Solution manual - Chapters 810

Jens Zamanian

March 24, 2014

1 Chapters 8-10

Kronig-Penney model

We want to solve the Schrodinger equation for the potential

V (x) =

0 0 < x < a, a+ b < x < 2a+ b, . . .V0 b < x < 0, a < x < a+ b, . . .

(1)

which can be seen in Fig. 1. The lattice constant of this potential is given by a+b. The Schrdinger

Figure 1: The Kronig-Penney potential.

equation in the

0(x) = ~2

2m

d2

dx20(x), (2)

with solutions(x) = A sinx+B cosx, (3)

where =

2m/~. In the regions where V = V0 the equation reads

(x) = ~2

2m

d2

dx2(x) + V0(x), (4)

1

with solutions(x) = C sinhx+D coshx, (5)

where =

2m(V0 )/~. Alternative solutions are ex and ex.In the region 0 < x < a we have the solutions (3) and in the region b < x < 0 we have the

solutions (5). In order to find solutions for all values of x we must match the solutions and theirderivative at for example x = 0. The continuity gives

D = B, (6)

where we have used that sinh 0 = 0 and cosh 0 = 1 (and the same for the sine and cosine functions).The derivatives are given by

(x) = A cosx B sinx (7)(x) = C coshx+ D sinhx, (8)

which yields

C =A

, (9)

when matching at x = 0. We then get that solutions must be on the form

(x) = A sinx+B cosx 0 < x < a (10)

(x) =A

sinhx+B coshx b < x < 0. (11)

In this case the lattice vectors can be taken as X = n(a + b), where n is an integer as thesesvectors will point at the points where we have atoms (actually it will point at the right handside of the atoms, but this does not matter much). From Blochs theorem we know that wemust have

k(x+X) = eikXk(x). (12)

If we use this for x = b and for X = a+ b (which gives x+X = a we get

A sin(a) +B cos(a) = eik(a+b)[A

sinh(b) +B cosh(b)

], (13)

which may be written

A

[sin(a) + eik(a+b)

sinh(b)

]+B

[cos(a) eik(a+b) cosh(b)

]= 0. (14)

where we have used that cosine and hyperbolic cosine are even functions and that sine and hy-perbolic sine are odd. The Bloch theorem for the derivatives gives (x+X) = eikX(x) which(after some calculations) gives

A[ cos(a) eik(a+b) cosh(b)

]+B

[ sin(a) + eik(a+b) sinh(b)

]= 0. (15)

The latter two equations only have nontrivial solution for A and B if sin(a) + eik(a+b) sinh(b) cos(a) eik(a+b) cosh(b) cos(a) eik(a+b) cosh(b) sin(a) + eik(a+b) sinh(b) = 0. (16)

You may calculate this by hand, but here it is clearly useful to use algebra software like Mathe-matica. You can also plugg this into Wolfram Alpha (www.wolframalpha.com). The result is

(eik(a+b) + eik(a+b)) + 2 cos(a) cosh(b) (2 2)

sin(a) sinh(b) = 0. (17)

2

Now cos[k(a+ b)] = (eik(a+b) + eik(a+b))/2 and we may write this as

cos[k(a+ b)] = cos(a) cosh(b) +2 2

2sin(a) sinh(b), (18)

or finally

k =1

a+ barccos

[cos(a) cosh(b) +

2 22

sin(a) sinh(b)

]. (19)

This equation may in principle be solved for k for a given k, but this is a hard task.We consider the special case where the potentials approaches delta peaks. This is done by

taking the limit V0 and b 0 in such a way that V0b = const. When we do this we have

b =

2m(V0 k)

~b 0, (20)

because b goes to zero faster thanV0 goes to infinity (because we keep V0b constant). We hence

havecos(a) cosh(b) cos(a). (21)

Furthermore we get

2 22

sin(a) sinh(b) 2

2sin(a)b

2b

2sin(a), (22)

where we have used that sinh(x) x for small x. Note also that 2b V0b which is a constant(define W0 =

2ab/2. We then have the limiting equation

k =1

aarccos

[cos(a) +

W0a

sin(a)

]. (23)

Now for a given value of W0 we may solve for k a as a function of ak/pi using either numericalor algebra software. Pluggin the equation into Mathematica yields the results of Fig. 2

-1.0 -0.5 0.0 0.5 1.00

50

100

150

200

-1.0 -0.5 0.0 0.5 1.00

50

100

150

200

Figure 2: (a)2 k plotted as a function of ak/pi for W0 = 1 and W0 = 5. It is seen that thebandgap increases when W0 increases.

