Exploring the Nature of Photons: Position, Speed, and Emission/Absorption

[bahamagreen]

Just wondering...

If photons do not have a position operator

- is it because theory math does not provide it, or because theory math forbids it?

- how is speed (the difference between two positions divided by time) determined?

- are emission and absorption inferred from measuring energy changes only?

- do emissions and absorptions have positions (are they events)?

- how are an emission and absorption attributed to the same photon?

 

[vanhees71]

One can mathematically show that a massless quantum field with spin $\geq 1$doesn't allow to define a self-adjoint operator that has all the mathematical properties for a position observable is fulfilled. For massive fields that's always possible.

For photons, it's no problem to define energy and momentum operators. Note that this also must be so from the inner workings of relativistic QFT: To make sense you always must be able to define the operators that allow to define how the various quantities entering the formalism behave under (proper orthochronous) Poincare transformations (i.e., translation in space and time, rotation, and change from one inertial frame of reference to another moving with constant speed against the former, i.e., boosts). The generators of the time translation and the space translations are energy and momentum. Thus, as for any sinsible physical system described by QFT also for the electromagnetic field you can define energy and momentum operators, and a free photon's energy-momentum eigenmode has energy and momentum as a classical massless particle would have, i.e., $E=c |\vec{p}|$ (where classical massless particles are not very well definable; what I mean here is to use the energy-momentum relation for a massive particle $E=c \sqrt{\vec{p}^2 + m^2 c^2}$ and take the formal limit $m \rightarrow 0$). The "speed of the photon" (in some specific sense) is then $v=c^2 |\vec{p}|/E=c$.

Emission and absorption are interactions of the electromagnetic quantum field with some charged particles, e.g., an atom. Atoms can be pretty well localized (nowadays it's amazing what can be done with all kinds of traps to put and hold atoms and other particles at a pretty well defined place). Now to absorb a photon by this atom you have to use one with pretty the right energy (i.e., also with a pretty well defined momentum). You have of course to have a momentum that the photon coming from the photon source (which for this purpose must indeed be pretty well localized) to aim well with the photon towards the atom in its trap. Then with some probability (which can be calculated using the quantum rules). The emission and absorption processes are pretty well defined events, and that's also how the "speed of photons" can be measured. It's not different than with a classical electromagnetic wave: You send a flash of light (i.e., an electromagnetic wave packet) to a mirror and measure how long it takes for the wave to be reflected back to you, which you can only do by detecting it. Then you divide distance by the measured time, and you'll get (to very good accuracy) $c$, the speed of light in a vacuum.

If you do this (gedanken) experiment with one photon, it doesn't make sense to say "the very same photon I've sent to the mirror is detected when it's reflected back to my detector". Photons are indistinguishable, and the photon interacts with the mirror. You can as well describe the "reflection of the photon" (which is a rough picture based on the behavior of classical em. waves) as an annihilation process of the photon by interaction with some charged particles in the mirror which then get excited and emit a new photon with a momentum pretty well determined by the laws of reflection.

That's how in fact all interactions of photons with matter are described by quantum electrodynamics: A photon collides with some charged particle and then anything might happen with a certain probability which is allowed according to the conservation laws valid for the photon. E.g., the scattering of a photon with an electron may again lead to just a photon and an electron with energies and momenta changed such that the conservation of energy and momentum are fulfilled. In the formalism (called perturbation theory) to the lowest order the photon and the electron are destroyed and then recreated as a new photon and a new electron in some sense, but in the essence what you really only can observe is that there was happening an elastic scattering of a photon with an electron resulting again in a photon and an electron (which is also known as Compton scattering).

 

[bahamagreen]

Emission and absorption are interactions of the electromagnetic quantum field with some charged particles, e.g., an atom. Atoms can be pretty well localized... ...The emission and absorption processes are pretty well defined events, and that's also how the "speed of photons" can be measured.

Thanks, I'm still not clear how "interactions of the electromagnetic quantum field" (which for photons would not have position operators?) can be events (which would have positions).

If photons don't have position operators (and I'm thinking that extends to their interactions?), how can interactions be well localized? Are the photon's position revealing interactions themselves localized events, but the position of the photon itself at the time of these interactions somehow without position?

Sorry to be obtuse, just seems to me that if a photo does not have a position operator, neither should the photon's interactions, since interaction localization should reveal the photon's location.