Periodic boundary conditions

1. Nov 28, 2005

StatusX

I'm taking solid state, and again and again we use the periodic boundary conditions, that the wavefunction should be unchanged by displacements of the length of the sample, L (assume 1D for simplicity). The argument was that the surface is so far away that it shouldn't have an effect on the properties in the interior, so we are free to use boundary conditions that are mathematically convenient. But this argument doesn't work for me, because if the surface is so far away as to be irrelevant, why should 'L' enter into things at all? Why should the properties be different deep inside a 1 cm sample from inside an otherwise identical 2 cm sample?

2. Nov 28, 2005

Physics Monkey

They shouldn't be different! The intensive thermodynamic quantities like density, pressure, etc are indeed independent of the size of your system. Think about it for second, does the Fermi energy depend on the size of your system? What about the density of states per unit volume? It is only when you calculate extensive variables like total energy or total magnetization that the volume of your system enters (as it should). Even when dealing with such extensive quantities, the dependence on volume is trivial: it is always volume times some intensive quantity. For macroscopic systems the dependence on volume is exactly as simple as you would expect.

Of course, things can get interesting when this picture eventually begins to break down.

Last edited: Nov 28, 2005
3. Nov 29, 2005

marlon

Err, do you know the concept of Bloch Wave functions?

The plane wave wavevector (or Bloch wavevector) k (multiplied by Planck's constant, this is the particle's crystal momentum) is unique only up to a reciprocal lattice vector, so one only needs to consider the wavevectors inside the first Brillouin zone.

The Brillouin zone is the Fourier Transform of the Wigner Seitz Cell. This is the smallest primitive unit cell that ALSO exhibits the same symmetry as the crystal itself. Keep in mind that the coordinates in the Brillouin zone are MOMENTA not x,y and z-coordinates. So this length L has a rather special meaning.

marlon

4. Nov 29, 2005

StatusX

Maybe that's all true, but my point is that the reason I was told, that the surface should not have an effect, is not a good enough reason. I guess the real reason is something like "if the sample is large, there needs to be more allowed k values to give the same intensive properties," but this was never made clear to me.

5. Nov 29, 2005

marlon

i don't quite understand what you are saying here. Could you clarify, please

thanks

marlon

6. Nov 29, 2005

nbo10

There can be very strange surface effects. But for most anything you do for classes these effects can be ingored

7. Nov 29, 2005

StatusX

I mean, these boundary conditions mean the allowed k values are spaced by 2$\pi$/L, and I wasn't sure why such a property should depend on L.

8. Nov 29, 2005

Physics Monkey

You would expect the spacing to depend on L. You know the set of inequivalent states is always contained in the 1st Brillouin Zone and the volume of the zone is fixed by the lattice parameters i.e. it is independent of system size. But it is perfectly clear that a bigger sample should have more states, right? Since you have to stuff all these extra states in the same volume in k space, your states must be closer together for a larger system.

9. Dec 1, 2005

Modey3

StatusX,

Think of a 1 dimensional lattice. A small lattice has fewer modes of vibration than a larger lattice. You can prove that to yourself.

Modey3

10. Dec 1, 2005

armandowww

...if I could express myself too... In one of many proofs of Bloch Theorem, boundary conditions by Born and von Karmann are simply essential. Without them you are not able to execute the Fourier expansion of psi on the "big" crystal... That proof is very important because you can see, once you've done it, how to build up wavefunction u (with the same periodicity of potential) and the appearance of n, the "clever" band index, which explain why dispersion relations are never single-valued.
Moreover, in quantum mechanics periodical boundary conditions are enough to emphasize the hamiltonian hermitian properties... without request meticolously that psi must vanish on crystal surface.
Strictly speaking, in the independent electron approximation, periodicity for your potential is unreal because 10^23 is not the mathematical infinite... but we are physicists... so, please try to satisfy yourself with some kind of trick.:tongue:

