Photons and Linewidths
Most recent answer: 05/17/2013
- Aman Kanojia (age 22)
India
This is a brilliant question. You're absolutely right that if the photon had an exactly defined energy, then its wavefunction would have to be infinitely spread out. So it must have some spread in energies. How can energy be conserved then if it takes away the energy lost in a transition between two well-defined energy levels?
The secret is that the energy levels of the atom or molecule are not absolutely sharply defined. In particular the excited state (e.g. 2p for hydrogen) is not quite a true energy eigenstate, i.e. state with exactly defined energy. You can tell that must be right because exact energy eigenstates never do anything, so one would not emit light! In the full quantum mechanical expression for the energy, including the quantum version of the electromagnetic field, those excited states actually represent a narrow band of energies. The emitted photons share that range, thus conserving energy. That allows them to have a finite extent in space.
In many cases, a less fundamental solution applies. The atoms may be colliding and so forth, which already makes the situation more complicated and creates an energy range. Even in the ideal case, however, there's some little linewidth to those transition lines, keeping things consistent.
Mike W.
(not yet checked by Lee, who's traveling)
(published on 05/17/2013)
Follow-Up #1: why do excited-state electrons emit photons?
- broccoli (age 22)
singapore
That's a very nice but very hard question. We just answered one that's closely related, so I've put it in that thread. Please follow-up if you want more explanation.
Mike W.
(published on 05/27/2013)