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:

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

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 $$\frac{\hbar^2 k^2}{2m}$$:

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

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

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

Hope that helps.

 

[keith-k11]

Thanks very much for this solution.

 

[paranormal]

The ground state energy of a free electron fermi gas is not simply the sum of all the energies of the states within the fermi sphere because of the Pauli exclusion principle. This principle states that no two electrons can occupy the same quantum state simultaneously. Therefore, in a fermi gas, as the electron states fill up, the energy levels become increasingly crowded and it becomes more difficult for electrons to occupy higher energy states. This results in a lowering of the overall energy of the system.

In your calculation, you are assuming that all the electrons are at the same energy level, which is not the case due to the Pauli exclusion principle. The correct expression for the ground state energy takes into account the distribution of electrons among energy levels, which is given by the Fermi-Dirac distribution. This distribution shows that the energy levels are not evenly spaced, with more electrons occupying lower energy levels and fewer electrons occupying higher energy levels.

Therefore, the correct expression for the ground state energy of a free electron fermi gas is:

E = \frac{3}{5} N \epsilon_f

Where N is the number of electrons and \epsilon_f is the fermi energy, which is related to the fermi wavevector k_f by:

\epsilon_f = \frac{\hbar^2 k_f^2}{2m}

This expression takes into account the distribution of electrons among energy levels and gives the correct result for the ground state energy. I hope this helps to clarify your understanding.