Simple Proves of Blochs Theorem

The Prove

We first give a very short prove for a special case which is take from the "Kittel". It treats the one-dimensional case and is only valid if y is not degenerated, i.e. there exists no other wavefunction with the same k and energy E.

We consider a one-dimensional ring of lattice points with the geometry as shown in the picture.

This is of course just a representation of a one-dimensional crystal consisting of N atoms with spacing a and periodic boundary conditions.

The potential V thus is periodic in a, we have V(x) = V(x + s · a) with s = integer

The decisive thought invokes symmetry arguments. Since no particular coordinate x along the ring is different in any way form the coordinate (x + a), we expect that the value of any wave function y(x), will differ at most by some factor C from the value at (x + a), i.e.

y(x + a)	=	C · y(x)

If we now proceed from (x + a) to (x + 2a), or to x + Na, we obtain

y(x + 2a)	=	C² · y(x)

y(x + Na)	=	C^N · y(x)	=	y(x)

because after N steps we are back at the beginning.

We thus have C^N = 1 and C must be one of the N roots of 1, i.e.

C	= exp	i · 2ps N

With s = 0, 1, 2, 3,...,N – 1

We now have y(x + a) = y(x) · exp(i2ps/N) and this equation is satisfied if

y(x)	=	u_k(x) · exp	i · 2p · s · x N · a

With u_k(x) = u_k(x + a), i.e. for any function u that has the periodicity of the lattice.

Try it:

	y(x + a)	=	u_k(x + a) · exp	i · 2p · s · (x + a) N · a

	y(x + a)	=	u_k(x) · exp	i · 2p · s · x N · a	· exp	i · 2p · s N	=	y(x) · exp	i 2p · s N

If we introduce k = 2ps/Na we have Blochs theorem for the one-dimensional case.
q.e.d.

The Problem

This "prove", however, is not quite satisfactory. It is not perfectly clear if solutions could exist that do not obey Blochs theorem, and the meaning of the index k is left open. In fact, we could have dropped the index without losing anything at this stage.

It does, however, give an idea about the power of the symmetry considerations.

A very similar prove is contained in Alonso-Finn. It uses a slightly different approach in arguing about symmetries.

Again, we consider the one-dimensional case, i.e. V(x) = V(x + a) with a = lattice constant.

But now we argue that the probability of finding an electron at x, i.e. |y(x )|², must be the same at any indistinguishable position, i.e.

[y(x)]²	=	[y(x + a)]²

This implies

y(x)	=	C · y(x + a)

\|C\|²	=	1

We thus can express C as:

C	=	exp (i · k · a)

for all a and k. At this point k is an arbitrary parameter (with dimension 1/m). This insures that |C| = exp (ika) · exp –(ika) =1

We thus have

y(x)	=	exp(ika) · y(x + a)

And this is already a very general form of Blochs theorem as shown below.

Writing it straight forward for the three-dimensional case we obtain the general version of Blochs theorem

y_k(r + T)	=	exp (ik · T) · y_k(r)

with T = translation vector of the lattice and r = arbitrary vector in space.

The index k now symbolizes that we are discussing that particular solution of the Schrödinger equation that goes with the wave vector k

The generalization to three dimensions is not really justified, but a rigorous mathematical treatment yields the same result. The more common form of the Bloch theorem with the modulation function u(k) can be obtained from the (one-dimensional) form of the Bloch theorem given above as follows:

Multiplying y(x) = exp(ika) · y(x + a) with exp-(ikx) yields

exp – (ikx) · y(x)	=	exp – (ikx) · exp( ika) · y(x + a)	=	exp (ik · [x + a]) · y(x + a)

This requires unambiguously that exp-(ikx) · y(x) must be periodic with the periodicity of the lattice, or exp-(ikx) · y(x) = u(x) with u(x) being periodic as before.

This again gives Blochs theorem:

y(x)	=	u(x) · exp – (ikx)

Once more, no index k at y or u is required. We also did not require specific boundary conditions. The meaning of k, however, is left unspecified. Of course, the plan wave part of the equation makes it clear that k has the function of a wave vector, but it has not been explicitly introduced as such.

To index

2.1.4 Periodic Potentials