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)