# A Derivation of DOS

1. Apr 8, 2016

### Array

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 text books use k=2π/L

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

So where is this factor 2 coming from?

Thanks,

Array

2. Apr 8, 2016

### drvrm

the solution of Schrodinger’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 [Broken]

Last edited by a moderator: May 7, 2017
3. Apr 8, 2016

### Array

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