Consider a free Fermi gas in two dimensions, confined to a square area A=L2

(a) Find the Fermi energy (in terms of Nand A), and show that the average energy of the particles is F2.

(b) Derive a formula for the density of states. You should find that it is a constant, independent of .

(c) Explain how the chemical potential of this system should behave as a function of temperature, both when role="math" localid="1650186338941" kTFand when Tis much higher.

(d) Because gis a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for μas a function of N. Show that the resulting formula has the expected qualitative behavior.

(e) Show that in the high-temperature limit, kTF, the chemical potential of this system is the same as that of an ordinary ideal gas.

Short Answer

Expert verified

(a) Fermi energy is, 12NFand the average energy relation is given below.

(b) Density of states is, g=NF

(c) As the density state is constant, so the behavior of chemical properties is the same as the Fermi-Dirac distribution i.e. it will decrease continuously.

(d) Value of μis, μ=kTlneFkT-1

(e) Since, μ=-kTlnAN×2lQ2. Therefore, we can say that the chemical potential of this system is the same as that of an ordinary ideal gas.

Step by step solution

01

Part(a) Step 1: Given information

We have been given that,

A=L2

02

Part (a) Step 2: Simplify

In two dimensional case,

N=20nmax0π/2ndndθ=π2nmax2

We know that,

F=h2nmax28mL2=h28mA×2NπF=h24πmAN

Now, we can find the total energy U,

U=20nmax0π2ndndθ=π2nmax2U=0Fπ8mAh28mAh212dU=0F4πmAh2dU=0FNFdU=12NF

So, we can easily get average energy of the particles is UN=F2

03

Part (b) Step 1: Given information

We need to find out the density of states.

04

Part (b) Step 2: Simplify

From part (a),

Density of states is,

g=NF

That is independent of

05

Part (c) Step 1: Given information

We need to find out behavior of chemical properties

06

Part (c) Step 2: Simplify

Since the density of state is constant, the behavior of the chemical potential is the same as that of the Fermi-Dirac distribution. As we know that Fermi-Dirac distribution has some symmetrical property but when the energy difference from μis higher than the chemical potential, it gets broken because the occupancy is zero when <0. So, the electrons with energy less than the Fermi energy are excited to a higher level than the Fermi energy with broken symmetry, which means that the number of electron loss with energy <Fis lower than the number of electron gain with energy >F.

To be sensible with the difference in no. of electrons, chemical potential should move i.e. Fermi-Dirac distribution has to move totally. As the number of electrons with energy >Fis larger than <F, so Fermi-Dirac distribution has to move on the left side with fixed F, which means that the chemical potential decreases. When the temperature slightly grows up, then the difference in the number of electrons also grows up. So, the chemical potential will decrease continuously and at some point, it will have a negative value.

07

Part (d) Step 1: Given information

We need to find out value ofμ

08

Part (d) Step 2: Simplify

Since g()is a constant, we can analytically integrate the equation,

N=0g1e-μkT+1d=NF0e--μkT1+e--μkTd=NF-kTln1+e--μkT=0==NFkTln1+eμkT

From this,

μ=kTlnFkT-1=FwherekTFkTlnFkTwherekTF

The second case is approximately and it will be negative value.

09

Part(e) Step 1: Given information

We need to find out that the chemical properties of this system is the same as that of an ordinary ideal gas

10

Part (e) Step 2: Simplify 

From the answer of part (d),

μ=kTlnFkT=kTlnAN×4πmkTh2=-kTlnAN×2lQ2

It seems to be the same form with the equation of chemical potential of an ordinary ideal gas.

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Most popular questions from this chapter

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is

(a) 1eVless than μ

(b) 0.01eVless than μ

(c) equal to μ

(d) 0.01eVgreater than μ

(e) 1eVgreater thanμ

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At t=0the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are Natoms, the total magnetization is typically~2μeN, where µa is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure 7.30. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h21rand therefore reduces the magnetization by ~2μe. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of 1λ.. (in the limit of long wavelengths). Therefore, since=hfand p=hλ.. for any "particle," the energy of a magnon is proportional

In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small.

to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write =p22pm*, wherem* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.24×1029kg, about14times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.

(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by

NmV=2π2m×kTh2320xex-1dx.

Evaluate the integral numerically.

(b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(O)-M(T))/M(O).Write your answer in the form (T/To)32, and estimate the constantT0for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cv/Nk=(T/Ti)32, where Tidiffers from To only by a numerical constant. EstimateTifor iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470k.)

(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section 8.2we will consider a different two-dimensional model in which magnetization does occur.)

Carry out the Sommerfeld expansion for the energy integral (7.54), to obtain equation 7.67. Then plug in the expansion for μto obtain the final answer, equation 7.68.

In a real semiconductor, the density of states at the bottom of the conduction band will differ from the model used in the previous problem by a numerical factor, which can be small or large depending on the material. Let us, therefore, write for the conduction band g(ϵ)=g0cϵ-ϵcwhere g0cis a new normalization constant that differs from g0by some fudge factor. Similarly, write gat the top of the valence band in terms of a new normalization constant g0v.

(a) Explain why, if g0vg0c, the chemical potential will now vary with temperature. When will it increase, and when will it decrease?

(b) Write down an expression for the number of conduction electrons, in terms of T,μ,candg0cSimplify this expression as much as possible, assuming ϵc-μkT.

(c) An empty state in the valence band is called a hole. In analogy to part (b), write down an expression for the number of holes, and simplify it in the limit μ-ϵvkT.

(d) Combine the results of parts (b) and (c) to find an expression for the chemical potential as a function of temperature.

(e) For silicon, g0cg0=1.09andg0vg0=0.44*.Calculate the shift inµ for silicon at room temperature.

Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation 7.121 numerically, without approximating it as an integral.*

(a) Write equation 7.121 for this system as a sum over energy levels, taking degeneracy into account. Replace Tandμwith the dimensionless variables t=kT/hfandc=μ/hf.

(b) Program a computer to calculate this sum for any given values of tandc. Show that, for N=2000, equation 7.121 is satisfied at t=15provided that c=-10.534. (Hint: You'll need to include approximately the first 200 energy levels in the sum.)

(c) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.

(d) Now reduce tto 14 , and adjust the value of cuntil the sum again equals 2000. Plot the number of particles as a function of energy.

(e) Repeat part (d) for t=13,12,11,and10. You should find that the required value of cincreases toward zero but never quite reaches it. Discuss the results in some detail.

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