# Reciprocal Lattices

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Is there an easy way to understand and/or visualize the reciprocal lattice of a two or three dimensional solid-state lattice? What is the significance of the reciprocal lattice, and why do solid state physicists express things in terms of the reciprocal lattice rather than the real-space lattice?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user David Hollman
Good question, actually. The answer is not immediately obvious.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Noldorin

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The significance of the reciprocal lattice is tied with diffraction of waves on a crystal.

How do we determine the crystalstructure of some material? We usually do by bombarding a tiny piece of crystal with X-rays or neutrons or another type of wave of appropriate wavelength and properties. We then look at the diffraction pattern. Now, to make a long story short, the pattern that appears will essentially be a pattern in the reciprocal space.

http://en.wikipedia.org/wiki/X-ray_crystallography

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Raskolnikov
answered Nov 29, 2010 by (260 points)
This is pretty brief, but basically right. It's all to do with the idea of Bragg diffraction. k-space (reciprocal space) becomes obviously helpful when considering it from that perspective.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Noldorin
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To understand why reciprocal space is important it is perhaps useful to illustrate related concept of frequency space.

Anytime one is analyzing waves (be it sound waves, EM waves or any other kind) one can exploit the translational symmetry of the laws of motion. The dual quantity to position is momentum (for massless fields like EM waves this also corresponds to frequency) and because of the said symmetry everything becomes whole lot easier. Instead of working with general wave (which can be a pretty difficult beasts) one instead works with monochromatic (i.e. single frequency) waves. For this special kind of waves the differential equations simplify just to algebraic equations, so the problem becomes easily tractable.

Now, the method explained in the previous paragraph is known as Fourier analysis and Fourier transform and is very general. Put simply, anytime you have some nice symmetry you can use Fourier analysis to move to the dual space where the problem will simplify greatly. When applied to lattices (which have a lot of translational symmetry) one obtains the concept of a reciprocal lattice.

Mathematical note:

From mathematical point of view one exploits that the system is essentially described by integrable functions on some locally compact abelian group $G$. By Peter-Weyl theorem we know that the space of such functions is parametrized by irreducible representations of $G$ which form a dual Pontryagin group $\hat{G}$.

For example when working with periodic functions one is actually working with the functions defined on the circle $S^1$. Now the irreps of $S^1$ are parametrized precisely by integers. So in this case we get the decompositions of a periodic function $$f(x) = \sum_{k \in \mathbb{Z}} f_k \rho_k(x)$$ where $f_k$ are the Fourier modes of $f(x)$ and $\rho_k(x) = \exp(-i k x)$ are irreps of $S^1$. In general the irreps of an LCA group are given by some exponential so this explains the ubiquity of exponentials in physics.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Marek
answered Nov 29, 2010 by (635 points)
Why the down-vote?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Marek
Your answer, while mathematically detailed, is off the mark. It doesn't tackle @david's original question which was: " an easy way to understand and/or visualize the reciprocal lattice". Why do we need such technicalities as the Peter-Weyl theorem to explain what a dual lattice is?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
@cadet: He also asked what is the significance of the reciprocal lattice and why we don't just work in real-space instead. I think I answered that. As for the mathematics, it was just added as a note (and marked as such) for anyone that might be interested why all of this stuff works. Do you I understand you correctly that adding mathematical notes makes the answer so much worse it deserves to be down-voted?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Marek
It seemed to me that you were using a "hammer to break an egg". And again to explain the "significance" or the "why" of reciprocal lattices we hardly need to invoke language about locally compact groups and the peter-weyl theorem, which, in any case, is decipherable by a small minority. And while there are many talented individuals, such as yourself, on this site who have that capacity, there are many more who would feel intimidated rather than helped by how you framed your answer. Or, perhaps I made an error of judgment. If so, I apologize. Feel free to return the favor anytime :)

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
@cadet: Fair enough, I see your point. As I said already under some other question, I never know at what level to formulate my answers (and actually it's probably harder to explain at low level; that takes a good amount of experience in teaching). But seeing that there are other physical answers already I thought it would be useful for someone to see also the more mathematical point of view. Not sure if my thinking was correct.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Marek
up-voted. dual and reciprocal lattices are not the same thing, and mathematical or not, this answer does a better job of explaining why physicists would ever want to work in reciprocal space.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user wsc
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Its questions like this one that keep me coming back to this site !

Is there an easy way to understand and/or visualize the reciprocal lattice of a two or three dimensional solid-state lattice?

YES ! The reciprocal lattice is simply the dual of the original lattice. And the dual lattice has a simple visual algorithm.

1. Given a lattice $L$, for each unit cell of $L$ find the point corresponding to that cell's "center of mass" (see below).
2. Connect each such "center of mass" to its nearest neighbors.
3. The resulting lattice is the dual of $L$.

To find center of mass of unit cell (we consider 2d case, generalizes to arbitrary dimension):

1. Draw the perpendicular bisectors of the edges which bound the unit cell.
2. For regular lattices these lines should intersect at a single point in the interior of the cell. This point is the "center of mass" of the cell.

Performing these simple steps you find that the dual of a square lattice is also a square lattice, and that the triangular and hexagonal lattices are each others duals ! You can see a nice illustration of this fact here.

