Does a photon attract or repel matter?

Question

So, I know experimentally that matter and light interact. And I can calculate cross sections in Quantum Electrodynamics. Yet a fundamental intuition about how light and matter interact is completely missing in my understanding.

If a photon were (for a brief instant) next to a proton, would they attract? Repel? How about an electron? Is there some kind of potential $V(x_2-x_1)$ which can approximate the interaction between light in a particle-like state and matter? If we can do it for protons and electrons, which are both particles and waves, why can't we also do it for photons?

Maybe this is going to be categorized as too broad. If that's so, let's just narrow it down to: Does a photon (a gaussian wavepacket of light) attract a proton, or repel it? If the answer depends on the distance, is there must be an optimal distance between them like in the Van der Waals potential?:

I'm not sure the necessary insight is even anything to do with quantum mechanics. Think about how a classical electromagnetic wave interacts with a point charge. That's more or less it. It's not an attraction or a repulsion. The charge feels a force that depends on the Lorentz force law. — , Dec 29 '19 at 20:41
The quantum mechanical part is because I was asking about light in a particle-like state (a single photon). I didn't expect that this could be expressed completely classically. But I don't really know. — doublefelix, Dec 29 '19 at 20:54
If you want to think about it in terms of QFT, maybe another way of seeing that it doesn't make sense to conceptualize it this way is the following. We can draw a world-line of one electron, and a world-line of another electron, and then we can construct a Feynman diagram by letting a photon be emitted by one and absorbed by the other. But you can't take an electron and a photon and connect them with a particle X being emitted by one and absorbed by the other. Regardless of whether X is a photon or an electron, you'll have at least one vertex that isn't a legal vertex in QED. — , Dec 29 '19 at 21:56
I had a hard time visualizing what you were saying with X - but I think the phenomenon of compton scattering is enough to see that the particles really do interact/scatter. And the corresponding diagrams are of course also not 0 for their scattering interaction. So I don't see why the question doesn't make sense to ask. Of course in QFT we have no way of describing a gaussian wavepacket for photons that I am aware of (bc position space isnt available), but that seems more a flaw of the formalism than a physical effect; light is seen to come in discrete chunks in experiment. — doublefelix, Dec 29 '19 at 23:36
It's ridiculous to say that two particles which can scatter do not interact. If the feynman diagrams aren't mediated by a boson, then so be it. Maybe diagrams don't have to be mediated by a boson to be an interaction. I did not feel I understood the "bosons must mediate" thing, but after writing this question & seeing the related question in the comments: https://physics.stackexchange.com/questions/522318/why-do-forces-have-to-be-mediated-by-bosons-or-do-they?noredirect=1#comment1179158_522318 I begin to question whether that is really a valid principle. — doublefelix, Dec 31 '19 at 13:09

score 1 · Answer 1 · answered Feb 09 '21 at 21:21

I think this question needs to be asked very carefully to get a precise answer. But, the zeroth order answer should be clear: the photon has no electric charge or magnetic moment, so it will neither attract nor repel another electric charge.

One interpretation of this question, given in the comments to another answer, is to ask about the trajectory of a photon passing near an atom.

Well, in fact, even here we need to be careful. When I hear "trajectory" I think of a one dimensional line moving through space; and it makes sense to think of the impact parameter (distance of closest approach) of such a line with an atom. But an EM wave, or a photon wavepacket, is not (generally) of this form.

If the EM wavelength (1 over the photon momentum) is larger than the size of the atom, then the atom will look approximately like a point particle, compared to a traveling plane wave. First, note there is no useful sense in which we can think of the EM wave / photon as having a 1 dimensional trajectory and ask for its distance with respect to an atom in this limit. We should think of a wave extending over space, scattering off of a point source. This is the limit where we expect more or less elastic scattering to apply. There is an interaction, but we describe this interaction in terms of the scattered wave produced at infinity, and not a trajectory.

If the EM wavelength (1 over the photon momentum) is smaller than the size of the atom, then we can think of a trajectory, if we imagine building a wavepacket of high energy radiation. But in this limit, where by definition the photon has a larger energy than the binding energy of the atom, we certainly do not expect an elastic scattering process to occur. Instead, the EM radiation will be absorbed by the atom and re-emitted with a different energy with some probability, or possibly even ionize the atom. If the photon is super-well localized and not on a collision course with the atom, it will have a small probability to interact; the trajectory won't be "dragged toward" or "pushed away from" the atom.

Another possible interpretation of this question is whether one can derive an "effective potential" analogous to the Coulomb potential for photon-charged particle interactions. To address this, one should remember that how Coulomb potential can be derived from quantum field theory: one starts with the full relativistic scattering amplitude, takes the non-relativistic limit, and matches with the scattering amplitude with a potential in non-relativistic quantum mechanics. The problem is that there is no way to take the non-relativistic limit for a photon.

