# Simple Proofs of Bloch's Theorem

The Proof

| We first give a very short proof for a special case which is taken from the book of Kittel ("Quantum Theory of Solids"). It treats the one-dimensional case and is only valid if y is not degenerate, 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 x with period length a, i.e. 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 from 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) | = | C 2 · y (x) |  |
--- | --- | --- | --- | ---
y(x + Na) | = | CN · y (x) | = | y(x)
|  | because after N steps we are back at the beginning.
|  | We thus have CN = 1 and C must be one of the N roots of 1, i.e.
C | = exp | i · 2p s N
--- | --- | ---
|  | With s = 0, 1, 2, 3, ..., N – 1
| We now have y(x + a) = y(x) · exp(i2p s/N) and this equation is satisfied if
y (x) | = | uk(x) · exp | i · 2p · s · x N · a
--- | --- | --- | ---
|  | With uk(x) = uk(x + a), i.e. for any function u that has the periodicity of the lattice.
|  | Try it:
| y(x + a) | = | uk(x + a) · exp | i · 2p · s · (x + a) N · a
--- | --- | --- | --- | ---
| y(x + a) | = | uk(x) · exp | i · 2p · s · x N · a | · exp | i · 2p · s N | = | y (x) · exp | i 2p · s N
--- | --- | --- | --- | --- | --- | --- | --- | --- | ---
|  | If we introduce k = 2p s /Na we have Bloch's theorem for the one-dimensional case. q.e.d.

The Problem

| This "proof", however, is not quite satisfactory. It is not perfectly clear if solutions could exist that do not obey Bloch's 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 proof is contained in the relevant Alonso–Finn book ("Quantum and Statistical Physics"). 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 + a) | = | C · y(x)
--- | --- | ---
|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 ensures that |C|2 = exp (ika) · exp (–ika) = 1
|  | We thus have
y(x + a) | = | exp(ika) · y(x)
--- | --- | ---
|  | 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 Bloch's theorem:
yk(r + T) | = | exp (ik · T) · yk(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 shows unambiguously that exp(–ikx) · y(x) = u(x) is periodic with the periodicity of the lattice.
|  | And this, again, gives Bloch's 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 plane wave part of the expression makes it clear that k has the role of a wave vector, but it has not been explicitly introduced as such.

With frame

2.1.4 Periodic Potentials

© H. Föll (Semiconductors - Script)