The thermal interpretation of quantum physics

[vanhees71]

Bell's intention behind introducing the word "beable" was to talk about properties that have values whether or not they are observed. So in classical physics, fields, and particle positions and momenta are beables. In QM, it seems that measurement results are beables, and that's about it.

Well, yes. But as the Bell tests show, within QT there are then no beables. The only way out of this were to find a deterministic theory as successful as QT, which (again according to Bell's famous analysis) must be non-local. One may speculate, whether such a theory exists, but as long as there is none, it's not more than speculation.

 

[DarMM]

But that's obviously wrong and not how QT is used in practice. If I measure $S_z$ accurately, I don't get the value $-1/4$ but one of the possible values $\pm 1/2$ (supposed we deal with spin-1/2 particles).

Certainly, but as I said before the TI explains why you get $\pm \frac{1}{2}$ despite the true value being $-\frac{1}{4}$

 

[stevendaryl]

Well, yes. But as the Bell tests show, within QT there are then no beables.

Not even measurement results? A "beable" is what actually exists. So no beables means nothing exists. Or maybe, in a solipsistic sense, nothing exists outside our own minds (but then how do the minds exist?)

 

[A. Neumaier]

The definition of states and observables I gave are not pseudo-mathematical speculations but common practice in physics.

You simply call physics what you believe, and philosophy what you disgree with and pejoratively call speculations. But this is not the common notion of either physics or philosophy.

Your definitions are not common practice; they are absent in the much more frequent expositions of variants of the Copenhagen interpretation, and used only by those working with the minimal interpretation. This shows that they are a matter of philosophy, not of physics per se. I didn't call it specuations but philosophy.

Moreover, since in the sense you use these notions, neither preparation procedures nor observables are mathematical objects, forming equivalence classes of them with respect to an ill-defined equivalence relation is pseudo-mathematical.

 

[A. Neumaier]

What's sure is that this is not as simple as you seem to imply with your "thermal interpretation" by just identifying the expectation values with observables.

I am sure of the opposite, though I prefer Bell's word beable for uncertain but definite properties, not to increase confusion.

Again you gave no specific argument against it but only a story about Schrödinger and Bohr. But the thermal interpretation is quite different from Schrödinger's, who didn't have a notion of quantum fields and hence got into trouble.

If I measure $S_z$ accurately, I don't get the value $−1/4$ but one of the possible values $\pm 1/2$. The measurement accuracy is not due to the state of the measured system but due to the construction of the measurement device.

The thermal interpretation accepts only that it is an accurate measurement of the silver position but denies that this is an accurate measurement of the spin. It asserts instead that the silver position measures the true value $-1/4$ of the spin with a large error of $|-1/4\pm 1/2|\ge 1/4$.

There is no way to determine experimentally what should be called the true value. It is a theoretical convention made by the interpretation, and tradition and I differ in the choice of convention.

The convention in the thermal interpretation has the advantage that it loses nothing about the agreement with the experimental record but eliminates statistics from the foundations, makes thereby quantum mechanics much less mysterious and much less different from classical mechanics, and solves the measurement problem which you (unlike Peres, the most consequent of the defenders of the minimal interpretation) do not even perceive as one.

 

[stevendaryl]

The convention in the thermal interpretation has the advantage that it loses nothing about the agreement with the experimental record but eliminates statistics from the foundations, makes thereby quantum mechanics much less mysterious and much less different from classical mechanics, and solves the measurement problem which you (unlike Peres, the most consequent of the defenders of the minimal interpretation) do not even perceive as one.

Okay, but it is the discrete values of measurement results that led to the development of quantum mechanics in the first place, as a way to explain why those values.

 

[A. Neumaier]

what's observable is the corresponding cross-section

But like probabilities, a cross section can be written as a q-expectation; so in fact you agree that at least some q-expectations are observable!

 

[stevendaryl]

You simply call physics what you believe, and philosophy what you disgree with and pejoratively call speculations. But this is not the common notion of either physics or philosophy.

