# Nearly free electron model - band gap

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1. Mar 4, 2015

### unscientific

For a wavefunction at the Brillouin boundary we have:

$$\langle k|H|k\rangle = \epsilon_0 (\vec k)$$
$$\langle k'|H|k'\rangle = \epsilon_0 (\vec k+\vec G)$$
$$\langle k'|H|k\rangle = V_G = \frac{1}{L^3} \int e^{i(\vec k - \vec k') \cdot \vec r} V(r) d\vec r$$
$$\langle k|H|k'\rangle = V_G^*$$

Using degenerate perturbation theory, we diagonalize the hamiltonian and find the eigenvalues (energies) to be:
$$E_{\pm} = \epsilon_0 (\vec k) \pm |V_g|$$

This would mean that for every $k$ right on the boundary we have a bandgap of $2|V_G|$.

Since $V_G = \frac{1}{L^3} \int e^{i(\vec k - \vec k') \cdot \vec r} V(r) d\vec r$, shouldn't its magnitude be fixed at $2|V_G$? However, in my notes for a periodic perturbation $V_G = \tilde V cos(\frac{2n\pi}{a})$ the gaps are increasing with $k$ (The bandgap at $k=\frac{2\pi}{a}$ seems to be twice as big as the bandgap at $k=\frac{\pi}{a}$.

2. Mar 9, 2015

### Greg Bernhardt

Thanks for the bump greg. I think the potential here depends on $k$, even though not stated. The constant band-gap thing only works when the potential is periodic like $V_G = \tilde V cos(\frac{2n\pi}{a})$.