chapter 7 solid state calculations

8/2/2019 Chapter 7 Solid State Calculations

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Chapter 7: Energy Bands (Part II)

Ryan John A. CuberoDepartment of Physics

De La Salle University - Manila2401 Taft Avenue, Manila

0.1 Wave Equation of Electron in a Periodic Potential

In this section, we shall treat in detail the wave equation for a general potential at general values of

k.

First, we let U(x) be the potential energy of an electron in a linear lattice of lattice constant a.

Note that the potential energy is invariant under a crystal lattice translation, i.e., U(x) = U(x + a).

This implies that the potential in consideration is periodic in nature. Since the potential U(x) is

periodic, one can expand the potential as a Fourier series in the reciprocal lattice vectors G, i.e.,

U(x) =G

UGeiGx (1)

where the values of the coefficients UG for actual crystal potentials tend to decrease rapidly with

increasing magnitude of G. For a bare coulomb potential, UG decreases as 1/G2.

Now, we choose our potential U(x) as a real function such that

U(x) =G

UGeiGx

=G

UGeiGx

+ U0e0

+G

UGeiGx

=G

UGeiGx + U0 +

G

UGeiGx assuming the crystal is symmetric about x = 0.

=G

UG(eiGx + eiGx) assuming further that U0 = 0

= 2G>0

UG cos(Gx) via the definition of the cosine function in terms of exponentials (2)

This potential form will be used in calculations on the preceeding chapters.

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Now, an electron in a crystal can be described by the wave equation

H = (3)

where H is the Hamiltonian ang is the energy eigenvalue. Note here that , the solutions, are called

eigenfunctions (or orbitals; Bloch functions). In a one-elctron approximation, i.e., we disregard the

interaction between electrons, where the wavefunction (x) describes the motion of one electron in the

potential of the ion cores and in the average potantial of the other conduction electron, one obtains

H(x) =

p2

2m+ U(x)

(x)

=

p2

2m+G

UGeiGx

(x)

=

The wavefunction (x) may be expressed as a Fourier series summed over all values of the wave

vector permitted by the boundary conditions, so that

(x) =k

Ckeikx (4)

where k is real. Inspecting at the boundaries, x = 0 and x = L, one has to assert that = 0 for a

general potential U(x). Hence,

(x) =k

Ckeikx

=k

Ck(cos(kx) + i sin(kx)) (5)

At x = 0, (0) = 0. This can only be satisfied when the coefficients of cos(kx) is zero. Moreover, at

x = L, the condition that (L) = 0 can only be satisfied if

sin(kL) = 0

kL = 2n

k =2n

L(6)

where n is any integer, positive or negative. Take note that the wave eigenfunction may not be

periodic in the fundamental translation a. The translational properties are to be determined by the

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Bloch theorem

k(r) = uk(r)exp(ik r) (7)

Moreover, not all k = 2nL

enter the Fourier expansion of (x). If one wavefunction, k already

contains the wave vector k, then the rest will be of the form k + G, i.e., k+G.

Note that unlike in the phonon problem where there are no components of the ion motion outside

the first zone, this situation (the electron problem) is like an x-ray diffraction problem because EM

field exists everywhere within the crystal and not only in the ions.

Hence,

p2

2m+G

UGeikx

(x) = (x)

2

2m

d2

dx2+G

UGeikx

k

Ckeikx =

k

Ckeikx

k

2Ck2m

d2

dx2eikx +

G

k

UGCkei(k+G)x =

k

Ckeikx

k

2Ck2m

k2eikx +G

k=k+G

UGCkGei(k)x =

k

Ckeikx

k

2

Ck2m k2eikx +

G

k

UGCkGeikx = k

Ckeikx (8)

wherein this condition can only be met when

2

2mk2Ck +

G

UGCkG = Ck

2

2mk2

Ck +

G

UGCkG = 0 (9)

If we let k =2k2

2m, then one has

(k )C(k) +G

UGC(k G) = 0 (10)

which is known as the central equation which then describes the wave equation in a periodic lattice.

However, this equation is formidable because of the infinite number of C(k G) to be determined.