One could ask what happens with a massive, neutral particle -- say a neutron, or imagine the photon really did have a tiny mass. In this case, you can go through the above exercise. You will find that neutral particles will have a zero effective potential in the non-relativistic limit.

score 1 · Answer 2 · answered Feb 09 '21 at 21:26

In photonic crystals, there are bandgaps because there is a difference in energy for standing electromagnetic waves, depending on where the nodes are. The energy is lower when the antinodes (regions with stronger electric fields) are in matter with larger $\varepsilon_r$.

So there is an interaction between light and matter. I am a simple experimentalists and would not dare to say even where the photons are in such a case. Quantum electrodynamics can only treat very simple cases. It cannot really deal with solids or even with larger atoms.

score -2 · Answer 3 · answered Dec 30 '19 at 00:45

-2

You are not specifying matter, but you are asking about a massless photon and a massive proton. And you are not asking about an actual scattering, you are just saying that the massless photon and the massive proton are next to each other (at certain distance).

You are saying that if a massless photon were for an instant next to a massive proton, what you are not specifying is, the instant. Photons are massless particles, travel at speed c when measured locally, in vacuum, and whatever the relative speed of the massive proton is, from its reference frame as per SR, the massless photon always travels at speed c. You cannot specify a reference frame or an "instant" in the temporal dimension where both the massless photon and the massive proton would be at rest relative to each other.

If you are asking about EM attraction or repulsion, then the answer is no. The photon does not EM repel or interact any object in the universe, the photon is EM neutral.

If you asking about gravitational attraction, then the answer is yes. Massless photons do have stress-energy, just like massive protons, they do have their own gravitational (static) field, and both do bend spacetime.

Do photons bend spacetime or not?

answered Dec 30 '19 at 00:45

Árpád Szendrei

28,452

2

I never expected them to be at rest w.r.t. each other, I only said "for an instant" actually to make it clear that I understood the photon would quickly be somewhere else. I also know that the photon is charge neutral, and the interaction need not be named EM attraction. But clearly, since the presence of a photon affects the motion of an electron (compton scattering), it is either attractive or repulsive - possibly depending on the distance. Maybe I should have formulated my question in terms of electrons bc of compton, but to me I'm just as interested in either answer. – doublefelix Dec 30 '19 at 00:51
@doublefelix so are you asking about scattering? – Árpád Szendrei Dec 30 '19 at 00:52
Sure, you can think of it that way. But I'm not just interested in what momentum the (say, electron) ends up with in the scattering limit. I'm interested in the nature of the interaction at close range. I think this is perfectly well-defined, even in a quantum mechanical regime, where you might replace "position of the particle" with "expected value of the position of the particle" if utmost clarity was desired. – doublefelix Dec 30 '19 at 00:55
@doublefelix am I right that you are thinking of the photon and the proton as classical objects? In that case it does not work, because the photon is not localized until it is absorbed. You cannot think of the photon when it flies near the proton as a classical object. The photon is a QM object, and it is only localized when it is absorbed. Are you asking about elastic or inelastic scattering? – Árpád Szendrei Dec 30 '19 at 00:58
Every particle is localized to some extent - in QM the wave function may spread to infinity, but 99% of the probability density is focused in some region. QFT doesn't have position space as far as I know, but experimentally it's still true. That is why when you do scattering experiments, depending on the situation, there may be a small spread of angles on which the particle is allowed to arrive (consistent with conservation of E and p). It is not arbitrarily unlocalized to all space. This roughly defines a position for the particle, and I'm asking how that position moves. – doublefelix Dec 30 '19 at 01:08
The only completely non-localized particle would be a momentum eigenstate, which is known to be an unphysical idealization anyway – doublefelix Dec 30 '19 at 01:09
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@doublefelix "This roughly defines a position for the particle, and I'm asking how that position moves." This defines a trajectory for the particle maybe showing where it will be absorbed eventually. I think I understand now what you are asking. When the photon interacts with the atom, where will that trajectory move. So you are basically asking how physically the atom will change the trajectory of the photon as the photon passes by the atom (when scattering). You are asking whether they attract or repel as the photon passes by? – Árpád Szendrei Dec 30 '19 at 03:29
Exactly! You phrased it even better than I. – doublefelix Dec 30 '19 at 09:50
Why the downvote? – Árpád Szendrei Jan 17 '20 at 22:06
I have now upvoted your comment, as I should have, since it phrased the question clearly. I was not the downvoter of your answer, but I also did not upvote the answer, as it does not answer my question. – doublefelix Jan 17 '20 at 22:10
@doublefelix can you please tell me exactly how I did not answer your question and I will try to edit my answer accordingly. – Árpád Szendrei Jan 18 '20 at 02:34
My question is exactly as in your comment that I +1'd. But nobody seems to know the answer. – doublefelix Jan 18 '20 at 11:37

Does a photon attract or repel matter?

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