See post 11, what I understand from it (just a layman talking here...) is that the properties of the photons (like wavelenght, intrinsic speed or direction) would not be affected by the interaction with the virtual particles (that pop into and out of existence from the zpe) but that only a small delay in transition would occur.
Photons interact with particles and fields through scattering.
(1) Photons being elastically or inelastically scattered by particles such as electrons.
This which can be treated as a two particle collision where the electron can either be initially at rest or moving.
The photon does not slow down for either elastic or inelastic collisions and travels at the speed of light c before and after the collision.
The photon energy is defined by the equation E = hc/λ and if the collision is inelastic the photon loses energy due to an increase in its wavelength λ but it doesn't slow down.
(2) Photons being scattered by the electric field surrounding a charged particle such as an electron.
In this case the electromagnetic field surrounding the electron can be treated as an electromagnetic vacuum which is a field in the lowest energy state or the zero point energy.
This field undergoes vacuum polarization which results in the formation of virtual particle/antiparticle pairs.
The scattering of a photon in the electromagnetic field is explained by photons being scattered by these virtual particles in a process known as
Delbrück scattering.
This is quite the opposite to your explanation in post #11; photons are not captured and released after the virtual particle pair disappears and as with scattering involving particles the velocity of the photon remains the same before and after collisions.
Furthermore only high energy photons can interact since virtual particle/antiparticle pairs have high energies as they have a short lived Δt value as defined by the Heisenberg energy time relationship ΔEΔt ≥ h/4pi
Scattering of photons by virtual particles is always an elastic collision.