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Homework # 7 Chapter 7 Kittel Phys 175A Dr. Ray Kwok SJSU

Problem 7-1 Nabel Alkhawlani
Square lattice, free electron energies: a) show for a simple square lattice (two dimensions) that the kinetic energy of a free electron at the a corner of the first zone is higher than that of an electron at midpoint of a side face of the zone by a factor of 2. b) what is the correspondence factor for a simple cubic lattice (three dimensions)? C) what bearing might the result of b) have on the conductivity of divalent metals?


This is the 2 - Drecipical square lattice with the first brillouin zone.

By taking the brillouin zone

Kr = 2 Kx 2

2

h 2h 2 2 ε kr = (kr ) = (kx) = 2ε kx 2m 2m

For 3-D, the beruilluin zone

Kr = 3Kx

ε kr = 3ε kx


in case of the divalent metal, if the band gap energy is higher than the difference of the kinetic energy between the midpoint of a side face and the corner, then some electrons will prefer moving to the second zone than moving to the corners of the first zone. In that case the divalent metal will become a conductor/

Chapter 7: Problem #2 Free electron energies in reduced zone: Consider the free electron energy bands of an fcc lattice in the approximation of an empty lattice, but in the reduced zone scheme in which all ks are transformed to lie in the first Brillouin zone. Plot roughly in the [111] direction the energies of all bands up to six time the lowest band energy at the zone boundary at k = (2π/a)(1/2,1/2,1/2). Let this be the unit of energy.















Ch. 7; Eq. (1): ε = (ħ2/2m)(kx2 + ky2 + kz2) Ch. 2; Eq. (21): k' = k + G Ch. 2; Eq. (15): G = v1b1+v2b2+v3b3 Useful textbook pages: 32, 38, 223 For fcc lattice the reciprocal lattice vectors are: b1 = (2π/a)(-x+y+z) b2 = (2π/a)(x-y+z) B3 = (2π/a)(x+y-z) Find G: G = (2π/a)(-v1x + v1y+v1z +v2x – v2y + v2z +v3x+v3y-v3z)

















k = (2π/a)(1/2,1/2,1/2) εmax=(ħ2/2m)k2 = (6)(ħ2/2m)(2π/a)2(1/4+1/4+1/4) εmax= (9/2)(ħ2/2m)(2π/a)2 K = G + k' = (2π/a)(1/2x+1/2y +1/2z) + (2π/a)(-v1x + v1y+v1z +v2x – v2y + v2z +v3x+v3y-v3z) K2 = ((2π/a)((1/2 – h + k + l)x + (1/2 + h – k + l) + (1/2 + h + k – l)))2 ε000 = (1/2x + 1/2y + 1/2z)2 = 3/4, yes ε001 = (3/2x + 3/2y – 1/2z)2 = 19/4, no ε010 = (3/2x - 1/2y + 3/2z)2 = 19/4, no













ε011 = (5/2x + 1/2y + 1/2z)2 = 7/2, yes ε110 = (1/2x + 1/2y + 5/2z)2 = 7/2, yes ε101 = (1/2x + 5/2y + 1/2z)2 = 7/2, yes ε111 = (3/2x + 3/2y + 3/2z)2 = 27/4, no ε00-1 = (-1/2x - 1/2y + 3/2z)2 = 11/4, yes ε0-10 = (-1/2x + 3/2y - 1/2z)2 = 11/4, yes ε-100 = (3/2x - 1/2y - 1/2z)2 = 11/4, yes ε0-1-1 = (-3/2x + 1/2y + 1/2z)2 = 11/4, yes ε-10-1 = (1/2x - 3/2y + 1/2z)2 = 11/4, yes ε-1-10 = (1/2x + 1/2y - 3/2z)2 = 11/4, yes




ε-1-1-1 = (-1/2x - 1/2y - 1/2z)2 = 3/4, yes 4

3.5

3

2.5 000 00-1,0-10,-100 011,101,110 -1-1-1 0-1-1,-10-1,-1-10

2

1.5

1

0.5

0 0

1/2

Kittel, 7.3 Kronig-Penney Model a)

