Free electron model

Masatsugu Suzuki Department of Physics, SUNY at Binghamton (Date: January 13, 2012)

Abstract As an example we consider a Na atom, which has an electron configuration of (1s)2(2s)2(2p)6(3s)1. The 3s electrons in the outermost shell becomes conduction electrons and moves freely through the whole system. The simplest model for the conduction electrons is a free electron Fermi gas model. In real metals, there are interactions between electrons. The motion of electrons is also influenced by a periodic potential caused by ions located on the lattice. Nevertheless, this model is appropriate for simple metals such as alkali metals and noble metals. When the Schrödinger equation is solved for one electron in a box, a set of energy levels are obtained which are quantized. When we have a large number of electrons, we fill in the energy levels starting at the bottom. Electrons are , obeying the Fermi-Dirac statistics. So we have to take into account the Pauli’s exclusion principle. This law prohibits the occupation of the same state by more than two electrons. Sommerfeld’s involvement with the quantum electron theory of metals began in the spring of 1927. Pauli showed Sommerfeld the proofs of his paper on paramagnetism. Sommerfeld was very impressed by it. He realized that the specific heat dilemma of the Drude-Lorentz theory could be overcome by using the Fermi-Dirac statistics (Hoddeeson et al.).1 Here we discuss the specific heat and Pauli paramagnetism of free electron Fermi gas model. The Sommerfeld’s formula are derived using Mathematica. The temperature dependence of the chemical potential will be discussed for the 3D and 1D cases. We also show how to calculate numerically the physical quantities related to the specific heat and Pauli paramagnetism by using Mathematica, based on the physic constants given by NIST Web site (Planck’s constant ħ, Bohr magneton B, Boltzmann constant kB, and so on).2 This lecture note is based on many textbooks of the solid state physics including Refs. 3 – 10.

Content: 1. Schrödinger equation A. Energy level in 1D system B. Energy level in 3D system 2. Fermi-Dirac distribution function 3. A. 3D system B. 2D system C. 1D system 4. Sommerfeld’s formula 5. Temperature dependence of the chemical potential 6. Total energy and specific heat 7. Pauli paramagnetism

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8. Physical quantities related to specific heat and Pauli paramagnetism 9. Conclusion

1. Schrödinger equation3-10 A. Energy level in 1D system We consider a free electron gas in 1D system. The Schrödinger equation is given by

p2 2 d 2 (x) H (x)   (x)   k    (x) , (1) k 2m k 2m dx2 k k where

 d p  , i dx

and  k is the energy of the electron in the orbital. The orbital is defined as a solution of the wave equation for a system of only one electron:one-electron problem.

Using a periodic boundary condition:  k (x  L)  k (x) , we have

ikx  k (x) ~ e , (2) with 2 2 2  2   2   k  k   n , 2m 2m  L  2 eikL  1 or k  n , L where n = 0, ±1, ±2,…, and L is the size of the system.

B. Energy level in 3D system We consider the Schrödinger equation of an electron confined to a cube of edge L.

p2 2 H     2    . (3) k 2m k 2m k k k

It is convenient to introduce wavefunctions that satisfy periodic boundary conditions. Boundary condition (Born-von Karman boundary conditions).

 k (x  L, y, z)  k (x, y, z) ,

 k (x, y  L, z)  k (x, y, z) ,

 k (x, y, z  L)  k (x, y, z) .

2

The wavefunctions are of the form of a traveling plane wave. ikr  k (r)  e , (4) with

kx = (2/L) nx, (nx = 0, ±1, ±2, ±3,…..), ky = (2/L) ny, (ny = 0, ±1, ±2, ±3,…..), kz = (2/L) nz, (nz = 0, ±1, ±2, ±3,…..).

