# Boundary conditions for crystals

As students on solid state physics, we are all taught to use the periodic boundary condition, taking 1D as an example: $\psi(x)=\psi(x+L)$ where $L$ is the length of the 1D crystal.

My question is:

1. Why this boundary condition is acceptable? Only because we think we are dealing with the bulk, and the surface are not relevant too much?

2. How about other boundary condations, such as zero boundary condtion, grain boundary condition etc? If ignoring their complexity, can they get the same answer as the periodic boundary condition?

3. How about other artificial boundary conditions, for example, if I let $\psi(x)=2 \psi(x+L)$, could I get the same energy band?

• I have wondered this as well. There are not just surface effects that can occur with different boundary conditions. Consider an Ising magnet on a Moebius strip. The ground state necessarily has a freely propagating kink. Taking the thermodynamic limit does not change this. It almost seems like laziness to consider periodic boundary conditions to be the absolute. :P – Ryan Thorngren Sep 12 '12 at 9:22
• It is because we are considering just the bulk, and because you're students, we prefer to teach you model systems to practice the basics on. There is a huge amount of physics which is not taught at all, but which you can learn quite easily yourself once you have the fundamentals. And yes, that includes surface effects and disorder; and yes, massive changes occur. But no, you're not going to see them dealt with in mathematical glory for quite some time. – genneth Sep 12 '12 at 21:46

Your last question has a definite "no" answer. If you allow a boundary condition like $\Psi(x+L) = 2\Psi(x)$ ($L$ being the lattice constant I presume in your head), then you will never find a normalizable wave function in an infinite space... that's bad, isn't it ?
You probably meant $\psi(0)=\psi(L)$, $0<x<L$. Such a boundary condition follows from the symmetry you suppose the crystal has.