Proof that the thermal interpretation of QM is wrong

In summary, the conversation discussed the thermal interpretation of quantum physics and how it is believed to solve the measurement problem. However, the speaker presented a proof that the thermal interpretation is in fact wrong and cannot solve the problem. They argued that looking at the problem from the perspective of the whole Universe, rather than just the cat, simplifies the analysis and shows that the superposition of an alive and dead cat is actually stable. This means that the beables, or observable properties, in the thermal interpretation cannot solve the Schrodinger cat paradox. They also addressed the possibility of interference terms and explained why they are negligible in this scenario. Ultimately, the speaker's proof shows that the thermal interpretation cannot resolve the measurement problem in quantum physics.
  • #36
A. Neumaier said:
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.
 
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  • #38
Demystifier said:
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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  • #39
A. Neumaier said:
This is obviously wrong because Bohmian mechanics is reversible.
Now you are doing the same error as Loschmidt did on the Boltzmann H-theorem.
 
  • #40
Demystifier said:
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...
 
  • #41
A. Neumaier said:
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?
 
  • #42
Demystifier said:
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.
Demystifier said:
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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  • #43
Demystifier said:
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.
 
  • #44
vanhees71 said:
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.
 
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  • #45
A. Neumaier said:
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.
 
  • #46
Demystifier said:
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.
 
  • #47
Demystifier said:
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.
 
  • #48
A. Neumaier said:
In the TI, ##\rho## is a beable,
Then TI is MWI in disguise. :-p
 
  • #49
A. Neumaier said:
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?
 
  • #50
Demystifier said:
Then TI is MWI in disguise. :-p
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.
 
  • #51
Demystifier said:
If so, then how to know in general which superpositions are allowed and which are not?
The state of the universe determined everything. A quantum system is a subsystem of the universe. Hence it is at time ##t## in the state determined by the corresponding reduced density matrix at time ##t##. These are the allowed states of quantum systems.

If you are able to prepare a particular spin state (a superposition, say), it means that at the time of preparation, the state of the universe is such that the reduced density matrix of the spin is in this state. (See post #22 for full details.)

In a similar way, all preparable and all measurable information give little bits of information about the state of the universe, and from this information, one can draw theoretical conclusions.
 
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  • #52
Demystifier said:
I'm not sure what assumption are you talking about?

Let me take a specific example. Suppose you have a collection of ##N## particles confined to a box of size ##L^3##, the points ##(x,y,z)## such that ##0 \leq x \leq L##, ##0 \leq y \leq L##, ##0 \leq z \leq L##. But initially, the wave function only has non-negligible support in a small corner ##0 \leq x \leq L/10##, ##0 \leq y \leq L/10##, ##0 \leq z \leq L/10##. So the actual particle distribution is uniform throughout the larger volume, while ##|\psi|^2## is uniform in the smaller volume, and neglible outside it. So there is initially a discrepancy between the actual and calculated particle distributions. Are you saying that eventually the discrepancy will disappear with time? (Or maybe we need to add some interactions?)

I assume that like Newtonian physics, if you wait long enough, the Poincare recurrence time, eventually the system will return arbitrarily closely to the initial state, with a big discrepancy between ##|\psi|^2## and the actual particle distribution? But that would seem to contradict the argument that if the two are ever in sync, they will remain in sync. I'm confused.
 
  • #53
A. Neumaier said:
The state of the universe determined everything. A quantum system is a subsystem of the universe. Hence it is at time ##t## in the state determined by the corresponding reduced density matrix at time ##t##. These are the allowed states of quantum systems.

If you are able to prepare a particular spin state (a superposition, say), it means that at the time of preparation, the state of the universe is such that the reduced density matrix of the spin is in this state. (See post #22 for full details.)

In a similar way, all preparable and all measurable information give little bits of information about the state of the universe, and from this information, one can draw theoretical conclusions.

This is actually what's lacking with RQM although almost similar, dumb-down version of it . Absolute relativization of quantities, namely the relativization of the possession of values (or definite magnitudes) to interacting physical systems is mute with respect to realism. It doesn't care any of the information in comparison to TI. It stay true to available formalism--Special Theory of Relativity. “The real events of the world are the "realization" (the "coming to reality", the "actualization") of the values q, q′, q″, … in the course of the interaction between physical systems. This actualization of a variable q in the course of an interaction can be denoted as the quantum event q.”

https://arxiv.org/ftp/arxiv/papers/1309/1309.0132.pdf
 
  • #54
julcab12 said:
This is actually what's lacking with RQM although almost similar, dumb-down version of it . Absolute relativization of quantities, namely the relativization of the possession of values (or definite magnitudes) to interacting physical systems is mute with respect to realism. It doesn't care any of the information in comparison to TI. It stay true to available formalism--Special Theory of Relativity.
What is RQM?

Relativistic quantum theory is quantum field theory. In the standard model nothing is relative except the local gauges and the choice of the Lorentz frame.
 
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  • #55
A. Neumaier said:
What is RQM?
Relational Quantum Mechanics, the interpretation of QM (and QFT) proposed by Rovelli. It is inspired by relativity theory, but is logically independent on it.
 
  • #56
A. Neumaier said:
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.
How can ##\rho## (assuming it's what's called the statistical operator in the standard interpretation) be a "beable", if it depends on the picture of time evolution chosen? The same holds for operators representing observables.

