A Are quantum fields real objects in space?

[Lord Jestocost]

According to some versions of Copenhagen interpretation, the Moon does not exist when nobody looks at it. For instance, Wheeler said that “no phenomenon is a real phenomenon until it is an observed phenomenon.”

Regarding Wheeler’s statement, P. C. W. Davies and Julian R. Brown say in “The Ghost in the Atom: A Discussion of the Mysteries of Quantum Physics”:

“It means that, on its own an atom or electron or whatever cannot be said to 'exist' in the full, common-sense, notion of the word.” (italics in the original)

Or, as N. David Mermin (in “Making Better Sense of Quantum Mechanics”) has recently read some of Bohr’s quotations:

“Both quotations state that physics is not so much about phenomena, as it is about our experience of those phenomena.”

What you term “moon” is first and foremost a mental image, which is in your mind and not in the external world, at the end an encodement of a set of potentialities or possible outcomes of measurements. The Copenhagens merely warn people not to mistake this map for the territory.

 

[Demystifier]

The Copenhagen’s merely warn people not to mistake this map for the territory.

Actually, we have three things that should be distinguished:
1) The map (mathematical formalism of QM).
2) The territory (the physical objects existing irrespective of our observations).
3) Our vision of the territory (the measurement outcomes).
I view Copenhagen as a statement that we should talk seriously about 1) and 3), but not about 2). Realists, on the other hand, view 2) as the central object of research, while 1) and 3) are just the research tools.

 

[bhobba]

Well, he claims that there isn't anything according to Copenhagen.

I will not get into a discussion on what nothing means, form your own view of that, but clearly there is something under what most would think as 'something'. Consider a particle in a box and you measure its position. If there is no particle you will never - doesn't matter how many times you measure a position - you will never get an an answer. If there is a particle you will always get an answer. It modifies the usual idea of something somewhat - but most would say there is something there - we just do not know its properties until measured and can only predict probabilities. What's going on when not measured Copenhagen as far as I know from the various versions I have read says - we do not know - not there is nothing. Certainly the formalism doesn't say anything.

Thanks
Bill

 

[Lord Jestocost]

1) The map (mathematical formalism of QM).
2) The territory (the physical objects existing irrespective of our observations).
3) Our vision of the territory (the measurement outcomes).

I, personally, would beware of restricting the "Territory" to "physical objects".

 

[slufoot]

they are as real as anything.if it has an effect on any thing it is real

 

[martinbn]

I will not get into a discussion on what nothing means, form your own view of that, but clearly there is something under what most would think as 'something'. Consider a particle in a box and you measure its position. If there is no particle you will never - doesn't matter how many times you measure a position - you will never get an an answer. If there is a particle you will always get an answer. It modifies the usual idea of something somewhat - but most would say there is something there - we just do not know its properties until measured and can only predict probabilities. What's going on when not measured Copenhagen as far as I know from the various versions I have read says - we do not know - not there is nothing. Certainly the formalism doesn't say anything.

Thanks
Bill

That is my view and understanding.

 

[martinbn]

Mathematically it makes perfect sense to have a Hamiltonian as an object by its own. Physically, for a version of classical mechanics without particle trajectories see my https://link.springer.com/article/10.1007/s10702-006-1009-2

I might take a look, but I am already skeptical. You are switching between the notions. The point was whether you can have a version without particles (without objects no matter the name), now you are talking about a version without particle trajectories.

The Bell theorem says that if there is something, then this something obeys nonlocal laws. And yet, if I remember correctly, in other threads you deny nonlocality. My point is that it is inconsistent to accept that both (i) there is something and (ii) this something obeys local laws.

I have no problem with "nonlocality" except for the name. I also don't see how Bell's theorem says that if there is something it obeys nonlocal laws! It says that if those things that exist out there posses quantities that have values at all times then the laws are nonlocal.

How about the following quotes of Mermin, taken from https://en.wikipedia.org/wiki/Relational_quantum_mechanics#History_and_development :
David Mermin has contributed to the relational approach in his "Ithaca interpretation."[8] He uses the slogan "correlations without correlata", meaning that "correlations have physical reality; that which they correlate does not", so "correlations are the only fundamental and objective properties of the world".

That is more of the same. The fact that spins of two particles are correlated and the sipns don't have values before measurements doesn't imply that there are no particles.

 

[martinbn]

There is certainly nothing you are reading correctly.

