In the case of degenerate, isotropic bands in one dimension, Lwdin second-order perturbation theory says that the

In the case of degenerate, isotropic bands in one dimension, Löwdin second-order perturbation theory says that the energies of the bands are given by the eigenvalues of the following matrix:

where i and j run over the range of degenerate states with energy E_n, and l is summed over all other states not degenerate with these. The matrix element p_ii = 〈i|p|i〉 = 0, but p_ij = 〈i|p|j〉 can be nonzero for crystals without inversion symmetry.

Suppose that a band arises from two degenerate atomic states, and the matrix element p₁₂ = p₂₁ is nonzero. Ignore the coupling to other states (e.g., assume the energies of other states are far fromE_n). Find the 2×2 matrixH_ij for this system. Use Mathematica or a similar program to plot the eigenvalues of this matrix as functions of k. Set E₁ = E₂ = 0, p₁₂ = 0.5, and h̄ = m = 1.

Does this result violate the theorem deduced in Section 1.6, that the first derivative of the energy bands vanishes at the Brillouin zone center? To explore this, suppose that there is a small interaction between the bands; in particular, set E₁₁(0) = ΔE = −E₂₂(0), where ΔE = 0.01 or some other small number. Plot the eigenvalues of this matrix. What happens where the bands crossed? What is the slope of the bands there? What will happen in the limit ΔE → 0?

This type of band structure is called a camelback structure and is common in crystals without inversion symmetry.

Transcribed Image Text:

Hij(k) = En (0)8ij + ħkpij m + ħ²k² -Sij + 2m ħ²k² m² PilPlj En(0) - E₁(0) (1.9.48)