# Density of states

1. Aug 14, 2010

1. The problem statement, all variables and given/known data
According to the non-relativistic quantum mechanics of a particle of mass m in a cubic box ov volume V = L3, the single particle energy levels are given by

Ek=$\hbar$2k2/2m
where k is the magnitude of the wavevector K = (kx,ky,kz) and where the components of k are quantised as kx=$\pi$nx/L etc with nx=0,1,2,...

a)Show that the density in the k-space of the single particle states that are available to electrons is given by

$\rho$=Vk2/$\pi$2

2. Relevant equations

3. The attempt at a solution
In my notes I have the derivation of this, however it shows it to be 2$\pi$2 in the denominator, as do other online sources.

Is there something I'm missing that means in this situation the factor of can be taken out?

(There are further parts to this question that ask to prove certain things that only hold true without the factor of 2 so it's not an error)

2. Aug 14, 2010

### javierR

Keep in mind that the allowed states of a particle depend also on the spin of the particle...in quantum mechanics courses you start off with spin-0 particles for simplicity. In the question, they are asking about electrons, which have spin 1/2. How does this change the counting of the number of states in a given region of k-space?

3. Aug 15, 2010

Thank you, multiplying by the the factor (2s+1) where s=1/2 was what I needed.

I thought I had the next parts but I was incorrect.

b)Hence show that the total kinetic energy of a non-relativistic gas of N electrons at T=0 is

E=(V/5$\pi$2)(2me/$\hbar$2)3/2EF5/2

c)Furthermore show that Ef is given by

2/2me(3π2N/V)2/3

Therefore,

KF=((3π2N/V)1/3

I am able to prove part C.

I show that N=(L3/3π2)KF2

Rearranging for KF and substituting into the first equation in my first post gives the answer.

However, I'm sure sure how to approach part B, all my attempts so far have the wrong number in front of the first pi and the wrong power above Ef at the end.