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Phonon, anyone?

  1. Jan 27, 2005 #1
    I have never encountered phonon before, so I am going to make a list of statements, and please be generous to point out any mistakes. Thank you.

    1. There is no new physics behind the idea of phonons: it is just an alternative way of doing things.

    2. When one says that there are, say, 5 phonons, it can be one of the following case: one atom oscillating with 5(h bar)w, 2 atoms with a sum of 5(h bar)w, and so on.

    3. Propagation of phonons means nothing more than the exchange of energy between oscillators.
  2. jcsd
  3. Jan 28, 2005 #2
    What do you mean here ??? Phonons are the QM-interpretation of energy-values. The atoms in a lattice can vibrate and this motion is expressed in terms of oscillators. A phonon is nothing else then an energy-quantum of such oscillators, just like the photon is an energy-quantum of EM-radiation (eg light)

    yes indeed. The phonon is ONE energyquantum [tex]E_{n} = (n + \frac{1}{2}) \hbar \omega[/tex]
    But beware that two harmonic oscillators will have an influence on each other's energy-levels. This is due to their mutual interaction.

    Yes, when a phonon is absorbed by an oscillator, this oscillator is in a higher energy-state. to lower the energy, just emit a phonon.

  4. Feb 1, 2005 #3
    Phonons are not single atoms oscillating. Phonons are excitations of the normal modes of a crystal, which in principle is of infinite size (no surfaces). Some phonons have very long wavelengths - in a truly infinite crystal there are phonons with frequencies in the audio range and wavelength of many meters.

    Propagation should the same as propagation of a classical wave on a string.
  5. Feb 1, 2005 #4


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    Nope. The "other" way of doing this does not contain the same set of information. Once one invokes phonons, then all (and I mean ALL) the rules of QFT comes in. This includes the self-energy terms that reveal themselves via the perturbative expansion. You do not get, for example, the strong and weak coupling description of superconductivity using the "other" description.

    [tex]5 \hbar \omega[/tex] is NOT 5 phonons. If you must equate the two different pictures, phonons are the normal modes of oscillations.

    You left out momentum and spin (even though phonons have spin of zero). It's hard to know what you mean by "propagation" here. Phonons (and any other QFT particles) are exchange particles that carry energy, momentum and spin. They don't really "propagate". They are the replacement of the classical field. However, their formalism (QFT) provides MORE fundamental information than the classical field formalism. In that sense, they are not necessarily simply an "alternative" description.

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