The quantum mechanical vacuum (i.e. the vacuum of a typical QFT) viewed as the zero-point state of the system has an energy (of course, here enters the Vacuum Catastrophe and all the mess that follows).

The question: if we have a ray of light (a null geodesic) traveling through this vacuum will the ray of light scatter? In a perfect classical vacuum there is no scattering of a light ray, but in this quantum vacuum there is a ground state field that has energy, and therefore it is conceivable that the photon-field would scatter off (interact with) the ground state energy field. Is this incorrect? IF this is not a well-posed question, please help me understand how to make it so.

Note: I realize that the light ray could interact with itself with nonzero probability and thus could scatter off itself - is this distinguishable from the light scattering off of the "energy in the vacuum"?

  1. Nothing "travels" through the quantum mechanical vacuum. The quantum mechanical vacuum is a quantum state with no definite location (manifest in that it is a Lorentz-invariant state, which no state with any sort of definite location could be). If there's a single light ray in the universe, the quantum state of the universe is the state of a bunch of photons in the coherent state that represents the light ray, and no vacuum state at all. The "light ray" state can't interact with the "vacuum" state because they do not exist simultaneously at all.

  2. Even if we do not get pedantic about what a quantum state is and how it can interact, it is also a simple fact that quantum mechanics conserves energy-momentum (at least its expectation value) as strongly as classical mechanics does. The light ray suddenly changing its direction would violate conservation of energy-momentum, which is allowed neither classical nor quantumly.

  3. The only interaction that would be allowed here is photon-photon scattering of two different photons in the light ray. However, it is not evident to me that scattering (a process predicated on approximating the in- and out-coming particles as "free" can happen in this form between the "components" of a coherent state such as a laser beam - the number of photons in such a coherent state is generically indeterminate to begin with, making the application of the particle picture questionable.

  • $\begingroup$ For point 3.: The constraint on the elastic photon-photon scattering cross section set by PVLAS seems to imply new physics, since the result is larger than what the Standard Model predicts. arxiv.org/abs/0805.3036 They used magnetic fields rather than coherent lasers; is that more convincing to you? $\endgroup$ – N. Steinle Jan 15 at 18:43

An upper limit can be set by observing stars. Since no such scattering has ever been observed it is safe to say that the vacuum does not scatter light.

  • $\begingroup$ This seems like more of a comment than an answer. $\endgroup$ – N. Steinle Jan 7 at 17:15
  • $\begingroup$ Nope, it is an answer, more specifically, it is a "no". $\endgroup$ – my2cts Jan 7 at 22:10
  • $\begingroup$ @N.Steinle This answer ("no") is in the correct place. There are several discussions around the SE network about whether brief answers are more appropriate comments, and the consensus is that they aren't. Here's a recent one. $\endgroup$ – rob Jan 7 at 23:04

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