# How do Gamma point phonons work?

1. Jan 3, 2007

### sphericalCat

Hi folks,

I'm having lots of trouble wrapping my head around this. For a start, how can you have a vibration along a [0 0 0] vector? I mean, where does it point?? And how can it have a finite momentum or frquency? (Does it?)

Help would be greatly appreciated, but, please, speak slowly and clearly, and use short words - I am a chemist!

2. Jan 3, 2007

### Gokul43201

Staff Emeritus
A Gamma Point phonon is not a vibration that points "along a [0 0 0] vector." For one thing (among several), the Gamma Point is a point in reciprocal space (k-space), not a direction vector in real space. You need to understand how a phonon dispersion (or band structure) comes about. I suggest you take the standard approach of learning about vibrational modes in a 1D chain and then extending that to a 3D lattice. Any introductory text on Solid State Physics will cover this (eg: Kittel or Ashcroft & Mermin).

I also remember seeing a thread here ages ago that linked to several Java applets that provided neat visualizations of physics concepts - one of these demonstrated 1D phonon modes with associated dispersion relations. I can't seem to find it now.

If you are asking if a phonon mode can have "non-zero" frequency (or energy) at the Gamma point (ie: at zero wave-vector, or infinite wavelength), the answer is "yes, it can". This you will see, when you advance to step 2 - phonon modes in a 1D chain with a basis of 2 (atoms per lattice site).

3. Jan 3, 2007

### sphericalCat

Thanks for the pointer. I have looked at the textbooks before, but now, at least, I know which bits to concentrate on.

One quick question about what you said:

Am I wrong in thinking that, in the current context, the k-points correspond to real-space lattice vectors?

4. Jan 3, 2007

### Dr Transport

Yes,

k-space is reciprocal space not real-space, they are not the same.

5. Jan 5, 2007

### sphericalCat

That much I know, but I was under the impression, that half the point of reciprocal space was to express the orientation vectors, describing periodic planes in real space, as discrete points in k-space. No?

6. Jan 6, 2007

### Gokul43201

Staff Emeritus
Yes, this is true. The lattice points in reciprocal space represent the normals of a corresponding family of planes in real space.

Perhaps you are getting confused between the terminologies "k-space" (which represents all points in reciprocal space) and the "reciprocal lattice" (which is a subset of the reciprocal space, restricted to points of the form $G=2\pi(n_1 \mathbf{a^*}+n_2 \mathbf{b^*}+n_3 \mathbf{c^*})$ with integer n's)?

7. Jan 7, 2007

### sphericalCat

Ok, this all makes sense so far. By the way, as you suggested, I've gone back to the textbook, and things are, generally, beginning to make a bit more sense.

Here's where I come unstuck: any given orientation vector can describe three orthogonal planewaves, right? Now, presumably, you can express most transverse modes as longitudinal modes along another axis, right? What happens when you get to the Gamma point? You, presumably, still have three orthogonal modes. How does the notation differentiate between them?

Come to think of it, there's a more basic problem that I have. As you said, every point in k-space describes a normal to some family of planes, so we are talking of actual palpable directions in real space, right? I can pick any three arbitrarily small numbers, and, so long as long as at least one of them is not zero, the resulting set of Miller indices will describe some set of planes of fixed orientation. Again, what happens when you get to the Gamma point? So far as I can tell, (0 0 0) points everywhere at once. A set of planes infinitely far removed from each other can still, presumably, have a fixed orientation. How do you pin it down?

8. Jan 7, 2007

### Gokul43201

Staff Emeritus
If I understand you correctly your question is how [0,0,0] is a lattice point in the reciprocal lattice, when there's no such thing as a (0,0,0) plane in the real lattice. Before I attempt to answer this question directly, let me note that the usefulness of the construction of the reciprocal lattice to physicists is better seen through its construction from lattice points in the real lattice through the requirement that $e^{iK \cdot R}=1$.

The construction of the reciprocal lattice from normal vectors to planes in the real lattice and their interplanar spacings is used in some textbooks to teach you what a reciprocal lattice is, but I don't know what the usefulness of this visualization is. I haven't even actually devoted any thought to reconciling the two contructions with each other. Nevertheless, from the point of view of this second construction, I believe the point [0,0,0] in the reciprocal lattice is no more than the origin, from which other reciprocal lattice points are contructed.

In the band pictures, you must note that the [0,0,0] point in k-space is no different from any $[n_1a^*,n_2b^*,n_3c^*]$ point (due to the periodicity of the reciprocal lattice, or putting it more bluntly, since $e^{i(k+K) \cdot R}=e^{ik \cdot R}e^{iK \cdot R}=e^{ik \cdot R}$), and it is for this reason, that the reduced representation is used.

Last edited: Jan 7, 2007
9. Jan 8, 2007

### sphericalCat

Thanks, once again, for the reply. Again, all makes sense, but... what i want to know is, basically, this:

1) Any given k-point can describe three normal modes, which, depending on symmetry, may or may not be degenerate. Is there a customary way of labelling these individually. My book (Hook and Hall) doesn't seem to have much to say on this.

2) Can you always express a transverse mode as an longitudinal mode along an orthogonal direction? If you can, then, once again, what do you do once you get to the Gamma point, where you still have the three orthogonal components but no specified axis to relate them to.

