[alancalverd]

Even then, you need to remember that "observation" is introduced as an auxiliary and explanatory concept in physics, rather like "cross section" in engineering. Nobody is going to cut a ship in half to make it work, but drawing what it would look like if cut in half, is very helpful in explaining how it works.

The natural world does what it does, and we petty beings try to understand and model it as though we were observers to every action.

To go back a couple of paragraphs, the properties of longitudinal compression waves are pretty much independent of frequency, so they are all "sound" whether we, bats or transducers can sense them, but the interaction of electromagnetic radiation with matter is strongly dependent on its frequency so we assign specific names to parts of the spectrum,  and the tiny bit that excites transient chemical responses in the eye is called light.

[Bill S]

Thanks Alan. I like the "cross section" analogy; it's worth keeping in mind when thinking about QM.

As far as terminology is concerned, I guess we are back to Humpty Dumpty!

That should work, as long as each of us is aware of how the other is using a term.

[Colin2B]

Sometimes one sees the assertion that light travels as a wave, but is “detected” as a photon.  I was looking for opinions on that idea.

Simple view:
At the source an atom cannot emit photons with any old energy. When an electron moves between higher to lower energy levels in the atom an em wave is emitted with the specific energy for that particular transition equal to the difference in energy between those levels. The energy of this wave is =hv and is called a photon. Clearly this is the wave that travels to the detector. This process has a symmetry, just as the movement of a source electron creates an oscillating em field (wave) then at the detector end the oscillating field can move an electron.
One of the key points in understanding photons is to understand the quantum nature of the detector. When an em wave over a certain energy hits a detector the photoelectric effect releases an electron which triggers a photomultiplier and a counter goes click. If the wave does not have that energy then the counter doesn't go click.
When Einstein wrote his paper on the photoelectric effect they didn’t know about the quantum nature of the electrons in the photoelectric metal so the view was that the photons were particles that transferred their momentum to the electron kicking it free, just like billiard ball interactions, and this led Einstein to conclude that light is a stream of particles, photons.
However, we now understand much more about the quantum nature of the detector where each electron is in a quantum potential well - defined by the binding energy. If the incoming energy is large enough to expel the electron from the well it leaves, if not it dissipates its energy in collisions with nearby electrons and atoms and remains trapped in the potential well, ie not emitted. So what you are measuring is not the quantum nature of light, but the quantum nature of the detector. Crucially, in QED you don’t have to assume that the incoming light is quantised for the photoelectric effect - the quantisation can be derived separately. Even Einstein stated in his paper on photoelectric effect that his working was heuristic.
So in QED and QFT the energy is transferred by a disturbance of the em field (as described by Maxwell’s equations) but detected as a photon. I think it was Pete who said it isn't a photon until it's detected.
A wave by any other name ........ 

By the way you may have missed this:

You're confusing the term observer with person. The observer could be a system of clocks and rods and detectors at the location at the time of the event. It may be a person but that's not necessary.