From WolfWikis

Phonons

Transverse and longitudinal

Before we allow our hydrogen atoms to fly away let us consider the vibrational states of our 1D wire of 3D objects. We can do that in a classical way, but it is easily extended into a quantum mechanical picture. Let us just assume that the interaction between two neighbors is a simple spring of the harmonic oscillator type. That means that any deviation (lengthening or shortening) of the ideal neighbor to neighbor distance will increase the potential energy and lead to a restoring force F that tries to put the hydrogen atoms back into place.

If we consider the forces acting upon atom #n we have two contributions from its left and its right neighbor:

F = F_left+F_right = -k_spring[u_n-u_n-1]-k_spring[u_n-u_n+1]

The meaning of u depends a little on whether we allow the atoms to move in the direction of the chain or in the directions perpendicular to it, resulting in two different kinds of motions, a transverse and a longitudinal one, but in either case we are essentially talking about what happens to the bond length between neighbors.

According to Newton we can describe the force as F = M.a, where M is the mass of atom #n and a is its acceleration, i.e. a= ∂²u_n/∂t². This means we end up with a system of classical differential equations:

M.∂²u_n/∂t²= k_spring[u_n+1+u_n-1-2u_n]

Fortunately we already know what kind of functions to try and solve this system with, because in its equilibrium state the chain has full translation symmetry. The solutions are very Bloch-like:

u_n = A exp[i(qna-ωt)]; where we have written the spatial wave vector as q and the angular (temporal) frequency as ω and 'a' is the unit translation vector that takes one atom into its neighbor in the equilibrium state of our chain.

There is a close relationship between the temporal and spatial frequencies as we can see if we substitute the solution back into our equations:

Mω² = -k_spring(exp[iqa]+exp[-iqa] -2)= 4k_springsin²(qa/2).

A bit of math will give:

ω= 2√[k_spring/M].|sin(qa/2)|

This expression links the temporal frequency ω with the spatial one (q). As we said before the speed of your car was the link between the regular spacing of lamp posts (spatial frequency) and how often you see them pass by (the temporal one). We can therefore conveniently introduce the ratio of the two frequencies as a new variable:

The (phase) velocity v= ω/q = 2√[k_spring/M].a.sin(qa/2)/(qa/2) which approximates to ≈ √[k_spring/M].a for qa<<1

For low values of q, i.e. for waves with a long wavelength the phase velocity is essentially constant √[k_spring/M].a therefore.

How phonon bands split into acoustic and optical phonon branches

Cell doubling

So far we have assumed that each unit cell contains a single atom, so that a unit translation takes an atom into its neighbor. In that case the (longitudinal) phonon at qa= +1/2 represents a motion where two adjacent atoms either move towards each other or apart. As long as this is a dynamic process, this is simply one of the ways our chain can vibrate. But suppose the distortion becomes static. In that case we get the pair formation we talked about in the previous topic. Such a phonon that becomes a static permanent distortion of the translation symmetry of the chain is said to go soft.

In case the atoms are hydrogens they may well fly away as diatomic molecules, but suppose there are other bonds to keep the chain intact. What would happen to our phonons?

From a symmetry point of view we have doubled our unit cell, which would introduce a new Brillouin edge at what was qa=1/4. The phonon modes outside the new boundary would become aliases of the ones inside.

The one at the old edge that went static (soft) now transforms as the totally symmetric irrep at k=0 and is no longer distinct. Towards the new zone boundary the states will mix and split up. The result is two branches: a lower one called acoustic and an upper one called optical.

Notice that acoustic phonons result from long-wavelength waves (qa=small) for which the content of the individual cells are more or less in phase which each other.

For optical phonons the phases of atoms in a cell are often opposite in sign. This means that the length of a chemical bond has to vary. In fact the optical phonon at qa=0 that resulted from the 'phonon gone soft' at q_olda= +1/2 represents a stretching of the bond length between the red and the blue atom in our diagram in all cells simultaneously. It stands reason that this will require more energy (have an higher frequency ω) than a wave where the red-and-blue molecules move aside a little in a wave pattern spanning many cells.

Classical or quantum mechanical?

Of course we have mixed two approaches here: the above derivation was entirely classical rather than quantum mechanical, but then we did invoke mixing of states.

However, if we do a new classical derivation, this time with two atoms in the unit cell, e.g. giving them a different mass (red and blue in the diagram) we arrive at a system of coupled differential equations that solves into two bands as shown. Vice versa we can take our classical phonon bands and turn them into a quantum mechanical model precisely the way the classical spring was turned into the well-known harmonic oscillator model. We simply apply the same principles for each mode separately. ^[1] The energy becomes quantized in units of ћω for each phonon mode.

Being quantized we could look upon phonons as (pseudo-)particles, much like we do for photons. The only important thing to keep in mind is that phonons are bosons. That means that each mode can contain an arbitrary number of the same phonons, rather than being subject to the Pauli exclusion principle. In that sense a diagram of phonon bands may look like an electronic band structure but the latter has a Fermi level up to which all states are full. For phonon bands there is no such thing

Exciting phonons

Thermal phonons

Phonons can be generated in a variety of ways. First of all at finite temperatures there are always phonons present, because of thermal motion.

Infrared radiation

It is also possible to excite them with radiation, but there is a bit of a problem. Nature has two important laws:

Conservation of energy E

Conservation of momentum p

As the k-vector of a phonon represents its momentum the dispersion diagram represents energy on the vertical axis and momentum on the horizontal. For a photon we also know the relationship between these two quantities. It is actually a very simple linear relationship:

E = h.ν

p = E/c --> E = c.p

p = (h/c).ν

Thus the photon E versus p relation can be added to the dispersion diagram as a straight line ('working line') with a known slope. Whenever this line intersects the phonon bands both conservation laws are fulfilled and absorption can occur. However, the horizontal scale of the diagram is in reciprocal angstroms, whereas the vibrational energy absorbed typically requires an photon in the infrared region, thus with a wavelength of microns. This means that the working line is almost vertical and only phonons very close to k=0 can be excited. In other words for IR absorption we need not worry about the dispersion of the phonon bands.

However, when we are dealing with e.g. a rough surface it is possible to combine scattering with absorption, in that case the scattering event possesses a k-vector of its own (momentum transfer) and transitions in other parts of the phonon band may become possible.

Neutrons

With a beam of neutrons we have a lot more control over both their energy and their momentum transfer and this is why inelastic neutron scattering is often used to map out the phonon bands of a single crystal

Electronic 'vibronic' transitions

It is possible to generate (or absorb) phonons during an electronic transition in a solid much like the vibronic transitions of a molecule. Again momentum matching plays a large role. An electronic transition for which the k-vector remains the same ('straight up' or direct transtions) do not require mediation of phonons but if the transition involves a change in k-vector a phonon must iether be created or absorbed during the process to make sure momentum is conserved. These are indirect transitions. Their intensity tends to be more temperature dependent because at higher temperatures more phonos are available for abosorption.

Other mechanisms

There are many other ways to generate a phonon, e.g. by mechanical means (hit with hammer), magnetic means or from the decay of other phenomena like plasmons.

1. ↑ Chapter 6.1.2 Basic semiconductor physics Chihiro Hamaguchi Springer ISBN 3-540-41639-0 2001