Ground state energy of free electron fermi gas

[Repetit]

Can someone explain to me why the ground state energy of a free electron fermi gas is not just:

$$E = 2 \int_0^{k_f} \frac{\hbar^2 k^2}{2m} 3k^2 dk$$

Where the factor of two is due to the fact that there are two electron states for each value of k. The idea is to add up all the energies of all states within the fermi sphere, but it does not give the correct result which is:

$$E = \frac{3}{5} N k_f$$

Where N is the number of electrons, and $$k_f$$ is the radius of the fermi sphere. What am I doing wrong? If you need more info please let me know.

Thanks in advance
René

 

[StatusX]

What is the integral defining N? (ie, the integral over the fermi sphere of dN)

 

[kavehsh]

There are a few things wrong with your formulas:

Your total energy integral should look like this:

$$<br /> E = 2 \int_0^{k_f} \frac{\hbar^2 k^2}{2m} \frac{V}{(2\pi)^3} 4 \pi k^2 dk<br />$$

The $$\frac{V}{(2\pi)^3}$$ is necessary because you need to account for the volume of k-space each state occupies. The $$4\pi$$ comes from converting the 3D integration in spherical coordinates into a 1D integration only in the radial direction (dk).

You'll see that you can also find the total particle number the same way, without the $$\frac{\hbar^2 k^2}{2m}$$:

$$<br /> N = 2 \int_0^{k_f} \frac{V}{(2\pi)^3} 4 \pi k^2 dk<br />$$

After performing the integrations, you can write E in terms of N, and you'll find that:

$$<br /> E = \frac{3}{5} N \frac{\hbar^2}{2m} (k_f)^2 = \frac{3}{5} N E_f<br />$$

Hope that helps.

 

[keith-k11]

Thanks very much for this solution.