Density of states at the Fermi energy

Masatsugu Sei Suzuki

Department of Physics

(Date: May 06, 12)

1. Density of states for the 3D system

dkkV

dkkV

dD 2

2

2

34

)2(2)(

.

The factor 2 comes from spin freedom. Using the dispersion relation

22

2k

m

ℏ ,

we get

dmV

dkkV

dD

2/3

22

2

2

2

2)(

ℏ

,

or

2/3

22

2

2)(

ℏ

mVD ,

where

2/3

22

2

2

ℏ

mV

.

Note that the number of free electrons is

2/3

003

2)( F

FF

ddDN

Then we have

FF

FF

N

D

2

3

3

2)(

2/3

2. The total number of free electrons below the Fermi energy:

3

3 3

4

)2(2 Fk

VN

or

3

23

1Fk

V

Nn

or 3/123 nk F

Then the Fermi energy is given by

3/222

22

322

nm

km

FF ℏℏ

.

3. Electronic heat capacity coefficient

Suppose that the heat capacity of atoms (1 mol) with the volume Vm can be measured as a function of temperature. The heat capacity consists of lattice contribution and electronic contribution. The electronic heat capacity can be expressed by

TTkDC BFel 22 )(3

1.

with

22 )(3

1BF kD

and

Fm

F

mVD

2/3

22

2

2)(

ℏ

where Vm is the molar mass. Then we get

)(35715.2

)(3

1

)(

3

1

22

22

F

A

BAF

A

BA

A

F

D

kND

kNN

D

where

A

FF

A

N

DD

)()(

(1/eV.atom) or (1/eV.at)

and

35715.23

1 22 BAkN (mJ eV/K)

((Mathematica))

___________________________________________________________________

4. Expression of )( F

AD

What is the expression for )( F

AD ? It is expressed by

erg

eVm

erg

eVm

N

V

erg

eV

N

DD FF

A

m

A

FF

A

2/3

22

2/3

22

2

2

12

2

1)()(

ℏℏ

where

Clear@"Global`∗"D;rule1 = 9kB → 1.3806504× 10−16, NA → 6.02214179 × 1023,

c → 2.99792× 1010, — → 1.054571628 10−27,

me → 9.10938215 10−28, mp → 1.672621637× 10−24,

mn → 1.674927211× 10−24, qe → 4.8032068× 10−10,

eV → 1.602176487× 10−12, meV → 1.602176487 × 10−15,

keV → 1.602176487× 10−9, MeV → 1.602176487 × 10−6,

fi → 10−8, J → 107=;

1

meV J

π2 NA kB2

3êê. rule1

2.35715

AA

m

N

M

N

V

is the volume per atom. M is the molar mass and is the density. Suppose that there are p

electrons per atom. Then the number density of electrons is given by

pn

1 eV = 1.602176487 x 10-12 erg

Then

erg

eVp

m

erg

eVp

m

erg

eVmD FF

A 2/33/2

2

2/33/2

23/4

3/12/3

22313454.0

32

2

1)(

ℏℏℏ

where

3/222 3

2

p

mF

ℏ

In other words, )( F

AD depends on the mass m, the atomic volume , and the number

of electrons per atom.

5. Electronic heat capacity coefficient

erg

eVp

mD F

A 2/33/2

273886.0)(35715.2

ℏ

When the mass is different from that of the free particle, we have

3/22/3*

142/33/2*

21069637.973886.0

p

m

m

erg

eVp

m

mm

ℏ (mJ/mol K2)

((Example-1))

Al:

Density: = 2.6989 g/cm3

Atomic weight = 26.9815386 g/mol

fcc; a = 4.05 A. p = 3 (trivalent)

AN

Ma

4

3

The Fermi energy:

3/222

32

nm

F ℏ

= 11.6524 eV

Since

p = 3

we get

23

31080641.1

43

an /cm3

Then we have

3/22/3*

141069637.9

p

m

m

For p = 3 and 4

3a ,

m

m

m

m *2/3

*

31289.13631169.0 (mJ/mol K)

((Example-2)) Pb

Density: = 11.342 g/cm3

Atomic weight = 207.2 g/mol

fcc; a = 4.95 A. p = 4.

AN

Ma

4

3

The Fermi energy:

3/222

32

nm

F ℏ

= 9.44945 eV

23

31031918.1

44

an /cm3

3/22/3*

141069637.9

p

m

m

For p = 4 and 4

3a ,

m

m

m

m *2/3

*

37585.2473886.0 (mJ/mol K)

4. Derivation of )( F

AD from measured

)( F

AD can be experimentally determined from the relation

35715.2

)/()(

2KmolmJ

D measuredF

A

[1/(eV at)].

(mJ/.mol K2) D(F) (1/eV at)

Al 1.35 0.573

Pb 2.98 1.264

Sn (white) 1.78 0.755

APPENDIX

Myer Chapter 13 problems

1. Estimate the strength of the electron-phonon interaction, V, for Al and Pb,

where

])(

1exp[

F

cVN

T

and N(F) is the density of states for a given spin at the Fermi energy.

