# Crystal momentum in a lattice.

Tags:
1. Oct 24, 2014

### emily1986

Background information:
The wave function for an electron in a crystal lattice is modeled by a Bloch wave. A Bloch wave is a function with the periodicity of the lattice multiplied times a complex exponential function. This exponential function has a wave vector k, called the crystal momentum, which can have any value. To my knowledge this k independent of the configuration of the lattice. Assume the potential of the atoms in the lattice are weak. If we take the potential to zero, the piece of the Bloch wave that holds the lattice periodicity disappears and we are left with the exponential (i.e. a plane wave).

Question:
Increasing the value of the crystal momentum by a reciprocal lattice vector does not change the wave function. But obviously, a function modulated by e^(i⋅16⋅π⋅r) would be much different than a function modulated by e^(i⋅2⋅π⋅r). The spacial frequency of the former is much higher. How could this not make a difference? I know it has something to do with breaking of symmetry, and a Bloch wave not being an eigenstate of the momentum operator. Perhaps the wavefunction is only sampled periodically so it doesn't make a difference? I'd like to see the mathematical justification, but couldn't find a clear explanation in Ashcroft or on the web. I'm mostly interested in how two functions with a different crystal momentum could mathematically be the same. Thank you.

2. Oct 25, 2014

### DrDu

3. Oct 25, 2014

### ehild

The crystal is periodic with the lattice vector, so you get the same situation if you shift the origin by a lattice vector a. The wavefunction has to be periodic, too, it is the same at x and at x+a.
In the "empty" crystal, the wavefunctions are simple waves in the form ψ=ei(kx-ωt) (in one-dimensional case). It does not change if you add integer multiple of 2pi/a to the wave vector.
But the frequency and the wavevector are related, and adding a reciprocal-lattice vector to k will change the angular frequency, producing a new band.

4. Oct 25, 2014

### emily1986

According to my understanding, the wave function is periodic, but does not have the same periodicity as the lattice. For example take a look at this one dimensional example.
The wave function is not the same at x and x+a, correct? The modulation of the exponential term screws up the periodicity. If we imagined a wave function with a larger value of k, we'd see a higher frequency of modulation. Yet it would still be physically the same, but why?

5. Oct 25, 2014

### ehild

You are right, I mixed things. The wavefunction in its Bloch form is the product of a periodic function U(x) and the wave part eikx. According to the periodic boundary condition, the wavefunction is periodic with L, the size of the crystal, which is the number of the cells N multiplied by the lattice parameter, a. This condition gives the allowed values of k= 2pi/a (n/N). n is integer.The wave vector is a "label" of the wave function, and the energies depend on it. The wavefunction itself has no physical meaning. Physically ψψ* has meaning, and it is the same everywhere in the "empty" crystal.The wave factor cancels. It does not matter how many ripples you get between the lattice points.

Last edited: Oct 25, 2014
6. Oct 25, 2014

### DrDu

In the link I gave, I should have specified more clearly to look at the following picture:
http://en.wikipedia.org/wiki/Empty_...iewer/File:1D-Empty-Lattice-Approximation.svg
You can clearly recognize the free electron parabolas, however, there exist shifted copies centered in each Brillouin zone. If you switch on an interaction, the parabolas will cease to be parabolas and the degeneracy points on the boundaries of the BZ will be lifted as a band gap forms. If the interaction is small, if an electron is accelerated and crosses the band gap, it will jump from the lower to the upper band (which corresponds to the motion of the free particle), but with increasing band gap, the electron will stay on the lower band. We speak of Umklapp (from German "flip over") scattering as the electron gets Bragg scattered from the lattice, transferring momentum to the lattice.

7. Oct 25, 2014

### DrDu

Not necessarily. Let's say you start from a Bloch wave is of the form $\exp(ikx)u(x)$. If you increase k to k+G, then $\psi=\exp( ikx +iGx)(x)$ where $u$ has the periodicity of the lattice. However in the case of an empty lattice the u are of the form $u=\exp(iG'x)$. So you can find another wavefunction with a different function u' and $G'=-G$, so that the wavefunction will remain unchanged.