What Specifies Energy Bands in Solid?

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SUMMARY

The energy bands in solids are specified by the band index n and the electron wave vector k, as established in Ashcroft and Mermin's work. For each band index n, the wave vector k runs through all values that satisfy periodic boundary conditions, leading to an infinite family of solutions. The boundary conditions, determined by k, influence the logarithmic derivative of the wave function, which is crucial for solving the Schrödinger equation within a single cell. This establishes a definitive relationship between k and the energy bands in solid-state physics.

PREREQUISITES
  • Understanding of band theory in solid-state physics
  • Familiarity with the Schrödinger equation
  • Knowledge of periodic boundary conditions
  • Basic concepts of wave functions and their derivatives
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  • Study the implications of periodic boundary conditions in quantum mechanics
  • Explore the derivation of energy bands using the Schrödinger equation
  • Investigate the role of wave vectors in solid-state physics
  • Learn about the mathematical formulation of band theory in Ashcroft and Mermin's text
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Students and professionals in solid-state physics, quantum mechanics researchers, and anyone interested in the theoretical foundations of energy bands in materials.

mkbh_10
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n is the band index and k is the electron wave vector. Now is energy band specified when for each n , k runs through all the values available that specify periodic boundary conditions ?

But in Ashcroft Mermin , it is also given that for a given k , there exists an infinite family of solutions labelled by n which is the band index.

I am confused here about what specifies the band.
 
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Look at it like this: k determines the boundary conditions for an elementary cell, namely the value of the logarithmic derivative of the wavefunktion ##\psi'/\psi (a)=\psi'/\psi(0) \exp(ika)##.
Now the Schroedinger equation in a single cell, with the boundary condition specified by k, has an infinity of solutions labelled by n.
 
So the second one is correct ans.
 

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