In Problem $7.28$you found the density of states and the chemical potential for a two-dimensional Fermi gas. Calculate the heat capacity of this gas in the limit role="math" localid="1650099524353" $kT\ll {\epsilon }_F$· Also show that the heat capacity has the expected behavior when $kT\gg {\epsilon }_F$. Sketch the heat capacity as a function of temperature.

We have been given that the density of states and the chemical potential for a two-dimensional Fermi gas which we have found earlier in Problem $7.28$is$g\left(\epsilon \right)=\frac{N}{{\epsilon }_F}$.

We need to find the heat capacity of this gas in the limit $kT\ll {\epsilon }_F$and show that the heat capacity has the expected behavior when $kT\gg {\epsilon }_F$. Also we need to sketch the heat capacity as a function of temperature.

The total energy is given by the following integral:

$U={\int }_0^{\infty }\epsilon g\left(\epsilon \right){\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$ (Let this equation be (localid="1650103209259" $1$))

We have found the energy density earlier in Problem $7.28$such as

localid="1650102270353" $g\left(\epsilon \right)=\frac{N}{{\epsilon }_F}$

Substitute the value of $g\left(\epsilon \right)$in equation (localid="1650103517684" $1$), we get:

$U=\frac{N}{{\epsilon }_F}{\int }_0^{\infty }\epsilon {\overline{n}}_{FD}\left(\epsilon \right)d\epsilon$

Integrating by parts, we get:

$U=\frac{N}{{\epsilon }_F}\frac{{\epsilon }^2}{2}{\overline{n}}_{FD}{|}_0^{\infty }-\frac{N}{{\epsilon }_F}{\int }_0^{\infty }\frac{{\epsilon }^2}{2}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$ (Let this equation be ($2$))

If we substitute $\epsilon =0$, the integral will become zero due to the dependence of the term on ${\epsilon }^2$and the first term vanishes.

The equation ($2$) reduces to :

$U=-\frac{N}{{\epsilon }_F}{\int }_0^{\infty }\frac{{\epsilon }^2}{2}\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }d\epsilon$ (Let this equation be ($3$))

We know that ${\overline{n}}_{FD}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$(e-\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}$

Taking derivative with respect to $\epsilon$, we get:

localid="1650103912932" $\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=\frac{\partial \left[{\displaystyle \frac{1}{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1}}\right]}{\partial \epsilon }$

$\frac{\partial {\overline{n}}_{FD}}{\partial \epsilon }=-\frac{1}{kT}\frac{e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}}{{(e^{{\displaystyle \raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}+1)}^2}$

Let $\raisebox{1ex}{$(\epsilon -\mu )$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.=x$, then we can write as:

$dx=\frac{d\epsilon }{kT}$

We also need to change the boundaries of integration, so:

$\epsilon \to \infty$ $x\to \infty$

$\epsilon \to 0$ $x\to -\frac{\mu }{kT}$

As $kT\ll \mu$, we can put $-\infty$as the lower limit of integration.

The integral in equation ($3$) will become:

$U=\frac{N}{2{\epsilon }_F}{\int }_{-\infty }^{\infty }{\epsilon }^2\frac{e^x}{{(e^x+1)}^2}dx$ (Let this equation be (localid="1650114310341" $4$))

Now expanding ${\epsilon }^2$about $\mu$using Taylor series, we get:

$$\begin{gathered} {\epsilon }^2={\mu }^2+2\mu (\epsilon -\mu )x+(\epsilon -\mu )x^2 \\ {\epsilon }^2={\mu }^2+2\mu \left(kTx\right)+{\left(kTx\right)}^2 \end{gathered}$$

By substituting the value of in equation ($4$), we get three integrals say $I_1,I_2,I_3$

The equation ($4$) will become:

where,

$$\begin{gathered} I_1={\mu }^2{\int }_{-\infty }^{\infty }\frac{e^x}{{(e^x+1)}^2}dx \\ {I}_2=2kT\mu {\int }_{-\infty }^{\infty }\frac{x{e}^x}{{(e^x+1)}^2}dx \\ {I}_3={\left(kT\right)}^2{\int }_{-\infty }^{\infty }\frac{x^2{e}^x}{{(e^x+1)}^2}dx \end{gathered}$$

Simplifying the integrals $I_1,I_2,I_3$, we get:

$I_1={\mu }^2$

The second integral is odd integral and the limits of integration are from to , so this integral is simply zero.

$I_2=0$

$I_3={\left(kT\right)}^2\left[\frac{\pi 2}{3}\right]=\frac{{\left(kT\pi \right)}^2}{3}$

Substitute the values of in equation ($5$), we get:

$U=\frac{N}{2{\epsilon }_F}\left[{\mu }^2+\frac{{\left(kT\pi \right)}^2}{3}\right]$

Let $\mu ={\epsilon }_F$, we get:

$$\begin{gathered} U=\frac{N}{2{\epsilon }_F}\left[{\epsilon }_F^2+\frac{{\left(kT\pi \right)}^2}{3}\right] \\ U=\frac{N{\epsilon }_F}{2}+\frac{N{\left(kT\pi \right)}^2}{6{\epsilon }_F} \end{gathered}$$

The heat capacity is the partial derivative of with respect to the temperature, that is:

$$\begin{gathered} C_V=\frac{\partial U}{\partial T} \\ {C}_V=\frac{\partial \left[{\displaystyle \frac{N{\epsilon }_F}{2}}+{\displaystyle \frac{N{\left(kT\pi \right)}^2}{6{\epsilon }_F}}\right]}{\partial T} \\ {C}_V=\frac{N{\left(k\pi \right)}^{2}T}{3{\epsilon }_F} \end{gathered}$$

The graph of the heat capacity as a function of temperature is linear, where the slope of the line is $Slope=\frac{N{\left(k\pi \right)}^2}{3{\epsilon }_F}$.