3

Bloch energies

In the weak potential limit we have solutions

k(r) =1Vei(kK)r (24)

with energies

=~2(kK)2

2m, (25)

and where K is a reciprocal lattice vector. Since we have a square lattice, the reciprocal latticevectors are of the form

K =2pi

a(n1, n2) , (26)

where a = 5A is the lattice constant and ni are integers. Note that we may then label the energiesas nk and we see that for each k in the first Brillouin zone we have different bands where ndepends on the values of n1 and n2. Here we want to find the lowest few solutions for the vector

k = (0.5A1, 0). Straight forward calculations using (25) we get

n n1 n2 nk(eV)

0 0 0 0.951 1 0 2.172 0 1 6.973 1 1 8.204 1 0 11.8

...

(27)

The solutions (24) are in the form nk = eikrunk, where unk = V 1/2eiKr. For the three lowest

states we then have

u0k =1V

(28)

u1k =1Vei2pix/a (29)

u2k =1Vei2piy/a (30)

u3k =1Vei2pi(xy)/a. (31)

2D band structure

For a two dimensional square lattice with spacing a the reciprocal lattice vectors are K =(2pi/a)(n1, n2) where ni are integers. We want to plot the energy k = ~2(k K)2/2m alongthe lines W X, where the points are (0, 0), (pi/a, pi/a) and (pi/a, 0), see Fig. 1. For the part(0, 0) to (pi/a, pi/a) we may parametrise the k vectors as

k =pi

a(x, x), (32)

where x goes from 0 to 1. The energies on this range are then

(x) =~2

2m

(k(x)2 K)2 = ~2

2m

4pi2

a2

[(n1 x

2

)2+(n2 x

2

)2]=

2pi2~2

ma2

[(n1 x

2

)2+(n2 x

2

)2]. (33)

4

Figure 3: The 2D-reciprocal lattice with the points , W and X.

In the range W X we may parametrise k as

k =pi

a(1, 1 x), (34)

where again x goes from 0 to 1 and the corresponding energies

(x) =2pi2~2

ma2

[(n1 1

2

)2+

(n2 1 x

2

)2]. (35)

A plot of the resulting energy levels for the values ni = 1, 0, 1 can be seen in Fig. 1.

G W X

0.5

1.0

1.5

2.0

2.5

3.0Energy

Figure 4: Energies along the path W X.

5

Fermi surface of s-band

For a two dimensional lattice with spacing a and Bravais vectors R = a(n1, n2) we have that foran atomic s-level, the energy is given by1

E(k) = Es +

R (R)eikR

1 +

R (R)eikR , (36)

where Es is the s-level energy of the free atom and

= drU(r) |(r)|2 (37)

(R) =

dr(r)(rR) (38)

(R) = dr(r)U(r)(rR). (39)

For s-levels the overlap between wave functions at nearest neighbors is small so that , and are small corrections. This means that we may neglect the in the denominator (this can be seenby Taylor expanding the denominator for small and noting that the corrections are second orderin the small quantitates). In the 2D square lattice there are 4 nearest neighbors given by

R = a(1, 0) and R = a(0,1). (40)

We then getR

(R)eikR = (ax)eiakx + (ax)eiakx + (ay)eiaky + (ay)eiaky . (41)

Moreover, for a lattice with inversion symmetry (which we have in this case) we have (R) =(R). We also have that, since U(r) has the same symmetry as the lattice (ax) = (ay).This means that all the s above are the same. We finally use that 2 cos = ei+ei. Combiningall this we get

E(k) = Es 2 [cos(akx) + cos(aky)] . (42)A plot of the Fermi surface for Es < EF , Es = EF and Es > EF is seen in Fig. 5.

1This is obtained by using the equation (10.12) in A.M. noting that the s-levels are non-degenerate and it isonly needed to consider one nonzero bm at the time.

6

- 0 -

0

k_x

k_y

Figure 5: The Fermi surface for Es < EF (red), Es = EF (blue) and Es > EF (yellow).

7