Q. 7.28 - An Introduction to Thermal Physics

Question

Consider a free Fermi gas in two dimensions, confined to a square area $A=L^2$•

(a) Find the Fermi energy (in terms of $N$ and $A$), and show that the average energy of the particles is $\raisebox{1ex}{${\in }_F$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

(b) Derive a formula for the density of states. You should find that it is a constant, independent of $\in$.

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when $kT\ll {\in }_F$ and when $T$ is much higher.

(d) Because $g\left(\in \right)$ is a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for $\mu$ as a function of $N$. Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, $kT\gg {\in }_F$, the chemical potential of this system is the same as that of an ordinary ideal gas. 

Step-by-Step Solution

Answer

(a) Fermi energy is, $\frac{1}{2}N{\in }_F$ and the average energy relation is given below.

(b) Density of states is, $g\left(\in \right)=\frac{N}{{\in }_F}$

(c) As the density state is constant, so the behavior of chemical properties is the same as the Fermi-Dirac distribution i.e. it will decrease continuously.

(d) Value of $\mu$ is, $\mu =kT\ln \left(e^{\raisebox{1ex}{${\in }_F$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}-1\right)$

(e) Since, $\mu =-kT\ln \left(\frac{A}{N}\times \frac{2}{l_Q^2}\right)$. Therefore, we can say that the chemical potential of this system is the same as that of an ordinary ideal gas.  

Part(a) Step 1: Given information

We have been given that,

$A=L^2$

Part (a) Step 2: Simplify

In two dimensional case,

  $N=2{\int }_0^{n_{max}}{\int }_0^{\mathrm{\pi }/2}ndnd\theta =\frac{\mathrm{\pi }}{2}{n}_{max}^2$

We know that,

  $$\begin{gathered} {\in }_F=\frac{h^2{n}_{max}^2}{8m{L}^2}=\frac{h^2}{8mA}\times \frac{2N}{\mathrm{\pi }} \\ {\in }_F=\frac{h^2}{4\mathrm{\pi mA}}N \end{gathered}$$

Now, we can find the total energy $U$,

  $$\begin{gathered} U=2{\int }_0^{n_{max}}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}ndnd\theta =\frac{\mathrm{\pi }}{2}{n}_{max}^2 \\ U= {\int }_0^{{\in }_F}\in \left[\mathrm{\pi }\sqrt{\frac{8mA}{h^2}}\sqrt{\in }\sqrt{\frac{8\mathrm{mA}}{{\mathrm{h}}^2}}\frac{1}{2\sqrt{\in }}\right]d\in \\ U={\int }_0^{{\in }_F}\in \left[\frac{4\mathrm{\pi mA}}{h^2}\right]d\in \\ U={\int }_0^{{\in }_F}\in \frac{N}{{\in }_F}d\in \\ U=\frac{1}{2}N{\in }_F \end{gathered}$$

So, we can easily get average energy of the particles is $\frac{U}{N}=\frac{{\in }_F}{2}$

Part (b) Step 1: Given information

We need to find out the density of states.

Part (b) Step 2: Simplify

From part (a), 

Density of states is,

   $g\left(\in \right)=\frac{N}{{\in }_F}$

That is independent of $\in$

Part (c) Step 1: Given information

We need to find out behavior of chemical properties

Part (c) Step 2: Simplify

Since the density of state is constant, the behavior of the chemical potential is the same as that of the Fermi-Dirac distribution. As we know that Fermi-Dirac distribution has some symmetrical property but when the energy difference from $\mu$ is higher than the chemical potential, it gets broken because the occupancy is zero when $\in <0$. So, the electrons with energy less than the Fermi energy are excited to a higher level than the Fermi energy with broken symmetry, which means that the number of electron loss with energy $\in <{\in }_F$ is lower than the number of electron gain with energy $\in >{\in }_F$.

To be sensible with the difference in no. of electrons, chemical potential should move i.e. Fermi-Dirac distribution has to move totally. As the number of electrons with energy $\in >{\in }_F$ is larger than $\in <{\in }_F$, so Fermi-Dirac distribution has to move on the left side with fixed ${\in }_F$ , which means that the chemical potential decreases. When the temperature slightly grows up, then the difference in the number of electrons also grows up. So, the chemical potential will decrease continuously and at some point, it will have a negative value.

Part (d) Step 1: Given information

We need to find out value of $\mu$

Part (d) Step 2: Simplify

Since $g(\in )$ is a constant, we can analytically integrate the equation,

    $$\begin{gathered} N={\int }_0^{\infty }g\left(\in \right)\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\in -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}d\in \\ =\frac{N}{{\in }_F}{\int }_0^{\infty }\frac{e^{{\displaystyle \raisebox{1ex}{$-\left(\in -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{1+e^{{\displaystyle \raisebox{1ex}{$-\left(\in -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}d\in \\ = \frac{N}{{\in }_F}{\overline{)\left(-kT\right)\ln \left(1+e^{\raisebox{1ex}{$-\left(\in -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}\right)}}_{\in =0}^{\in =\infty } \\ =\frac{N}{{\in }_F}kT\ln \left(1+e^{\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}\right) \end{gathered}$$

From this,

     $$\begin{gathered} \mu =kT\ln \left(\frac{{\in }_F}{kT}-1\right) \\ =\left\{\begin{array}{l}{\in }_F where kT\ll {\in }_F\\ kT\ln \left(\frac{{\in }_F}{kT}\right) where kT\gg {\in }_F\end{array}\right. \end{gathered}$$

The second case is approximately and it will be negative value.

Part(e) Step 1: Given information

We need to find out that the chemical properties of this system is the same as that of an ordinary ideal gas

Part (e) Step 2: Simplify

From the answer of part (d),

$$\begin{gathered} \mu =kT\ln \left(\frac{{\in }_F}{kT}\right) \\ = kT\ln \left(\frac{A}{N}\times \frac{4\mathrm{\pi mkT}}{h^2}\right) \\ =-kT\ln \left(\frac{A}{N}\times \frac{2}{l_Q^2}\right) \end{gathered}$$

It seems to be the same form with the equation of chemical potential of an ordinary ideal gas.