The thermal interpretation of quantum physics

[A. Neumaier]

If that's true, then it could explain the Born rule in general, and violations of Bell inequalities in particular, by beables that satisfy local laws. Is that correct? If so, then I am very skeptical that it's true because it would be in contradiction with the Bell theorem.

No. The fields satisfy local causal laws but products of fields with different arguments are nonlocal and satisfy causality only in the sense of extended causality discussed in Subsection 4.4 of Part II.

Thus there are local beables such as $\langle \phi(x)\rangle$ and nonlocal beables such as $\langle \phi(x)\phi(y)\rangle$, where $x$ and $y$ can be at large spacelike distance, or rather smeared versions of these distributions.

Most of physics works under conditions where only local beables are probed directly, and where nonlocal beables only have a small correcting influence (linear response theory). But experiments testing Bell violations probe conditions in which the nonlocal beables are quite influential.

 

[Demystifier]

No. The fields satisfy local laws but products of fields with different arguments are nonlocal.

Thus there are local beables such as $\langle \phi(x)\rangle$ and nonlocal beables such as $\langle \phi(x)\phi(y)\rangle$, where $x$ and $y$ can be at large spacelike distance, or rather smeared versions of these distributions.

Most of physics works under conditions where only local beables are probed directly, and where nonlocal beables only have a small correcting influence (linear response theory). But experiments testing Bell violations probe conditions in which the nonlocal beables are quite influential.

But $\langle \phi(x)\phi(y)\rangle$ satisfies local laws. For instance, if $\langle \phi(x)\rangle$ satisfies the local law
$$(\Box_x +m^2) \langle \phi(x)\rangle=0$$
then $\langle \phi(x)\phi(y)\rangle$ satisfies the local laws
$$(\Box_x +m^2) \langle \phi(x)\phi(y)\rangle=0$$
$$(\Box_y +m^2) \langle \phi(x)\phi(y)\rangle=0$$
Interactions make it more complicated, but they are still local. The expected values in QFT satisfy local laws (that's why QFT is called "local"), so the expected values cannot serve as beables that explain violation of Bell inequalities.

 

[A. Neumaier]

But $\langle \phi(x)\phi(y)\rangle$ satisfies local laws. For instance, if $\langle \phi(x)\rangle$ satisfies the local law
$$(\Box_x +m^2) \langle \phi(x)\rangle=0$$
then $\langle \phi(x)\phi(y)\rangle$ satisfies the local laws
$$(\Box_x +m^2) \langle \phi(x)\phi(y)\rangle=0$$
$$(\Box_y +m^2) \langle \phi(x)\phi(y)\rangle=0$$
Interactions make it more complicated, but they are still local. The expected values in QFT satisfy local laws (that's why QFT is called "local"), so the expected values cannot serve as beables that explain violation of Bell inequalities.

You may call this local. But this sort of locality is not excluded by violations of Bell-type theorems!

 

[Demystifier]

You may call this local. But this sort of locality is not excluded by violations of Bell-type theorems!

I disagree. Of course, the local equations above are consistent with Bell-type theorems, but my point is that those equations cannot describe beables in the Bell sense.

 

[ftr]

Look around you. Everything flows, hence is represented by currents moving densities. Or is solid, hence is represented by stress fields deforming densities. On other scales it is not so different, in principle.

I think you misunderstood me. I am very much aware of the field point of view.
https://rreece.github.io/talks/pdf/2017-09-24-RReece-Fields-before-particles.pdf
I guess I was just asking in what sense you think they are "real" as in my post.

 

[A. Neumaier]

I disagree. Of course, the local equations above are consistent with Bell-type theorems, but my point is that those equations cannot describe beables in the Bell sense.

Why not? Where is the contradiction?

 

[ftr]

beables in the Bell sense.

What do you think Bell meant?

https://cds.cern.ch/record/980036/files/197508125.pdf

 

[A. Neumaier]

I think you misunderstood me. I am very much aware of the field point of view.
https://rreece.github.io/talks/pdf/2017-09-24-RReece-Fields-before-particles.pdf
I guess I was just asking in what sense you think they are "real" as in my post.

I don't understand your question.

There is just one way in which something is real - it exists and has properties independent of what we think about it. Our subjective views are approximations to the real.

 

[Demystifier]

@A. Neumaier , @DarMM , @ftr and others, I think I finally understand what, from my perspective at least, is the main problem with the thermal interpretation of QM. The problem is that it introduces a kind of beables that are the exact opposite of what Bell beables are supposed to be. Bell beables are supposed to be kinematically local (i.e. have definite values at spacetime positions) but dinamically non-local (i.e. satisfy non-local equations of motion). By contrast, the beables in the thermal interpretation are kinematically non-local but dynamically local. In that sense, the thermal interpretation is more similar to the many-world interpretation than to the Bohmian interpretation.

