Fermi temperature of a 1D electron gas

  • #1
DrClaude
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Homework Statement



Consider a one-dimensional metal wire with one free electron per atom and an atomic spacing of ##d##. Calculate the Fermi temperature.

Homework Equations



Energy of a particle in a box of length ##L##: ##E_n = \frac{\pi^2 \hbar^2}{2 m L^2} n^2##

1D density of states: ##g(E) = \frac{L}{\pi \hbar} \sqrt{\frac{2m}{\epsilon}}##

Fermi temperature: ##T_\mathrm{F} = \frac{E_\mathrm{F}}{k_\mathrm{B}}##

Fermi-Dirac occupancy: ##\bar{n}_\mathrm{FD} = \left[ e^{\beta (E - \mu)} + 1 \right]^{-1}##


The Attempt at a Solution



The solution is easy to find by considering the system at ##T=0##, where one simply needs to calculate the energy of the highest occupied level. For ##N## electrons, with ##2## electrons per ##n## state, i.e., ##N= 2 n_\mathrm{max}##,
$$
\begin{align*}
E_\mathrm{F} &= \frac{\pi^2 \hbar^2}{2 m L^2} n_\mathrm{max}^2 \\
&= \frac{\pi^2 \hbar^2}{2 m L^2} \left(\frac{N}{2} \right)^2 \\
&= \frac{\pi^2 \hbar^2}{8 m} \left(\frac{N}{L} \right)^2 \\
&= \frac{\pi^2 \hbar^2}{8 m d^2}
\end{align*}
$$
since ##N/L = 1/d##, so ##T_\mathrm{F} = \frac{\pi^2 \hbar^2}{8 m k_\mathrm{B} d^2} \left(\frac{N}{L} \right)^2 = \frac{\pi^2 \hbar^2}{8 m k_\mathrm{B}}##.

My problem comes when trying an alternate approach, by considering the system at ##T = T_\mathrm{F}##. In this case, one can calculate the average number of electrons using
$$
N = \int_0^\infty \bar{n}_\mathrm{FD} g(E) dE
$$
##\mu(T_\mathrm{F}) = 0##, so the equation becomes
$$
\begin{align*}
N &= \frac{L \sqrt{2m}}{\pi \hbar} \int_0^\infty \frac{E^{-1/2}}{e^{E / (k_\mathrm{B} T_\mathrm{F})} + 1} dE \\
&= \frac{L \sqrt{2m}}{\pi \hbar} \left(1 - \sqrt{2} \right) \sqrt{\pi } \zeta \left(1/2\right) \sqrt{k_\mathrm{B} T_\mathrm{F}}
\end{align*}
$$
from which I recover
$$
T_\mathrm{F} = \frac{\left(2 \sqrt{2}+3\right) \pi \hbar ^2}{2 \zeta^2 \left(1/2\right) k_\mathrm{B} m} \left(\frac{N}{L} \right)^2
$$
Comparing with the approach above, this result is ##\approx 3.5## times bigger.

I am confident of the first result above (and it is the same as I found in some on-line and off-line resources) and would appreciate some help in figuring out what is wrong in the second approach.
 

Answers and Replies

  • #2
TSny
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I don't think ##\mu(T_\mathrm{F}) = 0##.
 
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  • #3
DrClaude
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I don't think ##\mu(T_\mathrm{F}) = 0##.
That was my initial thought when a saw that solution from a student. We were confused by fig. 7.16 in Schroeder's textbook, which shows the chemical potential go to 0 at ##T_\mathrm{F}##. Looking more closely, that figure is related to problem 7.23 for a the case where ##\mu = 0## at ##T = T_\mathrm{F}##. I guess this is just a special case.

Thanks!
 
  • #4
TSny
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It is interesting that for the 3d case, ##\mu \approx 0## at the Fermi temperature. See second page of http://young.physics.ucsc.edu/112/mu_T.pdf . In the graph, you can see that ##\mu= 0## at a temperature slightly less than the Fermi temperature. See also equation (9) on that page.

##\mu## at a given temperature is whatever it has to be in order for ## N = \int_0^\infty \bar{n}_\mathrm{FD} g(E) dE ## to hold. For the 1d case, ##\mu## is apparently not close to zero at the Fermi temperature.
 
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  • #5
DrClaude
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This is indeed interesting. If I calculate
$$
N = \int_0^\infty \bar{n}_\mathrm{FD} g(E) dE
$$
for a 3D Fermi gas with ##\mu=0## and solve for the temperature ##T_0##, I find
$$
\frac{T_0}{T_\mathrm{F}} = 2 \left( \frac{2 \sqrt{2}+3}{\pi }\right)^{1/3} \left(\frac{2}{3 \zeta \left(3/2 \right)}\right)^{2/3} \approx 0.988734
$$
confirming that the temperature for which ##\mu=0## is almost the Fermi temperature.
 
  • #6
TSny
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Yes. Your result agrees with equation (9) of the link. That's nice.
 

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