The construction of particles in QFT

[ilper]

So you never heard of position operator in QM? If so, I would suggest you to first learn more about QM before trying to understand QFT.

Having position operator is not the same as having position. There is no position operator with eigenfunction Dirac delta function.

 

[Demystifier]

Having position operator is not the same as having position. There is no position operator with eigenfunction Dirac delta function.

True, but the same can be said for field operator in QFT.

 

[ilper]

No. QFT is not modeling the process as "things in space changing in time". It is modeling the process as a whole self-consistent 4-dimensional spacetime with self-consistent 4-dimensional objects in it. There is no "vanishing" of anything at the end.

That's for sure. But in order to grasp things physically as a mental picture it is not so bad. Zee in his book is adopting it. Note also that there is a nonrelativistic variant of QFT.
I am more interested in your note "Nothing is vanishing at the end." If so there is no collapse and one would expect the excitation to be measured in other points as well?

 

[A. Neumaier]

Apart from that I could not imagine what would that mean physically (an operator acting on every state and giving infinity) but math gives every imaginary possible.
Ok but what do we get when measured. Any number of particles with a subsequent probability, or what?

Measured is the expectation value of operators obtained by integrating the quantum field over a finite region of spacetime. This produces extensive numbers (with a computable uncertainty) roughly proportional to the integrated volume, not probabilities. Just like measuring the concentration of salt at various places in a lake produces extensive numbers, not probabilities.

 

[ilper]

Measured is the expectation value of operators obtained by integrating the quantum field over a finite region of spacetime. This produces extensive numbers (with a computable uncertainty) roughly proportional to the integrated volume, not probabilities. Just like measuring the concentration of salt at various places in a lake produces extensive numbers, not probabilities.

Ha, to get the expectation value you need the probabilities for every value from the eigenvalues of the operator. So it is what I thought - loosely speaking a couple of vectors from eigenvalues and probabilities in spacetime.

 

[A. Neumaier]

Ha, to get the expectation value you need the probabilities for every value from the eigenvalues of the operator. So it is what I thought - loosely speaking a couple of vectors from eigenvalues and probabilities in spacetime.

No. To predict the expectation value you just use the formula ##\langle A\rangle=Tr~\rho A##. The measurement does not produce numbers with probabilities for averaging; it produces directly the expectation value.

 

[Demystifier]

No. To predict the expectation value you just use the formula ##\langle A\rangle=Tr~\rho A##. The measurement does not produce numbers with probabilities for averaging; it produces directly the expectation value.

But it doesn't get rid of probabilities. To account for the possibility of different measurement outcomes, your ##\rho## needs to satisfy a stochastic equation, which, of course, involves probabilities.

 

[PeterDonis]

If so there is no collapse and one would expect the excitation to be measured in other points as well?

You're missing the point. When you define a model in QFT, you are defining in advance exactly where in spacetime measurements are being made; those are the "sinks". If you change where measurements are made, you change the model, and you have to re-do all your calculations with the changed model. Asking whether something could be measured at other places than where it is measured in a given QFT model makes no sense.

 

[ilper]

You're missing the point. When you define a model in QFT, you are defining in advance exactly where in spacetime measurements are being made; those are the "sinks". If you change where measurements are made, you change the model, and you have to re-do all your calculations with the changed model. Asking whether something could be measured at other places than where it is measured in a given QFT model makes no sense.

Does the light from a star know you will measure it on Earth? If you doesn't intend to measure it it will not shine?
If QFT is just a model so pure defined, where do come from this talks about fields being building blocks of Universe from David Tong (not sure about his name but I saw a lecture in Cambridge or Oxford on youtube).

 

[PeterDonis]

Does the light from a star know you will measure it on Earth?

You're missing the point again. When you use QFT to model a photon from a distant star being received on Earth, those two events (emission from the star, detection on Earth) are inputs to the model. You already know that those are the endpoints. You don't use QFT to find that out.

If you want to use QFT to find out where the light from the distant star might go, you use multiple models, choosing different endpoints for each one. For example, you could pick the Earth as the endpoint in one model and some planet in the Andromeda galaxy as the endpoint in another. Then those two calculations would give you the relative probability of a photon from the distant star being detected at those two locations.

where do come from this talks about fields being building blocks of Universe from David Tong

From the fact that the most fundamental theory we have of everything except gravity is a QFT--the Standard Model of particle physics. None of which changes anything I said.

 

[A. Neumaier]

But it doesn't get rid of probabilities. To account for the possibility of different measurement outcomes, your ##\rho## needs to satisfy a stochastic equation, which, of course, involves probabilities.

My statement was about measuring quantum fields. The only statistics appearing is the same as that for measuring classical fields as in pre-quantum times. Born's rule is not involved.

 

[ilper]

From the fact that the most fundamental theory we have of everything except gravity is a QFT--the Standard Model of particle physics. None of which changes anything I said.

In QM (intended to be TOE) things don't happen just for separate cases. And QFT is following QM making it relativistic.

 

[EPR]

In QM (intended to be TOE) things don't happen just for separate cases. And QFT is following QM making it relativistic.

Those 'separate' cases form the observed reality.

 

[ilper]

Those 'separate' cases form the observed reality.

