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\heading{E3520: Ideal Fermi gas in 2D space} 
\auname{Amit Omer and Naor Yossi}


{\bf The problem:}

\Dn


${N}$ fermions with mass ${\mathsf {m}}$ and spin ${\frac{1}{2}}$
are in a two dimensional box with area ${A}$.

\Dn
1)  \hspace{5 mm} Show that:\hspace{3 mm}$N\left(\beta,\mu\right)=A\frac{\mathsf{m}}{\pi}T \ln
\left(1+e^{\frac{\mu}{T}}\right)$


\Dn
Guideline: Use variables exchange ${X=e^{\beta \left(E-\mu
\right)}}$ and use the integral
${\int_{1}^{\infty}\frac{dx}{x\left(x+1\right)}=\ln
\left(\frac{1+X_{1}}{X_{1}}\right)}$.

\Dn
2)   \hspace{5 mm}Take ${T=0}$ and find ${N=A\frac{m}{\pi}\mu}$. Explain that result
intuitively by looking at ${g\left(E\right)}$.

\Dn
3)   \hspace{5 mm}Find the chemical potential ${\mu\left(T,\frac{N}{A}\right)}$ and
the fermi energy ${E_{F}\equiv \mu\left(T\rightarrow 0 \right)}$.

\Dn 
4)\hspace{6 mm}Show that in low temperatures
%
%
\[\mu\left(T\right)\approx E_{F}-Te^{-\frac{E_{F}}{T}}\]
%
%
\hspace{10 mm}While in high temperatures we get a result consistant with the
general expression you

\Dn
 \hspace{10 mm}developed in the Boltzmann's proximity frame.

\Dn
5)   \hspace{5 mm}Find ${E\left(\beta,\mu\right)}$ and
${{P}\left(\beta,\mu\right)}$ in temperature zero, and by
placing the expression for ${\mu\left(T=0,\frac{N}{A}\right)}$.

\Dn
\hspace{9 mm} Show that the following results
%
%
\[E=A\frac{\pi}{m}\frac{1}{2}\left(\frac{N}{A}\right)^{2},
{P}=\frac{\pi}{m}\frac{1}{2}\left(\frac{N}{A}\right)^{2}\]
%
%

\Dn
6)  \hspace{5 mm}Clarify yourself why in zero temperature
${{P}\propto\frac{1}{A^{2}}}$ while in high temperature
${{P}\propto\frac{1}{A}}$.


\Dn\Dn

\newpage
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{\bf The solution:}

\Dn

1)
\Dn
\[
Z={{\sum_{\{n_i\}}}}e^{-\beta\left(E_i-\mu\\n_i\right)}
\]
\Dn
\[
E_i=\epsilon_i{\cdot n_i},\hspace{5 mm} {n_i}=0,1\hspace{2 mm} - \hspace{2 mm} fermions,\,\,\,K_B=1\,\,\left(\beta=\frac{1}{T}\right)
\]
\Dn
\[
Z={{\sum_{\{x_i,p_i,\sigma_i,n_i\}}}}e^{-\beta{\cdot n_i}\left(\epsilon_i-\mu\right)}=
\prod_{\{x_i,p_i,\sigma_i\}}\left(1+e^{-\beta\left(\epsilon_i-\mu\right)}\right)
\]
\Dn
\[
N= T {\frac{\partial{\ln Z}}{\partial \mu}}={{\sum_{\{x_i,p_i,\sigma_i\}}}}\frac{1}{1+e^{\beta\left(\epsilon_i-\mu\right)}}
\]
\Dn
\[
\epsilon_i=\frac{p_i^2}{2m},\hspace{5 mm}spin \frac{1}{2}\,\, \rightarrow\,\, degeneracy\,\,2\,\,\left(no\,external\,magnetic\,field\right)
\]
\Dn
\[
=\frac{2}{\left(2\pi\right)^2}\cdot \int_A{dA}\int_0^{2\pi} {d\phi}\int_0 ^\infty{\frac{pdp}{1+e^{\beta\left(\frac{p^2}{2m}-\mu\right)}}}
\]
\Dn
\[
{x=e^{\beta \left(\frac{p^2}{2m}-\mu\right)}}
\]
\Dn
\[
=\frac{Am}{\pi\beta}\int_{e^{-\beta\mu}}^\infty \frac{dx}{x\left(1+x\right)}=\frac{Am}{\pi\beta}\ln \left(\frac{1+e^{-\beta\mu}}{e^{-\beta\mu}}\right)=\frac{AmT}{\pi} \ln {\left(1+e^{\frac{\mu}{T}}\right)}
\]

\Dn
\Dn
2)
\Dn
\[
T=0 \,\,\,\rightarrow \,\,\, e^{\frac{\mu}{T}}\gg 1
\]
\Dn
\[
N=\frac{AmT}{\pi}\ln{\left(1+e^{\frac{\mu}{T}}\right)}=\frac{AmT}{\pi}\ln{\left(e^{\frac{\mu}{T}}\right)}=\frac{Am\mu}{\pi}
\]
%
%
\Dn

We can look at it intuitively by saying that at T=0 all the energy levels are populated equaly 
(2 fermions at each level) until ${E_F}$, so that the density number is just constant.   
\Dn

for ideal fermions gas, spin ${\frac{1}{2}}$ and 2D:\,\,\,\,${g(E)=\frac{mA}{\pi}}$
\Dn

(according to section 6.2 in the lecture notes)
\Dn

so ${N=g\cdot E_F |_{T\rightarrow 0}=\frac{Am\cdot\mu}{\pi}}$ (as we got in the direct way).