I think it's pretty much only physicists who use "philosophy" as a pejorative for those ideas that they disagree with. An analogy in another area is "ideology" in politics. People pretty much only use "ideology" to describe those political ideas that they disagree with.

 

[A. Neumaier]

Okay, but it is the discrete values of measurement results that led to the development of quantum mechanics in the first place, as a way to explain why those values.

Yes, but this is a historical statement, and it lead into a history of over 90 years of problematic foundations.

If Born or Ehrenfest (who came with his theorem quite close to the thermal interpretation) would have had instead the idea of the quantum bucket and that a bucket measures everything in discrete units, even when the measured stuff is continuous, the history of quantum interpretations could have been very different.

If you have a hammer, everything looks like a nail...

 

[vanhees71]

I give up. Obviously there's no way to reach agreement. I gave several arguments about the very clear fact, which is independent of any interpretational issues, why the expectation values (which are formally the same as the quantity you call q-expectations) are NOT the observables. I gave physical reasons, I gave examples, I made the historical point of why this misconception was abandoned almost immediately when it came up. You just pick imprecisions which are unavoidable in colloquial conversations and deny this simple fact repeatedly. In this way it doesn't make sense to discuss!

 

[Demystifier]

@A. Neumaier I also have a question.

Consider the following 3 hermitian scalar field operators (at the same time): $\phi({\bf x})$, $\pi({\bf x})\equiv\dot{\phi}({\bf x} )$ and
$$A({\bf x})\equiv [ \phi({\bf x}) \pi({\bf x}) +\pi({\bf x}) \phi({\bf x}) ]/2$$
In the thermal interpretation, the corresponding expected values
$$\langle\phi({\bf x})\rangle , \;\; \langle\pi({\bf x})\rangle , \;\; \langle A({\bf x})\rangle$$
are all beables. But are all these beables equally fundamental?

If they are all equally fundamental, that there is an infinite number of fundamental beables at each point ${\bf x}$, because there are also beables corresponding to products of an arbitrary number of field operators. Isn't it strange that there is an infinite number of fundamental beables?

Or if they are not all equally fundamental, then one would expect that only $\langle\phi({\bf x})\rangle$ and $\langle\pi({\bf x})\rangle$ are fundamental, while $\langle A({\bf x})\rangle$ is a function of $\langle\phi({\bf x})\rangle$ and $\langle\pi({\bf x})\rangle$. But then how one would explain that
$$ \langle A({\bf x})\rangle \neq \langle\phi({\bf x})\rangle \langle\pi({\bf x})\rangle \; ?$$

 

[Pleonasm]

So, according to the thermal QM, every event (including all this great discussion) was pre-programmed by the Big Bang's primordial fluctuations?

Pretty much

 

[A. Neumaier]

Consider the following 3 hermitian scalar field operators (at the same time): $\phi({\bf x})$, $\pi({\bf x})\equiv\dot{\phi}({\bf x} )$ and
$$A({\bf x})\equiv [ \phi({\bf x}) \pi({\bf x}) +\pi({\bf x}) \phi({\bf x}) ]/2$$
In the thermal interpretation, the corresponding expected values
$$\langle\phi({\bf x})\rangle , \;\; \langle\pi({\bf x})\rangle , \;\; \langle A({\bf x})\rangle$$
are all beables. But are all these beables equally fundamental?

If they are all equally fundamental, that there is an infinite number of fundamental beables at each point ${\bf x}$, because there are also beables corresponding to products of an arbitrary number of field operators. Isn't it strange that there is an infinite number of fundamental beables?

What is fundamental depends on the point of view.

Note that values at single space-time points are ill-defined since quantum fields are distributions only. Thus one needs to consider open neighborhoods of points. But fields always have infinitely many degrees of freedom in every open neighborhood of a given space-time point; thus this cannot be a useful criterion for fundamentality.

Fermionic fields (e.g., el;ectron fields in QED) are unobservable (q-expectations vanish identically), but fermionic currents and fermionic pair correlators are observable.
In QCD, quark and gluon fields are considered fundamental since they appear in the action. However, the beeables are only the q-expectations of the gauge invariant expressions, i.e., certain renormalized polynomial expressions in these fields. One can order them in terms of complexity. Currents and current pair correlators are the most relevant (more easily observable) ones of these.

