The thermal interpretation of quantum physics

[DarMM]

Again expectation values are not the observables. I have no clue what you (or Bell) mean with the word "beables". It's obviously something philosophical

Not particularly it's just the basic fundamental things in your model/theory as opposed to the things that you observe. Like in General Relativity Riemann curvature is a beable, length in a given frame is an observable.

without giving a positive explanation which physcial meaning the quantum state should take in the new interpretation

To me it is very clear, the state $\rho$ assigns values to system properties via the map $\langle A \rangle_{\rho}$.

 

[vanhees71]

Interpretation is about the connection of the formal entities of the theory (for QT the Hilbert space, the statistical operators, and the operators representing observables) with physics. Of course, Arnold mathematically defines his symbols. I had the impression that the physical meaning behind the symbols is the usual probabilistic one. Then I learned in this discussion that Arnold wants to precisely abandon this probalistic interpretation. For me he has still not explained what the physical interpretation of the formalism (in his case obviously the q-expectation value algebra/analysis) is supposed to be.

For sure it is wrong to simply identify the expectation values with observables. This was known already since 1926 and lead Born to his probabilistic interpretation of the quantum state, and I've not seen any convincing argument, particularly no empirical one, that this is not the correct interpretation of QT.

 

[DarMM]

For me he has still not explained what the physical interpretation of the formalism (in his case obviously the q-expectation value algebra/analysis) is supposed to be.

Very simple, it's a value.

So when we write down the following in regular QM:
$$
\langle S_z \rangle_{\rho} = -\frac{1}{4}\\
\langle S_z^{2} \rangle_{\rho} = \frac{1}{16}
$$

What that means is that there are properties $A$ and $B$ with values $A = -\frac{1}{4}$ and $B = \frac{1}{16}$.

I know I've just relabeled them, but I'm trying to express that in the thermal interpretation they are simply properties like momentum, not statistical expectations despite the notation of the standard views.

However the thermal interpretation then goes on to explain why we get discrete values and not the true values of these quantities when we perform experiments.

 

[A. Neumaier]

This is no answer to my request.

Again expectation values are not the observables. I have no clue what you (or Bell) mean with the word "beables". It's obviously something philosophical, but I want to understand the physics. Philosophy of science is some a posteriori analysis of science but not science, and I'm interested in the science.

It is very clear that nothing that depends on the state of the system, which is interpretationally defined as an equivalence class of preparation procedures can be the observable. An observable is interpretationally defined as an equivalence class of measurement procedures, and the measurement procedure doesn't depend on the state of the measured system but on the construction of the measurement appratus.

In classical mechanics, particles exist. States define their properties (i.e., the beables of classical mechanics) and are given by the exact positions and momenta of the particles, some of which can be approximately measured. Fields are coarse-grained approximate concepts. This is the standard interpretation of classical mechanics. Experimental physics is about how to do the measurements, and under which conditions which measurements are how accurate.

Nothing else is needed. In particular, there is no need for pseudo-mathematical philosophy about states as equivalence class of preparation procedures or observables as equivalence class of measurement procedures. This is specific to a statistical interpretation, which has no place in the foundations.

In quantum field theory, fields exist. States define their properties (i.e., the beables of quantum field theory) and are given by the exact q-expectations of the fields and their normally ordered products, some of which can be approximately measured. Particles are coarse-grained approximate concepts. This is the thermal interpretation of quantum physics. Experimental physics is about how to do the measurements, and under which conditions which measurements are how accurate.

Nothing else is needed. In particular, there is no need for pseudo-mathematical philosophy about states as equivalence class of preparation procedures or observables as equivalence class of measurement procedures. This is specific to a statistical interpretation, which should have no place in the foundations.

 

[A. Neumaier]

For sure it is wrong to simply identify the expectation values with observables. This was known already since 1926 and lead Born to his probabilistic interpretation of the quantum state

You are completely wrong about the early history of quantum mechanics!

Please give explicitly the details of the argument that would prove that it is wrong to identify expectation values with observables.

 

[stevendaryl]

Again expectation values are not the observables. I have no clue what you (or Bell) mean with the word "beables"

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.

 

[vanhees71]

You asserted this WRONGly several times, without giving any rational justification. I am not doing the same as Schrödinger, who didn't think in terms of quantum fields.

We agree to disagree. I give up. I've given a very simple reason that expectation values are not the observables in QT. This is independent of any interpretational opinion but an empirical fact.

Perhaps one more example (from QFT!) helps: If you measure the invariant-mass spectrum of a resonance (physicists slang) what you really measure are cross sections of a scattering process (since resonances are no asymptotic free states). Take e.g. the process $\pi^+ + \pi^- \rightarrow e^+ + e^-$. You find a pronounced peak around 770 MeV invariant mass (to be very precise, you find a quite broad peak of around 150 MeV width and a quite narrow one on top). Theoretically this can be very well described by a model by Sakurai, called vector-meson dominance, where you assume that the hadronic electromagnetic current is proportional to the light-vector-meson fields (the $\rho$ and $\omega$ for the peaks discussed here).