The components of the wavevector k are the quantum numbers, along with the quantum number ms of the spin direction. The energy eigenvalue is 2 2  (k)  (k 2  k 2  k 2 )  k 2 . (5) 2m x y z 2m

Here

 p (r)    (r)  k (r) . (6) k i k k k

So that the plane wave function  k (r) is an eigenfunction of p with the eigenvalue k . The ground state of a system of N electrons, the occupied orbitals are represented as a point inside a sphere in k-space. Because we assume that the electrons are noninteracting, we can build up the N- electron ground state by placing electrons into the allowed one-electron levels we have just found.  ((The Pauli’s exclusion principle)) The one-electron levels are specified by the wavevectors k and by the projection of the electron’s spin along an arbitrary axis, which can take either of the two values ±ħ/2. Therefore associated with each allowed wave vector k are two levels:

k, , k, .

In building up the N-electron ground state, we begin by placing two electrons in the one- electron level k = 0, which has the lowest possible one-electron energy  = 0. We have

L3 4 V N  2 k 3  k 3   (2 )3 3 F 3 2 F where the sphere of radius kF containing the occupied one-electron levels is called the Fermi sphere, and the factor 2 is from spin degeneracy. The electron density n is defined by

N 1 n   k 3   V 3 2 F 3

The Fermi wavenumber kF is given by

2 1/ 3 kF  3 n . (9)

The is given by

2  2 / 3   3 2n . (10) F 2m

The Fermi velocity is

k  1/ 3 v  F  3 2n . (11) F m m ((Note)) The Fermi energy F can be estimated using the number of electrons per unit volume as -15 2/3 2/3 F = 3.64645x10 n [eV] = 1.69253 n0 [eV], -3 22 where n and n0 is in the units of (cm ) and n = n0×10 . The Fermi wave number kF is calculated as 7 1/3 -1 kF = 6.66511×10 n0 [cm ]. The Fermi velocity vF is calculated as 7 1/3 vF = 7.71603×10 n0 [cm/s].

EF eV 14 12  10 8 6 4 2 n0 1 cm 3 5 10 15 20 25

22 -3 Fig.1 Fermi energy vs number density n (= n0×10 [cm ]).  

2. Fermi-Dirac distribution function3-10 The Fermi-Dirac distribution gives the probability that an orbital at energy  will be occupied in an ideal gas in thermal equilibrium

1 f ( )  , (12) e (  ) 1 where  is the chemical potential and  = 1/(kBT).

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(i) lim   F . T 0

(ii) f() = 1/2 at  = .

 ( ) (iii) For  - »kBT, f() is approximated by f ( )  e . This limit is called the Boltzman or Maxwell distribution.

(iv) For kBT«F, the derivative -df()/d corresponds to a Dirac delta function having a sharp positive peak at  = .

F E 1

0.8 

0.6

0.4

0.2

E 0.9 0.95 1.05 1.1  Fig.2 Fermi-Dirac distribution function f() at various T (= 0.002 – 0.02). kB = 1. (T = 0) = F = 1.

120

100

80

60

40

20

0.9 0.95 1.05 1.1  Fig.3 Derivative of Fermi-Dirac distribution function -df()/d at various T (= 0.002 – 0.02). kB = 1. (T = 0) = F = 1.

3. Density of states3-10 A. 3D system There is one state per volume of k-space (2/L)3. We consider the number of one- electron levels in the energy range from  to +d; D()d

L3 D( )d  2 4k 2dk , (13) 2 3

5 where D() is called a density of states. Since k  (2m / 2 )1/ 2  , we have dk  (2m/ 2 )1/ 2 d /(2  ) . Then we get the density of states

3 / 2 V  2m  D( )     . (14) 2 2  2 

A Here we define D ( F ) [1/(eV atom)] which is the density of states per unit energy per unit atom.

D( ) D ( )  F , (15) A F N where

 F 3 / 2  F 3 / 2 V  2m  2 V  2m  3 / 2 N  D( )d   d   . (16)  2  2   2  2  F 0 2    0 3 2   

Then we have

3 DA ( F )  . (17) 2 F This is the case when each atom has one conduction electron. When there are nv electrons A 9 per atom, D (F) is described as

3nv DA ( F )  . (18) 2 F

A For Al, we have F = 11.6 eV and nv = 3. Then D (F) = 0.39/(eV atom).

