Density of states in k-space for particle in a cubic box

[madorangepand]

Homework Statement

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

E_k=$\hbar$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

Homework Equations

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)

 

[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?

 

[madorangepand]

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)(2m_e/$\hbar$²)^3/2E_F^5/2

c)Furthermore show that Ef is given by

ℏ²/2m_e(3π²N/V)^2/3

Therefore,

K_F=((3π²N/V)^1/3




I am able to prove part C.

I show that N=(L³/3π²)K_F²

Rearranging for K_F 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.