sci.physics.electromag

On Oct 31, 11:29 am, p.kins...@ic.ac.uk wrote:
> Thomas wrote:
> > When you speak of (single) photons, you can't speak of energies or
> > wavelengths.
>
> Yes, you can. You can define photons to be (single) excitations
> of a quantized EM field mode; these modes are solutions to
> the classical Maxwell's equations. If, for example, the modes
> are those in an optical cavity (resonator), then they have a
> specific energy and a specifc wavelength.
>
He may have been right concerning virtual photons. I don't think
the number, energy, or wavelength of virtual photons are good quantum
numbers. Remember the uncertainty relations related to photon
emission. The energy bandwidth of emitted radiation times the lifetime
of the excited state is equal or greater than Planks constant. The
energy bandwidth of the emitted photon corresponds to the uncertainty
in energy of the virtual photons.
Consider the absorption of a single real photon by an atom,
resulting in a transition of energy E_t. The Feynmann diagram
associated with the dipole interaction shows a straightforward
interaction, with no virtual photons. The diagram associated with the
quadropole interaction shows a transition to a still higher
intermediate state, involving two virtual photons with an uncertain
energy. The diagram associated with an octopole interaction involves
four virtual photons... There is more than one Feynmann diagram
necessary to describe a real transition precisely because there is an
uncertainty in the energy of a virtual photons involved. There is also
an uncertainty in the number of virtual photons involved. Why would
you need perturbation theory if virtual photons had a precisely
determined energy?