All these look quite nonfundamental according to your notion of fundamentality.

Or if they are not all equally fundamental, then one would expect that only $\langle\phi({\bf x})\rangle$ and $\langle\pi({\bf x})\rangle$ are fundamental, while $\langle A({\bf x})\rangle$ is a function of $\langle\phi({\bf x})\rangle$ and $\langle\pi({\bf x})\rangle$. But then how one would explain that
$$ \langle A({\bf x})\rangle \neq \langle\phi({\bf x})\rangle \langle\pi({\bf x})\rangle \; ?$$

In the thermal interpretation, q-expectations are most naturally classified by their slowness - the most easily observable ones are those where the high frequency dependence on space-time coordinates is most negligible. Surely there is no dependence in the sense that the q-expectations of products factor. Thus your last fact needs no explanation - equality would need it!

 

[Demystifier]

In the thermal interpretation, q-expectations are most naturally classified by their slowness - the most easily observable ones are those where the high frequency dependence on space-time coordinates is most negligible.

It looks as if, in thermal interpretation, QFT is just an effective theory, not a fundamental one. Is that right?

 

[A. Neumaier]

It looks as if, in thermal interpretation, QFT is just an effective theory, not a fundamental one. Is that right?

No.

The thermal interpretation interprets the fields, independent of whether or not they are fundamental. For definiteness I assumed that QFT is fundamental but this is not essential; see Footnote 13 in Section 4 of Part II. Thus it would also apply to string theory, or to a lattice theory of which QFT would be an effective approximation.

 

[Demystifier]

No.

The thermal interpretation interprets the fields, independent of whether or not they are fundamental. For definiteness I assumed that QFT is fundamental but this is not essential; see Footnote 13 in Section 4 of Part II. Thus it would also apply to string theory, or to a lattice theory of which QFT would be an effective approximation.

I guess my problem then is that I cannot digest that $\langle \phi(f) \rangle$ is a fundamental ontology, because the test function $f$ is too arbitrary.

 

[A. Neumaier]

I guess my problem then is that I cannot digest that $\langle \phi(f) \rangle$ is a fundamental ontology, because the test function $f$ is too arbitrary.

Fundamental is whatever the theory regards as fundamental, and in QFT these simply are the $\phi(f)$.

But you can take $f(x)=e^{-(x-\mu)^TM^{-2}(x-\mu)/2}$ where $M$ is a diagonal matrix whose diagonal entries contain spatial and temporal resolutions since any other q-expectation of the form you mention is a linear combination of these.

 

[Demystifier]

Fundamental is whatever the theory regards as fundamental, and in QFT these simply are the $\phi(f)$.

But you can take $f(x)=e^{-(x-\mu)^TM^{-2}(x-\mu)/2}$ where $M$ is a diagonal matrix whose diagonal entries contain spatial and temporal resolutions since any other q-expectation of the form you mention is a linear combination of these.

It's OK for a theory viewed as a practical instrumental tool, but for me it's not OK for a theory that is supposed to say something about beables (ontology).

 

[A. Neumaier]

It's OK for a theory viewed as a practical instrumental tool, but for me it's not OK for a theory that is supposed to say something about beables (ontology).

Well, beables are what the interpretation declares to be beables. Bell's requirement for beables was just that they are definite properties of the quantum system, independent of whether they are observed.

There is no need to have some beables to be more fundamental than others, or that beables must have a particularly intuitive meaning. It is enough that the easily measurable beables (such as observable smeared currents) have such a meaning.

 

[Demystifier]

Well, beables are what the interpretation declares to be beables. Bell's requirement for beables was just that they are definite properties of the quantum system, independent of whether they are observed.

There is no need to have some beables to be more fundamental than others, or that beables must have a particularly intuitive meaning. It is enough that the easily measurable beables (such as observable smeared currents) have such a meaning.

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.

 

[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.