# Is emission/absorption of a photon lossy?

I recall vaguely that energy is absorbed/radiated in packets called quanta. Quanta were what are now known as photons.

What I'm curious about - Is absorption/radiation vis-a-vis photon lossy? Do the total number of photons exactly match the energy acquired/released?

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I'm flummoxed. I can't make out the right answer. Would anybody care to make a recommendation please? – Everyone Sep 6 '12 at 12:55

It is not a lossy process. For there to be a loss at one place would require there to be a gain somewhere else. When the atom releases the photon there would be recoil of the atom that would contain some energy, but 100% of the energy lost by the atom would be gained by the photon.

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Absorption is often lossy. A photons energy is inversely proportional to wavelength, if that energy is larger than the bandgap of the material absorbing the photon then the excess energy goes into heat. In Silicon, as an example, if a green photon is absorbed and if the electron hole pair is allowed to recombine, there is a photon emitted that has a wavelength that corresponds to the bandgap of Si of 1.12 eV.

Emission is not lossy.

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I'm not sure I would say that absorption is lossy. The energy gain of the absorber is always equal to the energy of the incident photons. I'm guessing you're thinking of fluorescence or phosphoresence where the emitted light has a lower frequency than the absorbed light, but describing this as a lossy process is a bit misleading. – John Rennie Sep 1 '12 at 8:29
It is not at all misleading. There is always conservation of energy. If you apply that uniformity the answer is always NO - there is never any loss. So why ask the question? I assumed the person didn't want an tautological answer and that had some fundamental understanding and was asking amore interesting question. So I'll be more explicit and rephrase the question as " is Absorption and re-emission lossy or not". Keep in mind that if you have emission then there must be an electron involved - reflection is something different (even though it does involve electrons). – placeholder Sep 1 '12 at 20:51
Does that mean the photon would give it's energy even in the event the 'target' is incapable of absorbing energy at the photon's frequency? – Everyone Sep 11 '12 at 19:19
Think about what you mean by "incapable of absorbing". That means the 'target" is transparent to that wavelength. In an ideal world the answer is no, in real life imperfections will lead to some loss. And if there is an index of refraction change (almost certainly is) then a certain number of photons will be reflected and a certain amount transmitted at the boundary.-> fresnel eqn's. – placeholder Sep 12 '12 at 16:28
Or if the material is rough (think of powdered glass) then the photon energy will go purely into heat. – placeholder Sep 13 '12 at 2:42

The emission and absorption can be lossy, but it does not have to be.

While in the overall picture energy is conserved other processes can either increase or decrease the photon energy. In practice this is used for cooling atoms with a laser beam. This doppler cooling uses photons to reduce the speed of gas atoms, the incoming photon energy is split between the kinetic energy of the atom to slow it down, the remaining energy is emitted as another photon with less energy.

As a side note: in the future we might be able to cool larger objects with this method, as the photons can be used to 'destroy' phonons and therefore cool a solid transparent material.

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Photon number does not have to be conserved in light-matter interaction processes. For example, take the light storage experiments in hot/cold atomic vapor, where a pulse of light is converted to atomic coherence (definite phase relationship between the ground states of the atomic system). Dissipation in the form of phonons is also possible.

Ultra-Slow Light and Enhanced Nonlinear Optical Effects in a Coherently Driven Hot Atomic Gas

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