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Once the Cs in the central equation are determined, then one can express k(x) as

k(x) =G C(k G)e

i(kG)x

=G

C(k G)eiGxeikx

= uk(x)eikx (11)

where one defines uk(x) =

G C(k G)eiGx. Because uk(x) is a Fourier sum over the reciprocal

lattice vectors, it is invariant under a crystal lattice translation T, i.e.,

T(uk(x)) = uk(x + T)

=G

C(k G)eiG(x+T)

=G

C(k G)eiGxeiGT

= uk(x)eiGT

= uk(x)(1)

= uk(x) (12)

Hence, uk(x + T) = uk(x). This is the general proof of Blochs theorem valid even when k(x) is

degenerate

0.2 Crystal Momentum of an Electron

Earlier, we used the wave vector k to label the Bloch function due its several properties:

1. Under a crystal translation, r to r + T, one has

K(r + T) = uk(r + T)eik(r+T)

= uk(r)eikTei

kr

= eikTk(r) (13)

Hence, one can think of eikT as a phase factor by which Bloch function is multiplied whenever a

translationT is made in the crystal lattice. Moreover, one can also think of eikT as an eigenvalue

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of the crystal translation operator, T, i.e.,

T

k(r) = k(r +T)

= eikTk(r) (14)

so that k is a suitable label for the eigenvalue.

2. If the lattice potential vanishes, i.e., U = 0 such that the central equation boils down to

(k )Ck = 0 (15)

so that all C(k G) = 0 except C(k). Hence,

uk(x) =G

C(k G)eiGx

= C(k)e0

= C(k) (16)

which is a constant. Hence, the wave function simplifies into a plane wave solution as expected

for a free electron.

3. Note that for a normalized wave function

(x) =k

Ckeikx (17)

Hence, calculating for the average momentum,

p = (x)|p|(x)

=

(x)

i

d

dx

(x)

=

(x)

i

(ik)(x)

= (x)|k|(x)

= k(x)|(x)

= k (18)

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which is the crystal momentum of an electron.

The quantity k enters in the conservation laws that gover collision process in crystals, that is,

selection rules for transitions (allowed transitions). For instance, when an electron k absorbs a

phonon, p, the selection rule is k + p = k + G, i.e., the electron is scattered from a state k to k

with G, a reciprocal vector.

0.3 Solutions of the Central Equation

The central equation

(k )C(k) +G

UGC(k G) = 0 (19)

represents a set of simultaneous linear equations that connect the coefficients C(kG) for all recirpocal

lattice vectors G. Since C(k) and C(k G) are independent of each other, the equations will be

consistent if the determinant of the coefficient vanishes.

We now write out the equation for an explicit problem. Now, let g denote the shortest G. Moreover,

we suppose that the potential energy U(x) contains only a single Fourier component U = Ug = Ug.

Hence,

(k )C(k) +G

UGC(k G) = 0

(k )C(k) + UgC(k g) + UgC(k + g) = 0

U C(k + g) + (k )C(k) + U C(k g) = 0 (20)

Thus, with 5 sets of central equations, the determinant can be written now as

k2g U 0 0 0

U kg U 0 00 U k U 00 0 U k+g U0 0 0 U k+2g

(21)

Take note that the determinant is, in principle, infinite. However, it will often be sufficient to set equal

to zero the portion shown above. The solution of the determinant gives a set of energy eigenvalues

nk where n is an index for ordering the energies and k is the wave vector we used to label the C(k)s.

Here, k will often be taken in the first zone. However, if we choose a k different from the original

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by some reciprocal lattice vector, we will still obtain the same sets of equation in a different order but

with the same energy spectrum.

0.4 The Kronig-Penney Model in Reciprocal Space

In this section, we solve for the Kronig-Penney model of a periodic delta function potential as an

example to the use of the central equation an exactly solvable model which has already been discussed

in the previous sections.

