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\exnumber{3515}
\heading{Ideal Fermi gas in 1D space} 
\auname{Roy Apsel}


{\bf The problem:}

\Dn

Consider ${N}$ electrons that are kept between 
the plates of a capacitor. 
%
\[
V\left(x,y,z\right)=\left\{\begin{array}{ll}
\frac{1}{2}m\omega^{2}\left(x^{2}+y^{2}\right) &\textrm{ $0\leq z \leq L$}\\
\infty & \textrm{else}\\
\end{array}\right.
\]
%
The system is in thermal equilibrium at zero temperature. 
Find the force that the gas exerts of the plates assuming 
that it can be treated as one-dimensional. 

Write the condition on ${N}$ for having this assumption valid.

Tip: Find first the one particle states, 
and illustrate them using a schematic drawing. 
Express your results using  ${N,L,m,\omega}$ only.

\Dn\Dn


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{\bf The solution:}

\Dn
 
The one-particle energy spectrum is given by
\begin{equation}
	\epsilon_{n_{x},n_{y}}(p_{z}) = \omega(n_{x}+n_{y}+1) + \frac{p_{z}^{\,2}}{2m}		\qquad\qquad(\hbar=1)
\end{equation}
Where $p_{z} = \frac{\pi}{L} n_{z}$.\\

It is assumed that in the range where the system can be treated as one-dimensional, the contribution of energy levels $\,n_{x},n_{y}>0\,$ can be ignored, i.e.
\begin{equation}
	\epsilon 	= 	 \omega + \frac{p_{z}^{\,2}}{2m}
\end{equation}

The one-particle energy range for which this limit is valid, is then anywhere between the ground state, and the 1st excited state on the XY plane
\begin{equation}
	\omega < \epsilon < 2\omega
\end{equation}
The occupation of particles in this energy range occurs over the $\epsilon_{n_{z}}=\frac{1}{2m}\left(\frac{\pi}{L} n_{z}\right)^2$ levels, generally ranging from $0$ to $\omega$.\\

At T=0, the Fermi distribution behaves like a step function, with energy states lower than $\epsilon_{F}=\mu$.\\
The relation between N and the chemical potential $\mu$ is then given by
\begin{equation}
	N = \sum_{i} \left<n_{i}\right> = 2\int \frac{dz\,dp_{z}}{2\pi} = \frac{2L}{\pi} \sqrt{2m} \, (\mu-\omega)^{1/2}
\end{equation}

Combining (3) \& (4), (with $\omega < \epsilon_{F} < 2\omega$), gives a condition on N for the validity of this limit
\begin{equation}
	N < \frac{2L}{\pi} \sqrt{2m\omega}
\end{equation}

At a general temperature T, the force $P$ exerted on the plates can generally be found by calculating the grand canonical potential $\Omega$, and by using the relation $PL=-\Omega$.\\
At T=0 it is also possible to find $P$ through the total energy of the system, which is given by
\begin{equation}
	E = \sum_{i} \epsilon_{i} \left<n_{i}\right> = 2\int \frac{dz\,dp_{z}}{2\pi} \,\epsilon(p_{z}) = \omega N + \frac{1}{3} \left( \frac{\pi}{2L \sqrt{2m}} \right)^2 N^3
\end{equation}

The force exerted on the plates is then
\begin{equation}
	P   =  -\frac{\Omega}{L} = -\frac{(E-\mu N)}{L}
			= -\left( \frac{\partial E}{\partial L} \right)_{N}
		    		=   \frac{\pi^2}{12m} \left( \frac{N}{L} \right)^3
\end{equation}



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