|San José State University|
& Tornado Alley
Felix Bloch in his Reminiscences of Heisenberg and the early days of quantum mechanics explains how his investigation of the theory of conductivity in metal led to what is now known as the Bloch Theorem.
When I started to think about it, I felt that the main problem was to explain how the electrons could sneak by all the ions in a metal so as to avoid a mean free path of the order of atomic distances. Such a distance was much too short to explain the observed resistances... To make my life easy, I began by considering wave functions in a one-dimensional periodic potential. By straight Fourier analysis I found to my delight that the wave differed from the plane wave of free electrons only by a periodic modulation.
This was so simple that I couldn't think it could be much of a discovery, but when I showed it to Heisenberg he said right away: 'That's it!' Well that wasn't quite it yet, and my calculations were only completed in the summer when I wrote my thesis on "The Quantum Mechanics of Electrons in Crystal Lattices." [F. Bloch 1976, p.26]
Statement of the Theorem: Let r be any vector in a lattice. Let ψ be a single electron solution to the Schroedinger equation
where U((r+R)=U((r) for all r belonging to the lattice. Then there exists a wave vector k in the reciprocal lattice and a periodic function uk(r), such that uk(r+R)=uk(r), such that ψ is of the form
This means that ψ is a plane wave exp(ikr) modulated by the function uk(r).
Not all wave functions satisfy the Bloch Theorem. For example, if the wave function is for a lattice with boundaries then it is not of the Bloch form. The wave function of two or more interacting electrons is not of the Bloch form.
is periodic with the periodicity of the lattice. Let R be any element of the lattice and let G be any element of the reciprocal lattice. This means that GR is an integer multiple of 2π so
But exp(iGR)=1 so
Thus ψk(x) is of the Bloch form.
ψk(x) will be an eigenfunction of the momentum operator if the momentum operator commutes with the Hamiltonian operator. This requires that
for any function f(x).
While it is true that
in general it is not true that
is the same as U((x)(∂/∂x)f(x). This would be true if U((x) is a constant. If U((x) is piecewise constant then
would the same as U((x)(∂/∂x)f(x) at every point where the derivative is defined for both quantities.
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