# A Proof that the thermal interpretation of QM is wrong

#### 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.

#### 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.

#### Demystifier

2018 Award
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 this sense, the Schrodinger's famous thought experiment prepares the cat in a superposition of dead and alive. I still don't see how TI can possibly prevent it.

#### stevendaryl

Staff Emeritus
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.

#### 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.

In a similar way, all preparable and all measurable information give a little bit of information about the state of the universe, and from this information, one can draw theoretical conclusions.
In this sense, the Schrodinger's famous thought experiment prepares the cat in a superposition of dead and alive.
No; only the atom is prepared in a superposition of undecayed and decayed. Schrödinger arrives at your conclusion only by assuming that the subsystem consisting of a decaying atom and the cat is isolated, which is never the case.

Only the whole universe is isolated. The subsystem is open, interacting with the bottom of the box and with the air in it. Increasing the size of the subsystem does not help. This openness is sufficient to make the system dissipative; the quantum H-theorem (whose statistical mechanics derivation you conceded to be acceptable as proof) provides a proof of this. Thus the arguments of Section 5 of my Part III apply and produce a definite final state for the atom (undecayed or decayed) and the cat (alive or dead).

The validity of the quantum H-theorem also shows that increasing entropy does not prove that the system must end up in experimentally relevant times in an equilibrium state, thus making Valentini's and your conclusion of Bohmian mechanics soon being in quantum equilibrium as unwarranted.

#### stevendaryl

Staff Emeritus
No; only the atom is prepared in a superposition of undecayed and decayed. Schrödinger arrives at your conclusion only by assuming that the subsystem consisting of a decaying atom and the cat is isolated, which is never the case.

Only the whole universe is isolated.
Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.

#### stevendaryl

Staff Emeritus
Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.
In terms of density matrices, the evolution would produce something like this:

$\rho = p_1 \rho_{alive} + p_2 \rho_{dead} + \rho_{cross}$

where $\rho_{cross}$ represents the cross terms like $|dead\rangle\langle alive|$ and $|alive\rangle\langle dead|$ ("alive" and "dead" referring to states of the whole universe, not just the cat).

If you ignore the cross-terms, this can be interpreted as the cat having a probability of $p_1$ of being alive and a probability of $p_2$ of being dead. With the cross-terms, it's a little hard to say.

Regular quantum evolution is not going to change the probabilities $p_1$ and $p_2$. So I don't think that the universe will ever evolve into a state where the cat is definitely dead.

#### A. Neumaier

Right, but I don't think that really changes anything, does it? Instead of a cat that is in a superposition of alive and dead, you have the whole universe in a superposition of a universe with a dead cat and a universe with a live cat.
Superposition makes no sense for density operators.

We have a universe with an evolving density operator, and at the preparation time we have a binary subsystem whose reduced density operator is prepared as $\psi^S=\psi_S\psi_S^*$ with a superposition $\psi_S\in C^2$, and another subsystem describing a binary live/dead variable for a cat in local equilibrium, whose reduced density operator $\psi^C$ is prepared as $\psi_C\psi_C^*$ with $\psi_C={1\choose 0}$, where $A=\pmatrix{1 & 0\cr 0 & 0}$ is the
degree of aliveness. The question is what happens to the reduced density matrix $\psi^C$ under the unitary evolution applied to the density operator of the universe. The claim is that only two states are stable under small perturbations.

#### A. Neumaier

In terms of density matrices, the evolution would produce something like this:

$\rho = p_1 \rho_{alive} + p_2 \rho_{dead} + \rho_{cross}$

where $\rho_{cross}$ represents the cross terms like $|dead\rangle\langle alive|$ and $|alive\rangle\langle dead|$ ("alive" and "dead" referring to states of the whole universe, not just the cat).

If you ignore the cross-terms, this can be interpreted as the cat having a probability of $p_1$ of being alive and a probability of $p_2$ of being dead. With the cross-terms, it's a little hard to say.

Regular quantum evolution is not going to change the probabilities $p_1$ and $p_2$. So I don't think that the universe will ever evolve into a state where the cat is definitely dead.
It is more complicated; only the q-expectation of $A$ of the cat needs to evolve such that a binary decision can be made. I'll provide some analysis in a separate thread, but not today; I need time to work out a reasonably intuitive form and to type the details.

#### Demystifier

2018 Award
It is more complicated; only the q-expectation of A of the cat needs to evolve such that a binary decision can be made. I'll provide some analysis in a separate thread, but not today; I need time to work out a reasonably intuitive form and to type the details.
I'm looking forward to see that. This might clarify some important things about TI.

#### 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

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