The peak has a finite width. In the formalism it's due to a complex pole in the corresponding connected four-point function of the scattering process, which is in your formalism a q-expectation value; and this is not an observable to begin with, because what's observable is the corresponding cross-section, but that's semantics in this case; the physics argument is the following: The invariant-mass width of this peak is 150 MeV. The inverse of this width is the life-time of the resonance (called $\rho$ meson). Now to measure this width the mass resolution of the detector must be much better than this width to really resolve this peak (the more for the much narrower $\omega$ meson and the $\phi$ meson around 1 GeV mass). The resolution of the measurement device is not restricted by the standard deviation of the invariant mass, i.e., it is independent of the quantum state the system is prepared in, and that's why the statement that quantum-mechanical expectation values (or, as in this case, quantities like the cross section in this example derived from it) are not the observables of the QT formalism.

 

[vanhees71]

You are completely wrong about the early history of quantum mechanics!

Please give explicitly the details of the argument that would prove that it is wrong to identify expectation values with observables.

I have given this very simple argument several times. For a real-world example from QFT see #487.

Also, where is my history wrong? There's a famous story found in at least one biography of Bohr that Schrödinger visited the Bohr institute and was staying with Bohr's family. Bohr involved him in such heated discussions about this very issue that Schrödinger got sick from the stress and Bohr's wife had to keep Bohr out of Schrödinger's bed room to give him some rest from the discussions ;-)).

 

[vanhees71]

I have given this very simple argument several times. For a real-world example from QFT see #487.

Very simple, it's a value.

So when we write down the following in regular QM:
$$
\langle S_z \rangle_{\rho} = -\frac{1}{4}\\
\langle S_z^{2} \rangle_{\rho} = \frac{1}{16}
$$

What that means is that there are properties $A$ and $B$ with values $A = -\frac{1}{4}$ and $B = \frac{1}{16}$.

I know I've just relabeled them, but I'm trying to express that in the thermal interpretation they are simply properties like momentum, not statistical expectations despite the notation of the standard views.

However the thermal interpretation then goes on to explain why we get discrete values and not the true values of these quantities when we perform experiments.

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). The measurement accuracy is not due to the state of the measured system but due to the construction of the measurement device. That's the very point I raised several times, and this is very clear from everyday practice of applications of the QT formalism to real-world experiments in the labs.

In other words the expectation values are not the "true values". That makes the probabilistic interpretation of the quantum state necessary (at least I've not seen any convincing alternative interpretation, where the quantum state is not probabilistically interpreted). The conclusion is that in this case $S_z$ is indetermined, and only probabilities can be given.

Here we even have incomplete information to begin with, i.e., we have only $\langle S_z \rangle$ and $\langle S_z^2 \rangle$. Then you can only "guess" the corresponding statstical operator. One idea to get one is to use information-theoretical arguments a la Shannon, von Neumann, and Jaynes, according to which the statistical operator with the "least prejudice" is the one that maximizes entropy. The result is the operator-valued Gaussian
$$\hat{\rho}=\frac{1}{Z} \exp(-\alpha \hat{S}_z - \beta \hat{S}_z^2), \quad Z=\mathrm{Tr} \hat{\rho}.$$
You can verify this guess by measuring (accurately!) $S_z$ on an ensemble of equally prepared particles and use statistical methods to "test this hypothesis".

 

[vanhees71]

In classical mechanics, particles exist. States define their properties (i.e., the beables of classical mechanics) and are given by the exact positions and momenta of the particles, some of which can be approximately measured. Fields are coarse-grained approximate concepts. This is the standard interpretation of classical mechanics. Experimental physics is about how to do the measurements, and under which conditions which measurements are how accurate.

Nothing else is needed. In particular, there is no need for pseudo-mathematical philosophy about states as equivalence class of preparation procedures or observables as equivalence class of measurement procedures. This is specific to a statistical interpretation, which has no place in the foundations.

In quantum field theory, fields exist. States define their properties (i.e., the beables of quantum field theory) and are given by the exact q-expectations of the fields and their normally ordered products, some of which can be approximately measured. Particles are coarse-grained approximate concepts. This is the thermal interpretation of quantum physics. Experimental physics is about how to do the measurements, and under which conditions which measurements are how accurate.

Nothing else is needed. In particular, there is no need for pseudo-mathematical philosophy about states as equivalence class of preparation procedures or observables as equivalence class of measurement procedures. This is specific to a statistical interpretation, which should have no place in the foundations.

In classical physics fields are beables too. If you use the word in this sense, it's obvious that only determinstic theories can describe beables. QT in the standard interpretation then doesn't describe beables, and you may well be motivated to find a determinstic theory as successful in describing the empirical facts as QT. 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.

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