((Solution))

])(

1exp[

F

cVN

T

For Pb, Tc = 7.2 , = 105 K.

For Al, Tc = 1.2 K, = 428 K.

We assume that

)(2

1)( F

A

F

A DN ,

VA is in the units of [eV at].

c

F

AA

F

AA

FT

DVNVVN

ln

1)(

2

1)()( .

or

c

F

A

A

TD

V

ln)(2

1

.

Then we have

(mJ/.mol K2) D(F) (1/eV at) AV (eV.at)

Al 1.35 0.57 0.60

Pb 2.98 1.26 0.59

________________________________________________________________________

2.

Calculate the London penetration depth in Al at 0 K under the assumptions (a)

that all the valence electrons are responsible for the superconducting property, and

(b) that only electrons in an energy interval kB/2 in the neighborhood of the Fermi

energy contribute to the supercurrent. To what fraction of the valence electron

concentration does the latter case correspond?

((Solution))

Al fcc with the lattice constant a = 4.05 A.

Then the number density:

n0 =3 x 4/a3 = 18.06 x 1022/cm3.

The Fermi energy is

F = 11.6524 eV.

The London penetration depth is

20

2

4 en

mcL =125.046 Å.

The energy interval is

Bk2

1 = 0.018441 eV.

(mJ/.mol K2)

DA(F) = 0.57 (1/eV at)

)( F

A

ADNn = 6.33014 x 1020/cm3.

The revised penetration depth is

2

2

4'

en

mcL 667.19 Å

n

n0.0352351

((Mathematica))

((Note)) The brief solution is shown in the Myer's book. Our result is slightly different.

L =127 Å. 'L =1250 Å. n

n0.035%.

________________________________________________________________________

3. The critical field is described as

Clear@"Global`∗"D;rule1 = 9kB → 1.3806504× 10−16, NA → 6.02214179× 1023,

c → 2.99792× 1010, — → 1.054571628 10−27,

me → 9.10938215 10−28, qe → 4.8032068× 10−10,

eV → 1.602176487× 10−12, meV → 1.602176487× 10−15,

keV → 1.602176487× 10−9, MeV → 1.602176487× 10−6,

fi → 10−8, n0 → 18.06× 1022, Θ → 428, a1 → 4.05 fi,

DA1 → 0.57=;

λL =1

fi

me c2

4 π n0 qe2êê. rule1

125.046

∆E =1

2kB ΘêeV êê. rule1

0.0184411

n1 = NA DA1 ∆E ê. rule1

6.33014×1021

λL2 =1

fi

me c2

4 π n1 qe2êê. rule1

667.916

n1

n0êê. rule1 êê N

0.0350506

)1( 2

2

0

c

ccT

THH .

Obtain an expression for Hc0 in terms of Tc and (Cs - Cn) at Tc. Given that the

latter quantities are 1.2 K and 1.6 mJ/K mol, estimate Hc0 for Al.

((Solution))

At T = Tc = 1.2 K,

TcTcc

TTVSNdT

TdHTCC

c

2

)(

4|)(

Parabolic law:

)1()(2

2

0

c

ccT

THTH .

20

2)(

c

cc

T

TH

dT

TdH

Then we have

c

c

c

cc

TTVNST

H

TH

TCC

c

2

02

2

0

4

4|)(

or

cTTVNScc CCTH |)(2

0

For Tc = 1.2 K, cTTVNS CC |)( = 1.6 mJ/mol K.

We note that the units of 20cH is mJ/mol. The unit of 2

0cH should be Oe2 = Gauss2 =

erg/cm3. Therefore we need to calculate the volume of 1 mol of Al. Using the molar mass

M (= 26.98 g/mol) and the density (= 2.70 g/ cm3), we get the molar volume as

M

VM = 9.99 cm3.

Then we get

24342

0 10604.099.9/106.12.1 OecmergHc

or

780 cH Oe.

_______________________________________________________________________

4. It is found, for a certain superconducting alloy, that Hc1 is 400 Oe and that

the magnetization has fallen to half its (negative) value at Hc1 when the

applied field is 500 Oe. Assuming the Abrikosov model for the mixed state,

find the distance between the centers of the flux vortices.

((Solution))

At Hc1 = 400 Oe. -4M1=400 Gauss.

At H = 500 Oe. 124 MHMHB =500-200 = 300 Gauss.

The most stable arrangement of vortices is triangular.

Ba20

2

3 ,

with

7

0 100678.2 Gauss cm2

Then we get

107

02 1096.73003

100678.22

3

2

Ba cm2

or

a = 2820 Å.

Fig. Quantum fluxoids forming triangular lattice.

