PeterDonis said:
Sort of. You can do experiments with very faint light sources in which individual dots show up on a detector; the dots in such experiments can be interpreted as the detection of single photons by the detector. In most cases, however, such light sources do not emit a state of the quantum electromagnetic field that would normally be interpreted as a "single photon" state. Such a state would be a Fock state, and sources that emit Fock states are extremely hard to make. Most light sources emit coherent states, which have no definite photon number.
Such "faint light sources" are no single-photon sources but just coherent states of very low intensity, i.e., it's mostly the vacuum state but contains also ##N##-photon states for all ##N \in \mathbb{N}_0##.
True single-photon states are not so simple to have, but for some decades one has such sources. One standard way is to use parametric down-conversion, where you use a laser and certain sorts of birefringent crystals (like beta-barium borat, BBO), where for certain directions behind the crystal you get entangled photon pairs, i.e., one photon from the laser interacts with the crystal in such a way that two photons are created such that energy and momentum is conserved ("phase matching") as well as the polarization of the two photons are entangled. Depending on the type of conversion you get singlet or triplet polarization states. You can use this as a "heralded single-photon source", i.e., you use one photon as a "trigger", announcing the other photon (the "idler") to be present in the corresponding other direction determined by the phase-matching condition. If you also measure the polarization state of the trigger photon you also know the polarization state of the idler photon. In this way you get a true one-photon source.
PeterDonis said:
No. Photons are not little billiard balls. They aren't even close.
I couldn't agree more. Photons do not even have a position observable in the usual sense. To get an intuition for photons you should just think of them as electromagnetic waves and the intensity of the electromagnetic wave as the probability density to detect a photon at a place determined by the position of the detector. The most distinct property of a single-photon state in comparison to a classical em. wave is that it can be absorbed as a whole or not at all. There's no way to absorb "half a photon". That's the only sense you have "particle-like features" of single-photon states, but it's not possible to define a position of a single photon.
Another case are "cavity photons". There you have a box made of conducting walls. As classical electrodynamics tells you, in this case you have some boundary conditions for the electromagnetic field to fulfill, leading to discrete frequencies and wave numbers, out of which the most general radiation field can be built. These are all "standing waves", due to the boundary conditions. A photon is now a certain state of the corresponding quantized field. This photon is, of course "localized" within the cavity, i.e., the probability to detect it outside the cacity is 0, but it has no position within the cavity either. It's just within the cavity but not in any sense like a point particle, bouncing back and forth between the walls of the cavity.
I retired a few years ago after spending 50 years in IT. This is all for pure enjoyment. Physics Forums allows me to pick the brains of those with more formal education and experience in physics than myself.
