Let’s say you had a perfect vacuum, a laser emitter at Point A, and a receiver at Point B. Could an observer standing outside of the beam detect it without receiving (or scattering or otherwise interfering with) part of the beam or its reflection?
Something like this:
Is there any way to see/detect the beam of light “passing by” without interfering with it?
I don’t know the answer, but the cat is clearly at the wrong end of that laser.
How would you know unless you could detect the laser? Hmm? HMM? 
The answer is no, even when it is in atmosphere most of the time. The beam on my 200W laser cutter is invisible even in the dark. It does become visible when traveling through smoke, though.
And you’ve tried every possible experiment allowed by the laws of physics with it?
I just got my eyes and ears, man. Oh! And I got a magnifying glass like the guy in the picture. You want me to go get it?
On my phone; this’ll be skeletal. And I’m going on very old & cobwebbed knowledge.
Light is just EMR of a particular freq. As such it’s got an E & B field. Beam coherence of the photons is nice, but those fields aren’t cohered.
You ought to be able to sample them at a distance & make decisions based on what you sniff.
I’ll set aside the QM / philosophical discussion about whether that sampling changes, and thereby interferes with, the photon beam. IOW:
If a photon falls down in a forest, is the entire universe (subtly) affected?
Any obvious way of doing this, you’ll absorb photons.
Above, @wolfpup cites Quantum Non-Demolition experiments that measure a single photon without absorbing it. In addition, Weak Measurement can be used to measure quantum systems (including photons) by weakly coupling to the quantum system and only slightly perturbing the system, not enough to destroy its state.
The quantum non-demolition (QND) measurements are rather different from what I believe @Reply is talking about, in that those measurements are carried out directly in line with the beam. A photon under inspection reflects off the QND device during the measurement process. It’s not traveling in an unimpeded fashion.
A photon has no charge so \vec E=\vec B=0. Until the photon interacts with (absorbed by) your instrument or some third system that your (presumably electromagnetically based) instrument is measuring, there is no detection.
It is useful to understand that photons are fundamental particles (gauge bosons) that only interact per quantum mechanics. So, any Gedankenexperiment that has a classical observer observing a photon traverse a path is physically nonsensical. We can only infer the interaction of a photon through the observable interactions with normal matter, i.e. hadronic (for our everyday experience, limited to baryonic) matter. Photons can theoretically interact with virtual particles such as a fermion-antifermion pair, another photon, or even itself, but of course an observation of this interaction would fundamentally interfere with it, making the ‘virtual’ particles or any kind of self-interaction to be a ‘real’ particle interaction that also interacts with other particles. There is no way to observe a quantum system that does not involve an interaction with the observer in some way.
Stranger
D’oh! Thank you.
Is an electron emitted via the photoelectric effect the beam “bouncing” for the purposes of the question?
The article I cited above on QND appears to require registration in order to properly view the PDF, but here’s a simplified explanation (a fascinating phenomenon, even if only tangentially related to the OP) … free, but limited views per month:
https://www.science.org/content/article/spying-photon-without-harming-it
The photoelectric effect destroys the initial photon, so presumably that’s out of bounds for the OP.
Certainly the classical \bf{E} and \bf{B} fields aren’t directly applicable for a single photon, but a photon is an excitation of the quantized, relativistic extension of that field concept, a 4-vector often denoted A_\mu. While this is close to the classical potentials than the classical fields, those are intimately related, so I would say that a single photon “has” (or even “is”) an electromagnetic field, despite having zero charge. In fact, the opposite definition holds better in quantum land: a (charged) electron has no electromagnetic field in QED, but it talks to the electromagnetic field of the (uncharged) photon. (Classically, we’d say the electron “has” a field; and there are no photons.)
This work led to the 2012 Nobel Prize in physics. There is no photon beam in this case, but it’s very neat stuff.
This experiment, in brief: A photon is trapped in a “cavity”, which is to say there is a single excitation of the EM field in a small localized spot in the apparatus. A beam of atoms is prepared, with each atom in a special quantum state and with velocities chosen just so so that when the atom passes through the cavity, the atom’s quantum mechanical phase is detectably shifted compared to the case where there is no photon in the cavity. And then when the next atom passes through and shows the same thing, you can say that the (stationary, standing-wave) photon in the cavity was detected multiple times by the atom beam without killing the photon.
Modern QND measurements have been able to switch things around, with the atom in a trap and a photon traveling toward it, but those are the ones in which the traveling photon is reflected. The principle is very similar, with the photon either reflecting before entering the atom trap or after, and if it reflects after, then the atom’s phase will be perturbed. And, this can be achieved without loss of the (reflected) photon.
With no disrespect meant to the several far more learned participants in this thread than I, I think we can say in this case it’s
Q(M)uibbles all the way down.
Deep stuff; very deep. I am suitably humbled.
Right, that’s why I only said that the obvious ways of doing it don’t work. There are still clever and subtle ways to (arguably) do it (with quantum ibbles all over the place, of course).
But also applauded, at least by me, as your qualitative description of “sniffing around” at the photon’s field is close to the mark of how @wolfpup’s cited measurement works, even if with some roles flipped around and with a blind eye turned to your “at a distance” part. The atoms that pass through the cavity – and thus through the photon – had their initial quantum state specially prepared to provide a good “nose” for the the correspondingly specially prepared cavity. The atoms’ energy levels are affected by the radiofrequency photon’s electric field as they pass through, and the smell of that field sticks around in the atom as it leaves in the form of a measurable phase shift.