________________________________________________________________________

5. A thick specimen of type I superconductor is known to have a critical field of

500 Oe. It is found that a film of 5 x 10-5 cm thickness has a critical field of

550 Oe. What value would you expect for the critical field of a sample of only

10-6 cm thickness? You may assume that the penetration of field into the

superconductor is as given by the London theory and that the penetration

depth is independent of magnetic field and you may neglect demagnetization

effects.

((Solution))

Type-I superconductor

Hc = 500 Oe.

From Problem ((10-2)) Kittel 8th edition,

cc HH

12'

where is the thickness and l is the penetration depth.

Suppose that

c

c

HH '

where is constant.

For = 5 x 10-5 cm, Hc' = 550 Oe.

Then we get

)105(500

550' 5 c

c

H

H= 5.5 x 10-5.

Using the relation

c

c

HH )105.5(' 5 ,

we can estimate

6

5

10

500)105.5('

cH = 27. 5 kOe

________________________________________________________________________

6. In a tunneling experiment using a junction of Sn and Al at 2 K, the I vs V

characteristic as shown in Fig.1 was obtained. What can be deduced from

this value?

((Solution))

Tc = 3.72 K for Sn

Tc = 1.14 K for Al

At T = 2 K, Sn is in the S-state, but Al is in the N-state. Then Sn-Al junction is a junction

of S-I-N, where I is an insulator. In this case, V = 0.5 meV corresponds to e/ , where

2 is the energy gap of Sn.

25.02 = 1.0 meV.

This value is close to the accepted value 1.15 meV.

cBTkK 5277.3)0(2 1.12478 meV (BCS theory).

((Note))

1 meV = 1.602176487 x 10-15 erg

________________________________________________________________________

7. An electric current is passing through a junction that consists of Pb and Al

separated by a very thin insulating layer. The plot of tunnel current against

applied voltage at T = 0.5 K is shown schematically in Fig.2(a), the maximum

and minimum being at voltages V1 = 11.8 x 10-4 V and V2 = 15.2 x 10-4 V,

respectively. Explain the shape of the curve and derive values for the energy

gaps of superconducting Pb and Al. At what temperature the maximum and

minimum to disappear to yield a curve as shown in Fig.2(b)? You may

assume that the energy gap of a superconductor does not change appreciably

in the temperature range from 0 K to (1/2)Tc.

((Solution))

Pb: Tc = 7.20 K

Al: Tc = 1.20 K

At T = 0.5 K, Pb and Al are in the S-state. 21 and 22 are the energy gap of

superconductor S1 (Pb) and the superconductor S2 (Al).

eV 21

2

,

eV 21

1

(1> 2)

Then we get

52.1221 eV meV, 18.1121 eV meV,

or

35.11 meV for Pb, 17.02 meV for Al,

For 1.2K≤T<7.2 K,

Pb is in the S-state, while Al is in the N-state.

35.113

eV mV.

_____________________________________________________________________

8. Estimate the penetration depth and coherence length for pure Sn, where

critical temperature Tc = 3.7 K,

density = 7.3 g/cm3,

atomic molar mass M = 118.7 g/mol

effective mass m* = 1.9 m (m; mass of electron)

((Solution))

Suppose that each atom has one electron. The electron concentration n is

M

N

M

N

V

M

V

Nn AAA = 3.70359 x 1022/cm3.

The London penetration depth is evaluates as

2

2*

4 ne

cmL =380.622 Å.

The Fermi velocity is given by

3/12

**)3( n

mk

mv FF

ℏℏ = 6.28321 x 107 cm/s

Then the coherenth length

cB

FFF

Tk

vvv

5277.3

2

)2(

20

ℏℏℏ

= 2344.1 Å.

with

cBTk5277.32 1.12318 meV (BCS theory).

((Mathematica))

_____________________________________________________________________

Clear@"Global`∗"D;rule1 = 9kB → 1.3806504× 10−16, NA → 6.02214179× 1023,

c → 2.99792× 1010, — → 1.054571628 10−27,

me → 9.10938215 10−28, qe → 4.8032068× 10−10,

eV → 1.602176487× 10−12, meV → 1.602176487× 10−15,

keV → 1.602176487× 10−9, MeV → 1.602176487× 10−6,

Tc → 3.70, M → 118.7, fi → 10−8, ρ → 7.3,

m1 → 1.9 me=;

n1 =ρ NA

Mê. rule1

3.70359×1022

λL =1

fi

m1 c2

4 π n1 qe2êê. rule1

380.622

vF =—

m1I3 π2 n1M1ê3 êê. rule1

6.28321×107

ξ0 =1

fi

2 — vF

3.5227 π kB Tcêê. rule1

2344.1

3.5227 kB Tc

meVêê. rule1

1.12318

REFERENCES

H.P. Myers, Introductory Solid State Physics (Taylor & Francis, London, 1990).

C. Kittel, Introduction to Solid State Physics, 8th edition (John Wiley & Sons, New

York, 2005)