What is the QUANTUM mechanical "explanation" for the "red shift?"

Question

On another thread, users have asked for an explanation of the "red shift" of photons (the apparent loss of energy of photons due to the expansion of the universe.) All they ever got was a GENERAL RELATIVITY explanation.

So, I'll rephrase the question: if a so-called "photon" represents a discrete QUANTUM of energy exchanged between two or more atoms as a result of the the electron shell of one atom changing its energy level, then what happens to the energy of those photons that "arrive" at a lower energy level (lower "frequency") as a result of red shift?

Let me try and clarify my question based on my understanding of how "photons" work. Let say atom A releases a quantum of electromagnetic energy equal to 3 units. One atom B absorbs 1 of those energy units, and atom C absorbs 2 units. We could then say that atom A emitted two "photons." If, however, atom D absorbed all 3 energy units, then we would have to say that Atom A emitted one "photon." I think this I where people get confused--photons aren't "real" PARTICLES you can count like sheep. They're just a mathematical description of chunks of energy that get "exchanged" AS THOUGH that energy were being delivered as chunks.

So, back to my question: if the energy of EACH "photon" being received is less than it was when it was "emitted," where does that energy go? Does it become "more" photons? I know, I know, in RELATIVITY the work of bending space time robs the "photons" of some of their energy (precisely the way the bending of leaf springs in a dragster robs the axle of some of the energy directed toward the wheels.) BUT our question is: What is the QUANTUM explanation? And please, don't just point the relativistic answer we already got.

If you don't HAVE a quantum explanation, just say so.

Thanks!

Jon Custer: Please read my question. I did not say there was ANYTHING "wrong" with the Relativistic explanation. I said, we have repeatedly asked for a QUANTUM MECHANICAL explanation, and all we get is Relativity. There is NOTHING "wrong" with coffee, but I ORDERED a chai latte. — Tommy Jonq, Dec 05 '16 at 17:48
A light ray isn't simply made up of photons. Attempting to describe the red shift by concentrating on the properties of individual photons is a pointless exercise. — John Rennie, Dec 05 '16 at 17:48
FQQ: I'm sure the photons appreciate you sticking up for them, but you didn't answer my question. Did you READ my question? — Tommy Jonq, Dec 05 '16 at 17:49
John Rennie: "Attempting to explain is pointless..." ...Because WHY? I mean, if YOU don't WANT to explain, then why comment at all? If you don't know how, then why not just say I don't know? Why do you assume NOBODY knows just because you don't know? Does ANYBODY "moderate" these discussion? — Tommy Jonq, Dec 05 '16 at 17:54
@TommyJonq The moderators are probably very upset with you for not understanding this site in general, for example by responding to comments as if they were answers. You probably do not want them to "moderate" the discussion as it could end in adverse consequences for you and the disrespectful way you conduct your conversation (see: "did you READ my question?" and "if YOU don't WANT to explain, then why comment at all?") etc. — CR Drost, Dec 05 '16 at 18:15
If the photons are not just mediating an interaction and are rather present in the asymptotic state of a scattering experiment, I sure can count them like sheep with PM tubes! If you are considering two atoms, under semi-classical gravity, at different gravitational potentials "exchanging" a photon, then you would rather have that the allowed transition will be changed. There would not be disappearing energy because the mass of the excited bound state in the "receiving" atom would be higher/lower accounting for the "missing" energy. — G. Bergeron, Dec 05 '16 at 18:18
There's probably an answer to the question that could be given in terms of "in" states and "out" states in the context of QFT in curved spacetimes. But I'm not sure that this would satisfy the questioner, since the properties of the curved spacetime would still be integral to how the "photons" would behave. Really, there's no way around general relativity in answering this question... — Michael Seifert, Dec 05 '16 at 19:10
With asymptotic in/out states, you are right, but if the photon is absorbed by atoms, then one can use simple gravitational potential difference to explain dynamics (AFTER the transition). I think QM has been shown to mix well with simple gravitational potential using cold neutrons bouncing off a plate. Of course, I'm thinking of low curvature regions... — G. Bergeron, Dec 05 '16 at 19:46

score 6 · Answer 1 · answered Dec 05 '16 at 18:01

Redshift is a kinematical effect, not a dynamical one. Therefore, the explanation of redshift is the same in classical mechanics and in quantum mechanics.

For example, the derivation of the red-shift formula in non-relativistic classical mechanics is valid in non-relativistic quantum mechanics. Similarly, the derivation in (special) relativistic classical mechanics is valid in (special) relativistic quantum mechanics. Finally, the derivation of the formula in curved space-times in classical mechanics is valid in quantum mechanics in curved space-times.

score 0 · Answer 2 · answered Dec 05 '16 at 17:59

0

Quantum mechanics is a scientific theory. As covered in my answer here, theories do not in general pin everything down but rather they create a space for modeling phenomena.

Any quantum mechanical model of redshift will say "those photons smoothly lost energy on their trajectory, therefore they must have changed frequency to satisfy the Einstein-Planck relation $E = h f.$" That is because the latter property is part of the theory of quantum mechanics. However quantum mechanics itself is not going to argue with you about why this happened.

Now, given what we know about general relativity, we can use that to inform our quantum mechanical model, saying "oh, this cannot be a case of one photon becoming two photons to conserve energy; energy is explicitly not conserved here, there is just one photon whose wavelength is changing." You seem to be very distraught by this idea, that we might try to use one theory to inform another, however it is necessary in this case because you're asking about a fundamentally gravitational phenomenon (redshift) and we do not have a theory of quantum gravity which easily can be made to model such things. So any quantum model must incorporate ideas about this phenomenon from relativity: otherwise we will simply not describe the phenomenon.