10. Jan 9, 2007

### Gokul43201

Staff Emeritus
I think a quick recap of what a phonon wave-vector is may be useful. If the following is trivially basic, it is because I'm not sure what your level of reading and understanding of this area is, mostly since you are from another area.

Let's look at different vibrational modes in a 1D lattice (along the x-direction) of basis 1 (meaning there's only one type of atom in the lattice) and lattice parameter d (where o is a real space lattice point, and the arrows represent the amplitude and direction of displacement of the lattice points from their equilibrium positions, at some specific point in time):

Code (Text):
o---> o-->  o->   o   <-o  <--o <---o  <--o   <-o     o     o->   o-->  o--->
This mode has a wavelength, $\lambda= 12d$, making the magnitude of its wave-vector (i.e., its wavenumber), $k=2\pi/\lambda = \pi/6d$. Furthermore, we can see that this is a longitudinal mode as the displacements are along the same direction as the propagation (i.e., the x-direction). For a 1D lattice, you can have an infinite number of transverse modes, with the displacements along any of the equivalent normal directions to the x-axis. Any transverse mode thus has an infinite degeneracy. What distinguishes a transverse mode from a longitudinal mode for a given wavelelength, is the energy of that mode.

What this snapshot does not convey, is the direction of propagation of the mode. Well, since it's a 1D lattice, you know that the propagation must be along the x-direction, but you don't know if it's along the positive x or the negative x. You can only know that by taking successive snapshots at infinitesimally separated intervals in time. In essence, this series of snapshots conveys all the information given in the solution to the wave equation for the phonon mode.

However, keep in mind that simply knowing the value of the wave-vector $\vec{k}$, for a particular phonon mode tells you nothing about whether the mode is transverse or longitudinal - it only tells you which way the phonon is propagating and how long its wavelength is.

Next, notice what you have with two longitudinal modes in the same 1D lattice with wavelengths $\lambda =2d~,~~ \lambda = 2d/3$.

The first one, with $\lambda=2d~$ looks like this (I've scaled it to look at fewer points):

Code (Text):
o-->                  <--o                        o-->                     <--o
For the second one, with $\lambda=2d/3~$, I'll thrown in additional representative points (x) at every half-wavelength (or every d/3), which are not real lattice points, but to show what the amplitude would have been, were an atom located at such a point:

Code (Text):
o--> [color=red]<--x        x-->[/color] <--o        [color=red]x--> <--x[/color]       o--> [color=red]<--x           x-->[/color] <--o
Notice that by looking at the displacements of the real lattice points (or atoms), there is no way to distinguish the above two modes, with clearly differing wavelengths (and hence, different wave- vectors). The wavenumbers of these modes are $~k_1=\pi/d~$ and $~k_2 = 3\pi/2d = \pi/d + 2\pi/d~$. The point of this exercise was to illustrate (visually) that modes differing by a wave-vector of magnitude $~2\pi/d~$ and directed along a basis vector (in other words, waves differing by a wave-vector equal to a reciprocal lattice vector), are identical to each other. This is the motivation for the reduced representation - that you can describe the behavior of all the modes, simply by describing the behavior of modes within any single interval of size $~2\pi/d~$.

Hopefully, the above explanation has convinced you that this is not possible. Transverse and longitudinal modes are fundamantally different (from the relationship between individual displacements and propagation direction). So, it's meaningless to try to "express a transverse mode as a longitudinal mode along an orthogonal direction". While in both cases, the displacement of the atoms happens in the same direction, their wave-vectors are actually different (normal to each other), making them solutions to different wave equations.

Last edited: Jan 9, 2007
11. Jan 15, 2007

### sphericalCat

Just, finally, finished reading your reply from beginning to end. Thank you very, very much. It has cleared most of the mess in my head. Still a bit confused abot what happens at the Gamma point, where there is no finite wavelength or fixed direction of propagation. Do you just define directions of displacement relative to some arbitrary crystallographic axis?

12. Jan 15, 2007

### Gokul43201

Staff Emeritus
In a cubic lattice (for instance) the k-space point (0,0,0) is no different from the k-space point $(2n \pi/a,2m \pi/a,2 l \pi/a)$ (n,m,l integers). The phonon with wavenumber 0 (or infinite wavelength) is no different from a phonon with wavenumber $2n \pi /a$ (or wavelength=a/n). In a 1D lattice, this mode has all the lattice points at the same displacement (or no phase difference) at any instant (i.e., it's an oscillation of the entire chain).

Code (Text):
o --> o --> o --> o --> o --> o --> o --> o -->
Clearly, for a lattice of basis 1 (as depicted above), the phonon with k=0 has no potential energy (since there's no relative displacement), hence, $\omega(k=0)=0$ for a Gamma point phonon in a lattice of basis 1. If the basis is greater than 1 (i.e., more than 1 atom per Bravais lattice point), you can then have relative motion between the two atoms (that make up the basis, see below) even at k=0, giving rise to the optical phonon branch that has non-zero energy at the Gamma point.

Code (Text):
o --> <-0  o --> <-0   o --> <-0  o --> <-0  o --> <-0  o --> <-0
(o = atom 1, 0 = atom 2)