# Phonon-Phonon Coupling (Different Wave Vectors)

 P: 5 In a perfect periodic crystal, do lattice vibrational modes corresponding to different wave vectors (k-vectors) interact with each other? I'm asking with reference to anharmonic lattice-dynamics calculations of a time-independent nature. For example: consider a linear, monatomic chain with a primitive basis of one atom. Build a supercell containing 3 primitive cells. Calculate lattice vibrations using periodic boundary conditions at the two edges of the supercell (lattice constant is now triple the primitive lattice constant). The primitive cell had 1 degree of freedom, and the supercell will have 3 (3-1 translation = 2 vibrational dof). One of the supercell's vibrational modes will correspond to wave vector $k=\frac{1}{3}$, and the other to $k=\frac{2}{3}b$, where $b$ is the reciprocal lattice vector for the chain. $k$ and $b$ are defined relative to the primitive cell. Note that in speaking of modes in the supercell which "correspond to" particular wave vectors, I could have more formally said that there is a set of wave vectors which are commensurate with the supercell. Further clarification: if we had a linear, diatomic chain then the 3xsupercell would have 2 vibrational modes each for the wave vectors: $0b, \frac{1}{3}b$ & $\frac{2}{3}b$, and one could imagine mode-coupling between the two vibrational modes for $k=\frac{1}{3}b$, say. Does one also include coupling between modes corresponding to different wave vectors? The Question: In a calculation including mode-coupling (evaluating the potential energy surface of a nucleus as a function of 2 vibrational mode coordinates $V=V(Q_1,Q_2)$ - $Q_1,Q_2$ are vibrational normal coordinates - rather than just $V=V(Q_1)+V(Q_2)$), is it correct to couple the modes from different wave vectors? If not, is there a physical/mathematical reason why one refrains from coupling different wave vectors? It seems reasonable to me, as I just imagine coupling modes of different wave vectors as superposing two plane waves. To be clear: I'm trying to avoid a discussion of multi-phonon scattering, because I'm only interested in an "equilibrium" snapshot in time / the "mean" vibrational state of the perfect crystal.
 Sci Advisor P: 3,628 As the quadratic part of the Hamiltonian is diagonalized yielding the phonons, the lowest order of phonon-phonon coupling is trilinear, i.e. it involves 3 phonons. That's an effect of anharmonicity, i.e. also an equilibrium effect. With a diatomic basis, I don't see why you don't also consider k=0 for the optical phonons. So you could have k1=1/3, k2=2/3 and k3=0 (in units of b). I don't quite understand your mode coupling example.
P: 5
 Quote by DrDu With a diatomic basis, I don't see why you don't also consider k=0 for the optical phonons. So you could have k1=1/3, k2=2/3 and k3=0 (in units of b). I don't quite understand your mode coupling example.
You're correct; for completeness I should include k3=0. I omitted it in the diatomic case, because it is just a translation in the monatomic case. I apologise if that made things difficult to follow. I'll edit the post.

 Quote by DrDu As the quadratic part of the Hamiltonian is diagonalized yielding the phonons, the lowest order of phonon-phonon coupling is trilinear, i.e. it involves 3 phonons. That's an effect of anharmonicity, i.e. also an equilibrium effect.
What I mean by coupling: imagine a potential energy surface constructed by moving the nuclei along the $N$ normal mode coordinates $\left\{Q_i\right\}, i=1..N$, in a molecule with $N$ vibrational degrees of freedom. If the modes are considered completely independent (no coupling), then we might write:

$V(Q_1,Q_2,...,Q_N) \approx \sum_{i=1}^{N}V(Q_i)$
$V(Q_i) := V(Q_i,\left\{Q_l\right\}) , Q_l = 0, \forall l \ne i, l \in \left\{1..N\right\}$

If we consider pairwise coupling, then:

$V(Q_1,Q_2,...,Q_N) \approx \sum_{i<j}V(Q_i,Q_j)$
$V(Q_i,Q_j) := V(Q_i,Q_j,\left\{Q_l\right\}) , Q_l = 0, \forall l \ne i,j, l \in \left\{1..N\right\}$

