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I understand what are virtual photons and the difference between the real and virtual photons.

However, I am not able to clearly distinguish the difference between the 'phonons and virtual photons'. Both are involved in non radiative energy transfer.

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Could you give some context? Where did you read virtual phonons? – innisfree Jun 19 '14 at 7:51
@innisfree He didn't mention virtual phonons. – jinawee Jun 19 '14 at 9:20
@jinawee oops ok. – innisfree Jun 20 '14 at 11:04
up vote 3 down vote accepted

A photon is an excitation of a quantum field, which is classically $A_{\mu}$ governed by the Lagrangian,


which we may quantize canonically by the usual procedure, i.e. expanding the field as a plane wave, promoting the Fourier coefficients to operators, etc. A virtual photon does not necessarily satisfy,


i.e. the relativistic dispersion relation, as it is off shell. In Feynman diagrams, it appears as an internal line, for which we associate the usual propagator in the Feynman-'t Hooft gauge,

$$\frac{-i \eta_{\mu\nu}}{k^2-m^2}$$

Consider now a solid, which for our purposes is simply a lattice. Although the positions of the individual atoms are stuck, the sound waves due to the vibrations of the atoms may be treated with the same formalism of photons in statistical physics. We imagine sound waves in solids composed as discrete packets - these are the phonons.

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Thank you JamalS – QuantumGirl Jun 19 '14 at 14:19
Thank you @JamalS. The Problem I had was, the virtual photons also have longitudinal components as phonos. When the energy is transferred from donor to acceptor, it can emit a real photon or a virtual photon, (or the energy can be dissipated as heat-phonons). I was thinking of a way to differentiate the energy transfer mediated by virtual photons vs phonons. How a particular molecule decide whether to release a phonon or a virtual photon? – QuantumGirl Jun 19 '14 at 14:45
@user50724: Photons arise, in the end, from transitions to different energy states by the electrons of the constituent atoms of the molecule. A phonon we attribute to the vibration of the molecule - it is a quasiparticle, because the solid may behavely almost like a set of free particles. – JamalS Jun 19 '14 at 14:48

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