# Interpretation Born-Von Karman boundary conditions

The cyclic Born-Von Karman boundary condition says that if we consider a one dimensional lattice with length $L$, and if $\psi(x,t)$ is the wavefunction of an electron in this lattice, then we can say that $\psi(x+L,t) = \psi(x,t)$ for every $x$. Applying this boundary condition leads to correct solutions for $\psi$. This boundary condition can also be generalized to three dimensional lattices and can be applied when working with phonons instead of electrons.

I wonder if there is a reasoning why these boundary conditions can be applied. They appear to me as elegant boundary conditions, but I don't see any reason why these conditions could be applied. Can we derive the Von Karman boundary conditions or are they just an experimental result?

• Strictly speaking, we are ignoring surface effects altogether when we do this, and treating the lattice as infinite. Then, they key point is that the potential energy $V(x)$ felt by the electrons has this discrete translation symmetry. An arbitrary wave-function need not have this symmetry, but because the discrete translation operator $T$ commutes with the Hamiltonian, we can choose a basis of energy eigenstates that do. – lionelbrits Nov 20 '13 at 0:09
• @lionelbrits: Yup, thanks for the add-on! I didn't mention the notion of being able to choose a simultaneous eigenbasis of $H$ and the symmetry group $G$ since $[H,g]=0$ for $g\in G$ because I wasn't sure how much detail the asker wanted, but you're definitely right about that being the key concept. – DumpsterDoofus Nov 20 '13 at 0:45