((Note)) Average energy

 F 3 / 2  F 3 / 2 V  2m  2 V  2m  5 / 2 E  D( )d   3/ 2d    2  2   2  2  F 0 2    0 5 2   

Then we have

3 / 2 2 V  2m  5 / 2    F E 5 2 2  2  3  3 / 2   F N 2 V  2m  3/ 2 5    F 3 2 2  2  ______

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Here we make a plot of f()D () as a function of  using Mathematica.

D E f E 1

0.8   

0.6

0.4

0.2

E 0.2 0.4 0.6 0.8 1 1.2 1.4  Fig.4 D()f() at various T (= 0.001 – 0.05). kB = 1. (T = 0) = F = 1. The constant a of D() (= a  ) is assumed to be equal to 1.

B. 2D system For the 2D system, we have

L2 D( )d  2 2kdk . (19) 2 2

Since d  (2 / 2m)2kdk , we have the density of states for the 2D system as

mL2 D( )  , (20) 2 which is independent of .

C. 1D system For the 1D system we have

1/ 2 L 2L  2m  1 D( )d  2 2dk     1/ 2d (21) 2   2  2

Thus the density of states for the 1D system is

1/ 2 L  2m  D( )     1/ 2 . (22)   2 

4. Sommerfeld’s formula When we use a formula

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L3 F(k)  dkF(k) . (23)  3  k 2 the total particle number N and total energy E can be described by

2L3 N  2 f ( )  dkf ( )  dD( ) f ( ) , (24)  k 3  k  k 2 and

2L3 E  2  f ( )  dk f ( )  dD( )f ( ) . (25)  k k 3  k k  k 2

First we prove that

 f ( ) 1 7 g( )[ ]d  g()  k 2T 2 2 g (2) ()  k 4T 4 4 g (4) ()  B B   6 360 31 127  k 6T 6 6 g (6) ()  k 8T 8 8 g (8) () 15120 B 604800 B 73 1414477  k 10T 10 10 g (10) ()  k 12T 12 12 g (12) ()  ...(26) 3421440 B 653837184000 B

Here we note that

 f ( )   g( )[ ]d   f ( )g( ) |  g'( ) f ( )]d  g'( ) f ( )]d . (27)  0   0  0 0

We define

 ( )  g'( ) or g( )  ( ')d ' . (28) 0

Then we have a final form (Sommerfeld’s formula).

  1 7 f ( )( )d  ( ')d '  k 2T 2 2'()  k 4T 4 4 (3) ()   B B 0 0 6 360 31 127  k 6T 6 6 (5) ()  k 8T 8 8 (7) () 15120 B 604800 B 73 1414477  k 10T 10 10 (9) ()  k 12T 12 12 (11) ()  ... (29) 3421440 B 653837184000 B

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5. T dependence of the chemical potential We start with

N   dD( ) f ( ) where

3 / 2 V  2m  D( )      a  2 2  2  and

3 / 2 V  2m  a    2 2  2 

  1 2a 1 a N  f ( )D( )d  D( ')d '  k 2T 2 2D'()   3 / 2  k 2T 2 2   B B 0 0 6 3 6 2 

But we also have  F  (T  0) . Then we have

 F 2a N  D( )d   3 / 2 .  F 0 3

Thus the chemical potential is given by

2a 2a 1 a  3/ 2   3 / 2  k 2T 2 2 , 3 F 3 6 B 2  or

2  kBT 2    F [1 ( ) ] (3D case). (30) 12  F

For the 1D case, similarly we have

2  kBT 2    F [1 ( ) ] (1D case). (31) 12  F

We now discuss the T dependence of  by using the Mathematica.

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 1.4

1.3

1.2

1.1

1

T 0 0.05 0.1 0.15 0.2 0.25 0.3

Fig.5 T dependence of chemical potential  for the 1D system. kB = 1. F = (T = 0) = 1.

 1.1

1

0.9

0.8

0.7

0.6

0.5 T 0 0.05 0.1 0.15 0.2 0.25 0.3

Fig.6 T dependence of chemical potential  for the 3D system. kB = 1. F = (T = 0) = 1.