There is your chance to clear things up.

 

[Demystifier]

I have no problem with "nonlocality" except for the name. I also don't see how Bell's theorem says that if there is something it obeys nonlocal laws! It says that if those things that exist out there posses quantities that have values at all times then the laws are nonlocal.

Then we agree more than I thought. Since you are fine with nonlocality, can you just remind me what exactly do you not like about the Bohmian interpretation? It's important for this thread because Bohmian interpretation is made precisely with the intention to say what is real in space.

 

[DarMM]

I have no problem with "nonlocality" except for the name. I also don't see how Bell's theorem says that if there is something it obeys nonlocal laws! It says that if those things that exist out there posses quantities that have values at all times then the laws are nonlocal.

Just for accuracy, that's not quite what it says. Possessing quantities at all times is the assumption technically called "Decorrelating explanation" in the assumptions of the theorem. One can retain it, but give up one of the other assumptions aside from locality.

 

[martinbn]

Then we agree more than I thought. Since you are fine with nonlocality, can you just remind me what exactly do you not like about the Bohmian interpretation? It's important for this thread because Bohmian interpretation is made precisely with the intention to say what is real in space.

I am fine with nonlicality in the sense in which QM is nonlocal. I am not fine with nonlocality when it means infinite speed of propagation. The main problem I have with BM is that it is very unclear (in fact quite confused) about the ontology of the wave function. Ah, and it is nonrelativistic.

 

[Demystifier]

I am fine with nonlicality in the sense in which QM is nonlocal. I am not fine with nonlocality when it means infinite speed of propagation.

In BM there is no infinite speed of propagation.

The main problem I have with BM is that it is very unclear (in fact quite confused) about the ontology of the wave function.

You can think of wave function $\psi(x,t)$ in BM as something analogous to the principal function $S(x,t)$ in the Hamilton-Jacobi formulation of classical mechanics. I think it reduces the confusion a lot.

Ah, and it is nonrelativistic.

That's a separate topic on which I was writing (on this forum and elsewhere) a lot.

 

[Lord Jestocost]

...but clearly there is something under what most would think as 'something'.

One should add, however, that this 'something' cannot be objectified, so it's of inscrutanable nature.

 

[stevendaryl]

The answer to the question in the title is clearly "No!" In general, a wavefunction is a function on configuration space, not physical space. When you have two particles, for instance, $\psi(x_1, x_2, t)$ is the probability amplitude of finding the first particle at location $x_1$ and the second particle at location $x_2$. In contrast, a physical field (such as the electric field) gives a value at each point in physical space.

[edit]This is a poor response to the original question, which was about fields, not wave functions.

What I would say about fields is that in QM they are operators, not values.

 

[bhobba]

One should add, however, that this 'something' cannot be objectified, so it's of inscrutanable nature.

Of course - it has exactly the same status as Weinberg pointed out but his critics forgot he carefully qualified - reality. Such fundamental things are notoriously hard to pin down. Its just most people would say if every-time you measured position you got an answer you would say something is in there - what is meant by something - blackout.

Thanks
Bill

 

[atyy]

The answer to the question in the title is clearly "No!" In general, a wavefunction is a function on configuration space, not physical space. When you have two particles, for instance, $\psi(x_1, x_2, t)$ is the probability amplitude of finding the first particle at location $x_1$ and the second particle at location $x_2$. In contrast, a physical field (such as the electric field) gives a value at each point in physical space.

The field has an operator-valued value at each point in physical space.

 

[bhobba]

The answer to the question in the title is clearly "No!" In general, a wavefunction is a function on configuration space, not physical space. When you have two particles, for instance, $\psi(x_1, x_2, t)$ is the probability amplitude of finding the first particle at location $x_1$ and the second particle at location $x_2$. In contrast, a physical field (such as the electric field) gives a value at each point in physical space.

That is true and got my like - but I think the OP was referring to the Quantum Fields of QFT which are quantum operators assigned to every point - not state AKA wave-functions. In QFT the state space is a fock space.

Thanks
Bill

 

[stevendaryl]

That is true and got my like - but I think the OP was referring to the Quantum Fields of QFT which are quantum operators assigned to every point - not state AKA wave-functions. In QFT the state space is a fock space.

Yes. It's sort of confusing in quantum field theory that the field operators seems like noncommuting versions of the wave functions of non-relativistic quantum mechanics. They really are not analogous. There more like the position and momentum operators of non-relativistic quantum mechanics.

