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Your replies to my question about being able to see a photon, from the side (answer, unanimous, “no”) have raised in me some additional questions. Would it reasonable to think that in consequence, the photon, from any angle but directionally “in front”, is a black body? I ask because I am accustomed to think of the photon as a “light” particle, when perhaps it might be more correct to think of it as a “black” particle. When an observer sees light from a photon, isn’t it true that in fact, what is actually being seen is the photon itself? When the photon enters the eye of the observer, what happens (3.1) to the observer? (3.2)to the photon? 3.1 The photon has no mass, so the observer/photon combination remains at original observer mass. However, the photon has energy, what impact on the observer done this energy input have?
3.2 Upon entering the observer, does the photon disappear? If not what else can it do?

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You might be misunderstanding the term "black body" here. A black body is something that perfectly absorbs and emits thermal radiation. So it can't possibly be used to describe a photon. Can you clarify what you meant? –  dbrane Jan 20 '11 at 22:27
    
I don't see your previous question, but rhodopsin does react when struck by a single photon. We wouldnt notice it though likely because of noise suppression by the brain. –  Gordon Feb 6 '11 at 19:01
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Photons are not particles in the way that marbles or billiard balls are particles. You're never going to be able to "see" them by bouncing some other thing off them, because they will always move at exactly the speed of light. You could, in principle, reconstruct the path of a photon through some medium by looking at the perturbation of the medium caused by its passage, but in that case, it's debatable whether the photon that leaves the medium is really the "same" photon that entered, as the scattering process in the medium would involve repeated absorption and emission cycles. Which also points to one of the risks of talking about photons as discrete particles following a definite path, namely that they can be created and destroyed very easily through interactions with matter, and thus their identity is a little fluid.

When a photon strikes the eye of an observer, or the active surface in a detector, its energy goes into the medium. In an electronic detector, it usually knocks an electron loose in some manner, which then triggers some sort of cascade process that produces a measurable electronic pulse. In the eye, it causes some reconfiguration of light-sensitive molecules, which eventually produce a signal in the optic nerve that gets registered as a "flash" of light.

In the case of an atom absorbing a photon, I suppose it is true that the apparent mass of the atom in its excited state would be infinitesimally higher than the ground-state atom's, by an amount $E_\gamma / c^2$, but that's such a trivial effect that nobody bothers to deal with it. To give you an idea of the scale, a typical atomic transition in the visible light range has an associated energy of a few eV, while the rest mass of a hydrogen atom is about 1,000,000,000 eV. You would never notice the effect (unless you're talking about high-energy gamma ray photons, in which case the result of absorbing a photon in the nucleus is usually to blast the nucleus apart after a very short time, often creating a bunch of new particles in the process).

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Dear Chad, an appropriate answer, and +1 point. However, it's not right that you can't scatter light off light. SLAC has done it nicely, at least indirectly and probed the theoretically known electron-box 4-point-function of QED, see physics.stackexchange.com/questions/1361/… –  Luboš Motl Jan 20 '11 at 22:30
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Yeah, you can bounce light off light if you have a powerful enough source at high enough energy. I don't think it would be useful for "seeing" photons in the sense of the original question. –  Chad Orzel Jan 20 '11 at 23:25
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