 

[A. Neumaier]

what Bell beables are supposed to be

What Bell specifies about beables is in his book of reprints,

• J.S. Bell, Speakable ans unspeakable in quantum mechanics, Cambridge University Press, Cambridge 1987.

In Chapter 5 (p.41), we read:

it should again become possible to say of a system not that such and such may be observed to be so but that such and such be so. The theory would not be about 'observables' but about 'beables'. These beables need not of course resemble those of, say, classical electron theory; but at least they should, on the macroscopic level, yield an image of the everyday classical world

Chapter 7 is about the special class of ''local beables''; on p.53 we find the criterion:

We will be particularly concerned with local beables, those which (unlike for example the total energy) can be assigned to some bounded space-time region.

My bilocal beables satisfy this 'local beable' criterion as stated. But his subsequent analysis assumes that there are no beables spanning big regions (in Figure 3, p.55, intersecting M, N, and $\Lambda$, say). This strong form of locality is not satisfied, so that Bell's conclusions don't apply. Thus my bilocal beables are not local beables in the sense intended by Bell, though his definition is too vague to express this.

Thus Bell's analysis has the same sort of fault as earlier von Neumann's no-go theorem for hidden variables, that he silently assumes more than is warranted by the physics.

But nothing in Bell's work forbids beables to be nonlocal.
In Chapter 19 on ''Beables for quantum field theory'', Bell writes (p.174):

The beables of the theory are those elements which might correspond to elements of reality, to things which exist. Their existence does not depend on 'observation'.

Nothing else is required.

the main problem with the thermal interpretation [...] the beables in the thermal interpretation are kinematically non-local but dynamically local.

I don't understand why this should be a problem...

 

[Demystifier]

There is also a formal similarity between many worlds and thermal interpretation. In the 2-particle case, the many-world beable is a 2-particle wave function, which in the QFT language can be expressed as something like
$$\psi(x,y)=\langle 0|\phi(x)\phi(y)|\psi\rangle$$
This is quite similar to the bi-local beable in the thermal interpretation
$$\langle \phi(x)\phi(y) \rangle= \langle\psi|\phi(x)\phi(y)|\psi\rangle$$

 

[A. Neumaier]

There is also a formal similarity between many worlds and thermal interpretation. In the 2-particle case, the many-world beable is a 2-particle wave function, which in the QFT language can be expressed as something like
$$\psi(x,y)=\langle 0|\phi(x)\phi(y)|\psi\rangle$$
This is quite similar to the bi-local beable in the thermal interpretation
$$\langle \phi(x)\phi(y) \rangle= \langle\psi|\phi(x)\phi(y)|\psi\rangle$$

This is a very superficial and meaningless similarity. The many-world expression is linear in $\psi$ which poses all kinds of problems with interpreting superpositions. The thermal expression is quadratic in $\psi$ which makes a big difference - one cannot superimpose thermal beables!

 

[Demystifier]

Thus my bilocal beables are not local beables in the sense intended by Bell, though his definition is too vague to express this.

Thus Bell's analysis has the same sort of fault as earlier von Neumann's no-go theorem for hidden variables, that he silently assumes more than is warranted by the physics.

Well, I am not saying that your interpretation is wrong. I am just explaining what do I not like about it.

 

[A. Neumaier]

Well, I am not saying that your interpretation is wrong. I am just explaining what do I not like about it.

But you phrased it wrongly as not deserving the name beables:

beables that are the exact opposite of what Bell beables are supposed to be

 

[Demystifier]

But you phrased it wrongly as not deserving the name beables:

They deserve the name of beables, but not beables in the Bell sense. For instance, as you quoted, Bell required that "beables need not of course resemble those of, say, classical electron theory; but at least they should, on the macroscopic level, yield an image of the everyday classical world". I don't see how your bilocal beables, on the macroscopic level, yield an image of the everyday classical world.

 

[A. Neumaier]

They deserve the name of beables, but not beables in the Bell sense. For instance, as you quoted, Bell required that "beables need not of course resemble those of, say, classical electron theory; but at least they should, on the macroscopic level, yield an image of the everyday classical world". I don't see how your bilocal beables, on the macroscopic level, yield an image of the everyday classical world.

I don't see either how a Bohmian particle position yields on the macroscopic level an image of the everyday classical world. Only certain things computable from beables (in Bell's sense) are required to yield on the macroscopic level an image of the everyday classical world.

Note also that ''classical'' does not exclude ''nonlocal''. The mass of the Earth is a very nonlocal classical beable, and Bell mentioned explicitly the total energy of a bounded system as a nonlocal beable.

Thus the thermal bilocal beables are as much beables in Bell's sense as Bohmian particles. But unlike the latter, they are not local.

