I The thermal interpretation of quantum physics

[A. Neumaier]

Well, smearing is something closely related to measurements, so to me it doesn't make much sense to associate smearing with something that should not depend on measurements. For instance, in a measurement of a far galaxy one uses smearing over test functions which are light years wide, but it does not make sense to think that the galaxy itself does not have properties on much smaller scales invisible to us.

Beables can be formed with arbitrary smearing, and measured is a particular smearing. Thus the collection of beables is measurement independent. Which one you can measure depends on the detector and the precise measurement protocol.

Similarly, in traditional Bohmian mechanics, all particle positions are beables. Thus the collection of beables is measurement independent. Which ones you can measure also depends on the detector (since only those falling on the detector qualify).

 

[Demystifier]

Beables can be formed with arbitrary smearing, and measured is a particular smearing. Thus the collection of beables is measurement independent. Which one you can measure depends on the detector and the precise measurement protocol.

Similarly, in traditional Bohmian mechanics, all particle positions are beables. Thus the collection of beables is measurement independent. Which ones you can measure also depends on the detector (since only those falling on the detector qualify).

For the sake of comparison, let us consider a distorted version of Bohmian mechanics which says the following: Particle positions are beables for all kinds of particles, irrespectively of whether the particles are fundamental or not. It would mean that not only electrons and quarks have trajectories, but also that pions, protons, neutrons, nuclei, atoms and molecules have Bohmian trajectories. In such a distorted version perhaps even phonons would have Bohmian trajectories, or at least it would not be completely clear why phonon should not have a trajectory. Such a distorted version of Bohmian mechanics would be quite analogous to the idea that any smearing defines a beable in the thermal interpretation.

But such a distorted version of Bohmian mechanics is not satisfying (do I need to explain why?). For a similar reason, it does not look satisfying to me that any smearing defines a beable in the thermal interpretation.

Or let me ask a question. Suppose that $\phi(x)$ is the field operator for a non-fundamental field, such as pion field (see e.g. Bjorken Drell QFT book) or phonon field . Would you say that $\langle \phi(f)\rangle$ is a beable in this case?

 

[A. Neumaier]

Suppose that $\phi(x)$ is the field operator for a non-fundamental field, such as pion field (see e.g. Bjorken Drell QFT book) or phonon field . Would you say that $\langle \phi(f)\rangle$ is a beable in this case?

Of course, since it is a q-expectation. In the thermal interpetation, all q-expectations are beables. Moreover, pions and phonons can be detected, so these beables are even approximately measurable.

The water field is a more intuitive nonfundamental field of beables. Corresponding currents are routinely measured by engineers or even when we wash our hands. They even gave the name for the general concept of currents!

It would be very strange if the fields describing ordinary life experience were not beables!

 

[Demystifier]

Of course, since it is a q-expectation. All q-expectations are beables. Moreover, pions and phonons can be detected, so these beables are even approximately measurable.

A water field is a more intuitive nonfundamental field of beables. The corresponding currents are routinely measured by engineers or even when we wash our hands.

It would be very strange if the fields describing ordinary life experience were not beables!

OK, now I better understand what do you mean by a beable. But then I don't understand why do you associate beables only with fields? For instance, why wouldn't the expectation value $\langle {\bf x}\rangle$ of the position operator ${\bf x}$ in non-relativistic QM also be a beable? Are you saying that it is not measurable even in an approximate sense?

 

[DarMM]

OK, now I better understand what do you mean by a beable. But then I don't understand why do you associate beables only with fields? For instance, why wouldn't the expectation value $\langle {\bf x}\rangle$ of the position operator ${\bf x}$ in non-relativistic QM also be a beable? Are you saying that it is not measurable even in an approximate sense?

It would be in the Thermal Interpretation of Non-Relatvistic QM, however the motivations for the TI are much clearer and the results more general in the field case I would say.

 

[A. Neumaier]

OK, now I better understand what do you mean by a beable. But then I don't understand why do you associate beables only with fields? For instance, why wouldn't the expectation value $\langle {\bf x}\rangle$ of the position operator ${\bf x}$ in non-relativistic QM also be a beable? Are you saying that it is not measurable even in an approximate sense?

Because from a fundamental point of view we have only fields, and particles are approximate concepts. This is clear from the standard model, but also at higher levels. Even when a macroscopic system such as a cup of water is described nonrelativistically in first quantized form, it is imposssible to define single-particle position operators on the physical Hilbert space, because of indistinguishability of particles of the same kind. (One needs it to get the thermodynamic properties right!) Not even the physical Hilbert space of a pair of entangled electrons or silver atoms has a position operator for each particle, let alone that of a photon pair.

Thus physical quantum systems with position operators for each particle are the exception rather than the rule.

To get position operators for each particle, one either needs a nonphysical extension of the Hilbert space that artificially allows one to distinguish each particle. Or one needs a crystal structure that makes some of the particles distinguishable by approximately fixing their position. In the latter case, nuclei only have position operators; electrons still have none. And upon melting, even the nuclei lose their distinguishability and hence their position operators; so these cannot be fundamental things.

But should you consider a system with a position operator, such as a single massive particle or an anharmonic oscillator, then for this system, the q-expectation of position would be a beable.

 

[Demystifier]

But should you consider a system with a position operator, such as a single massive particle or an anharmonic oscillator, then for this system, mean position would be a beable.