Mathematics is for the separate cases. Physics is about the whole physics (nature).

 

[EPR]

Nature appears to be fundamentally probabilistic. This is the new physics.

 

[ilper]

Those 'separate' cases form the observed reality.

Nature appears to be fundamentally probabilistic. This is the new physics.

Well but you see what David Tong is saying that QF are building blocks. That does not fit with probability.

 

[ilper]

If you want to use QFT to find out where the light from the distant star might go, you use multiple models, choosing different endpoints for each one. For example, you could pick the Earth as the endpoint in one model and some planet in the Andromeda galaxy as the endpoint in another. Then those two calculations would give you the relative probability of a photon from the distant star being detected at those two locations.

You insisted earlier that in QFT nothing vanishes. But if I get the photon on Earth the poor creatures from Andromeda will not detect the excitation of the field nevertheless that in the multiple model it has reach them in order to calculate the relative probability. 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. (like Born 'solution' to QM paradoxes).

 

[ilper]

My statement was about measuring quantum fields. The only statistics appearing is the same as that for measuring classical fields as in pre-quantum times. Born's rule is not involved.

If the the statistic is as for classical fields you will receive a definite result and expectation value is superfluous. Then so are also operators.

 

[A. Neumaier]

If the statistic is as for classical fields you will receive a definite result and expectation value is superfluous. Then so are also operators.

The ensemble expectation values of the quantum fields are the quantum versions of the values of the classical fields. They don't have a statistical interpretation - one cannot prepare multiple copies of a quantum field at the same spacetime location to get its statistics!

 

[vanhees71]

Well but you see what David Tong is saying that QF are building blocks. That does not fit with probability.

Quantum fields use the concepts of quantum theory, and quantum theory is intrinsically probabilistic, and as far as we know for more than 90 years now Nature seems to be intrinsically probabilistic. Many people seem still to have some quibbles with this, because of philosophical prejudices. This brought, however, some of the most profound findings of modern physics, particularly in quantum optics, and this lead to the newest generation of engineering science, i.e., quantum-information technology.

You may still have your philosophical prejudices against a probabilistic nature, but the facts are the facts!

 

[ilper]

Quantum fields use the concepts of quantum theory, and quantum theory is intrinsically probabilistic, and as far as we know for more than 90 years now Nature seems to be intrinsically probabilistic. Many people seem still to have some quibbles with this, because of philosophical prejudices. This brought, however, some of the most profound findings of modern physics, particularly in quantum optics, and this lead to the newest generation of engineering science, i.e., quantum-information technology.

You may still have your philosophical prejudices against a probabilistic nature, but the facts are the facts!

So Quantum Fields are probabilistic fields and not some real stuff.
1. one has different probabilities for the eigenvalues of the operator of some physical quantities in spacetime.
2. after measurement they collapse (to vacuum state) as its cousin the wavefunction (to 0)
Do you agree?

 

[EPR]

What do you mean by 'real'?
Quantum fields are matter - what you observe in daily life(basically fermionic and bosonic fields).

Is an electron real? Certanly not classically real but a very measureable entity.

 

[vanhees71]

No, of course not.

In quantum theory you have a Hilbert space and an algebra of observables realized as self-adjoint operators on Hilbert space the eigenvalues of these operators are the values the observables described by them can take. Then there is a self-adjoint operator with trace 1, the statistical operator, that describes the state of the system and is determined by the preparation of the described system at the initial time of the experiment. With the eigenvectors of the observable operators and the statistical operator you can calculate the probability to find any of the possible values of an observable when measured, given the state of the system. There not more to this. In general you cannot say what happens to the system in a measurement. This you have to analyse given the concrete setup of the measurement. The socalled collapse of the state is an ad-hoc assumption of some kinds of the Copenhagen interpretation and inconsistent with relativistic local quantum field theory, according to which there cannot be any instantaneous actions at a distance by assumption. Fortunately the collapse assumption is not needed to describe nature with quantum theory.

In relativistic quantum field theory the observable operators are constructed with help of fundamental field operators (which usually do not directly represent observables). This is a mathematical concept to construct locally interacting field theories which are consistent with the relativistic space-time description and its implied notion of causality.

The most common experiment described by relativistic QFT are scattering experiments, which consist in the preparation of two particles (described by asymptotic free "in-states") which scatter at each other when coming close. What's then observed after the collision are again particles (described by asymptotic free "out-states") and their properties (mass- and momentum distributions etc.), i.e., you describe a reaction in collisions in terms of cross sections, which are transition probability rates per incoming-particle flux for a given reaction.

 

[ilper]

What do you mean by 'real'?
Quantum fields are matter - what you observe in daily life(basically fermionic and bosonic fields).

Is an electron real? Certanly not classically real but a very measureable entity.

I can not tell what is real but I can tell what is not. The probability is not real. Then as the fields are probabilities there are also not real. And the electron which is an excitation of probability is not real. Nothing is real.

 

[A. Neumaier]

I can not tell what is real but I can tell what is not. The probability is not real. Then as the fields are probabilities there are also not real. And the electron which is an excitation of probability is not real. Nothing is real.

Fields are not probabilities. You are thoroughly mistaken in your views.

 

[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.