91Ó°ÊÓ

Most spin-1/2 fermions, including electrons and helium-3 atoms, have nonzero magnetic moments. A gas of such particles is therefore paramagnetic. Consider, for example, a gas of free electrons, confined inside a three-dimensional box. The z component of the magnetic moment of each electron is ±µa. In the presence of a magnetic field B pointing in the z direction, each "up" state acquires an additional energy of $-{\mathrm{μ}}_{\mathrm{B}}B$, while each "down" state acquires an additional energy of $+{\mathrm{μ}}_{\mathrm{B}}B$

(a) Explain why you would expect the magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets studied in Chapters 3 and 6, for a given number of particles at a given field strength.

(b) Write down a formula for the density of states of this system in the presence of a magnetic field B, and interpret your formula graphically.

(c) The magnetization of this system is ${\mathrm{μ}}_{\mathrm{B}}\left(N_{↑}-N_{↓}\right.$, where Nr and N1 are the numbers of electrons with up and down magnetic moments, respectively. Find a formula for the magnetization of this system at $T=0$, in terms of N, µa, B, and the Fermi energy.

(d) Find the first temperature-dependent correction to your answer to part (c), in the limit $T≪T_{\mathrm{F}}$. You may assume that ${\mathrm{μ}}_{\mathrm{B}}B≪kT$; this implies that the presence of the magnetic field has negligible effect on the chemical potential $\mathrm{μ}$. (To avoid confusing µB with µ, I suggest using an abbreviation such as o for the quantity µaB.)

Step by step solution

01

Part(a) Step 1: Given information

We have to find out magnetization of a degenerate electron gas to be substantially less than that of the electronic paramagnets.

02

Part(a) Step 2:Simplify

Magnetising is very small.So, we can conclude that the a small fraction of electrons are free to filp their spin.

The fermi gas most of electrons can not flip their spin.

03

Part(b) Step 1: Given information

We have been given that $g\left(\mathrm{Ï\mu }\right)=g_0\sqrt{\mathrm{Ï\mu }}$

04

Part(b) Step 2:Simplify

Solution is as follows:

$g_0=\frac{\mathrm{Ï€}(8m{)}^{3/2}}{2h^3}=\frac{3N}{2{\mathrm{Ï\mu }}_F^{3/2}}$

05

Part(c) Step 1: Given information

We have been that $M={\mathrm{μ}}_B\left(N_{↑}-N↓\right)$

06

Part(c) Step 2: Simplify

It will be in the form of:

$M={\mathrm{μ}}_B^{2}B\sqrt{{\mathrm{Ï\mu }}_F}\frac{3N}{2{\mathrm{Ï\mu }}_F^{3/2}}$

$M=\frac{3N{\mathrm{μ}}_B^{2}B}{2{\mathrm{Ï\mu }}_F}$

07

Part(d) Step 1:Given information

We have to find the first temperature-dependent correction to answer to part (c) .

08

Part(d) Step 2:Simplify

The steps will be as follows:

${∫}_{-\mathrm{δ}}^{∞}\sqrt{\mathrm{Ï\mu }+\mathrm{δ}}{\stackrel{¯}{n}}_{\mathrm{FD}}\left(\mathrm{Ï\mu }\right)d\mathrm{Ï\mu }=-\frac{2}{3}{∫}_{-\mathrm{δ}}^{∞}(\mathrm{Ï\mu }+\mathrm{δ}{)}^{3/2}\frac{∂{\stackrel{¯}{n}}_{\mathrm{FD}}}{∂\mathrm{Ï\mu }}d\mathrm{Ï\mu }$

$M=\frac{3N{\mathrm{μ}}_B^{2}B}{2{\mathrm{Ï\mu }}_F}\left[1-\frac{{\mathrm{Ï€}}^2}{12}{\left(\frac{kT}{{\mathrm{Ï\mu }}_F}\right)}^2\right]$

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