# Periodic boundary conditions and Bloch's theorem

1. Feb 8, 2008

### Manchot

One thing that's always bothered me about Bloch's theorem is the periodic boundary conditions which are imposed on the system. Clearly, when dealing with an actual solid, the more natural choice would be to impose zero at the boundaries. I know that periodic conditions make the math easier, but honestly, the whole thing seems like a complete farce to me. Why bother trying to find the eigenstates of the system if the eigenstates you obtain aren't even the correct ones?

2. Feb 8, 2008

### f95toli

Remember that the size of a "typical solid" (i.e. the samples used in most experiments) is usually much,much larger than the interatomic spacing. Periodic boundary conditions simply assume that all boundaries (surfaces) of the samples are so far away that they can be neglected, and only the lattice (which is periodic) affects the properties of the material. This is a VERY good approximation for bulk materials but of course it breaks down in e.g. very thin films or nanowires where the presence of surfaces can have a large effect throughout the whole material.

3. Feb 8, 2008

### Manchot

I'm okay with it on physical grounds: that's not the problem. However, I've never really seen any justification on mathematical grounds.

4. Feb 8, 2008

### jpr0

If you take an infinite 2D system (so no boundaries) and find the eigenstates of the system (Bloch waves) you can determine, say, the dispersion relation as a function of the wave-vector, $\epsilon_{\mathbf{k}}$, which is a continuous variable in the Brillouin zone. If you now do the same analysis for a finite sized system, you'll see that $\epsilon_{\mathbf{k}}$ assumes exactly the same form (as determined by the lattice), except that $\mathbf{k}$ can now assume only discrete values within the Brillouin zone. As you tend the size of the system to infinity (when the length is much larger than the lattice constant) the discrete spectrum starts to look continuous (the spacing between allowed values of $\mathbf{k}$ becomes very small). The same applies for periodic boundary conditions. As f95toli pointed out, the boundaries play a lesser role for large systems, but become important for "finite" sized systems.

I suppose mathematically, it would be related to the fact that you can turn the following sum into an integral when the system size, $L^2$ becomes very large:

$$\frac{1}{L^2}\sum_{\mathbf{k}\in BZ} \to \frac{1}{\Omega}\int_{\Omega} d^2\mathbf{k}$$

($\Omega$ is the area of the BZ. You usually end up evaluating such summations - in the calculation of the Green's function for example, and other quantities).

Last edited: Feb 8, 2008
5. Feb 9, 2008

### Manchot

Fine, Bloch's theorem gives you most of the continuum states of the system. How does this help you find the bound states?

6. Feb 10, 2008

### jpr0

Bound states due to what? You mean if the periodic potential is an array of potential wells or something along these lines?

7. Feb 10, 2008

### Manchot

For the sake of simplicity, let's say that you have a one-dimensional periodic potential that is in a (very wide) infinite well.

8. Feb 10, 2008

### jpr0

Bloch's theorem gives you the eigenstates and dispersion relation for the infinite, periodic system. By taking a linear superposition of Bloch states with different values of $k$ (that satisfy the dispersion relation for a given energy) you can construct eigenstates which satisfy the hard-wall boundary conditions of an infinite potential well. The eigenenergies of these states correspond to evaluating the dispersion relation (as obtained in the infinite system) at discrete points in the brillouin zone (the allowed values of $k$ are determined by the boundaries). This is exactly the same procedure as just solving the Schroedinger equation in a box. You find eigenstates of the infinite system, and calculate the dispersion relation. You then take a linear superposition of these states (corresponding to $+k$ and $-k$) and find that the boundary conditions for the infinite well can be met provided $k$ is quantized.

9. Apr 11, 2009

### luman

A recent Springer book on a relevant subject