OK, now I understand what are beables in your theory, so now I can ask a question about dynamics. Let $H$ be the Hamiltonian of a closed quantum system and let $U(t)=e^{-iHt}$ be the corresponding operator of unitary evolution. Is the time dependence of $\langle \phi(t)\rangle$ always given by
$$\langle \phi(t)\rangle=\langle\psi |U^{\dagger}(t) \phi U(t)|\psi\rangle \; ?$$
In particular, if the closed system includes the measuring apparatus, is the time-dependence formula above valid during the measurement?

 

[A. Neumaier]

OK, now I understand what are beables in your theory, so now I can ask a question about dynamics. Let $H$ be the Hamiltonian of a closed quantum system and let $U(t)=e^{-iHt}$ be the corresponding operator of unitary evolution. Is the time dependence of $\langle \phi(t)\rangle$ always given by
$$\langle \phi(t)\rangle=\langle\psi |U^{\dagger}(t) \phi U(t)|\psi\rangle \; ?$$
In particular, if the closed system includes the measuring apparatus, is the time-dependence formula above valid during the measurement?

Only if $\hbar=1$, the system is isolated, the state is pure, and there are no classical external forces. Strictly speaking, the only fully isolated system is the whole universe; but in some approximate sense, you may consider many smaller systems as isolated. But a system containing a measuring apparatus is never in a pure state.

 

[Demystifier]

Only if $\hbar=1$, the system is isolated, the state is pure, and there are no classical external forces. Strictly speaking, the only fully isolated system is the whole universe; but in some approximate sense, you may consider many smaller systems as isolated. But a system containing a measuring apparatus is never in a pure state.

So let as study a mixed state $\rho$ of the full universe without classical external forces (in units $\hbar=1$). In this case we have the deterministic evolution
$$\langle \phi(t)\rangle = {\rm Tr}\rho(t)\phi$$
where
$$\rho(t)=U(t)\rho U^{\dagger}(t)$$
How does the thermal interpretation explain the appearance of the state reduction when measurement is performed?

By the state reduction I mean the apparently random transition
$$\rho \rightarrow \rho' = \frac{\pi\rho\pi}{ {\rm Tr} \pi\rho\pi}$$
where $\pi$ is the projection operator into the state of one of the possible measurement outcomes.

 

[A. Neumaier]

So let us study a mixed state $\rho$ of the full universe without classical external forces (in units $\hbar=1$). In this case we have the deterministic evolution
$$\langle \phi(t)\rangle = {\rm Tr}~\rho(t)\phi$$
where
$$\rho(t)=U(t)\rho U^{\dagger}(t)$$
How does the thermal interpretation explain the appearance of the state reduction when measurement is performed?

By the state reduction I mean the apparently random transition
$$\rho \rightarrow \rho' = \frac{\pi\rho\pi}{ {\rm Tr}~ \pi\rho\pi}$$
where $\pi$ is the projection operator into the state of one of the possible measurement outcomes.

There is no such state reduction. The state $\rho(t)$ of the universe never collapses, since the universe is an isolated system, hence satisfies a unitary, deterministic dynamics.

But the reduced state of the system plus apparatus is $\rho_P(t)=P\rho(t)P^*$ for some linear operator $P$ from the Hilbert space of the universe to the much smaller Hilbert space of (system plus apparatus), determined by the split of the universe into (system plus apparatus) and environment - the thermal interpretation analogue of a Heisenberg cut.

If (system plus apparatus) are only weakly coupled to the environment, $\rho_P(t)$ satisfies an approximate dynamical law in which stochastic and dissipative terms are present. See the derivation of the piecewise deterministic process (PDP) in B&P cited in Section 4.3 of Part III, which for photodetection in a low intensity light beam results (in the approximation derived using the standard projection operator techniques) in projections at random times where a detection event happens. This is what you were looking for, I guess.

 

[ftr]

Because from a fundamental point of view we have only fields

Do these fields represent some other entities as in classical fields or are you saying that is what reality is made of, i.e. some numbers and relations between them. Or what?

 

[Haelfix]

Only if $\hbar=1$, the system is isolated, the state is pure, and there are no classical external forces. Strictly speaking, the only fully isolated system is the whole universe; but in some approximate sense, you may consider many smaller systems as isolated. But a system containing a measuring apparatus is never in a pure state.

What purifies that state?

 

[A. Neumaier]

What purifies that state?

Possibly cooling to absolute zero, but this is impossible.

 

[A. Neumaier]

Do these fields represent some other entities as in classical fields or are you saying that is what reality is made of, i.e. some numbers and relations between them. Or what?

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.

 

[Demystifier]

There is no such state reduction. The state $\rho(t)$ of the universe never collapses, since the universe is an isolated system, hence satisfies a unitary, deterministic dynamics.

But the reduced state of the system plus apparatus is $\rho_P(t)=P\rho(t)P^*$ for some linear operator $P$ from the Hilbert space of the universe to the much smaller Hilbert space of (system plus apparatus), determined by the split of the universe into (system plus apparatus) and environment - the thermal interpretation analogue of a Heisenberg cut.

If (system plus apparatus) are only weakly coupled to the environment, $\rho_P(t)$ satisfies an approximate dynamical law in which stochastic and dissipative terms are present. See the derivation of the piecewise deterministic process (PDP) in B&P cited in Section 4.3 of Part III, which for photodetection in a low intensity light beam results (in the approximation derived using the standard projection operator techniques) in projections at random times where a detection event happens. This is what you were looking for, I guess.

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.

 

[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.