I would be fine with a one dimensional lattice for the purpose of answering this question. I am trying to figure out what more general theorem (if any) gives Bloch's theorem as the number of unit cells-->infinity.

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    $\begingroup$ Welcome to Physics.SE! For the benefit of users who read a question just because the title sounded interesting it is useful to link to a source for, in this case, Bloch's theorem. I'm going to add a Wikipedia link for you. $\endgroup$ – dmckee --- ex-moderator kitten Feb 19 '12 at 20:53

Bloch's theorem generalizes nicely to a finite size crystal if we take periodic boundary conditions (pbc). If we have pbc than the a translation by one unit cell is still a symmetry of the system and so Bloch's theorem will apply.

The only difference will be that the quasimomentum $q$ will only be allowed to take certain discrete values since the wavefunciton must periodic over the entire system. Specifically $qL = 2\pi$ where $L$ is the length of the system. This is all exactly analogous to taking a free particle and putting it in a periodic box.

Periodic boundary conditions are not physical of course (except for rings of atoms) but as with the particle in a box they give the essential structure. In this case the point is that the quasimomentum is still a good quantum number locally, but the energy spectrum is now discrete with energy spacing like $1/L$.

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    $\begingroup$ That is what I thought at first, but a college brought up the fact that periodic boundary conditions are not a good approximation near edges. So I was wondering whether there was a further generalization that naturally treated edges different (or the states changed as you approach the edges.) $\endgroup$ – Alex Eftimiades Feb 20 '12 at 15:05
  • $\begingroup$ To understand the detailed behavior near the edges you're going to have solve the Schrodinger equation near the edges. No way around it that I know of nor do I suspect there can be any. This is is done all the time in the literature. You can still apply Bloch's theorem in the directions perpendicular to the surface of course. $\endgroup$ – BebopButUnsteady Feb 20 '12 at 15:37
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    $\begingroup$ Then at what point does Bloch's theorem stop applying? How many unit cells is "enough" for Bloch's theorem to apply? $\endgroup$ – Alex Eftimiades Feb 20 '12 at 21:33
  • $\begingroup$ Bloch's theorem stops applying when you destroy translational symmetry. This symmetry is crucial for the theorem. But, if you take homogeneous Dirichlet boundary conditions instead of periodic ones, you still can try to use linear combinations of Bloch states with the same energy and opposite quasimomenta to satisfy these boundary conditions. $\endgroup$ – Ruslan Nov 14 '14 at 14:56

You may think this way: take a perfect infinite crystal where Bloch theorem perfectly work and add potential which makes real crystal finite. Next question you may ask how this potential is "seen" by quasiparticles which have been obtained from infinite crystal consideration. This procedure is perfectly self-consistent and is applicable in all cases.

Also, using this approach you may get some idea when Bloch theorem stops working. However, there is no definite answer to this question: it depends on which value you are interested in. If you are interested in momentum conservation, fourier spectrum of this confining potential should be small at wavevectors you consider. If you consider energy, levels spacing due to quantum confinement should be small compared to some characteristic energy, etc.

  • $\begingroup$ I suppose I am mainly interested in what kind of how quasimomentum would be allowed in such a system. Would it vary with position on the lattice? $\endgroup$ – Alex Eftimiades Feb 23 '12 at 1:00
  • $\begingroup$ What do you mean by "how allowed"? Following the procedure I outlined you may define this value in any case. The next question is whether it will have any meaning/be useful in calculations. Answer to this question depends on a system you consider. $\endgroup$ – Misha Feb 23 '12 at 4:20
  • $\begingroup$ Ok then, I guess that's all there is to it. I checked the first answer only because it was more directed to my question (rather than the comment.) I appreciate your answer too and voted it up. $\endgroup$ – Alex Eftimiades Feb 23 '12 at 11:16

For one dimensional crystal, when we have periodic boundary condition, we may think that the atoms are arranged in a circle, and so each lattice point will be equal to other lattice point and the translation symmetry is conserved.

And for three dimensional crystal, due to periodic boundary condition, each atom will feel the same in the lattice. They are in a finite crystal and then equal to the atom one crystal away from the atom. So the translation symmetry by Bravais lattice is still conserved. So the Bloch theorem still applies here.

  • $\begingroup$ This just reads like a paraphrase of points from existing answers. Doesn't really add anything new. $\endgroup$ – paisanco Oct 2 '15 at 2:24
  • $\begingroup$ Well, I just want to say that we can think of the atoms arranged in a circle for 1D crystal, which makes the conservation of translation symmetry a little more clear. $\endgroup$ – Sheng Oct 2 '15 at 4:26

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