Energy band in K space VS real space

[jackychenp]

Hi All,

There is a simple question in my mind.
A band with energy Ek has dispersion in k space. Then what it looks like in the real space?

 

[cgk]

The bands only exist in k-space, since the effective mean field one-particle Hamiltonian (Fock operator/Kohn-Sham operator), of which the e(k) are the eigenvalues, is diagonal in k-space, but not in real space.

If you transform the eigenstates of this operator (the crystal orbitals) into real space, you get the Wannier orbitals, which look closely like normal atomic orbitals (in particular, they are identical in each unit cell). But these Wannier orbitals do not diagonalize the effective one-particle Hamiltonian anymore, so there is no e(r) relation in this sense.

 

[Gordianus]

Let´s assume a perfect crystal and no scattering processes. Consider an electron and constant external electric field. If the electron were free, it would accelerate at uniform rate. However, the electron moves in a periodic structure. The E vs k relationship gives us an important information: at a given value of k, the slope of the curve is proportional to the electron's speed. Thus, although the external field is constant the electron moves in a complex way given by the E-k curve.

 

[jackychenp]

Hi cgk,

Do you mean integrate in the whole Brillioun Zone to get Wannier orbitals? In that case, one wave function psi(k) is just part of orbitals.

The bands only exist in k-space, since the effective mean field one-particle Hamiltonian (Fock operator/Kohn-Sham operator), of which the e(k) are the eigenvalues, is diagonal in k-space, but not in real space.

If you transform the eigenstates of this operator (the crystal orbitals) into real space, you get the Wannier orbitals, which look closely like normal atomic orbitals (in particular, they are identical in each unit cell). But these Wannier orbitals do not diagonalize the effective one-particle Hamiltonian anymore, so there is no e(r) relation in this sense.