A Proof that the thermal interpretation of QM is wrong

[Demystifier]

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

https://arxiv.org/abs/quant-ph/0112005

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]

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]

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.

 

[Demystifier]

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.

 

[Demystifier]

I am not saying that the thermal interpretation is completely wrong. I think it contains some right ideas. But my proof shows that something is still missing. In a sense, I think one could say that Bohmian mechanics is the thermal interpretation done right. For more details see https://pdfs.semanticscholar.org/7a...11.1117284887.1555316195-736102029.1555316195

 

[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.

 

[Demystifier]

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]

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.

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.

 

[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]

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.

 

[Demystifier]

In the TI, $\rho$ is a beable,

Then TI is MWI in disguise.

 

[Demystifier]

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.

 

[A. Neumaier]

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.

 

[stevendaryl]

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.

 

[julcab12]

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

 

[A. Neumaier]

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.

 

[Demystifier]

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.

 

[vanhees71]

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 $10^{-5}$.

 

[A. Neumaier]

How can $\rho$ be a beable

I answered your posts in the other thermal interpretation thread since they have nothing to do with Demysitifier's alleged proof, hence are here off-topic.

 

[*now*]

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.

 

[A. Neumaier]

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.)

 

[PeterDonis]

Moderator's note: A number of posts about the Schrodinger's Cat thought experiment and how the thermal interpretation analyzes it have been moved to a new thread:

https://www.physicsforums.com/threads/schrodingers-cat-and-the-thermal-interpretation.971699/