P613 HW # 2 Solutions JM Densmore

1. The Free and Independent Electron Gas in Two Dimensions

(a) In a 2D system with periodic BC you can solve the SE with momen-tum kx = 2π/Lnx and ky = 2π/Lny. The number of allowed valuesof k space inside a volume Ω will be

ΩV

4π2

The volume in k space is the area of a circle with radius kf . Takinginto account the number of spins for the electron

n =k2

f

2π(1)

(b)

1n

= πr2s (2)

rs =√

2kf

(3)

(c) In 2D the density of particles in range k + dk is 2πkdk

n =∫

2πkdk

4π2f(εk) (4)

Changing to an integral over energy we find

n =m

πh2

∫f(ε)dε (5)

(6)

g = mπh2 ε > 0

g = 0 ε < 0

(d) Since g is constant g′ = 0 all the terms in the BS expansion are equalto zero. So we find

n =∫ εf

0

m

πh2 dE + (µ− εf )m

πh2 (7)

n = n(µ− εf )m

πh2 (8)

µ = εf (9)

(e) From 2.67 we have

n =m

πh2

∫1

eβ(ε−µ) + 1dε

1

This integral can be looked up in a table. Using the electron densityand the fermi energy it is easy to show

εf = µ +1β

ln(1 + e−βµ) (10)

(f) µ(T ) does not have an analytic expansion about T≈ 0, which is whythe BS expansion does not work in 2D.

2. Entropy of a free electron gas

(a) We’ll start with the partition function, Z, for a free electron gas

Z =∑ni

e−β(niεi−niµ) (11)

=∑n1

e−β(n1ε1−n1µ) ∗∑nN

e−β(nN εN−nN µ) (12)

For Fermi-Dirac statistics n can be either 0 or 1. The sums are easilyevaluated.

Z = Π(1 + e−β(εi−µ)) (13)

The entropy S is

S = −∂Ω∂T = kB(lnZ + βε) (14)

With the energy defined as

ε = − ∂

∂βlnZ (15)

=∑ (εi − µ)e−β(εi−µ)

1 + e−β(εi−µ)(16)

and

ln(fi) = ln

(e−β(εi−µ)

1 + e−β(εi−µ)

)(17)

= −β(εi − µ)− ln(1 + e−β(εi−µ)) (18)

from (4), (6), & (8) it is easy to show that

S = −kb

∑

k

[fi ln fi + (1− fi) ln(1− fi)]

(b) We can use integration by parts to integrate

s = −kB

∫dEg(E)(f ln(f) + (1− f) ln(1− f))

2

hE =∫

g(E)dE

∂

∂E(f ln(f) + (1− f) ln(1− f)) = f ′ ln f + f ′ − f ′ln(1− f)− f ′

= f ′ ln(f

1− f) (19)

ln( f

1− f

)= −β(ε− µ) (20)

The entropy becomes

s = βkB

∫ ∞

0

h(ε)(ε− µ)(−∂f

∂ε

)(21)

Taylor expanding h(ε)

h(ε) =∞∑

l=0

1l!

h(l)(ε)(ε− µ)l

(l + 1) = 2n ⇒ l = 2n− 1

s =1T

∞∑n=1

h(2n−1)(µ)(2n)∫ ∞

−∞

(ε− µ)2n

2n!(−f ′) (22)

The above integral is just (kBT )2nC2n. For n =1, µ ≈ εf .

s =π2

3k2

BTg(εf ) (23)

The electron specific heat is

cv = T∂S

∂T/n

cv =π2

31n

k2BTg(εf )

3. Pauli paramagnetism This problem was solved on the quiz, but hereI will use a different approach. The energy of a electron whose magneticmoment is parallel (”+”) or antiparallel (”-”) to H is given by

ε± = p2

2m ± µBH = E0 ± µBH (24)

where E0 = p2/2m. The energy levels of the system are populated accord-ing to the fermi-dirac distribution function.

f(ε) =1

eβ(ε−µ) + 1(25)

3

The density of levels is given by 4πVh3 p2dp. The magnetization / volume is

M

V=

4πµB

h3

∫ ∞

0

dpp2[f(ε+)− f(ε−)] (26)

Define A = -H if ”+”, A = H if ”-”. We’ll change the integral in p to anintegral in E.

√2m3/2

∫ ∞

0

√EdE

eβ(E+α) + 1=

23α3/2[1 +

π2

8(

1βα

)2 + ...] (27)

with α = −µbA + µ then f(+) ⇒ α = µ + µBH; f(−) ⇒ α = µ− µBH

M/V =8πµB(2m3)1/2

3h3

((µ + µBH)3/2[1 +

π2

8( kBT

µ + µBH

)2] +

(µ− µBH)3/2[1 +π2

8( kBT

µ− µBH

)2]

)(28)

After expanding in powers of H and keeping only the leading terms

M

V=

8πµ2B(2m3µ)1/2H

h3

(1− π2

24(kbT

µ

)2 + ...

)(29)

µ is the chemical potential. In the low temperature limit it is just thefermi energy εf . You should be able to convince yourself that this thesame result as the quiz. χ = µ2

Bg(εf ) ≈ 10−6

Using the above results, making sure to use the correct units. We findthat RW = 1.

7.2 Measure ∆K, then use ∆k = 2π/∆L to calculate the length of the box.7.3See problem 1.

7.4 Make a plot of εf vs n, which comes out to be a linear plot.7.5 vsom ≈ 107, vdru ≈ 104. εsom ≈ 10 eV; εdru ≈ 10−2. εsom/εdru ≈ 102.

4