

Answer 2

That substitution can be directly derived from tight-binding approximation. Please refer section 2 of this article, you may find the solution by inserting Eq. (2.1) into (2.2).