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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?

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In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to associate the much larger symmetry group of lattice translation operators with the system.

Doing so, one obtains a system which is considerably more mathematically convenient than the otherwise rigorously-correct picture of a misshapen blob of atoms. As such, the Born-von Karman boundary conditions can be considered an approximate model to the behavior of real materials. That said, it's a pretty good approximation for macroscopically-sized materials.

That said, I don't have an explicit proof of why this approximation doesn't cause the math to explode and give wrong answers, so someone better versed than me could probably elaborate on this.

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    $\begingroup$ 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. $\endgroup$
    – lionelbrits
    Commented Nov 20, 2013 at 0:09
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    $\begingroup$ @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. $\endgroup$
    – DumpsterDoofus
    Commented Nov 20, 2013 at 0:45

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