And so on in that fashion for higher order mode couplings. Of course, that's for a molecule. For a supercell in a periodic solid, I'm asking whether or not we should consider mode couplings when the modes correspond to different wave vectors. That is, should we write (for pairwise coupling):

$V(Q_1,Q_2,...,Q_N) \approx \sum_{k}\sum_{i^{k}<j^{k}}V(Q_i^{k},Q_j^{k})$
$V(Q_i^{k},Q_j^{k}) := V(Q_i^{k},Q_j^{k},\left\{Q_l\right\}) , Q_l = 0, \forall l \ne i^k,j^k, l \in \left\{1..N\right\}$

? (where the superscript $k$ indicates that supercell mode corresponds to wave vector $k$ - the sum over $k$ includes all wave vectors commensurate with the supercell) That would be not coupling modes from different wave vectors. If I were to couple different wave vectors, I would retain the molecular formula, only with periodic boundary conditions.

I hope that clarifies my question :)

If not, what the above equations mean (in words):

For no coupling, one constructs a potential energy surface with displacements of nuclei along one vibrational normal coordinate at a time, with all other vibrational coordinates held at their "equilibrium" values. For pairwise coupling, one allows two vibrational coordinates to vary at the same time (take all pairs). In the $N$ mode coupling situation, one would construct the entire potential energy surface by allowing all $N$ coordinates to vary at the same time. Then the approximation would become an equality.

