# A Proof that the thermal interpretation of QM is wrong

#### A. Neumaier

Hi Arnold, is this an okay place for me to ask you some clarifying questions on the papers?

#### martinbn

@Demystifier I don't understand the argument. To me it seems that you just repeat what the measurement problem is, and that's it. I don't see how it is related to any interpretation.

Also, is there any interpretation that solves the measurement problem without introducing new problems?

#### vanhees71

Gold Member
Well, and I'm even more stupid in not seeing, where a measurement problem (in the sense of a physical problem!) might be. QT is the most successful theory of physics we have, which means we have a mathematical formalism and at least one interpretation which is not contradicting any basic principles, the statistical minimal interpretation, which enables predictions with this formalism about real-world measurements in the lab, and these predictions up to now always agreed with these real-world measurements. From a scientific point of view there's nothing else you can expect from a physical theory.

Now QT contradicts our intuition, but that's not very surprsing since our intuition is based on what our senses can detect, and that's a pretty limited "window" to the world around us. With our eyes we can detect about one octave of the electromagnetic spectrum, and it's tailored by evolution to let us survive in our environment and not to observe photons, atoms in the microscopic or the cosmic microwave background and black holes. We are also by far not sensitive enough to "feel" gravitational waves with our senses directly. So we apprehend with our senses only a tiny bit of the reality around us, but we are able to construct measurement devices that extend our ability to learn about nature, and with this technological progress of observability we detected a lot of things not directly available by our senses (e.g., $\gamma$ rays, X-rays, microwaves, radiowaves, extending our senses to the (almost) complete electromagnetic spectrum), and this led us first to the discovery of the atomistic structure of the matter around us and then way beyond (of all the elementary building blocks only the "first family" of fermions make up the matter we directly observe around us), and it's pretty save to guess that the "particle zoo" is still not complete.

So it's not very surprising that our "common sense" fails us in describing things which are beyond our direct experience correctly, and that's why the modern theories of physics have been enforced on us by extendended observational possibilities. So I don't understand, why one is surprised about something called "measurement problem", although from a rational point of view there's none. To the contrary, it's surprising how far we could get with our theoretical understanding after such and extension of observational possibilities. I'm sure and hope very much that there's still much more to be discovered and maybe one day we'd need even more comprehensive and "better" theories.

#### stevendaryl

Staff Emeritus
Well, and I'm even more stupid in not seeing, where a measurement problem (in the sense of a physical problem!) might be
The statement of the minimal interpretation involves the separation of the world into the measuring device and the system being measured. It treats the measuring device using rules that do not apply to the system being measured. In particular, it assumes that after the measurement interaction is completed, the measurement is in a definite "pointer state", with the probabilities of the different pointer states being given by the Born rule. This is a rule that applies to measurement devices, but apparently does not apply to the system being measured. Since the measuring device is itself a quantum system, there should not be any rules that apply to measurement devices that aren't derivable from rules for other systems.

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#### Demystifier

2018 Award
@Demystifier I don't understand the argument. To me it seems that you just repeat what the measurement problem is, and that's it. I don't see how it is related to any interpretation.
It's related to the thermal interpretation in the use of the assumption that $\langle O\rangle$ are beables. For that purpose, it's useful to compare it with Bohmian beables (particle trajectories) to see how Bohmian beables resolve the measurement problem, while thermal beables don't.

Also, is there any interpretation that solves the measurement problem without introducing new problems?
No. A new problem in the thermal interpretation is a weird ontology (beables). But that's not what I object to in my proof.

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#### Demystifier

2018 Award
Then how does Bohmian mechanics explain it - I have never seen this before.

You cannot invoke decoherence since the decoherence arguments also apply to the thermal interpretation, which you just ''proved'' wrong!
Yes I can invoke decoherence and no I didn't prove decoherence wrong. What I proved is that decoherence is not enough. Bohmian and thermal interpretation agree that decoherence needs to be enriched with appropriate beables in order for decoherence solve the measurement problem. What I have shown is that beables of thermal interpretation are not really appropriate for that purpose.

#### Demystifier

2018 Award
Well, and I'm even more stupid in not seeing, where a measurement problem (in the sense of a physical problem!) might be.
You are not stupid, you just use a narrower definition of "physical". Within your definition of "physical", there is no problem. One must take a wider perspective to see where the problem is. Whether we call it "physical", "philosophical" or "metaphysical" is a matter of semantics.

#### A. Neumaier

What I have shown is that beables of thermal interpretation are not really appropriate
Only for a fake universe whose state is a mixture of a universe with a life cat and one with a dead cat. Your result is essentially assumed in your hypothesis!

The TI assumes instead a universe in a Gibbs state, corresponding to approximate local equilibrium, which can never be of your form.

It is like disproving Bohmian mechanics by starting with positions not in quantum equilibrium!

#### Demystifier

2018 Award
Only for a fake universe whose state is a mixture of a universe with a life cat and one with a dead cat. Your result is essentially assumed in your hypothesis!
But von Neumann equation allows such states. An interpretation should explain why such mathematically possible solutions do not realize in nature.

The TI assumes instead a universal in a Gibbs state, which is never of your form.
Gibbs state is a state in a thermal equilibrium. Our Universe obviously is not in thermal equilibrium.

#### A. Neumaier

But von Neumann equation allows such states. An interpretation should explain why such mathematically possible solutions do not realize in nature.

