Can EM (radio) wave oscillation interact with an antenna without photon absorption? EM wave interaction with an antenna has been covered, e.g.: How do we visualise antenna reception of individual radiowave photons building up to a resonant AC current on the antenna?
Most answers focus on antenna atoms individually absorbing a photon's energy. But photons have oscillating EM fields. When light passes through water it loses velocity (but not energy we are told) because of the interaction of photon EM fields with electron electric fields. Might this play any role in the interaction of radio waves with a metal antenna? Can the photon EM oscillation induce its frequency on the antenna's delocalized valence electrons? Does the antenna absorb EM energy this way and then re-radiate so there is no net loss to the original EM field (but if the antenna heats up there is loss...).
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My antenna example/question introduces too many (conduction) interactions. My real interest centered on the reduced velocity of the photon in glass or water? Is that absorption and reemission as HF suggests? Or does the photon's electric field interact with electron charge in the medium?
anna v writes: " there is nothing oscillating in the photon except the quantum mechanical probability of detecting it. A photon just has energy, spin and momentum. The energy of the photon is E=hν where ν is the frequency...."
To characterize photon energy as an inert, quiescent, non-oscillatory quantity is to embrace the original concept of kinetic energy (Coriolis, Watt, Lord Kelvin) from the mid-19th century! Have we come no further? Physicists will make the photon into whatever is convenient for the question at hand, particle, field or mathematical operator. In fact, the photon – no mass, no trajectory - is a 'particle' only by analogy: it terminates at a point with momentum.
 A: 
But photons have oscillating EM fields

Wrong,  there is nothing oscillating in  the photon except the quantum mechanical probability of detecting it.  A photon just has energy, spin and  momentum. The energy of the photon is $E=hν$ where $ν$ is the frequency of the classical EM wave a quantum superposition of a large number of photons will make.
This is just a special case to see how different photons and classical light are:

On the left is the build up of a polarized classical electromagnetic wave, the red arrow depicting the electric field maximum. The middle describes the photon, which individually has only forward and backward spin, nevertheless in confluence of many photons it builds up the classical wave.
The classical Maxwell mathematics for elecrtomagnetic waves is accurate when used with classical electricity mathematics.
When one discusses atoms and valence electrons, one is in the frame of quantum mechanics. Radio wave photons  are very low energy photons and can only scatter with  conduction band electrons in a conductor.  The photons scatter   electromagnetically  with the conduction band electrons , giving a small part of their energy and  creating a current, the loss of energy for the photon is minimal. The mathematics is such that the probability of this scattering has the frequency of the radio wave in its formulation. (even though the classical wavelength of radio waves is from centimeters to kilo meters) It needs a study of quantum electrodynamics to see how this happens (example).
A: Yes, indeed it can. If the antenna is not tuned to the photon's wavelength then the photon may bounce off it. A simple example of this is provided by a waveguide with a poorly-connected end plate. Some of the forward power (EM photons) will excite the end plate and radiate out beyond it, while others will bounce off it and travel back the way they came, as what is known as reflected power. For a waveguide in the form of a coaxial cable, the ratio between the forward and reflected waves is referred to as the voltage standing wave ratio or VSWR (pronounced "vizwah").
This is no cheesy interpretation thing which vanishes when the oscillating wave collapses on measurement, it is eminently measurable and survives the measurement process. In other words, it is a real, concrete phenomenon.
On the other hand, one may argue that actually the forward photon was absorbed and the reverse photon re-emitted. This is actually the standard theoretical model. But it is just one of those cheesy and unfalsifiable interpretation things, it adds nothing to the classical picture, no new predictions, nothing like that. So how much credence you give it is a philosophical issue not a physical lab-test one.
A: I will focus on what you wrote in the title

Can EM (radio) wave oscillation interact with an antenna without photon absorption?

Production of a radio wave
To understand what a radio wave is made of, you have to remember how it was generated:

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*on the surface of a metal rod, many electrons are accelerated back and forth

*during these accelerations, they emit photons

*for each period of the oscillation, the number of emitted photons increases to a maximum and then goes back to zero.

The technically useful thing about this is that the electrons, which are all accelerated in the same direction, all emit their photons with the same directions of their electric and magnetic field components:

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*the electric field is directed parallel to the electric field potential of the rod

*the magnetic field component is directed perpendicular to both the electric field and the direction of propagation

What we get is a stream of a periodically swelling number of photons from zero to a maximum and back to zero. During each half period the directions of the electric and the magnetic field component are changing.
How does that helps to transfer informations from the emitting to the receiving antenna?
Receiving a radio wave
With the above said the answer to your question is the following. But, as Guy Inchbald said, „it is just one of those cheesy and unfalsifiable interpretation things, it adds nothing to the classical picture, no new predictions, nothing like that. So how much credence you give it is a philosophical issue not a physical lab-test one.“
The receiving antenna is hit by numerous photons, since all bodies that are not at absolute zero Kelvin permanently emit (and receive) photons. These photons that hit the rod interact with a surface electron and are absorbed and re-emitted with different directions and frequencies. This is noise for the antenna.
The photons of the transmitter of a radio wave act together (as a group) and the electrons on the surface of the receiving antenna are periodically influenced by the common field components of the incoming photon stream.
The electrons do not dissipate the incoming energy (the noise) in all directions, instead the electrons start to move along the rod!
Regarding your question: Photons are indivisible particles from their emission to their absorption. They transfer energy - and you get a signal at the receiver. Or they do not interact with anything and continue their propagation.