 Sci Advisor P: 3,628 Phonon-Phonon Coupling (Different Wave Vectors) This certainly gives you couplings of q's with different k.
 P: 5 DrDu, that was my first thought as well. To complicate matters, I've found a respectable paper where this appears to be contradicted, and I've been having trouble with my calculations also (results tend to become unphysical when I do couple different k-vectors). This is why I'm asking, because there seems to be more to it. Reference: Physical Review B 78, 224108 (2008) "Effect of Quantization of Vibrations On The Structural Properties of Crystals" Therein, under the heading, "Choice of Vibrational coordinates in the VSCF" - VSCF meaning Vibrational Self-Consistent Field methodology for calculating vibrational frequencies - the following was stated, without justification: "... We now analyze the question of which vibrational coordinates are optimal for the VSCF approximation in the case of the four-atom cell. The three modes are zone-center optical phonon, which exhibits a double well, and the zone-boundary acoustic and optical phonons, which are practically harmonic. In principle, any linear combination of these modes is possible. Nevertheless, since the modes at different k-vectors don't mix, we fix the double-well mode and optimize the choice of modes in the two-dimensional boundary subspace." That was with reference to a 1 dimensional F-H (hydrogen bonded) chain, on which VSCF calculations were performed. The "modes" mentioned therein, are represented as eigenvectors in the VSCF; the eigenvectors come from ordinary harmonic approximation (dynamical matrix diagonalisation) calculations. The 4 atom cell is the periodic unit, because the structure is chosen like this: ... F - H - - - - F - H - - - - F - H - - - - F -H ... Perhaps they mean something different by the mixing of modes than what I mean by the coupling of eigenvectors, though it sounds exactly the same. Perhaps there is some reason that the different wave vectors don't mix for a 1 dimensional wave that is obvious to everyone else? Or maybe there is actually an important general principle here. Sorry for the long-winded explanations. If you think I should start a new thread, I'd be happy to, but I'm struggling to make this question more concise. Thanks a lot for reading all this. I'm very grateful you've taken the time.
 Sci Advisor P: 3,628 Probably I was wrong or misled by your notation. I would split the index of the Q into two indices where the first one labels the unit cell (or molecule). So let's try to clear this up. We can introduce ##q_{kl}=\sum_j Q_{jl} \exp i j k ## or ##Q_{jl}=1/2 \pi \int dk q_{kl}\exp -ikj##. Now your quadratic potential is ##V=\sum_{iljm} V_{iljm} Q_{il} Q_{jm} ## where ## V_{iljm} \propto \delta_{ij}##. Doing the summations you get something like ##V=\sum_{klm}\tilde{V}_{klm}q_{kl}q_{-km}=\sum_{klm}\tilde{V}_{klm}q_{kl}q^*_{km}## which is diagonal in k, due to the ##\delta_{ij}##. This can be generalized to potentials which depend only on j-i, i.e.translation invariant couplings between different molecules.
 P: 5 Thanks very much for your reply, which is an introduction to a completely new idea for me. I've done my best to follow it, but I'm not confident I've understood entirely. Here's what I have: $q_{kl}=\sum_{j}Q_{jl}exp(Ijk)$ $Q_{jl} = \frac{1}{2\pi}\int dkq_{kl}exp(-Ikj)$ where $q_{kl}$ is some discrete Fourier transform of $Q_{jl}$, which (in turn), is the eigenvector for vibrational mode $l$ of a primitive cell, broken up into the primitive cell indices $j$. The index $l$ runs over all vibrational modes in a primitive cell, and the index $j$ runs over all primitive cells in the supercell. $k$ is the wave vector (1 D wavenumber in this case) and $exp(Ijk)$ is just the 1 D version of the phase factor, since $k$ is expressed in fractional coordinates of the reciprocal lattice vectors (reciprocal to the primitive cell real space lattice vectors). I've used capital I as $\sqrt{-1}$. All good so far? You then write the quadratic potential as the summation over $i,j,l,m$ of $V_{ijlm}Q_{il}Q_{jm}$, where $V_{ijlm}$ is ...? Is this a 2nd order partial derivative, or is it $V(Q_{il},Q_{jm})$ with all other $Q_{ab} \ne Q_{il},Q_{jm}$ held at equilibrium? The Kronecker delta: because you've split up the eigenvectors into their primitive cell components, simultaneous displacement along eigenvectors belonging to different primitive cells breaks primitive cell translational invariance (though not supercell translational invariance), and hence the contribution to the potential must be zero for $i \ne j$. I'd really like clarification here, because I feel I'm starting to stretch a bit. Simplifying gives: $V = \sum_{ilm}V_{ilm}Q_{il}Q_{im}$ where $V_{ilm} := V_{ilim}$ Now jumping to your formula, I can see that (by analogy with earlier discrete Fourier transform): $q_{kl}q_{-km} = \sum_{j}Q_{jl}exp(Ijk)Q_{jm}exp(-Imk)$ and hence: $Q_{il}Q_{im} = \int dkq_{kl}q_{-km}exp(-Ik(i-m))$ giving: $V = \sum_{ilm}\int dk V_{ilm}q_{kl}q_{-km}exp(-Ik(i-m))$ Since it is only the difference between cells $i$ and $m$ which is important, one can use $m' := i-m$ and, since the supercell is a finite approximation to the infinite lattice, one must replace the integral over $k$ with a summation over wave vectors commensurate with the supercell. $V = \frac{1}{N_k}\sum_{klm'} \tilde{V}_{klm'} q_{kl}q_{-km'}$ with $\tilde{V}_{klm'} := Vexp(-ikm')$ and $N_k$ the number of wave vectors commensurate with the supercell. Returning from $m'$ back to $m$ (we remember our reference cell $i$ or consider it shifting by a phase factor), I get something proportional to your final expression: $V = \frac{1}{N_k}\sum_{klm} \tilde{V}_{klm} q_{kl}q_{km}^{*}$ which is indeed diagonal in $k$. What this means: we have built a pair-coupled potential due to the vibrations of different primitive cells in a supercell, without mixing wave vectors, whilst maintaining translational invariance?
 Sci Advisor P: 3,628 This sounds all quite reasonable. Take in mind that my post was only a crude sketch, so it is up to you to work out all the 2π's and sums instead of integrals. V_ilmn was meant to be the Taylor coeficient for the second order term.

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