For the delta-function potential and with P 1 – (x^2/2) + … (P/Ka) (Ka – (Ka)^3/6) + (1 – (Ka)^2/2) = 1 After some algebra… K^2 = ((6P)/(P + 3))/a^2 e = (h^2 K^2) / (2m), and since P x – (x^3/6) + … Cos (x) => 1 – (x^2/2) + … (P/Ka) (Ka – (Ka)^3/6) + (1 – (Ka)^2/2) = -1 There is a solution at Ka = pi, and another at Ka = pi + (some small number, x) (P/(pi +x)) (-x + x^3/6) – 1 + x^2/2 = -1 Drop the x^3 term, and use (pi + x) = pi…

Kittel, 7.3b cont. After some algebra… x(-2P/pi + x) = 0 Then, [e(x=0) – e(x=2P/pi)] is the energy gap and we have a band gap energy of… e(gap) = (2h^2 P) / (ma^2).

Problem 4: Potential Energy in the Wanshan Li Diamond Structure A: a) summing potential Q: For diamond structure (Fig 22,p17) with G=2A, energy V from atoms, r N ⋅s r r where A is a basis U (r ) = Σ V (r − R j )[1] j vector in reciprocal where r, location of lattice in conventional cubic cell electron (field point); a) Find Fourier coefficient Rj, location of atoms of electric potential (source points); s, energy, UG number of atoms per b) Show no energy gap at unit cell; N, number of kx=A. unit cells.

Problem 4: Nontrivial math (2/3) Extension from Equation (22), p170 r r r U (r ) = Σ U G exp(iG • r ) G

N .s r r r r 1 UG = Σ V ( r − R j ) exp(−iG • r )dv v unit∫_ cell j

r r r r 1 =Σ V (r − R j ) exp(−iG • r )dv ∫ j v unit _ cell N .s

From Fourier analysis,

r r' r r' 1 ' =Σ V ( r ) exp( − i G • ( r + R )) dv j ∫ j v unit _ cell

r r r 1 UG = U (r ) exp(−iG • r )dv ∫ v unit _ cell

N .s r r 1 r r' ' r' = Σ exp(−iG • R j ) ⋅ V ( r ) exp( − i G • r )dv ∫ j v unit _ cell

N .s

r r 1 r r' ' r' = Σ exp(−iG • R j ) ⋅ V (r ) exp(−iG • r )dv j v all _∫space s r r ≡ C ⋅ Σ exp(−iG • R j )[2] s

Plug [1] into previous equation, and simplify….

j

Problem 4:Central equation (3/3) Position of atoms in basis (p16), r Choose r

R1 = 0 r 1 1 1 R2 = a ( , , ) 4 4 4

2π A= (1,0,0) a

Plug into [2], and solve

U G = C (1 − 1) = 0

b) From central equation (22),p172  λk − ε det   UG

  = 0 λk − G − ε  UG

With UG=0, k=A ε ± = λ A = λ− A

That means, energy gap E gap =| ε + − ε − |= 0

Mike Tuffley Ch. 7 Prob. 5

qed

Problem 6 Consider a square lattice in two dimensions with the crystal potential

Apply the central equation to find the energy gap at the corner point (π/a, π/a) of the Brillouin Zone. It will suffice to solve a 2 x2 determinantal equation.

In a square lattice of spacing a, the reciprocal lattice vector is [Eqn 2.15]:

where n and m are integers. The point (π/a, π/a) lies on the edge of the Brillouin zone boundary in both the kx and ky direction. This satisfies the condition

We are therefore justified in using the determinantal equation:

because Solving the determinant:

The energy gap is therefore 2|UG|

Use 2 dimensional Fourier series to determine UG U(x,y) is periodic with period a in both dimensions. It can be expressed as a fourier series by [Eqn 2.9]:

Drop sine terms because they will evaluate to zero in the next step. Invert to get UG

for n=m=1 The energy gap is therefore 2UG = 8U where U is a constant.