What is the significance of the reciprocal lattice, and why do solid state physicists express things in terms of the reciprocal lattice rather than the real-space lattice?

As mentioned by others this has to do with fourier transforms. In solid-state physics we want to understand the excitations (waveforms) that a certain material, whose structure is given by some lattice $L$, can support. For a lattice only certain momenta are allowed due to its discrete structure. These allowed momenta correspond to the vertices of the dual lattice! For more see the wikipedia page or check out the first couple of chapters of little Kittel or Ashcroft and Mermin.

                                Cheers,


First of all, it is incorrect that reciprocal lattice vectors in 3D have dimensions $1/L^2$. Consider a 3D lattice with basis vectors $\{a_i\}$. The reciprocal lattice has basis vectors given by

$$b_i = \frac{1}{2V} \epsilon_i{}^{jk} \, a_j a_k$$

in index notation, with summation convention. A more familiar way to write this is in vector notation:

$$\mathbf{b}_i = 2\pi \frac{\mathbf{a}_j \times \mathbf{a}_k}{\mathbf{a}_i \cdot (\mathbf{a}_j \times \mathbf{a}_k)}$$

where $(i,j,k)$ are cyclic permutations of $(1,2,3)$. We can see that

$$\dim[\mathbf{b}_i] = \frac{\dim[\mathbf{a}]^2}{\dim[\mathbf{a}]^3} = \frac{1}{L}$$

and in terms of the lattice spacing $a$, $\vert\mathbf{b}\vert \sim \frac{1}{a}$. In fact, this is a basic fact true in any dimension.

We can also understand the normalization of the reciprocal lattice vectors by the factor $\mathbf{a}_i \cdot (\mathbf{a}_j \times \mathbf{a}_k)$ as being nothing more than $V$ - the volume of the unit cell. Why? So that the transformation between the lattice and reciprocal lattice vector spaces is invertible and the methods of Fourier analysis can be put to use.

For all regular lattices AFAIK the "dual" and "reciprocal" lattices are identical. For irregular lattices - with defects and disorder - this correspondence would possibly break down.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
answered Nov 29, 2010 by (1,985 points)
@space_cadet: Thanks for a great answer. This is what I thought the reciprocal lattice was, but I wasn't sure. The biggest difficulty I actually have is trying to apply this concept to graphene. I remember a lot of this type of stuff from my undergrad x-ray crystallography class, but most of those unit cells have atoms on the corners (unlike graphene's unit cell; you can see where the confusion comes from). I still don't understand why the unit cells in reciprocal space have different edge lengths. I'll see if I can summarize my confusion in a question and post a link in another comment

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user David Hollman
This is a great answer. However, surely that would mean for a cubic lattice of sides length $a$, the dual has sides of length $\sqrt{3a^2}$ whereas Kittel gives is as $2\pi/a$ without fully explaining why we would need the extra $2\pi$ factor ("... [the $2\pi$ factors] are are not used by crystallographers but are convenient in solid state physics.") ...

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Brendan
This is totally incorrect: the dual lattice and the reciprocal lattice are not the same thing. This leads to Brendans confusion about the reciprocal lattice constants, which must have units of inverse length.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user wsc
@wsc then why don't you tell us what the difference is between the two notions? To clear Brendan's confusion, each reciprocal lattice vector is multiplied by a factor of $2\pi/V$ where $V$ is the volume of each unit cell. For a 3D lattice, this implies that the reciprocal lattice vectors have dimension $1/a$, $a$ being the lattice constant of the main lattice. I simply didn't mention this fact in the main answer. Doesn't change anything I've said.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
actually in 3d that would leave you with units of [length]$^{-2}$. The point of reciprocal space is that we have a whole separate vector space: all vectors $G$ such that for any lattice vector $R$, $exp(iG\cdot R)=1$ and thus you can bring in the machinery of Fourier decomposition. The real-space dual lattice doesn't, for example, explain the notion of Brillouin zones, and why for an infinite but (discretely) homogeneous system we can characterize everything with a finite region of reciprocal space.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user wsc
Its cool @wsc. I try to approach the question from the viewpoint I think it would be easiest for the answerer to understand the concept. That does not mean its the approach and that is where criticism such as yours helps guide me in the right direction. Cheers!

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
This is just wrong. For example, the reciprocal space of a 2D triangular lattice is a 2D triangular lattice. Not hexagonal.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Steve B
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One problem I had in understanding reciprocal space is the origin in a real space unit cell is at infinity in reciprocal space and vise-versa. How could any real construct be mapped out in reciprocal space in a finite cell that included the origin?

However it is important to remember that we tend to deal with waves in reciprocal space. So for example, when crystallographers use x-rays, neutron diffraction etc. to obtain a pattern in reciprocal space, simply applying the reciprocal conversion (i.e. $2\pi/a$) does not obtain the positions of the atoms, but rather the wavelengths of a set of waves that will denote the positions of the atoms. A wave with wavenumber zero will have infinite wavelength (i.e. is not a wave [is this correct?])

This is basically saying what the others have said about the reciprocal space being the Fourier transform of the real space - waves do not have position that is so easily defined, but they do have well defined wavelength!

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Brendan
answered Jan 3, 2011 by (20 points)

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