Phonon Modes and Vanishing Eigenvalues in Periodic Lattices

[Derivator]

Dear all,

in these http://pages.unibas.ch/comphys/comphys/TEACH/SS04/course.pdf" lecture notes, the author says on page (0-120):
http://img15.imageshack.us/img15/615/capturena.png It is not obvious to me, why due to the translation invariance of the energy 3 eigenvalues of the D_IJ matrix have to vanish. Could someone explain it to me, please?

It is also not obvious to me, why the 3 rotations are not taken into account. Due to this sentence on page 123/124:

In contrast to the case of periodic boundary condition where the matrix D had 3 zero eigenvalues due to the 3 translations, the matrix D has now 6 zero eigenvectors (unless the molecule is diatomic in which case only 2 rotations exist).

I assume this has something to do with the periodic boundary conditions. But I also don't understand this.derivator

[1] http://pages.unibas.ch/comphys/comphys/TEACH/SS04/course.pdf

 

[Bill_K]

Derivator, I must say these lecture notes manage to take a really simple subject and make it sound complex! He's talking about the vibrations of a lattice with nearest-neighbor coupling between the particles. As usual with a coupled system you want to find its normal modes, and you do this by diagonalizing the coupling matrix. Each mode will have its own eigenvalue, which tells you the oscillation frequency. Well his point is that three of the modes can be easily identified. If all the particles move by the same amount in the same direction, their relative positions have not been changed and so the restoring force is zero. This corresponds to a translation of the entire system.

As far as the rotations go, you're right in that it has to do with the periodic boundary conditions. A rotation of the system is not allowed, since the displacement would increase indefinitely as you went away from the rotation axis, and this would violate the periodic boundary condition.

 

[Derivator]

Derivator, I must say these lecture notes manage to take a really simple subject and make it sound complex! He's talking about the vibrations of a lattice with nearest-neighbor coupling between the particles. As usual with a coupled system you want to find its normal modes, and you do this by diagonalizing the coupling matrix. Each mode will have its own eigenvalue, which tells you the oscillation frequency. Well his point is that three of the modes can be easily identified. If all the particles move by the same amount in the same direction, their relative positions have not been changed and so the restoring force is zero. This corresponds to a translation of the entire system.

...and because they move all by the same amount in the same direction, there are no relative movements, thus no vibrations, thus the frequency has to vanish?

As far as the rotations go, you're right in that it has to do with the periodic boundary conditions. A rotation of the system is not allowed, since the displacement would increase indefinitely as you went away from the rotation axis, and this would violate the periodic boundary condition.

sorry, i don' t understand this. If the entire system rotates, there should be no difference in the energy at all, or?

 

[SpectraCat]

...and because they move all by the same amount in the same direction, there are no relative movements, thus no vibrations, thus the frequency has to vanish?

Yes. Doesn't that make perfect sense?

sorry, i don' t understand this. If the entire system rotates, there should be no difference in the energy at all, or?

Well, yes .. but in the lattice case, remember you are using periodic boundary conditions to simulate a lattice of infinite extent. So, what happens if you rotate the lattice? Some of the atoms inside your PBC's get moved outside of them, and that is not allowed. However, there are 3 modes (probably the 3 lowest lying modes) that *would* correspond to rotations, if they were allowed. I am not sure if these have any physical significance .. I think they are just artifacts of the calculation method.

One other point. You posted the following quote in your OP:

In contrast to the case of periodic boundary condition where the matrix D had 3 zero eigenvalues due to the 3 translations, the matrix D has now 6 zero eigenvectors (unless the molecule is diatomic in which case only 2 rotations exist).

That part in red, while true, is not a complete description .. in fact *any* molecule with a linear structure (including diatomics), will only have two rotational degrees of freedom, and thus will have 3N-5 vibrational DOF.

 

[Derivator]

Well, yes .. but in the lattice case, remember you are using periodic boundary conditions to simulate a lattice of infinite extent. So, what happens if you rotate the lattice? Some of the atoms inside your PBC's get moved outside of them, and that is not allowed. However, there are 3 modes (probably the 3 lowest lying modes) that *would* correspond to rotations, if they were allowed. I am not sure if these have any physical significance .. I think they are just artifacts of the calculation method.

But in the case of periodic boundaries, according to the lecture notes, you reduce the degree of freedom only by 3, whereas in the case of free boundaries, you reduce it by 6.

If rotations are not allowed in the periodic lattice, why do you only reduce by 3 degrees of freedom (the translational degrees)?

Or is the point, that because periodic boundary conditions are not rotationally symmetric, angular momentum is not conserved and thus, the degree of freedom is reduced only by 3 and not by 6?