Phonons & Collective Oscillations: Locality Experiments

[NaOH]

I am struggling with the concept of phonons and if they should be considered (experimentally) to be treated as collective oscillations of a certain region instead of the whole crystal.

I know the title comes off as a little strange -- since phonons are not localised to anyone atom, but consider an experimental setup where we measure the phonon modes of a crystal, say neutron scattering, brillouin light scattering (BLS), or to lesser precision, raman spectroscopy.

In these techniques, the principle is that a particle, be it photon or neutron, scatters off the crystal, and excites/populates a phonon state.

This is where I start running into problems: as I understand, phonons are characteristic of the whole crystal, so it follows that the scattering event at point A (instantly?) causes the whole crystal to vibrate in such a way that is described by the created phonon... even if the process is not instantaneous, it would suggest that if I had a crystal big enough, the information that a phonon is created could potentially travel faster than the speed of light. Which is kind of problematic.

So I try to get out of this by saying that there is a certain "locality", or loci, around which I need to consider for collective oscillations around where the scattering event occurred to describe the phonon, from an experimental point of view, and take that the infinite lattice/whole crystal is an abstraction to simplify the problem. However, I am not sure (or know where else) to seek clarification on matters like these...

This question was kind of lodged behind my mind for almost a year whenever I think of MoS2, I think of http://www.researchgate.net/profile/BK_Tay/publication/237068085_From_Bulk_to_Monolayer_MoS2_Evolution_of_Raman_Scattering/links/00b7d52cb4f4fc9049000000.pdf and then this question. In the paper, raman spectroscopy is employed to study the energy dependence of phonon modes on the number of layers, and did so by initially having a multilayered MoS2 crystal, and then scraping off the layers, making something that looks like stairs (figure 1).
Therefore I reasoned (to myself) that if the phonon mode was indeed describing the whole crystal, then it should not matter where the laser is used, but indeed if there is a certain "locality" that is needed, then the layer dependence can be shown using the steps within the same MoS2 material.

Thank you.

 

[ZapperZ]

How about thinking of it this way: the interaction itself is local, but the phonon itself isn't!

This is because the interaction occurs where the particle of interest is located (even that concept in itself is a bit dubious). If we look at this within QFT picture, it is no different than an interaction of charge with virtual photons where these photons are the excitation of the EM field which is everywhere (non-local). Similarly, the phonon field is everywhere in the material and it is only the interaction that defines a "locality".

Zz.

 

[DrDu]

Phonons are not different in this respect from electrons. You can consider both localized and delocalized phonon states. In a crystal, only the latter are energy eigenstates while the former change with time.

 

[NaOH]

Thanks for the replies.

The second quantization formalism is a wonderful way to view this, but I have avoided referring to it in the topic post because I wanted a classical picture as well.

Piecing together the replies, is the following statement how I should interprete the scattering process, if the incident particle only interacts with one atom in the crystal?
Particle scatters off the crystal, slightly displacing an atom with the allowed energy (corresponding to a phonon). What happens to this displaced particle and how the energy is shared with it's neighbours and subsequently other parts of the crystal is not known (or at least not mapped out by BLS, neutron scattering, Raman scattering, etc.).

Phonons are not different in this respect from electrons. You can consider both localized and delocalized phonon states. In a crystal, only the latter are energy eigenstates while the former change with time.

I am not sure what you meant by delocalised phonon states and why they change with time? They are not superposed pure phonon states are they?

 

[DrDu]

I am not sure what you meant by delocalised phonon states and why they change with time? They are not superposed pure phonon states are they?

Any displacement $u(r_i,t)$ of an atom from its equilibrium position or combination thereoff represents a phonon. If only one atom is displaced, this is a localized phonon, but it will couple to other phonons and spread out with time. On the other hand you can form Bloch modes of these localized excitations and these will be eigenstates of energy and crystal momentum but will be delocalized over the whole crystal . I suppose you have only the latter ones in mind. You can express the localized phonons in terms of the delocalized ones and vice versa, the relation being basically a Fourier transform.

 

[NaOH]

Any displacement $u(r_i,t)$ of an atom from its equilibrium position or combination thereoff represents a phonon. If only one atom is displaced, this is a localized phonon, but it will couple to other phonons and spread out with time. On the other hand you can form Bloch modes of these localized excitations and these will be eigenstates of energy and crystal momentum but will be delocalized over the whole crystal . I suppose you have only the latter ones in mind. You can express the localized phonons in terms of the delocalized ones and vice versa, the relation being basically a Fourier transform.

Yes, my understanding of phonons is quite limited and in most undergrad texts, the phonon that are introduced only mention the kind which extend over the whole crystal. Where can I find more description about these "localised phonons"?

 

[DrDu]

I am not sure where to find such a definition, but this is only a rather unimportant question of nomenclature. What is important is the fact that you can represent a localized lattice deformation in terms of the extended phonons in terms of a Fourier transform. Hence a localized (in space and time) lattice deformation will be very broad in energy space. If the excitation goes on for longer (e.g. the driving electric field), the vibrational modes not in resonance with the excitation will die out. The energy distribution of the delocalized phonons becomes narrower and at the same time, the spatial extent of the deformation becomes larger and larger. This is not different from the excitation of water waves extending over a whole basin by a localized perturbation somewhere on the water surface.

 

[NaOH]

My concern with a wide energy space is that ultimately, I want to reconcile with spectroscopy and this will not allow me to observe the absorption peaks!

However, I suppose that I can find comfort that with an incident electric field that can drive for sufficiently long, the energy spectrum narrows and the absorption peaks will be obtained.

 

[DrDu]

My concern with a wide energy space is that ultimately, I want to reconcile with spectroscopy and this will not allow me to observe the absorption peaks!

However, I suppose that I can find comfort that with an incident electric field that can drive for sufficiently long, the energy spectrum narrows and the absorption peaks will be obtained.

Yes, this is the situation in classical spectroscopy with more or less monochromatic light.
However, these days you can do spectroscopy with exictation times in the femto and attosecond range and I suppose it is possible to watch the spreading of the initial phononic excitation.

 

[NaOH]

So I guess I should wrap things up with the math. The creation operator of a one phonon state is: $$\sum_{k} c_{k}a^{\dagger}$$
This means that one phonon, in general, is a superposition of pure phonon states which are the eigenstates of the normal mode.

This would mean that the scattering event creates a phonon, but with it an associated uncertainty of it's energy as well (maybe associated with line width).

However, constant $c_{k}$ is also a function of time, so while the interation time increases, the spread of $c_{k}$ decreases as well.