 

[stevendaryl]

The field has an operator-valued value at each point in physical space.

You're right. The OP was not talking about the wave function.

 

[DarMM]

The wavefunctions in QFT are of the form (in the Heisenberg picture):
$$\Psi(\phi), \quad \phi \in \mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right), \quad \Psi \in \mathcal{H} = \mathcal{L}^{2}\left(\mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right),d\nu\right)$$
with $\mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right)$ the space of tempered Schwarz distributions on a spacelike slice. Depending on the measure $d\nu$ the Hilbert space has a Fock decomposition. It doesn't for $d > 1$ for an interacting theory (without inifinite volume cutoff).

 

[stevendaryl]

The wavefunctions in QFT are of the form (in the Heisenberg picture):
$$\Psi(\phi), \quad \phi \in \mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right), \quad \Psi \in \mathcal{H} = \mathcal{L}^{2}\left(\mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right),d\nu\right)$$
with $\mathcal{S}^{'}\left(\mathbb{R}^{d-1}\right)$ the space of tempered Schwarz distributions on a spacelike slice. Depending on the measure $d\nu$ the Hilbert space has a Fock decomposition.

Yes. In going from non-relativistic quantum mechanics to quantum field theory, I originally found it confusing because I thought of $\phi$ as the analogy for the nonrelativistic wave function, when it's actually $\Psi$. The confusion is made worse by the fact that the Heisenberg equations for $\phi$ look (at least for free fields) like the Schrodinger equation for the single-particle wave function.

 

[ZMacZ]

My personal view on this it's either field XOR particle..
The only time this appears to be untrue is when the two combined cause effect at the same time..
But in that case it's a field separate from the particle effect..
I think a lot of people mistake a field for a particle effect that they can't explain given current physics knowledge..

But in any case, when particles are involved it's not a field, and when it's a true field, it's not particles..
That's the annoying part..the particles do something we can only explain using field theories..
Hence it's called particle field, which in and by itself is a misnomer of sorts..

Anyways, that's my 10 cents..

 

[bhobba]

I think a lot of people mistake a field for a particle effect that they can't explain given current physics knowledge..

In QFT its all explained by the mathematics of the field - it has a form similar to the so called second quantization formulation, and hence quantum particles
https://pdfs.semanticscholar.org/2fb0/4475228ff385a44a16e3ba42b432d3bf5b17.pdf

That's how particles emerge from fields in QFT,

Thanks
Bill

 

[Elroch]

It is safe to say that both wave particle duality and the collapse of the wave function are not present when quantum mechanics is modeled precisely (particles are just waves which are localised, "collapsing" is the strong dependence of the state on the information that is available.
Roughly speaking a quantum state which is defined over both space and time expresses a state of knowledge. It is this state of knowledge which is affected by things such as the detection of a particle at a specific location. With a two slit experiment, the Schroedinger equation for a particle without any knowledge of a detector consists of a field which diffracts outwards from the two slits and exhibits interference. But once detected, the past locations of the particle are highly constrained to the two paths from the slits to the detection. If we knew in advance where the array of detectors were and the physics indicated that it was highly likely that one of the detectors would detect the particle, the correct wave function to express our state of knowledge before detection would be the sum of the ones after each of the individual possible detections (with some appropriate weights). [This notion is not part of the Copenhagen interpretation, and I have not seen it in an introductory text - it owes more to Feynman's viewpoint. The Copenhagen viewpoint seems to be based on a desire to retain causality in the evolution of the wave function as a physical field. An argument against this is that the wave function is a state of belief depending on some set of relevant information, and this remains true when the information is associated with things in the future of the wave function (such as the location of a set of detectors].
By contrast with the two wave functions associated with passing through one or other of the slits, these wave functions associated with distinct detections interfere with each other very little and add classically, like probabilities. The thing that makes the detections have this dramatic effect on the wave function is that the paths that end in a detection have very high probability for reasons which may be best expressed in terms of them being high entropy. This makes the contribution from other paths very small, unlike when the detectors are not there.

 

[stevendaryl]

It is safe to say that both wave particle duality and the collapse of the wave function are not present when quantum mechanics is modeled precisely (particles are just waves which are localised, "collapsing" is the strong dependence of the state on the information that is available.

I don't see how that interpretation of quantum probabilities is consistent with what we know about quantum mechanics.