6. Total energy and specific heat Using the Sommerfeld’s formula, the total energy U of the electrons is approximated by

 (T ) 1 U  f ()D()d  D()d   2 (k T)2{D[(T)] (T)D'[(T)]}   B . 0 0 6

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The total number of electrons is also approximated by

  (T ) 1 N  f ( )D( )d  D( )d   2 (k T)2 D'[(T )].   B 0 0 6

Since N / T  0, we have

1 '(T )D[(T )]   2k 2TD'[(T)]  0 , 3 B or

1 D'[(T )] '(T)    2k 2T . 3 B D[(T )]

The specific heat Cel is defined by

dU 1 1 C    2k 2TD[(T )] {  2k 2TD'[(T)]} '(T )D((T ))}(T ) . el dT 3 B 3 B

The second term is equal to zero. So we have the final form of the specific heat

1 C   2k 2TD[(T )] . el 3 B

When (T)   F ,

1 C   2k 2D( )T . (32) el 3 B F

In the above expression of Cel, we assume that there are N electrons inside volume V (= L3), The specific heat per mol is given by

C 1 D( ) 1 el N   2 F N k 2T   2 D A ( )N k 2T . N A 3 N A B 3 F A B

A where NA is the Avogadro number and D ( F ) [1/(eV at)] is the density of states per unit energy per unit atom. Note that

1  2 N k 2 =2.35715 mJ eV/K2. 3 A B

A Then  is related to D ( F ) as

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1    2 N k 2D A ( ) , 3 A B F or 2 A  (mJ/mol K ) = 2.35715 D ( F ) . (33)

We now give the physical interpretation for Eq.(32). When we heat the system from 0 K, not every electron gains an energy kBT, but only those electrons in orbitals within a energy range kBT of the are excited thermally. These electrons gain an energy of kBT. Only a fraction of the order of kBT D(F) can be excited thermally. The total 2 electronic thermal kinetic energy E is of the order of (kBT) D(F). The specific heat Cel is 2 on the order of kB TD(F).

((Note)) A For Pb,  = 2.98, D ( F ) =1.26/(eV at) A For Al  = 1.35, D ( F ) =0.57/(eV at) A For Cu  = 0.695, D ( F ) =0.29/(eV at)

______7. Pauli paramagnetism 2 Sˆ ˆ B z ˆ The magnetic moment of spin is given by z  Bz (quantum   mechanical operator). Then the spin Hamiltonian (Zeeman energy) is described by

2 Sˆ ˆ ˆ B z ˆ H z B ( )B  Bz B, (34)  

e in the presence of a magnetic field, where the Bohr magneton µB is given by B   2mc (e>0).

(i) The magnetic moment antiparallel to H: Note that the spin state is  z   . The energy of electron is given by

   k  B H

2 2 with  k  ( / 2m)k . The density of state for the down-state

L3 V 2m D ( )d  4k 2dk  ( )3 / 2    H d ,  (2 )3 4 2 2 B or

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1 D ( )  D(   H ). (35)  2 B

Then we have

 1 N  D(   H ) f ( )d . (36)   2 B  B H

(ii) The magnetic moment parallel to H. Note that the spin state is  z   . The energy of electron is given by

   k  B H ,

L3 V 2m D ( )d  4k 2dk  ( )3 / 2    H d ,  (2 )3 4 2 2 B or

1 D ( )  D(   H ) . (37)  2 B

Then we have

 1 N  D(   H ) f ( )d . (38)   2 B  B H

The magnetic moment M is expressed by

   M   (N  N )  B [ D(   H ) f ( )d  D(   H ) f ( )d , (39) B   2  B  B BH BH or

  M  B D( )[ f (   H )  f (   H )]d  B B 2 0 (40)  f ( )   2 H D( )( )d   2 HD( ) B  B F 0 

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f ( ) Here we use the relation; ( )   (   ) (see Fig.3).  F The susceptibility (M/H) thus obtained is called the Pauli paramagnetism.