11. Dec 2, 2005

marlon

Well, the answer to that question is in my first post. The reason is : Fourier Transform. This is the way that the Brilluoin zone is defined. The fact that we go from a spatial coordinate base (Wigner Seitz Cell) to a momentum coordinate base (Brillouin Zone) is because of symmetry. Without going into this too deeply, the philosophy is that you try to define a region in which you calculate all interactions going on. This region needs to be as small as possible, containing the lowest possible amount of atoms so that the calculations require less computational effort. Thanks to the crystal symmetry you can define such a small region. You calculate the interactions in a few points (ie k-point sampling) and then, thanks to the symmetry, you automatically know these interactions in other "equivalent" points of the crystal.

marlon

12. Dec 2, 2005

Galileo

Marlon, I think StatusX wants a good physical argument for the use of periodic boundary conditions. That has nothing to do with Bloch's theorem, which tells you wavefunction properties in a periodic potential. The periodic boundary conditions here are applied without reference to any potential energy function. (Although I assume the context is the free electron fermi gas model in a box)

To StatusX, all I can add is that you need boundary conditions to solve your problem and there are two more or less 'natural' choices:
- The wavefunction is 0 at the ends of the box
- Periodic boundary conditions
Both will give identical number of states, density of states etc, but periodic boundary conditions will give you traveling waves which are more natural to use than standing waves. You want to study things like electron transport after all.

If L is large enough, the allowed values for k will be practically a continuum, so the particle really is free. But you will get a discrete set of eigenstates and is a lot easier to work with. It's just a mathematical trick.

13. Dec 2, 2005

marlon

Galileo,

I must say i disagree with what you state. One can use periodic boundary conditions because one iterates the unit cell when describing a crystal. In the momentum (crystal momentum) base (after Fourier Transformation) one iterates the Brillouin zone, thanks to the Bloch Theorem. That is why this is so important. If this was not possible, all ab initio calculations would be, let's say, quasi impossible. It is the aspect of symmetry that is central in this discussion.

True, i must admit that it still is not very clear to me what the original poster is actually asking about. Perhaps you could clarify this for me, if you wish

regards
marlon

14. Dec 2, 2005

Dr Transport

Think of a single Brillouin zone and calculate whatever, the band structure, absorption, density of states, etc..... If we have an ensemble of the zones, why should the properties of an ensemble be any different than a single zone. Periodic boundary conditions ensure that the properties of the material are the same regardless of where you are inside the crystal. All propertiues are calculated with regards to the unit cell, and as Galileo has stated there are two choices, periodic or zero at the boundaries.
If I measure the absorption in a crystal, it should be the same if I have a 1 cm3 sample or if I have a 100 cm3, $$\alpha = \frac{4\pi nk}{\lambda}$$ is the same for either sample.
Now if you are at a surface and calculating surface states, that is another matter all together.

If this doesn't convince you, think about a series of masses around a circle vibrating, if you go around one time you better get the same answer, hence periodic bc's.....

Last edited: Dec 2, 2005
15. Dec 2, 2005

Physics Monkey

As Galileo noted, Bloch's Theorem applies when the potential is periodic. Now, a finite crystal can't technically supply a periodic potential unless it has periodic boundary conditions. In this technical sense, it may be said that periodic boundary conditions are the natural way to reconcile a finite crystal with Bloch's Theorem. In other words, it wouldn't make mathematical sense to impose hard boundary conditions on the wavefunction when you technically have to assume periodic boundary conditions for the crystal to get Bloch's Theorem in a finite crystal.

Physically, of course, none of this matters. You can still quote Bloch's Theorem as an excellent approximation, put in the size of the crystal by hand, and take whatever boundary conditions you want, and get all the same results.

Just a thought.

16. Dec 3, 2005

Galileo

True, but it's completely possible (as done in many books) to discuss the (crude) free electron model without reference to atomic structure (since that doesn't matter in this model). So no unit cell, periodic potential crystal momentum etc are mentioned or needed. And yet periodic boundary conditions are imposed.
There is however a lucky relation between the two IIRC, but I`ll have to check my notes on that and I don't have access to them now.

17. Dec 4, 2005

marlon

Ofcourse, that is completely true.

I was talking about interacting many body physics, though. I assumed this since the OP was talking about periodicity in structure, which you do not have in the case of a free electron gas.

regards
marlon