Pressure in spin-1/2 fermions

  • #1

Homework Statement


I have to find the mean Energy $<E>$ and pressure of a system of N fermions with spin 1/2. The energy per particle is
\begin{equation}
\varepsilon = \frac{p^2}{2m}.
\endu{equation}

Homework Equations


The relevant equations are the degeneracy of the system:
\begin{equation}
g(\varepsilon)=\frac{4V (2m)^{3/2}\pi\varepsilon^{1/2}}{h^3}
\end{equation}
the mean occupation numbers of fermions
\begin{equation}
<n_\varepsilon>_{FD} = \frac{1}{e^{\beta(\varepsilon - \mu)}+1}
\end{equation}

The Attempt at a Solution


I calculated the mean energy doing
\begin{equation*}
\begin{split}
<E> &= \int_{0}^{\infty} \varepsilon g(\varepsilon) <n_\varepsilon>_{FD} d\varepsilon\\
&= \frac{4V (2mkT)^{3/2}\pi}{h^3} kT \int_{0}^{\infty} \frac{x^{3/2}}{z^{-1}e^x+1}dx.
\end{split}
\end{equation*},
and finally obtained:
\begin{equation}
<E> = \frac{V}{\lambda^3} kT \sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}}
\end{equation}
where z is the fugacity.
\begin{equation*}
z=\exp{\beta \mu}
\end{equation*}


Then I calculated the pressure making use of the grand partition function
\begin{equation}
\mathcal{Z} = \prod_\varepsilon 1+e^{-\beta(\varepsilon - \mu)}
\end{equation}
and doing this
\begin{equation*}
\begin{split}
\frac{PV}{kT} &= \ln \mathcal{Z} = \sum_{\varepsilon} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right)\\
&= \frac{4V (2m)^{3/2}\pi}{h^3} \int_{0}^{\infty} \varepsilon^{1/2} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right) d\varepsilon\\
&= \frac{4V (2m)^{3/2}\pi}{h^3} \left[\frac{2}{3} \varepsilon^{3/2} \ln \left(1+e^{-\beta(\varepsilon - \mu)}\right)\Big|_0^{\infty} - \frac{2\beta}{3} \int_{0}^{\infty} \frac{\varepsilon^{3/2}}{e^{\beta(\varepsilon - \mu)}+1} d\varepsilon \right]\\
&= \frac{8V (2m)^{3/2}\pi}{3h^3} \beta \int_{0}^{\infty} \frac{\varepsilon^{3/2}}{e^{\beta(\varepsilon - \mu)}+1} d\varepsilon
\end{split}
\end{equation*}
I finally obtained that the pressure is
\begin{equation}
P = \frac{2}{3} \frac{1}{\lambda^3} kT \sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}}
\end{equation}
Now, the problem is that, if I want to obtain the pressure at room temperature, I calculate using the canonical ensemble the chemical potential, and approximate the series
\begin{equation}
\mu = \frac{1}{\beta} \ln \left(\frac{N \lambda^3}{V}\right),
\end{equation}
\begin{equation}
\sum_{l=1}^{\infty} (-1)^{l+1}\frac{z^l}{l^{5/2}} \approx z
\end{equation}
I get that the pressure is:
\begin{equation}
P = \frac {2} {3} \frac{NkT}{V}
\end{equation}
That result really bothers me because of that factor of 2/3. I did the same process to obtain the pressure and energy for bosons and obtained the correct result.
I thought that the factor may arise because I am considering electrons and the spin may do a contibution.

If anyone could help me with this it would be very appreciated!
Thank you
 

Answers and Replies

  • #3
They go through the calculation in this paper:

https://www.phas.ubc.ca/~berciu/TEACHING/PHYS455/LECTURES/FILES/fermionsn.pdf

They get ##PV = \frac{2}{3} U##, not ##PV = \frac{2}{3} kT## (where ##U## is the energy).

I found my mistake. I was assuming that
\begin{equation*}
\Gamma(5/2)=\frac{\pi^{1/2}{4}
\end{equation*}
when it actually is:
\begin{equation*}
\Gamma(5/2)=\frac{3\pi^{1/2}{4}.
\end{equation*}
Then everything was correct.
Thank you for your response, and i'm gonna keep that paper. It's really interesting.
 

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