Gibbs state is a state in a thermal equilibrium. Our Universe obviously is not in thermal equilibrium.
Please read my corrected post. Gibbs states and local equilibrium are defined in Part III of my sequence of papers.

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#### Demystifier

2018 Award
It is like disproving Bohmian mechanics by starting with positions not in quantum equilibrium!
In Bohmian mechanics, if a (sufficiently complex) system is initially far from quantum equilibrium, it will soon evolve close to the equilibrium. This result is called subquantum H-theorem.

What if a quantum system starts evolution far from Gibbs state (as I assume in my proof)? Intuition says that it should soon decay towards a state close to the equilibrium. Yet my proof shows that it doesn't happen within the thermal interpretation.

2018 Award

#### A. Neumaier

In Bohmian mechanics, if a (sufficiently complex) system is initially far from quantum equilibrium, it will soon evolve close to the equilibrium. This result is called subquantum H-theorem.
This is obviously wrong because Bohmian mechanics is reversible.

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#### Demystifier

2018 Award
This is obviously wrong because Bohmian mechanics is reversible.
Now you are doing the same error as Loschmidt did on the Boltzmann H-theorem.

#### A. Neumaier

Now you are doing the same error as Loschmidt did on the Boltzmann H-theorem.
Loschmidt committed no error.

Boltzmann made an additional assumption, which produced dissipation. You need to make the same assumption. It is the assumption that you denied me to make. You measure with double standard....

#### Demystifier

2018 Award
Loschmidt committed no error.

Boltzmann made an additional assumption, which produced dissipation. You need to make the same assumption. It is the assumption that you denied me to make. You measure with double standard....
I'm not sure what assumption are you talking about?

#### A. Neumaier

In Bohmian mechanics, if a (sufficiently complex) system is initially far from quantum equilibrium, it will soon evolve close to the equilibrium. This result is called subquantum H-theorem.

What if a quantum system starts evolution far from Gibbs state (as I assume in my proof)? Intuition says that it should soon decay towards a state close to the equilibrium. Yet my proof shows that it doesn't happen within the thermal interpretation.
I'm not sure what assumption are you talking about?
An assumption about effective stochastic independence (lack of conspiracy) akin to Boltzmann's Stosszahlansatz. Valentini knew this, and wrote in the abstract of his paper
• Valentini, A. (1991). Signal-locality, uncertainty, and the subquantum H-theorem. I. Physics Letters A, 156(1-2), 5-11.
Antony Valentini said:
based on assumptions similar to those of classical statistical mechanics
After (5), he invokes [7] for the proof of the H-theorem. If you accept the approximation techniques going into this as a valid proof, you automatically accept dissipation (the H-theorem is just such a manifestation of this) and hence semiclassical behavior of q-expectations (independent of the TI), since - as apparent from many places, e.g. from B&P - these are based on precisely the same kind of techniques. Informally, he relies on coarse-graining, just as the TI (see Sections 4 and 5 of Part III of my series).

Also, after (16), Valentini assumes (''will be regarded here as proof'') that a monotonically increasing function bounded above by zero will reach zero. This is obviously false; a counterexample is $f(t)=-1-1/(1+t^2)$. Thus his proof is invalid. I didn't check the other details of his proof.

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#### A. Neumaier

But my proof shows that something is still missing.
What is missing in your caricature argument is the TI assumption that there is only a single universe, and all observed systems are subsystems of this single universe at appropriate times. Only this actual universe must have the properties claimed. Thus mixing or superposing universes is irrelevant to the TI.

#### A. Neumaier

I don't understand, why one is surprised about something called "measurement problem", although from a rational point of view there's none.
From your point of view there is none. But from several different rational points of view, at least those of Einstein, Schrödinger, t'Hooft, or Weinberg, there is a serious problem.

#### Demystifier

2018 Award
What is missing in your caricature argument is the TI assumption that there is only a single universe, and all observed systems are subsystems of this single universe at appropriate times. Only this actual universe must have the properties claimed. Thus mixing or superposing universes is irrelevant to the TI.
I don't think that I assume that. Sure, I assume the superposition $\rho=\rho_{\rm alive} + \rho_{\rm dead}$, but since $\rho$ is not a beable in TI, it does not mean that the Universe itself is in the superposition, nor that there are two Universes.

Moreover, if TI assumes from the start that such a superposition is impossible, then this assumption begs the question.

#### A. Neumaier

since $\rho$ is not a beable in TI
In the TI, $\rho$ is a beable, since it is uniquely determined by the collection of all q-expectations. It is just not easily macroscopically interpretable, and hence has a subordinate role in the interpretation.

#### A. Neumaier

I assume the superposition $\rho=\rho_{\rm alive} + \rho_{\rm dead}$,
and you assume unitary dynamics for each of these. Thus you relate properties in three different universes to each other. At most one of these is permitted in the TI.

2018 Award

#### Demystifier

2018 Award
and you assume unitary dynamics for each of these. Thus you relate properties in three different universes to each other. At most one of these is permitted in the TI.
If so, then how to know in general which superpositions are allowed and which are not?

#### A. Neumaier

Then TI is MWI in disguise.
It has similarities with MWI in that it is based on the unitary dynamics of the state of the universe.

It is very different in spirit since it has obvious unique outcomes, only needs a single world, and only uses the standard concepts of quantum theory.

"Proof that the thermal interpretation of QM is wrong"

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