# Explain Bloch theorem? Discuss its use in the Kronig-Penny model and hence in band theory.

Solution: The difference in electrical conductivities of conductor, semiconductor and insulators can be understood by extending the free electron model to take into account the interaction of the electrons with the position ion lattice. It would be more realistic to assume that the potential energy inside the lattice is periodic. This is responsible because of the periodic distribution of the lattice ion in a crystalline solid.

For a free particle with potential energy V =constant, the wave function Ψ(x,t) is given as

Ψ(x,t) = Aei(kx – ωt)

The space part of the wave function can be written as

Ψ(x) = e±ikx

For a potential with period L, V(x) = V(x + L) = V(x + 2L). L is spacing between ions. Bloch theorem states that for a particle moving in a periodic potential, the eigenfunction Ψ(x) are of the form

Ψ(x) = uk (x) e±ikx

These eigenfunctions are plane waves modulated by a function uk (x), where uk (x) has the same periodicity as the potential energy. The specific form of the function uk (x) will depend on the form of the function V(x).

Because the potential is periodic, one expects that the probability of finding a particle at a given x is the same as that of finding it at x + L.

Here,

Ψ*(x) Ψ(x)  = uk*(x) e±ikx. uk (x) e±ikx

= uk*(x). uk (x)

Therefore, when uk (x) = uk (x + L)

Ψ*(x) Ψ(x) = Ψ*(x + L) Ψ(x + L)

The Kroning-Penny Model

It is the treatment of energy levels of an electron experiencing one-dimensional periodic potential. Consider a periodic potential with period L =a + b as shown in figure(1). Where ‘a’ is the width of well and ‘b’ is width of barrier.

The potential function for the particle can be expressed as

V(x) = 0 for region I

V(x) = 0 for region II

For region (I) Schrodinger wave equation can be written as

$$\frac{d^2 Ψ_1}{dx^2} + \frac{2mE}{ℏ^2}Ψ_1$$ = 0

$$\frac{d^2 Ψ_1 }{dx^2} + \alpha^2 Ψ_1 = 0$$   ……………………… (1)

Where

$$\alpha^2 = \frac{2mE}{ℏ^2}$$

From Bloch theorem,

Ψ= u1eikx

$$\frac{d\Psi_1}{dx}$$ = $$e^ikx \frac{du_1}{dx} + iku_1 e_ikx$$

$$\frac{d^2 Ψ_1}{dx^2} = e^ikx \frac{d^2 u_1}{ dx^2} + ike^ikx \frac{du_1}{dx} + ike^ikx\frac{du_1}{dx} + \left( ik \right)^2 u_1 e^ikx$$

= e^ikx( $$\frac{d^2 u_1}{dx^2} + 2 ik\frac{du_1}{dx} – k^2 u_1$$)

Substituting the value of $$\frac{d^2 \Psi_1}{dx^2}$$ in equation (1)

= e^ikx ($$\frac{d^2 u_1}{dx^2} + 2 ik\frac{du_1}{dx} – k^2 u_1) + \alpha^2 u_1 e^{ikx} = 0$$

=$$(\frac{d^2 u_1}{dx^2} + 2 ik\frac{du_1}{dx} – k^2 u_1) + \alpha^2 u_1 = 0$$ …………..(2)

For region (II), Schrodinger wave equation can be written as

=$$\frac{d^2 \Psi_2}{dx^2} + \frac{2m(E – V)}{ℏ^2}\Psi^2 = 0$$

=$$\frac{d^2 \Psi_2}{dx^2} – \frac{2m(V – E)}{ℏ^2}\Psi_2 = 0$$

=$$\frac{d^2 \Psi_2}{dx^2} – \beta^2 \Psi_2 = 0$$

=$$\beta^2 = \frac{2m(V – E)}{ℏ^2}$$

And its Solution is $$\Psi_2 = u_2 e^ikx$$

Proceeding as above we get,

$$\frac{d^2 u_2}{dx^2} + 2 ik\frac{du_2}{dx} – \beta^2 u_2 – k^2 u_2 = 0$$ ……………. (3)

The solution of equation (2) and (3) are

$$u_1 = Ae^{(\alpha – k)x} +Be^{-i(\alpha + k)x}$$     ………………….. (4)

$$u_2 = Ce^{(\beta – ik)x} + De^{-(\beta + ik)x}$$  ………………………. (5)

We have the boundary conditions,

u1(x=0) = u2(x=0)

u’1(x=0) = u’2(x=0)

Since the potential is periodic with period, L = a + b i.e u(x + L) = u(x)

For, x = -b, u(-b + a + b) = u(-b)

u(a) = u(-b)

∴u1(a) = u2(-b)

u’1(a) = u’2(-b)

Using these boundary conditions and solving equations (4) and (5) we get

$$\frac{\beta^2 b}{2\alpha}sin\alpha L + cos\alpha L = cos kL$$

$$\frac{P}{\alpha L}sin\alpha L + cos\alpha L = cos kL$$ …………………. (6)

Here, $$P = \frac{\beta^2 bL}{2} = \frac{2m(V – E)}{ℏ^2}.\frac{bL}{2} = \frac{m(V – E)}{ℏ^2}bL$$

Therefore, $$P = \frac{mVbL}{ℏ^2} (for V >> E).$$

In arriving equation (6) we have made the use of one of the forms of Bloch functions, namely, $$\Psi = u(x) e^ikx$$. The same result will be obtained if we use the either form that is, $$\Psi(x) = u(x)e^-ikx$$ .

Figure 3: The solid lines show the dependence of the energy E on the wave vector k for an electron moving in the periodic array of rectangle potential wells. The dashed line represents the relation between the energy and the momentum for the free electron case.

When the particle is not free, that is, for an electron moving in one-dimensional array of potential, the dispersion relation is given by equation(6),

$$\frac{P}{aL}sin\alpha L + cos\alpha L = cos kL$$ ………………… (6)

Where, $$\alpha = \sqrt{\frac{2mE}{ℏ^2}$$

The plot between E and k on the basis of equation (6) is as shown by the solid lines of figure(3). Here the solid lines are the real solution: of equation(6) for which the values of E and k are defined. These solid lines correspond to allowed energy bands. For other energy intervals between two adjacent solid lines, the value of k is not defined, these are the values of E for which the left side of equations (6) is either greater than +1 or than -1. These energy intervals correspond to forbidden energy bands.