2  p  B D( F ) . (41)

Experimentally we measure the susceptibility per mol, p (emu/mol)

D( )    2 F N   2 N D A ( ) , (42) P B N A B A F

2 -5 whereB NA = 3.23278×10 (emu eV/mol) and DA(F) [1/(eV atom)] is the density of states per unit energy per atom. Since

1    2 N k 2D A ( ) , (43) 3 A B F

2 we have the following relation between P (emu/mol) and  (mJ/mol K ),

5 P  1.3714810  . (44)

((Exampl-1)) Rb atom has one conduction electron. 2 -5  = 2.41 mJ/mol K , P = (1.37x10 )×2.41 (emu/mol) 1 mol = 85.468 g -6 P =0.386×10 emu/g (calculation) ((Exampl-2)) K atom has one conduction electron. 2 -5  = 2.08 mJ/mol K , P = (1.37x10 )×2.08 (emu/mol) 1 mol = 39.098 g -6 P =0.72x10 emu/g (calculation) ((Exampl-3)) Na atom has one conduction electron. 2 -5  = 1.38 mJ/mol K , P = (1.37x10 )×1.38 (emu/mol) 1 mol = 29.98977 g -6 P =0.8224x10 emu/g (calculation)

The susceptibility of the conduction electron is given by

  P  L  P  P /3  2P /3 , (45) where L is the Landau diamagnetic susceptibility due to the orbital motion of conduction electrons. Using the calculated Pauli susceptibility we can calculate the total susceptibility:

Rb:  = 0.386×(2/3)×10-6 = 0.26×10-6 emu/g K:  = 0.72×(2/3)x10-6 = 0.48×10-6 emu/g

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Na:  = 0.822×(2/3)×10-6 = 0.55×10-6 emu/g

These values of  are in good agreement with the experimental results.6

8. Physical quantities related to specific heat and Pauli paramagnetism Here we show how to evaluate the numerical calculations by using Mathematica. To this end, we need reliable physics constant. These constants are obtained from the NIST Web site: http://physics.nist.gov/cuu/Constants/index.html

Planck’s constant,  =1.05457168×10-27 erg s -16 Boltzmann constant kB = 1.3806505×10 erg/K -21 Bohr magneton B = 9.27400949×10 emu 23 Avogadro’s number NA = 6.0221415×10 (1/mol) Velocity of light c = 2.99792458×1010 cm/s electron mass m = 9.1093826×10-28 g electron charge e = 1.60217653×10-19 C e = 4.803242×10-10 esu (this is from the other source) 1 eV = 1.60217653×10-12 erg 1 emu = erg/Gauss 1mJ = 104 erg

Using the following program, one can easily calculate many kinds of physical quantities. Here we show only physical quantities which appears in the previous sections.

9. Liquid 3He A 3He refrigerator uses 3He to achieve temperatures of 0.2 to 0.3 K. A dilution refrigerator uses a mixture of 3He and 4He to reach cryogenic temperatures as low as a few thousandths of a K. An important property of 3He, which distinguishes it from the more common 4He, is that its nucleus is a since it contains an odd number of spin 1/2 particles. 4He nuclei are bosons, containing an even number of spin 1/2 particles. This is a direct result of the addition rules for quantized angular momentum. At low temperatures (about 2.17 K), 4He undergoes a : A fraction of it enters a superfluid phase that can be roughly understood as a type of Bose-Einstein condensate. Such a mechanism is not available for 3He atoms, which are fermions. However, it was widely speculated that 3He could also become a superfluid at much lower temperatures, if the atoms formed into pairs analogous to Cooper pairs in the BCS theory of . Each , having integer spin, can be thought of as a boson. During the 1970s, , Douglas Osheroff and Robert Coleman Richardson discovered two phase transitions along the melting curve, which were soon realized to be the two superfluid phases of -3. The transition to a superfluid occurs at 2.491 mK on the melting curve. They were awarded the 1996 in Physics for their discovery. Tony Leggett won the 2003 for his work on refining understanding of the superfluid phase of helium-3.[23] 15

