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(i2ps/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 = 2ps /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)