 

[ftr]

I don't understand your question.

There is just one way in which something is real - it exists and has properties independent of what we think about it. Our subjective views are approximations to the real.

Ok. I will be bold, what is the ontology of field, just numbers. I think this has a bearing on what is meant by beable.

 

[julcab12]

I think we must ignore some bounds of what it means local or nonlocal. Everything is approximation. Much like in open sets; continuous infinitely small patches, but with blurred contours. Forget any classical picture of a particle like a ball or point, it is very rough approximation, forget wavefunction, it is less rough approximation, currently, we understand all particles as excitation of some field, imagine it as a wave of a sea, there is no sense of sailing in which point it is, or how wide it is. TI offers some 'temporary' relief to some weirdness of QM.

 

[A. Neumaier]

Ok. I will be bold, what is the ontology of field, just numbers. I think this has a bearing on what is meant by beable.

In the thermal interpretation, one says (by definition) that the q-expectations of operators acting on the state space of a quantum system (in particular of fields) exist objectively, define real (actual) properties (beables) of the system, thus defining an ontology. Having the ontology, one can draw conclusions, for example about what can be known (observed) about a quantum system, and how accurately.

In my view, there is nothing more to ontology than this.

 

[Demystifier]

I don't see either how a Bohmian particle position yields on the macroscopic level an image of the everyday classical world.

This may help: https://en.wikipedia.org/wiki/Pointillism

 

[A. Neumaier]

This may help: https://en.wikipedia.org/wiki/Pointillism

It just illustrates what I said before, that macroscopic beables with a classical meaning are composed of certain nonclassical beables. But the single Bohmian beable has no classical macroscopic meaning. Thus Bell cannot have meant that every beable must have a classical macroscopic meaning.

 

[Demystifier]

It just illustrates what I said before, that macroscopic beables with a classical meaning are composed of certain nonclassical beables. But the single Bohmian beable has no classical macroscopic meaning. Thus Bell cannot have meant that every beable must have a classical macroscopic meaning.

Of course not. He meant that collection of many microscopic beables must have a classical macroscopic meaning.

But the thermal interpretation, or any interpretation with bilocal beables, is different. A bilocal beable means that "having a dot here" and "having a dot there" is not the same as "having dots here and there". Maybe it's my lack of imagination, but I cannot imagine pointilism painting with such a property.

Anyway, pointilism painting is a great metaphor for Bohmian beables, I think we can agree on that.

 

[A. Neumaier]

pointilism painting is a great metaphor for Bohmian beables

Yes, but not for the beables in the thermal interpretatiton.

A bilocal beable means that "having a dot here" and "having a dot there" is not the same as "having dots here and there".

The thermal interpretation embodies the classical truth ''There is more to the whole than to the parts'', which is also observed in nonlocal entanglement experiments.

 

[Demystifier]

Yes, but not for the beables in the thermal interpretatiton.

The thermal interpretation embodies the classical truth ''There is more to the whole than to the parts'', which is also observed in nonlocal entanglement experiments.

Thanks for the discussion, now I think I understand what the thermal interpretation is.

 

[N88]

The thermal interpretation embodies the classical truth ''There is more to the whole than to the parts'', which is also observed in nonlocal entanglement experiments.

How is the word "nonlocal' defined in the TI? Is it required [or redundant and can be dropped] in the above phrase "nonlocal entanglement experiments"?

 

[A. Neumaier]

How is the word "nonlocal' defined in the TI? Is it required [or redundant and can be dropped] in the above phrase "nonlocal entanglement experiments"?

I used the word informally.
You may replace "nonlocal" by "long distance" in the above phrase, to get a more precise rendering of the intended meaning.

 

[N88]

I used the word informally.
You may replace "nonlocal" by "long distance" in the above phrase, to get a more precise rendering of the intended meaning.

Thank you. Is it too much to hope that TI can be formulated and discussed without reference to that (in my experience) "many-times confusing" word?

 

[A. Neumaier]

Thank you. Is it too much to hope that TI can be formulated and discussed without reference to that (in my experience) "many-times confusing" word?

In the three papers I am fairly careful with my words. Read Section 4 of Part II. But it doesn't make sense to avoid the word.

Nonlocal simply means referring to multiple, not infinitely close locations.

 

[Demystifier]

Noise from the environment causes it to quickly decay into one of the slow mode manifolds giving a discrete outcome not fully reflective of $\langle A\rangle$.

Where can I see a simple general quantitative explanation of why exactly that happens?
By simple I mean not longer than a couple of pages, by general I mean referring to a wide class of cases, by quantitative I mean containing equations (not merely verbal hand waving). Mathematical rigor is not required.

