Phonons are said to be the result of the quantization of crystal waves.Let the function [itex] u=u_0 e^{i(\vec{k}\cdot\vec{r}-\omega t)} [/itex] describe such a wave. Is it right to say that if we assume u to be the quantum mechanical wave function of a particle, that particle is a phonon? Thanks
The function u itself has nothing to do with quantization, it's just a general wave function. The quantization arises in situations when you have special boundry conditions that limit the actual values on ω or k, so that only certain frequencies for example are allowed. Now, if we look at one such allowed frequency mode (in the case this is quantized at all) , we can say that the smallest possible excitation is 1 phonon*. * though note that the mode could be in a superposition of being excited and not excited, so on average it could have smaller excitation energy, but if you have a device that measures the phonon number of suc a mode, then it would always register integer values.
Answer: no. What you have in a crystal is a bunch of atoms vibrating which are painful to describe as individual oscillators. It's far easier to describe collective oscillations of atoms, called modes (many atoms oscillating at once). It's even easier if you describe a bunch of atoms oscillating all at the same frequency - then it is a normal mode. It is normal because the atomic displacements in these oscillations (normal modes) diagonalise the dynamical matrix, thus forming a set of normal coordinates, or an orthonormal basis for describing your vibrations. This is a linear algebra thing. The diagonalised dynamical matrix (a mass weighted force-constant matrix - remember spring constants?) contains the angular frequencies of vibration along its diagonal. Now for phonons: It's very straightforward, but also terribly confusing. I'll follow Ashcroft and Mermin ("Normal Modes vs. Phonons" section of Chapter 23 entitled "General Form of the Lattice Specific Heat"). Basically we could talk about normal modes (which are specified uniquely by giving a wave vector and a branch index) being in the nth excited state, which means that we would use E = (n+1/2)hbar*omega. Alternatively, we can say there are "n phonons" of this type. It's just language. The reason people use that is because then it's a bit like the electromagnetic waves and photons. Photons are the quanta of the electromagnetic field (energy can only increase by jumps equal to the energy of a photon). Again, these come out of the math too (formulate Maxwell's equations in a vacuum - that is with no charge or current density - rearrange a bit and out pops a wave equation with velocity equal to the speed of light). The photon carries the electromagnetic information, and as the photoelectric experiments demonstrated, they are quantised in energy = hbar*omega. The phonon carries the "ionic displacement field" information - ions are nuclei of atoms in a crystal here. So you could "excite a phonon" by inelastically scattering a neutron off of an atom and the vibrational energy of the crystal goes up by hbar*omega. The zero point energy gives you the 1/2 bit - that's the energy when there are zero phonons. Conclusion: there is no phonon. It's a convenient construct of language. There are just vibrating atoms whose normal modes' quantised energy levels lend themselves nicely to a "corpuscular" (particle) description, even though they represent atomic oscillations under the influence of a lattice wave. So you can think of phonons kind of like energy/information packets which have memorised some transfer of energy into the atomic oscillations of the crystal (a neutron strike? a photon strike?), and a photon as some energy/information packets which have memorised some transfer of energy into oscillations of an electrical and magnetic field always perpendicular to each other, zooming through space at the highest speed we know of. Imagine having to say, "The last increase in energy experienced by the lattice wave with wave vector k and branch index..." - No. Just say, "That phonon there."