If you pass unpolarized light through a polarizing filter, it comes out polarized in the direction corresponding to the filter orientation. Viewing this as a matter of acquiring information about the photons is certainly not complete. If you have three filters oriented at angle $0^o$, $45^o$ and $90^o$, 12.5% of the unpolarized photons (1/8 of them) will pass through all three filters. That can't possibly be a matter of just learning the pre-existing polarization of the photons, because it is impossible to have a photon that will pass through a filter at both 0 and 90 degrees. Passing through the filter changes the polarization of the photons. So it's not simply a matter of updating information.

 

[TonyP0927]

Would you consider the quantum system itself to be real in Copenhagen?

The Copenhagen interpretation is more "real" because it's observable: Upon observation, all quantum possibilities collapse into one outcome. This has been tested with the double-slit experiment.
There's a problem though, and it has to do with the "quantum information paradox". Upon waveform collapse, the information about all of the other possible quantum states is lost, thus violating the conservation of information.
Everett's MWI offers a solution to this: All outcomes are equally real but exist in different branches of the multiverse, which we can not observe. The conservation of information states that information can not be destroyed but it does not mean it has to be accessible, so the MWI satisfies this.

 

[TonyP0927]

So the Moon (considered as a many-particle quantum object) is not real when nobody looks at it?

Oh it's real! The gravitational attraction is obvious, otherwise there would be drastic changes to tides if nobody looked at the moon.

You can think of the probability of a macroscopic object existing is equal to that the ~sum of the probabilities of enough particles being in the position they need to be to form the moon is essentially 100%.

 

[Elroch]

I don't see how that interpretation of quantum probabilities is consistent with what we know about quantum mechanics.

If you pass unpolarized light through a polarizing filter, it comes out polarized in the direction corresponding to the filter orientation. Viewing this as a matter of acquiring information about the photons is certainly not complete. If you have three filters oriented at angle $0^o$, $45^o$ and $90^o$, 12.5% of the unpolarized photons (1/8 of them) will pass through all three filters. That can't possibly be a matter of just learning the pre-existing polarization of the photons, because it is impossible to have a photon that will pass through a filter at both 0 and 90 degrees. Passing through the filter changes the polarization of the photons. So it's not simply a matter of updating information.

I did not use the incorrect notion that there is a "pre-existing polarization of the photons" that stays the same throughout the life of the photon. This is physically inaccurate: the polarisation of a photon changes when it interacts with a polarising filter.
Instead what you have is the following situation. After a photon reaches a polarising filter it either passes through and acquires a specific definite state of polarisation or it does not pass through. The probability of it passing through is determined by its previous (possibly definite, possibly different) state of polarisation.
So when you have detected a photon that has passed through a set of polarising filters, you know the state of polarisation it had in the space between each two polarising filters. This polarisation changes at each filter if the filters are not aligned.
So no problem there.

 

[stevendaryl]

I did not use the incorrect notion that there is a "pre-existing polarization of the photons" that stays the same throughout the life of the photon. This is physically inaccurate: the polarisation of a photon changes when it interacts with a polarising filter.
Instead what you have is the following situation. After a photon reaches a polarising filter it either passes through and acquires a specific definite state of polarisation or it does not pass through.

Okay, but "acquiring a definite state of polarization" is what is meant by "collapse". You were saying that it was about information. I don't see that it has anything to do with information.

 

[Elroch]

It's all about information. :) The information interpretation of quantum mechanics is entirely valid and I can detect no differences between it and my own way of thinking.
A wave function is a description of what is known about a particle which includes its position and momentum, but also all information about polarisation, spin etc., in full generality. (Also eg even flavor, in the case of neutrino oscillation).
In both the most common case of the discovery of the position of a particle by detection which is the typical example of "wave function collapse" and the case of a photon passing through a filter, what happens is an interaction between the particle and another quantum object which in the first case determines the position of the particle and in the second case determines its polarisation afterwards.
Position is an infinite dimensional property expressed in terms of a unique preferred basis of delta functions at every point, while polarisation is a 2-dimensional property which can be expressed equally well in terms of bases based on different chosen orientations: this is the distinct property that makes experiments with polarising filters different.
In both the case of position and polarisation, clearly the properties change as a wave function evolves. As I see it, the difference is that the former changes constantly, the latter changes only with certain interactions (hence astronomers can measure polarisation produced light years away).