In zero magnetic field, there are two distinct superfluid phases of 3He, the A-phase and the B-phase. The B-phase is the low-temperature, low-pressure phase which has an isotropic energy gap. The A-phase is the higher temperature, higher pressure phase that is further stabilized by a magnetic field and has two point nodes in its gap. The presence of two phases is a clear indication that 3He is an unconventional superfluid (superconductor), since the presence of two phases requires an additional symmetry, other than gauge symmetry, to be broken. In fact, it is a p-wave superfluid, with spin one, S=1, and angular momentum one, L=1. The ground state corresponds to total angular momentum zero, J=S+L=0 (vector addition). Excited states are possible with non-zero total angular momentum, J>0, which are excited pair collective modes. Because of the extreme purity of superfluid 3He (since all materials except 4He have solidified and sunk to the bottom of the liquid 3He and any 4He has phase separated entirely, this is the most pure condensed matter state), these collective modes have been studied with much greater precision than in any other unconventional pairing system. http://en.wikipedia.org/wiki/Helium-3

The atom 3He has spin 1/2 and is a fermion. Here we calculate the Fermi velocity, Hermi energy, and Fermi temperature for 3He at T = 0 K, viewed as a gas of noninteracting fermions. The density of the liquid is 0.081 g/cm3. We also calculate the heat capacity at low temperatures for T<

CV = 2.89 NkBT.

Liquid 3He as a fermion

spin I = 1/2. density  = 0.081 g/cm3, m = 3.160293 u

The number density n;

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Nm N    n   = 1.543 x 1028/m3 V V m

The Fermi energy is given by

2   (3 2n)2/3 = 0.3924 meV F 2m

The Fermi temperature is defined as

 F TF   4.55 K kB

The Fermi velocity is given by

2 v  F = 154.79 m/s F m

The heat capacity is given by

1 2 T C   NkB =1.0837 NkBT 2 TF

((Mathematica)) Liquid 3He

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Clear "Global`" ; rule1  NA  6.02214179  1023 , u  1.660538782  1027 , eV 1.602176487  1019 , kB  1.3806504  1023 , h  6.62606896  1034 , —  1.05457162853  1034 , gram  103,  cm  102, mol  NA,   0.081 gram cm3, m  3.160293 u ;

 n1  . rule1 m  

1.54351  1028  —2 F  3 2 n1 2 3 . rule1 2 m

6.28684  1023     F . rule1 eV 0.000392394  F TF  . rule1 kB 4.55353  2 F vF  . rule1 m 154.79  1 1 2 . rule1 2 TF 1.08373 

10. Conclusion The temperature dependence of the specific heat is discussed in terms of the free electron Fermi gas model. The specific heat of electrons is proportional to T. The Sommerfeld’s constant  for Na is 1.38 mJ/(mol K2) and is close to the value [1.094 mJ/(mol K2)] predicted from the free electron Fermi gas model. The linearly T dependence of the electronic specific heat and the Pauli paramagnetism give a direct evidence that the conduction electrons form a free electron Fermi gas obeying the Fermi- Dirac statistics.

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It is known that the heavy fermion compounds have enormous values, two or three orders of magnitude higher than usual, of the electronic specific heat. Since  is proportional to the mass, heavy electrons with the mass of 1000 m (m is the mass of free electron) move over the system. This is due to the interaction between electrons. A moving electron causes an inertial reaction in the surrounding electron gas, thereby increasing the effective mass of the electron.

______REFERENCES 1. L. Hoddeson, E. Braun, J. Teichmann, and S. Weart, Out of the Crystal Maze (Oxford University Press, New York, 1992). 2. NIST Web site: http://physics.nist.gov/cuu/Constants/index.html 3. A.H. Wilson, The Theory of Metals (Cambridge University Press, Cambridge, 1954). 4. A.A. Abrikosov, Introduction to the Theory of Normal Metals (Academic Press, New York, 1972). 5. N.W. Ashcroft and N.D. Mermin, Solid State Physics (Holt, Rinehart, and Wilson, New York, 1976). 6. C. Kittel, Introduction to Solid State Physics, seventh edition (John Wiley and Sons, New York, 1996). 7. C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, New York, 1980). 8 S.L. Altmann, Band Theory of Metals (Pergamon Press, Oxford, 1970). 9. H.P. Myers, Introductory Solid State Physics (Taylor & Francis, London, 1990). 10. H. Ibach and H. Lüth, Solid-State Physics An Introduction to Principles of Materials Science (Springer Verlag, Berlin, 2003).

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