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 $-{\mu }_{\mathrm{B}}B$, while each "down" state acquires an additional energy of $+{\mu }_{\mathrm{B}}B$

(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 ${\mu }_{\mathrm{B}}\left(N_{\uparrow }-N_{\downarrow }\right.$, 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 $T\ll T_{\mathrm{F}}$. You may assume that ${\mu }_{\mathrm{B}}B\ll kT$; this implies that the presence of the magnetic field has negligible effect on the chemical potential $\mu$. (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 $g\left(ϵ\right)=g_0\sqrt{ϵ}$

04

Part(b) Step 2:Simplify

Solution is as follows:

$g_0=\frac{\pi (8m{)}^{3/2}}{2h^3}=\frac{3N}{2{ϵ}_F^{3/2}}$

05

Part(c) Step 1: Given information

We have been that $M={\mu }_B\left(N_{\uparrow }-N\downarrow \right)$

06

Part(c) Step 2: Simplify

It will be in the form of:

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

$M=\frac{3N{\mu }_B^{2}B}{2{ϵ}_F}$

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:

${\int }_{-\delta }^{\infty }\sqrt{ϵ+\delta }{\overline{n}}_{\mathrm{FD}}\left(ϵ\right)dϵ=-\frac{2}{3}{\int }_{-\delta }^{\infty }(ϵ+\delta {)}^{3/2}\frac{\partial {\overline{n}}_{\mathrm{FD}}}{\partial ϵ}dϵ$

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

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