Fermion Pressure at Room Temperature

In summary, the conversation discusses finding the mean energy and pressure of a system of N fermions with spin 1/2. The relevant equations include the degeneracy of the system and the mean occupation numbers of fermions. The attempt at a solution involved calculating the mean energy and pressure using the grand partition function. However, the final result for the pressure was incorrect due to a mistake in assuming the value of the gamma function. The correct result was obtained after correcting this mistake.
  • #1
FranciscoSili
8
0

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
 
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  • #3
stevendaryl said:
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 going to keep that paper. It's really interesting.
 

Related to Fermion Pressure at Room Temperature

What is spin-1/2 fermion?

Spin-1/2 fermion is a type of quantum particle that has a spin quantum number of 1/2. It is one of the two types of fermions, the other being spin-1 fermion. Spin-1/2 fermions include particles such as electrons, protons, and neutrons.

What is pressure in spin-1/2 fermions?

Pressure in spin-1/2 fermions refers to the force exerted by these particles on their surroundings due to their spin properties. This pressure is a result of the Pauli exclusion principle, which states that two fermions cannot occupy the same quantum state. As a result, spin-1/2 fermions in a confined space will exert a pressure on each other.

How is pressure in spin-1/2 fermions measured?

Pressure in spin-1/2 fermions is measured using various techniques, such as ultracold atom experiments and neutron scattering. In these experiments, the behavior of the particles is observed and analyzed to determine the pressure they exert on each other.

What are the applications of studying pressure in spin-1/2 fermions?

Studying pressure in spin-1/2 fermions has various practical applications in fields such as condensed matter physics, materials science, and quantum computing. It helps us better understand the behavior of these particles and their interactions, which can lead to the development of new materials and technologies.

Are there any real-world examples of pressure in spin-1/2 fermions?

Yes, there are many real-world examples of pressure in spin-1/2 fermions. One example is the pressure exerted by electrons in a metal, which is responsible for the conductivity of the material. Another example is the pressure exerted by protons and neutrons in the nucleus of an atom, which holds the atom together.

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