So the answer to "where does that energy go? does it become "more" photons?" is: "no, you're not understanding the phenomenon properly, the phenomenon of redshift is explicitly one where energy disappears. We can probably phrase some quantum-mechanical models where the energy doesn't disappear but probably these will ultimately give the photon an effective mass in its interaction with the other fields which absorb the photon's energy: and thus the theory will fail because the photon is massless. It's part of this phenomenon that the energy disappears. Therefore you need to model this phenomenon non-conservatively, and quantum mechanics is not going to explain it, it's going to assume it."

answered Dec 05 '16 at 17:59

CR Drost

37,682

CR Drost: "we do not have a theory of quantum gravity." First of all, thank you. That answers my question. Second of all, Jesus H Christ, it took a dozen "experts," including you, multiple pages of comments on multiple question threads to say that? That answers my question: "QM doesn't HAVE an explanation [yet, if ever.] Thank you. That's all ANYBODY needed to say. – Tommy Jonq Dec 05 '16 at 18:07
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You're welcome. Yes, it takes us multiple pages of comments in multiple question threads to articulate not just the sound-bite that's useful to you but the broader context by which we know that you're asking the wrong question. That's because you're under the narcissistic misapprehension that our answers serve you simply because you posted the question: in fact this is a public forum and my answer was written to serve anyone who has the same general question as you, not just you in particular. In that sense I'm trying to salvage some sort of redeeming value out of this. – CR Drost Dec 05 '16 at 18:18
CR Drost: Now then, back to your QED assertion that the energy "disappears." Um, do you mean, it "disappears" so far as we are able to measure it from our observation point? That is to say, we are not going to observe a conservation of energy from where we are "standing?" Or, do you mean, it disappears from the universe all together? I was under the impression a long time ago in college that the "missing" work showed up as an increase in the observed strength of the object's gravitational effect on the intervening space time. – Tommy Jonq Dec 05 '16 at 18:23
Speaking of narcissism, have you noticed some people are simply incapable of saying "I don't know?" IF one of you precious geniuses who have so little time for us mortals had BEGUN his answer with "there is no QM explanation, and here's why," we wouldn't have to bother you with our "stupid" questions. And if we're such a waste of your precious time, then WHY are you doing it? – Tommy Jonq Dec 05 '16 at 18:26
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I don't think your questions are stupid, nor that this is a waste of my time. When I said your misapprehension was narcissistic (self-directed) I did not mean to say that you are a narcissist in general, just that the way you approached the problem was very "me-me-me"-focused leaving you ill-equipped to understand me, someone who does not care very much about you per se. As for why we don't post an authoritative answer of "I don't know", it's because (a) that's not authoritative and (b) that only helps you, not the broader community. – CR Drost Dec 05 '16 at 18:31
To get back to the physics, I mean that it simply disappears from the universe altogether. What you are remembering here is presumably redshift from a gravitational potential well; what you are asking about above is redshift from the expansion of space. See e.g. this link for more info, and Lubos Motl's post that it references. – CR Drost Dec 05 '16 at 18:33
CR Drost: Apology accepted, and again, I do thank you for trying to answer these questions. I think what I'm getting at is, why (sorry, sorry, sorry) do we (if I may be so bold) draw a distinction between gravity well and the general expansion of space we observe? Why can't they be the same thing? More to the point, why can't the universe be expanding BECAUSE it's fallen INTO a gravity well? ...(cont) – Tommy Jonq Dec 05 '16 at 18:35
...Given the "surface area" of the observable "edge" of our universe, that is, the event horizon (and yes, I remember the difference between the 3 dimensional volume we "see" and the surface area of the 4-dimensional manifold of space time) and the amount of EM radiation from near that boundary, would the missing energy, when converted into "mass" for the purposes of calculating gravity at E=mc2, perhaps "explain" why the expansion seems to be accelerating? The gravity pulling our universe apart is increasing, partly due to this red shift? – Tommy Jonq Dec 05 '16 at 18:43
I mean, when I worked in this stuff I was a condensed matter physicist so I am not the right person to ask about that. However it seems unlikely as (a) the equations probably don't look similar at all and (b) the gravitational redshift appears as things are coming out of potentials, hence you are trying to say "the universe is coming out of a gravitational potential, and it's coming out faster and faster, and it's accelerating because the potential is getting stronger and stronger because the mass is going into the gravitating object" -- but that should lead to decelerating expansion. – CR Drost Dec 05 '16 at 18:49
It is not true that the energy disappears from the universe altogether! The two atoms would be at different gravitational potentials and since the mass of the bound states changes when excited, there would be a gain/loss of gravitational potential energy. – G. Bergeron Dec 05 '16 at 19:43
G Bergeron: Thanks! Very helpful. Is that referring to the translating of electromagnetic energy into gravity waves? What I mean is, the change in G potential energy across the tensor is continuous, and not discrete, i.e. "quantum" in nature? Or am I lost? – Tommy Jonq Dec 06 '16 at 00:06
CR Drost: I see that I have gotten off topic (QM/QED) and into GR. I will go back and look at the notion that translation of EM energy into gravity potential should result in a "deceleration of expansion." And thereby get out of your hair ;-) Thanks again! Very helpful! – Tommy Jonq Dec 06 '16 at 14:36
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@TommyJonq This has nothing to do with gravitational waves. I am considering the before and after states where the two atoms will have gained or lost a tiny bit of mass. Since they are located at different gravitational potential, there would be an equivalent tiny change in energy, compensating for the redshift lost. And what tensor? – G. Bergeron Dec 08 '16 at 10:39

What is the QUANTUM mechanical "explanation" for the "red shift?"

2 Answers2