The Kronig-Penney model is described by a periodic delta function potential given by,

U(x) = Aas

(x sa) (22)

where A is a constant and a the lattice spacing. The sum is over all integers s between 0 and 1/a The

boundary conditions are periodic over a ring of unit length, which means over 1 /a atoms. Hence, by

taking the inverse Fourier expansion of U(x) = 2

G>0 UG cos(Gx), to obtain the Fourier component,

UG, one has

UG =10

U(x)cos(Gx)dx

= Aas

10

cos(Gx)(x sa)dx

= Aas

cos(Gsa)

= Aa

cos

1

a

Ga sin

1a

+ 1

2

Ga csc

Ga

2

= Aacos G sin

12a

Ga

sinGa2

= Aa(1)G

2Ga2

= Aa1

a

= A. (23)

We can now then write the central equation with k as the Bloch index as

(k )C(k) + An

C(k 2n/a) = 0 (24)

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where k =2k2

2mand the sum is over all integers n. Here, we want to solve for (k).

To do this, we define

f(k) =n

C(k 2n/a) (25)

so that the central equation for the Kronig-Penney model as given can be re-arranged as

C(k) = (2mA/2)f(k)

k2 (2m/2)(26)

Because the sum in the definition of f(k) is iver all coefficients of C, then for any n,

f(k) = f(k 2n/a) (27)

This allows us to write the expression for C(k) as

C(k 2n/a) = (2mA/2)f(k 2n/a)

k2 (2m/2)

= (2mA/2)f(k)

(k 2n/a)2 (2m/2)(28)

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Summing both sides,

n C(k 2n/a) =

n

(2mA/2)f(k)

(k 2n/a)2 (2m/2)

f(k) =

2mA

2

f(k)

n

1

(k 2n/a)2 (2m/2)

1 =

2mA

2

n

1

(k 2n/a)2 (2m/2)

2

2mA=n

1

(k 2n/a)2 (2m/2)

=n

1

(k 2n/a)2 K2

=n

12K

1

(k 2n/a) K 1

2K

1

(k 2n/a) + K

=n

1

2K

1

(2n/a) + (k K)

1

2K

1

(2n/a) + (k + K)

=1

2K

n

1

(2n/a) (k K)+

1

(2n/a) (k + K)

=1

2K

n

1

(2n/a) + (K k)+

1

(2n/a) (K+ k)

=1

2K

n

1

(2/a)((n) + (a/2)(K k))+

1

(2/a)((n) (a/2)(K+ k))

= a4K

n

1(n) + (a/2)(K k)

+n

1(n) (a/2)(K+ k)

=a

4K

cot

a2

(K k)

+ cot

a

2(K+ k)

=a

4K

cot

a2

(K k) cot

a2

(K+ k)

(29)

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By having x = Ka2 and y =ka2 , one can calculate further

2

2mA =

a

4K [cot(x y) + cot (x + y)]

= a

4K

1 + tan x tan y

tan x tan y+

1 tan x tan y

tan x + tan y

= a

4K

(1 + tan x tan y)(tan x + tan y) + (1 tan x tan y)(tan x tan y)

tan2 x tan2 y

= a

4K

2tan x + 2tan x tan2 y

tan2 x tan2 y

= a

4K

2tan x(1 + tan2 y)

tan2 x tan2 y

= a

4K2tan x sec2 y

tan2 x tan2 y=

a

4K

2sin x

cos x cos2 ysin2 xcos2 x

sin2 ycos2 y

= a

4K

2sin x

cos2 y sin2 xcosx

cos x sin2 y

= a

4K

2sin x cos x

cos2 y sin2 x cos2 x sin2 y

= a

4K

sin2x

(cos y sin x cos x sin y)(cos y sin x + cos x sin y)

=

a

4K sin2x

sin(x y)sin(x + y)

= a

4K

2sin2x

cos(x y + x + y) + cos(x y x y)

= a

4K

2sin2x

cos(2x) + cos(2y)

= a

4K

2 sin(Ka)

cos(Ka) + cos(ka)

(30)

Hence, we now have

2

2mA =a2

2Ka sin(Ka)

cos(ka) cos(Ka)

cos(ka) cos(Ka) =mAa2

2

1

Kasin(Ka)

P

Ka

sin(Ka) + cos(Ka) = cos(ka) (31)

which agrees with the results presented in the earlier sections of the book.

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chapter 7 solid state calculations

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