The construction of particles in QFT

[PeterDonis]

if I get the photon on Earth the poor creatures from Andromeda will not detect the excitation of the field

They will not detect that one particular photon, true. But there are zillions of photons that were emitted by the distant star, so the Andromedans will still see the star just fine.

The only decision one is left is that the field is a probability and the instant I get the photon the field (vanishes there) returning to vacuum state.

You continue to miss the point. Any one particular photon gets detected at one particular place. That's it. There is no "vanishing" of anything anywhere else. The two QFT models I described--one that has the photon detected on Earth, and one that has the photon detected on Andromeda--are to help you predict the relative probabilities of detection on Earth or Andromeda. They are not telling you that the photon goes to both Earth and Andromeda and then vanishes at one of the two. Predictions are not reality.

 

[ilper]

They will not detect that one particular photon, true. But there are zillions of photons that were emitted by the distant star, so the Andromedans will still see the star just fine.
You continue to miss the point. Any one particular photon gets detected at one particular place. That's it. There is no "vanishing" of anything anywhere else. The two QFT models I described--one that has the photon detected on Earth, and one that has the photon detected on Andromeda--are to help you predict the relative probabilities of detection on Earth or Andromeda. They are not telling you that the photon goes to both Earth and Andromeda and then vanishes at one of the two. Predictions are not reality.

Exactly the 1 photon and the excitation of the field are relevant here! You have the excitation reaching me and Andromeda and I get the photon! What happens to the excitation on Andromeda. Is it real or probability.
It is exactly what is the situation in QM with DSE, collapse of WF etc. Solved there because WF is probability. Unsolved if not!

 

[PeterDonis]

You have the excitation reaching me and Andromeda and I get the photon! What happens to the excitation on Andromeda.

You are still missing the point. There is only one excitation. It either reaches Earth or it reaches Andromeda. If it reaches Earth, it makes no sense to ask what happens to "the excitation on Andromeda" because there isn't any.

 

[A. Neumaier]

Exactly the 1 photon and the excitation of the field are relevant here! You have the excitation reaching me and Andromeda and I get the photon! What happens to the excitation on Andromeda. Is it real or probability.

The field is real and propagates everywhere. The photon is a detection event potentially triggered by the local field density at the detector.

It is like what happens to a water wave emanating from a boat. It travels everywhere and simply ends or is reflected at the shore, but where the circumstances are right it leaves a little trace. Randomly.

The mystery is only in how one tells the story...

 

[meopemuk]

It is like what happens to a water wave emanating from a boat. It travels everywhere and simply ends or is reflected at the shore, but where the circumstances are right it leaves a little trace. Randomly.

A single water wave can leave many little traces on different shores.

But the light (probability) wave emitted in a single atomic transition can be absorbed by only one detector. After the detector clicked, the probability wave collapses. This guarantees that there will be no second click in any other detector, and the energy emitted in the transition is absorbed only once.

So, there is no full analogy between light emission/absorption and water waves.

Eugene.

 

[ilper]

You are still missing the point. There is only one excitation. It either reaches Earth or it reaches Andromeda. If it reaches Earth, it makes no sense to ask what happens to "the excitation on Andromeda" because there isn't any.

Peter, have you read the thread from the beginning? I argued that the wavepacket can not be stabilized and it decays very soon so the excitation of the field which is a wavepacket will also reach Andromeda. The excitation in a linear theory like QM (and QFT which is continuation of it with SR) can not be a soliton. If the theory is not linear than the relativity would not be obeyed. (there is the no-copy theorem showing that linearity of QM saves the relativity in QM).
But even if you make a soliton and escape somehow no-copy theorem then you can not account for intereference in DSE with single photon.
I don't think QFT cares about this and it is swept silently under the carpet. If this problem was explained by QFT there wouldn't be the DSE wonder (at heart of QM - Feynman) and it be announced thriumphantly at the start of every QFT text. Its a pity I don't know why it is not discussed in QFT and many people are with wrong imaginations.
You see Neumeier thinks the excitation is everywhere but he think it is real. This is also wrong because it can not collapse as you know. (breaking SR of course).

 

[ilper]

The field is real and propagates everywhere. The photon is a detection event potentially triggered by the local field density at the detector.

It is like what happens to a water wave emanating from a boat. It travels everywhere and simply ends or is reflected at the shore, but where the circumstances are right it leaves a little trace. Randomly.

The mystery is only in how one tells the story...

As Eugene wrote if the excitation is real than the click of my detector would not destroy it and it must be detected on Andromeda too. If my detector has rendered it undetectable it ( my detector) must emanate signals everywhere faster than speed of light (immediate) to tell other detectors not to click.

 

[A. Neumaier]

A single water wave can leave many little traces on different shores.

But the light (probability) wave emitted in a single atomic transition can be absorbed by only one detector. After the detector clicked, the probability wave collapses. This guarantees that there will be no second click in any other detector, and the energy emitted in the transition is absorbed only once.

So, there is no full analogy between light emission/absorption and water waves.

Eugene.

The word like does not require a full analogy. Moreover, we were talking abou light from galaxies - this is very different from light emitted by an isolated atom, which you were describing.

 

[meopemuk]

The word like does not require a fill analogy. Moreover, we wer talking abou light from galaxies - this is ver different from light emitted by an isolated atom, which you were describing.

One atom emits one photon. One galaxy emits a zillion of photons. So, the difference is not in principle, but in numbers.

Eugene.

 

[A. Neumaier]

One atom emits one photon. One galaxy emits a zillion of photons. So, the difference is not in principle, but in numbers.

This would only be the case if the galaxy were a collection of noninteracting atoms. With interactions, this is not the usual situation. Typical light is approximately in a thermal coherent state, not in an N-photon state.

 

[Dale]

Due to the rampant personal speculation of the OP this thread is closed. Thanks to all who made a sincere effort to resolve it through instruction rather than moderation.