Why Cannot Photons Create Electron-Hole Pairs? My question might be dumb or considered trivial, but I will try to explain my confusion.
In a Particle Physics course, we were taught that only charged particles (such as electrons $e^{\pm}$, $\mu^{\pm}$, $\pi^{\pm}$, etc.) create so-called electron-hole pairs in Silicon, and thus we would never be able too see photons in the trackers of the LHC experiments, which use highly granular Silicon pixel or vertex detectors.
In a Laser Physics course, however, we were taught that photons can create electron-hole pairs, which would be the case in Avalanche Photodiode, where Si is used as well.
Honestly, I am totally confused. I understand the point of the Particle Physics professor, because at the end of the day, photons and neutrons (both electrically neutral particles) do NOT leave tracks in the trackers of CMS or ATLAS. That is for sure. But on the other hand, I feel like the Laser Physics Prof. also has a point; after all, when we have Laser light and Avalanche Photodiodes, we do get signals.
Thus, the question is: Can photons (or more generally electrically neutral particles) create electron-hole pairs in Silicon?
 A: This is a long handwaving  comment.
When a charged elementary particle goes through matter at high energy ,it hits electrons off the atoms it crosses, thus ionizing them. The electrons thrown off have their own story , see the tracks of a charged particle  in a hydrogen bubble chamber:

I chose this picture  because it shows that in the mass of liquid hydrogen that is the bubble chamber matter, charged tracks of  $K^-$ leave  visible lines, whereas the $K^0$  as neutral, leaves no track. Why, since the $K^0$ is also composite of charged quarks, as is the beam $K^-$ ?
The tracks are made out of tiny bubbles which are  secondary-electron "tracks" as the atoms  are hit by the primary charged particle, and ionize them. These are very low energy electrons that are again absorbed , but the crossection for the charged track to create them is much larger than for the neutral , which is practically zero. The scattering of a charged track with an electron is a first order diagram  in the calculation, whereas for a neutral track as the $K^0$  higher order electromagnetic diagrams are involved , which repress the crossection, as the electromagnetic coupling is 1/137. Obviously in the bubble chamber there are no secondary electrons from the neutral $K^0$  to leave bubbles
The high energy photons have small probability to scatter off the electrons of the neutral atoms, so as to leave bubbles in the recorded picture, I found  the crossection in carbon here, as an indication.

There are orders of magnitude between the charged and neutral case.
In the silicon tracking chambers it is the secondary electrons that can create the electron-hole pairs, and summed over the interval of the path in the detector , a signal can be gathered and recorded, that a charged track passed through there. The sum of the secondary electrons from neutrals , even if some may  exist, will be very much lower, and thus either the current induced is lost in the noise or the detector is calibrated to ignore it, since it will be random and not reliable.
Found this:

A charged particle traversing a depleted silicon volume generates electron-hole (e-h) pairs via ionization; the electrons or holes then travel guided by the field to electrodes connected to an amplifier. The field, signal drift, and charge collection for an over-depleted strip sensor are demonstrated in Figure3. The good energy resolution of silicon sensors (one e-h pair created per ~3.6 eV of deposited energy) together with their high density (390 eV/μm average energy loss for a minimum ionizing particle (MIP)) results in ~108(electron-hole pairs)/μm. Different to gas detectors, the average signal of 32000 (22800 most probable) holes in a standard 300 μm thick piece of silicon does not need sensor intrinsic amplification.

