Fermi energy of electron gas with electrostatic interaction

Question

I have been given the following exam question and am unsure how I would go about solving it:

Consider the case of a one-dimensional metal, consisting of a chain of $N$ positive charges $+q$ separated by a distance $2R$ and immersed in a neutralizing background of electrons with density per unit length $n_e$. The electrostatic energy due to the interaction of the electrons with the ion cores and among themselves is: $$\mathcal{E}_{el}=-\frac{\alpha q^{2}}{4\pi \epsilon_{0}}\frac{N}{R}$$

Assuming that the electrons form a non-interacting Fermi gas, calculate the Fermi energy. Write down an expression for the total kinetic energy of the electrons.

Now, ordinarily I would compute the Fermi energy as follows: First write down the density of states for a gas of Fermions:

$$g(\epsilon)=\frac{V m^{3/2}}{\sqrt{2}\pi^{2}\hbar^3}\epsilon^{1/2}$$

Then we have:

$$N = \int_{0}^{\infty}g(\epsilon)n_{F}(\epsilon)\:\mathrm{d}\epsilon = \int_{0}^{\infty}g(\epsilon)\Theta(E_{F}-\epsilon)\:\mathrm{d}\epsilon$$

Where $\Theta(\epsilon)$ is the Heaviside step function. We can thus calculate:

$$N=\int_{0}^{E_{F}}g(\epsilon)\:\mathrm{d}\epsilon = \frac{Vm^{3/2}}{\sqrt{2}\pi^{2}\hbar^{3}}\frac{2}{3}\epsilon_{F}^{3/2}$$

So we have the Fermi energy:

$$\epsilon_{F} = \left(\frac{N}{V}\frac{2\sqrt{2}\pi^{2}\hbar^{3}}{3m^{3/2}}\right)^{2/3}=\frac{2\hbar^{2}}{m}\left(\frac{\pi^{2}}{3}\right)^{2/3}n_{e}^{2/3}$$

To calculate the total energy, we have:

$$E = \int_{0}^{E_{F}}\epsilon g(\epsilon)\:\mathrm{d}\epsilon$$

But none of this takes into account the electrostatic energy, so I fear that I have terribly misunderstood something.

Thanks!

Answer 1

Kinetic energy in 1D, method 1. Free electrons. Assume no potential energy at the moment. Zero temperature. \begin{equation} n_e=\int_0^{E_F}g(\epsilon)d\epsilon=\int_0^{E_F}\frac{1}{\pi\hbar}\sqrt{\frac{m}{2\epsilon}}d\epsilon=\frac{\sqrt{2mE_F}}{\pi\hbar} \end{equation} \begin{equation} E_F=\frac{\pi^2\hbar^2n_e^2}{2m} \end{equation}

Kinetic energy, method 2. Number of filled states \begin{equation} N=\frac{kL}{\pi} \end{equation} where $L=\frac{N}{n_e}$ is the chain length. This gives the wave vector \begin{equation} k=\frac{\pi N}{L}=\pi n_e \end{equation} and thus the kinetic energy \begin{equation} E_{kin}=\frac{\hbar^2}{2m}k^2=\frac{\hbar^2\pi^2n_e^2}{2m} \end{equation}

similarly to method 1. The total (Fermi) energy is the sum of kinetic and potential (electrostatic) energy: \begin{equation} E_F=E_{kin}+E_{el}=\frac{\hbar^2\pi^2n_e^2}{2m}-\frac{\alpha q^2}{4\pi\epsilon_0}\frac{N}{R} \end{equation} Hope this make sense. In your solution, DOS is taken 3D, but the problem says the system is 1D metal. This, at least, should be corrected.