meaning of $k$ in $k\cdot p$ approximation in condensed matter physics

Question

Sometimes, I encounter such a practice in condensed matter literature:

One takes a $k\cdot p$ hamiltonian $H(k)$ and substitute $k$ for $\nabla_r$, and then he solves the equation: $H(\nabla_r)\psi=E\psi$. Here is some example:

1. In graphene, $v_F(\sigma\cdot k)\psi(k)=E\psi(k)$ is transformed to $v_F\sigma\cdot\nabla\psi(r)=E\psi(r)$.

2. In topological insulator (for example, quantum spin hall effect), $k_x$ in $H(k_x,k_y)$ is substituted to get $H(\partial_x,k_y)$ and $H(\partial_x,k_y)$ is used to solve edge state which is localized in $x=0$ line.

I wonder what justifies such a practice because here $k$ is lattice momentum, not the real one. Also, what do they get from solving, say, $v_F\sigma\cdot\nabla\psi(r)=E\psi(r)$. This solution seems to be too simple to represent a wave function in graphene considering the strong carbon potential.

Answer 1

Your equation $$ v_F (\sigma \cdot k)\psi(k) = E \psi(k) $$

seems to be the "momentum representation" via Fourier transform of the envelope function equation for graphene. In this case $k$ is just the wavevector introduced by the Fourier transform, not a "crystal wave vector" as in the Bloch ansatz, and the $\psi_k$-s are the Fourier coefficients for the corresponding envelope function, $\psi(x) = \int{dk \;\psi(k) \;e^{-ik \cdot x}}$. Note that the (periodic) envelope function is not the full wave function. It actually replaces the plane wave phase factor in the Bloch ansatz in order to obtain an extended wave function ansatz that can apply also to non-periodic systems. The envelope function is generally assumed to be slowly varying over a unit cell, and this is why reversing the Fourier transform retrieves a deceivingly simple equation for it. The simple form of the equation derives from a rather involved perturbation analysis as the lowest term, see for instance this review: The k · p method and its application to graphene, carbon nanotubes and graphene nanoribbons: the Dirac equation. Their eq.(190) is the equivalent $k \cdot p$ equation for graphene.

So, assuming that we do indeed deal with a "momentum representation" of the envelope equation, let's reverse the Fourier transform. Multiply both sides of the above equation by $e^{-ik \cdot x}$ and integrate on $k$: $$ v_F \int{dk \;(\sigma \cdot k)\;\psi(k) e^{-ik \cdot x}} = E \int{dk \;\psi(k) e^{-ik \cdot x}} $$

Introduce $$ \psi(x) = \int{dk \; \psi(k)\; e^{-ik \cdot x}} $$

and note that $$ i \;\nabla\psi(x) = \nabla \int{dk \;\psi(k)\; e^{-ik \cdot x}} = \int{dk\; k \;\psi(k)\; e^{-ik \cdot x}} $$

and $$ i\;\sigma \cdot \nabla\psi(x) = \int{dk \;(\sigma \cdot k) \;\psi(k) \;e^{-ik \cdot x}} $$

Therefore $$ i \;v_F\; \sigma \cdot \nabla\psi(x) = E \;\psi(x) $$ as required.

A similar procedure can be applied to a general Hamiltonian $H(k_x, k_y)$ acting on an envelope (wave) function $\psi(k_x, k_y)$, provided $H(k_x, k_y)$ is a polynomial of $k_x$, or has a suitable polynomial expansion in $k_x$. All that is needed is to realize that
$$ \int{dk_x \;k^n_x \;\psi(k_x, k_y)\; e^{-ik_x \cdot x}} = (i\partial_x)^n \int{dk_x \;\psi(k_x, k_y) \;e^{-ik_x \cdot x}} = (i\partial_x)^n \;\psi(x, k_y) $$