\newpage


3)
\Dn
\[
N=\frac{AmT}{\pi}\ln{\left(1+e^{\frac{\mu}{T}}\right)}\,\,\,\,\,\rightarrow \,\,\,\,\,\mu(T,\frac{N}{A})=T\cdot \ln{\left(e^{\left(\frac{N}{A}\right)\cdot\frac{\pi}{mT}}-1\right)}
\]
\[
E_{F}\equiv \mu\left(T\rightarrow 0 \right)
\]
\Dn
\[
e^{\left(\frac{N\pi}{AmT}\right)}\left[T\rightarrow 0\right]\gg 1
\]
\Dn
\[
\mu(T,\frac{N}{A})=T\cdot \ln{\left(e^{\left(\frac{N\pi}{AmT}\right)}\right)}=\frac{N\pi}{Am}
\]
\[
\mu\left(T\rightarrow 0 \right)\equiv E_{F}=\frac{N\pi}{Am}
\]

\Dn
\Dn
4)
\Dn

\underline{Low Temperatures:}
\Dn
\[
\frac{N\pi}{AmT}=\frac{E_F}{T}=\ln{\left(1+e^{\frac{\mu}{T}}\right)}
\]
\Dn
\[
\mu=T\cdot \ln{\left(e^{\left(\frac{E_F}{T}\right)}-1\right)}=T\cdot \ln{\left[{e^{\left(\frac{E_F}{T}\right)}\cdot\left(1-e^{\left(\frac{-E_F}{T}\right)}\right)}\right]}
\]
\Dn
\[
=T\left[\frac{E_F}{T}+\ln\left(1-e^{\left(\frac{-E_F}{T}\right)}\right)\right]
\]
\Dn
\hspace{10 mm}${e^{\left(\frac{-E_F}{T}\right)}\left[T\rightarrow 0\right] \rightarrow 0}$ \,\,\,\,\ and as we  know\,\,\,\,\, ${\lim\limits_{x\rightarrow0}{\ln\left(1+x\right)\approx x}}$
\Dn

\hspace{10 mm}therefore
\[
\Dn
\mu\approx T\left[\frac{E_F}{T}-e^{\left(\frac{-E_F}{T}\right)}\right]=E_F-T\cdot e^{\left(\frac{-E_F}{T}\right)}
\]

\newpage

\underline{High Temperatures:}
\Dn

\hspace{10 mm} ${n\lambda_T^2\ll 1\,\,\,\,\, therefore \,\,\,\,\, e^{\frac{\mu}{T}}\ll1}$
\Dn
\[
N=\frac{AmT}{\pi}\ln{\left(1+e^{\frac{\mu}{T}}\right)}\approx \frac{AmT}{\pi}\cdot e^{\frac{\mu}{T}}
\]
\[
\mu=T\cdot \ln\left(\frac{\pi N}{AmT}\right)
\]
\Dn
\hspace{10 mm}In the Boltzmann proximity frame ${\mu=T\cdot \ln\left(\frac{N\lambda_T^2}{Ag_0}\right)}$
\[
g_0=\sum_{spin}e^{-\beta\epsilon_{bound}}=2,\,\,\,\,\left(no\,external\,magnetic\,field\right)
\]
\Dn
\hspace{10 mm}${\lambda_T^2=\left(\frac{2\pi}{mT}\right)}$
\[
\mu=T\cdot \ln\left(\frac{2N\pi}{AmT}\cdot\frac{1}{2}\right)=T\cdot \ln\left(\frac{\pi N}{AmT}\right)
\]
\Dn
\hspace{10 mm}as we can see the result we derived is consistant with the Boltzmann proximity frame.   

\Dn
\Dn
5)
\Dn

\hspace{10 mm}according to section 6.2 in the lecture notes:
\Dn
\[
\frac{E}{A}=\frac{m}{\pi}T^2F_2\left(z\right)
\]
\Dn
\hspace{10 mm}for ${T\rightarrow 0}$ we get ${z\gg 1}$ and ${F_2\approx\frac{1}{2}\left[ln\left(z\right)\right]^2}$
\[
\ln\left(z\right)=\ln\left(e^{\frac{\mu}{T}}\right)=\frac{\mu\left(T\rightarrow 0\right)}{T}=\frac{E_F}{T}=\frac{\pi N}{AmT}
\]
\Dn
\[
\frac{E}{A}\approx\frac{m}{2\pi}T^2\left(\frac{\mu}{T}\right)^2= \frac{m}{2\pi}T^2\left(\frac{\pi N}{AmT}\right)^2
\]
\Dn
\[
E=\frac{\pi A}{2m}\left(\frac{N}{A}\right)^2
\]
\Dn
\hspace{10 mm}At ${T=0}$ we get ${E=F-TS=F}$ therefore ${P=-\left(\frac{\partial F}{\partial A}\right)_N=-\left(\frac{\partial E}{\partial A}\right)_N}$    
\[
P=\frac{\pi}{2m}\left(\frac{N}{A}\right)^2
\]
\Dn
\hspace{10 mm}For high temperatures we have ideal gas in Boltzmann approximation so that simply  ${P=\frac{NT}{A}}$
\newpage

6)
\Dn

The persure reflects the Dependence of the energy levels on the linear size of the box.
\Dn

${P=-\frac{\partial E_n}{\partial A}}$ averaged over n.

\Dn

${E_n\propto\frac{n^2}{L^2}=\frac{n^2}{A}}$ hence ${P\propto\frac{n_{typical}^2}{A^2}}$.
\Dn

for N fermion at zero temperature the typical value is ${n\sim N}$, therefore ${P\propto \frac{1}{A^2}}$.
\Dn

for high temperature T the typical value is determined from ${E_n\sim T}$\,\,      i.e.\\ 


${n^2\sim TA}$ and therefore  ${P\propto \frac{TA}{A^2}\propto \frac{1}{A}}$.



 






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\Dn



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