 

[cube137]

The official description of the thermal interpretation of quantum physics can be found in my just finished papers

Foundations of quantum physics I. A critique of the tradition,
Foundations of quantum physics II. The thermal interpretation,
Foundations of quantum physics III. Measurement.

They are also accessible through the arXiv at
arXiv:1902.10778 (Part I), arXiv:1902.10779 (Part II), arXiv:1902.10782 (Part III).

If you are short of time, start reading at post #260. DarMM gave in post #268 a nice summary of the thermal interpretation. The simplest quantum system, a qubit, was already described by Stokes 1852, in terms essentially equivalent to the thermal interpretation. Post #484 is my very short summary and post #479 contains links to my explanations of the connections between theory and experiments. Open problems related to the thermal interpretation are discussed in post #293.

Three reviews (Part I, Part II, Part III) are on PhysicsOverflow, together with some comments by me.

In Februray 20, 2016. I started the thread "Neumaier quantum 101" to get a handle of the thermal interpretation. It's nice to know Neumaier has even developed it further. I wrote this then:

https://www.physicsforums.com/threads/neumaier-quantum-101.858289/
"For example you have detectors in a circle and a photon was emitted from the center. Does the photon travel with a trajectory to a particular detector or does it travel as wave.. if wave.. what is the meaning it travels as wave function.. upon detection.. it only appears at one detector (the wave function collapses). In your view.. what really happened?"

Neumaier described it" It travels as a wave, as one knows since Huygens 1690. It doesn't appear on the detector; it disappears there, and the transmission of energy leaves random spots at a rate determined by the impinging energy density.".

But why only one spot appear and not two simultaneously? Won't it be possible to put the above under experimental test or scrutiny as it is the most graphic ramifications yet of the thermal interpretation? Before delving even to greater and densest depth. We must deal with the above first.

 

[Demystifier]

But why only one spot appear and not two simultaneously?

My post above is asking the same question in a different form.

 

[stevendaryl]

Where can I see a simple general quantitative explanation of why exactly that happens?
By simple I mean not longer than a couple of pages, by general I mean referring to a wide class of cases, by quantitative I mean containing equations (not merely verbal hand waving). Mathematical rigor is not required.

My feeling is that it doesn't happen. The wave function by linearity cannot evolve from a superposition of two possibilities to a choice of one possibility or the other.

There is a trick that can be performed that makes it almost seem as if this happens. If a microscopic superposition interacts with a macroscopic system (such as a measurement device), you can no longer describe the situation as a pure state of the microscopic system. So you form the density matrix and trace out the environmental degrees of freedom. The resulting reduced density matrix looks as if the system is in a mixed state, which can be interpreted as the system being in one or other possibilities with a particular probability for each. But that does not actually mean that the environment has selected one of the possibilities; if you enlarge the "system" to include the environment, you will see that it is still a pure state.

At least, that's the way things work in nonrelativistic quantum mechanics. @DarMM tells me that in QFT you can have the situation where you have essentially mixed-state density matrices that cannot be interpreted as arising from any pure state.

 

[Demystifier]

The wave function by linearity cannot evolve from a superposition of two possibilities to a choice of one possibility or the other.

I think here one has a different idea in mind. Suppose that initially a subsystem is in a superposition of the ground state and an excited state. When you couple it to the environment, the subsystem decays continually into the ground state only. This explains why do we finally find the subsystem in a definite state of energy, and not in a superposition or mixture of different energies. The problem with that, as I see it, is that in 50% cases of energy measurements we should finally find the subsystem in the excited state, which the decay above cannot explain.

 

[DarMM]

At least, that's the way things work in nonrelativistic quantum mechanics. @DarMM tells me that in QFT you can have the situation where you have essentially mixed-state density matrices that cannot be interpreted as arising from any pure state.

Yes, for the local algebras in QFT all states are mixed states, no pure states exist. In the sense that for all states $\omega$ and observables $A$, there always exists another pair of states $\omega_1$ and $\omega_2$ such that:
$$\omega\left(A\right) = \frac{1}{2}\omega_{1}\left(A\right) + \frac{1}{2}\omega_{2}\left(A\right)$$
So every state can always be thought of as "Classical Ingorance" of two others if one wishes.

However my view is that thinking of density matrices as ignorance of the "true" pure state is wrong even in non-relativistic QM as if they were you would expect the space of density matrices to be (isomoprhic to) $\mathcal{L}^{1}\left(\mathcal{H}\right)$ with $\mathcal{H}$ the quantum Hilbert space.

 

[Demystifier]

I guess the answer to my question above is the following. The dynamics of the open subsystem is described not only by a single Hamiltonian, but also by a series of Lindblad operators. The consequence, I guess, is that there are many (rather than one) stable states to which the system can finally decay. To which one it will decay depends on fine details of the initial state that in practice cannot be known exactly, so they play a role of "hidden variables".