Deriving DOS: Solve the Mystery of the 2x Factor!

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SUMMARY

The discussion centers on the derivation of the density of states (DOS) for a particle in an infinite box, specifically addressing the factor of 2 in the wave vector k. Participants clarify that while k is defined as π/L for individual solutions, textbooks use k=2π/L to account for the full k-space, which includes both positive and negative values of k. This distinction is crucial for accurately counting states in three-dimensional space, as each solution corresponds to a unique k-space vector. The conversation highlights the importance of understanding boundary conditions and the implications for state counting in quantum mechanics.

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  • Familiarity with Schrödinger’s equation and its applications in particle confinement.
  • Knowledge of k-space and its significance in quantum state representation.
  • Basic concepts of density of states in statistical mechanics.
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Students and professionals in physics, particularly those focused on quantum mechanics, solid-state physics, and materials science, will benefit from this discussion. It is particularly relevant for anyone studying the behavior of particles in confined systems and the mathematical foundations of density of states.

Array
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Hi guys,

I have a very particular question on the derivation of DOS.

For a particle in an infinite box k=π/L. However, when deriving the density of states, all textbooks use k=2π/L

Now you could argue that they account for spin degeneracy, but its not that! Because in the textbooks that happens (again) in a later step

So where is this factor 2 coming from?

Thanks,

Array
 
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Array said:
For a particle in an infinite box k=π/L. However, when deriving the density of states, all textbooks use k=2π/L

the solution of Schrödinger’s equation for particle in a 3D box
where Each solution can be uniquely associated with
a k- space vector ; k^2 = k(x) ^2 +k(y)^2 + k(z)^2 the boundary condition gives k =n.pi/L
The k has three axes k(x) ,k(y), k(z) the conditions at the walls are k(x) =n(x).pi/L .... and similarly for k (y) and K*z) and so on where n(x), n(y) and n(z) can take values as +/- 1, +/-2,. +/-3...
some of the texts might have used n(x) taking values 1.2,3,... so for proper counting the states they might have used 2.(pi)/L
see for details-http://web.eng.fiu.edu/npala/EEE6397ex/EEE_6397_Ch2_Energy%20Levels%20and%20Charge%20Carriers%20in%20Semiconductors_PART3.pdf
 
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Thanks for the link, I derive it the same way.

You're right. If they use 2π/L then they must take the entire k-space into consideration, because they didn't differ between +/-. Where as if you take the π/L if did already differ between positiv and negative solutions and you only take the respective segment in k-space?!

Thanks
 

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