What's a physical quantity (not knowing, what precisely "beable" means, I rather refer to standard language) are
$$P(t,a|\rho)=\sum_{\beta} \langle t, a,\beta|\rho(t)|t,a,\beta \rangle,$$
where ##|t,a,\beta## and ##\rho(t)## are the eigenvectors of ##\hat{A}## and ##\rho(t)## the statistical operator, evolving in time according to the chosen picture of time evolution. In the standard minimal interpretation ##P(t,a|\rho)## is the probability for obtaining the value ##a## when measuring the observable ##A## precisely at time ##t##.

Now, before one discuss or even prove anything concerning an interpretation, one must define, what's the meaning of this expression in the interpretation. I still didn't get, as what this quantity is interpreted in the thermal interpretation, because you forbid it to be interpreted as probabilities.

Also it doesn't help to discuss about a fiction like the "state of the universe". This is something which is not observable even in principle.

On the other hand, I think it's pretty save to say the universe, on a large space-time scale, is close to local thermal equilibrium, as defined in standard coordinates of the FLRM metric, where the CMBR is in local thermal equilibrium up to tiny fluctuations of the relative order of ##10^{-5}##.
 
  • #58
julcab12 said:
This is actually what's lacking ...
https://arxiv.org/ftp/arxiv/papers/1309/1309.0132.pdf
This is just an off-topic aside more suited to BTSM and regards whether this is despite considerations involving LQG, as I think with that it can be seen like QM with some adaptations as probabilistic relations between values of variables evolving together, not variables evolving just with respect to a single time parameter.
 
  • #59
Let me summarize what emerged from the discussion of this alleged ''proof''.

The thermal interpretation claims that Born's rule follows from the dynamics of the actual state of the actual universe.

The mixed state discussed in post #1 has nothing to do with the actual state of the universe, hence the proof there proves nothing about the thermal interpretation, except that its author didn't understand the claim. (See post #43 and subsequent ones.)
 
<h2>1. What is the thermal interpretation of QM?</h2><p>The thermal interpretation of QM is a theory that explains the probabilistic nature of quantum mechanics by attributing it to the thermal fluctuations of particles. It suggests that the uncertainty in the position and momentum of a particle is due to the random thermal motion of its constituent particles.</p><h2>2. What evidence supports the thermal interpretation of QM?</h2><p>One of the main pieces of evidence for the thermal interpretation of QM is the observed behavior of particles at the quantum level. The probabilistic nature of quantum mechanics can be explained by the random thermal motion of particles, and this has been observed in experiments such as the double-slit experiment.</p><h2>3. What are the main arguments against the thermal interpretation of QM?</h2><p>One of the main arguments against the thermal interpretation of QM is that it fails to explain certain phenomena, such as quantum entanglement and the wave-particle duality of particles. Additionally, the thermal interpretation does not provide a clear explanation for the collapse of the wave function, which is a fundamental aspect of quantum mechanics.</p><h2>4. Are there alternative interpretations of QM that contradict the thermal interpretation?</h2><p>Yes, there are several alternative interpretations of QM that contradict the thermal interpretation. These include the Copenhagen interpretation, the many-worlds interpretation, and the pilot wave theory. Each of these interpretations offers a different explanation for the probabilistic nature of quantum mechanics.</p><h2>5. What implications would arise if the thermal interpretation of QM is proven wrong?</h2><p>If the thermal interpretation of QM is proven wrong, it would have significant implications for our understanding of the quantum world. It would mean that the probabilistic nature of quantum mechanics is not caused by thermal fluctuations, and we would need to find a new explanation for this phenomenon. It could also lead to the development of new theories and interpretations of quantum mechanics.</p>

1. What is the thermal interpretation of QM?

The thermal interpretation of QM is a theory that explains the probabilistic nature of quantum mechanics by attributing it to the thermal fluctuations of particles. It suggests that the uncertainty in the position and momentum of a particle is due to the random thermal motion of its constituent particles.

2. What evidence supports the thermal interpretation of QM?

One of the main pieces of evidence for the thermal interpretation of QM is the observed behavior of particles at the quantum level. The probabilistic nature of quantum mechanics can be explained by the random thermal motion of particles, and this has been observed in experiments such as the double-slit experiment.

3. What are the main arguments against the thermal interpretation of QM?

One of the main arguments against the thermal interpretation of QM is that it fails to explain certain phenomena, such as quantum entanglement and the wave-particle duality of particles. Additionally, the thermal interpretation does not provide a clear explanation for the collapse of the wave function, which is a fundamental aspect of quantum mechanics.

4. Are there alternative interpretations of QM that contradict the thermal interpretation?

Yes, there are several alternative interpretations of QM that contradict the thermal interpretation. These include the Copenhagen interpretation, the many-worlds interpretation, and the pilot wave theory. Each of these interpretations offers a different explanation for the probabilistic nature of quantum mechanics.

5. What implications would arise if the thermal interpretation of QM is proven wrong?

If the thermal interpretation of QM is proven wrong, it would have significant implications for our understanding of the quantum world. It would mean that the probabilistic nature of quantum mechanics is not caused by thermal fluctuations, and we would need to find a new explanation for this phenomenon. It could also lead to the development of new theories and interpretations of quantum mechanics.

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