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Part 1: A photon's wavelength is red-shifted by expanding space. Since $E=hf$ the photon imparts some of it's energy to the curvature of space, slowing the expansion of space ever so slightly. Is that Correct?

Part 2: A photon is described by its wave function as it travels through space and the photon is in a superposition of multiple states (eg. it's polarity). The energy of the photon is therefore the sum of the energies of each state multiplied by the probability (or amplitude) of each state. When we measure the photon as it lands on our telescope a specific state is "selected". I assume that this also means that by measuring the photon a specific impact to the curvature of space is realized. I assume this means the photon is entangled with space. Is that correct?

Qmechanic
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Barry Saffron
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3 Answers3

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This question seems to be based on a lot of wrong assumptions:

  1. Naive energy conservation does not apply to General Relativity - formally, energy conservation follows from invariance under time translations, but the expanding universe is not invariant under translations, see e.g. this question and its answers. We cannot reason about energy being conserved in the context of the expansion of space.

  2. Photons don't have "wavefunctions", see this question and its answers. There is no good "probability to detect at a position" for inherently relativistic particles.

  3. Entanglement is a specific property of the combined quantum state of two or more quantum systems. "Space" is not a quantum system, so it is meaningless to assert that "space" is "entangled" with anything else. (It is possible that there are quantum states for "space" in the eventual theory of quantum gravity, but we do not have a universally accepted theory of quantum gravity yet.)

ACuriousMind
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    I disagree with 2, as far as particle physics goes. The photon is one of the elementary particles, and as to its wave function see this https://arxiv.org/abs/quant-ph/0604169 – anna v Sep 24 '22 at 02:25
  • @annav The paper you link agrees with me. Quote: "In particular, there is no position operator for a photon, leading some to conclude that there can be no properly defined wave function, in the Schroedinger sense, which allows localizing the particle to a point. On the other hand, it is known that even electrons, when relativistic, don’t have properly defined wave functions in the Schroedinger sense [2,3], and this opens our minds to broader definitions of wave functions." Claiming that I'm wrong because you have changed the definition of "wave function" doesn't strike me as very useful. – ACuriousMind Sep 24 '22 at 11:36
  • The standard model QFT and its accompanying Feynman diagrams assume point particles , and certainly are not the Schrodinger equation solutions that define the fields in the SM, but dirac and klein gordon and maxwell – anna v Sep 24 '22 at 18:45
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I think the essence of your question boils down to the following.

The photon (or rather the EM field) is quantized in a way we know. Gravity on the other hand, in the form of GR, is not. If gravity was really classical it could not be entangled with a quantum field. This is an easy exercise (and an interesting theorem). So the interpretation of your second statement depends on what you mean with gravity.

On the other hand variations of similar arguments are powerful "proof" that gravity, ultimately, needs to be quantized (otherwise, essentially, one would be able to violate Heisenberg uncertainty principle).

lcv
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  • I will post a reference for the third paragraph when I have time – lcv Sep 23 '22 at 23:23
  • I also kind of reinterpreted your question from a broader point of view. Let me know if you agree or not. – lcv Sep 23 '22 at 23:25
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Let's first look at a simpler scenario that one can have in a lab. Then afterwards, we can apply it to gravity.

When light (or a photon) reflects from a mirror, the change in propagation direction implies a change in the momentum of the photon. Momentum conservation requires that the mirror picked up the difference and therefore must acquire some small momentum kick from the photon. However, a photon produced in a lab will not be a perfect plane wave and therefore would consist of a superposition of multiple different plane waves with (slightly) difference momenta. When such a superposition reflects from the mirror, it will receive different kicks from the different plane waves in the superposition. Hence, it must become entangled with the photon. However, the effect is so small that it is generally safe to ignore it, as it is always done in the literature. Note, this has nothing to do with any mathematical formulation of the scenario and whether or not we can represent the mirror as a quantum state or not. It's just physics.

No apply this to a photon propagating past a planet. We know from physical observations that the gravity of the planet will change the propagation direction of the photon. Now we have the same situation as with the mirror in the lab: the photon must become entangled with the planet. Here the effect is even smaller than in the lab, and thus once again safely ignored.

Finally, we can consider the scenario that you describe. In principle, nothing prevents the photon from becoming entangled with gravity based on purely physical arguments. However, once again the effect would be so small that it would be extremely hard to measure it.

flippiefanus
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