# Momentum conservation in phonon-photon scattering event

I am following a discussion from Kittel´s Introduction to Solid State Physics in the subchapter ‚phonon momentum‘:

We have two different conservation laws of momentum in a crystal:

1. Total momentum is conserved (if we include all processes i.e. also the center of mass motion).
2. Crystal momentum (pseudo momentum) is conserved up until a reciprocal lattice vector $$K_{AB}=K_A+K_B+G$$, with $$K$$ the crystal momentum and $$G$$ a reciprocal lattice vector.

If a photon hits a crystal and gets reflected, the momentum change of the photon is according to bragg’s law (or Laue equation): $$k’=k +G$$ where $$k$$ is the initial momentum and $$k‘$$ is the outgoing momentum of the photon. $$G$$ is a reciprocal lattice vector. This means that the photon gives some momentum to the crystal, whose center of mass starts to move (since a momentum $$\hbar G$$ is transferred).

In the case that a phonon (lattice vibration) gets produced or absorbed in this scattering process, the conservation of momentum looks the following: $$k‘=k+G+K$$ where K is the crystal momentum of the phonon.

My question is:

We said crystal momentum $$K$$ is not conserved. By above equation, then also $$k‘$$ (the real momentum of the photon) is not conserved. How can this be?

As an example: Let’s say we have two photons interacting with the crystal producing two phonons A and B. $$k^‘_A=k_A+G+K_A$$ $$k_B^‘=k_A+G+K_B$$

Now let’s say both phonons meet at a later point in time and a scattering event happens. Since crystal momentum is not a conserved quantity, the conservation of crystal momentum looks like: $$K_{AB}=K_A+K_B+G$$ so for example, if $$K_A$$ and $$K_B$$ would be large enough, this would be an umklapp process and the total crystal momentum $$K_{AB}$$ would be smaller then $$K_A+K_B$$. So where is the real momentum of the initial photon going? It cannot (again) change the center of mass motion, because phonon-phonon interactions are not external ‚forces‘, which are the only ones that can change the center of mass motion.

Comment: I think the problem is that some of the real momentum of the photon is converted into crystal momentum of the phonon. But why is this even possible if it does not conserve total momentum?

Crystal momentum is a quantum number, characterizing states in a specific crystal band (whether we speak of electrons/holes or phonons). That is $$k$$ and $$k+G$$ describe the same quantum state/excitation. While the analogy between crystal momentum and real momentum is useful and not completely unfounded, we have to keep in mind that the photon actually interacts with the whole crystal - the difference between the change in the crystal momentum and the momentum of the photon is the momentum transferred to the lattice as a whole, which is not accounted in the Block waves solutions.
• But why is it then possible to state $k‘=k+G+K$. Interpreting K as crystal momentum (even though it is a quantum number), it is not conserved at all and therefore k‘ (the real momentum of the photon) is not conserved, which is strange. Jan 26 at 10:26
• @Lockhart it is not true that it is not conserved at all - due to the crystal periodicity it does conserve... up to a reciprocal lattice vector - Kittel definitely demonstrates it somewhere. But it is not a true momentum. Whould it be easier to think of $k'=k+G+K$ as a selection rule, like those in atoms (where photon momentum goes largely in accelerating atom - hence laser cooling)? Jan 26 at 10:31
• @Lockhart Let us not mix energy and momentum. Think of an atom absorbing a photon $\Omega, k$: energy goes into moving electron from one orbital to another: $\hbar\Omega=E_f-E_i$, whereas the momentum is transferred to the atom as a whole: $p'=p+\hbar k$ (for simplicity I ignore here the angular momentum and spin, which are affected by photon polarization.) Jan 26 at 10:45
• @Lockhart (if I correctly udnerstand your notation) you choose $k$ and $k'$ to lie within the first Brillouin zone, whereas K is given, which determines the value of $G$ - the momentum transferred to the crystal as a whole. Without $G$, $k'$ might turn out to be outside of the first Brillouin zone. Jan 26 at 10:54