Energy band in K space VS real space

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SUMMARY

The discussion centers on the relationship between energy bands in k-space and their representation in real space. It establishes that energy bands exist solely in k-space due to the diagonal nature of the effective mean field one-particle Hamiltonian, specifically the Fock operator/Kohn-Sham operator. Transforming the eigenstates of this operator into real space yields Wannier orbitals, which resemble atomic orbitals but do not diagonalize the Hamiltonian, indicating a lack of a direct e(r) relationship. The conversation emphasizes the complexity of electron motion in a periodic structure under an external electric field, as described by the E vs k relationship.

PREREQUISITES
  • Understanding of energy bands and dispersion in k-space
  • Familiarity with the effective mean field one-particle Hamiltonian, specifically the Fock operator and Kohn-Sham operator
  • Knowledge of crystal orbitals and their transformation to Wannier orbitals
  • Basic principles of electron motion in periodic structures and external electric fields
NEXT STEPS
  • Explore the concept of Wannier functions and their applications in solid-state physics
  • Study the Brillouin Zone and its significance in band theory
  • Investigate the implications of the E vs k relationship on electron dynamics in crystals
  • Learn about the Kohn-Sham equations and their role in density functional theory (DFT)
USEFUL FOR

Physicists, materials scientists, and researchers in condensed matter physics who are interested in the behavior of electrons in crystalline materials and the theoretical frameworks that describe these phenomena.

jackychenp
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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?
 
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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.
 
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.
 
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.

cgk said:
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.
 

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