# A Proof that the thermal interpretation of QM is wrong

#### Demystifier

2018 Award
Preface

After a lengthy discussion of the thermal interpretation of quantum physics in https://www.physicsforums.com/threads/the-thermal-interpretation-of-quantum-physics.967116/ , now I think I can prove that it is wrong, i.e. that it doesn't solve the measurement problem in a way it claims it does. Since the following is supposed to be a final proof, I don't want it to be lost among many other posts in the thread above. That's why I open a separate thread.

Introduction

Here I want to prove that the thermal interpretation, contrary to its claim, cannot solve the measurement problem. For definiteness I will present the measurement problem in the form of the Schrodinger cat paradox, but it can be presented in other forms as well. I will prove that the Schrodinger cat "paradox" is a true paradox within the thermal interpretation that does not have a solution within that interpretation.

Let $\rho^{\rm (cat)}$ be the density matrix describing the cat degrees of freedom. In principle, it is determined by the density matrix $\rho$ of the whole Universe as
$$\rho^{\rm (cat)}={\rm Tr}_{\rm no\,cat} \rho$$
where ${\rm Tr}_{\rm no\,cat}$ denotes the trace over all degrees of freedom except the degrees of freedom of the cat. Since $\rho^{\rm (cat)}$ describes an open system, it's dynamics is very complicated and nonlinear. Since the details of influence of the environment "no cat" degrees of freedom on the cat are not known in practice, the evolution of $\rho^{\rm (cat)}$ in practice can be described by stochastic equations. The thermal interpretation conjectures (without an actual proof) that this complicated, nonlinear and effectively stochastic evolution can explain why the superposition of an alive and a dead cat is unstable, so that the system exhibits a fast decay towards an either dead cat or alive cat. Here I prove that this conjecture is wrong.

The central idea of my proof is to consider the problem from the point of view of the whole Universe, instead from the point of view of the cat. Even though the whole Universe is in principle much more complicated than the cat, this actually simplifies the analysis because it is known that the whole universe evolves unitarily, given by the unitary evolution operator
$$U(t)=e^{-iHt}$$
where $H$ is the Hamiltonian of the Universe.

The proof

Let $\rho(t)$ be the density matrix of the whole Universe. In general, it evolves with time according to $\rho(t)=U(t)\rho(0)U^{\dagger}(t)$.

Now suppose that initially $\rho(0)=\rho_{\rm alive}$, where $\rho_{\rm alive}$ is the state of the Universe with an alive cat. The alive state is stable, i.e. the cat who is initially alive will stay alive for a long time. Hence we can write
$$U(t)\rho_{\rm alive}U^{\dagger}(t)=\rho_{\rm alive}(t)$$
where $\rho_{\rm alive}(t)$ is the state of the Universe with a cat alive during a long time. Similarly, if initially $\rho(0)=\rho_{\rm dead}$ then we have a dead cat for a long time, so we can write
$$U(t)\rho_{\rm dead}U^{\dagger}(t)=\rho_{\rm dead}(t)$$

But what if initially we have the superposition of a dead and an alive cat? It is certainly possible as an initial condition, but the question is what happens with such a superposition later? Is it stable or unstable? To simplify the analysis we shall assume that the initial superposition is incoherent, i.e. that
$$\rho(0)=\frac{1}{2}\rho_{\rm alive}+\frac{1}{2}\rho_{\rm dead}$$
without the interference term. (We shall show later that inclusion of the interference terms does not change the final results.) Hence the linearity of evolution for the whole Universe implies
$$U(t)\rho(0)U^{\dagger}(t)=\frac{1}{2}\rho_{\rm alive}(t)+\frac{1}{2}\rho_{\rm dead}(t)$$
This proves that the superposition is stable, i.e. that there is no decay to $\rho_{\rm alive}(t)$ or $\rho_{\rm dead}(t)$.

Now what about beables in the thermal interpretation? All beables in the thermal interpretation are of the form
$$\langle O(t)\rangle = {\rm Tr}O\rho(t)$$
where $O$ are hermitian observables. So if $O$ is a cat observable that describes some actual properties of the cat, we see that the actual property of the cat is
$$\langle O(t)\rangle = \frac{ \langle O(t)\rangle_{\rm alive} + \langle O(t)\rangle_{\rm dead}}{2}$$
which is neither $\langle O(t)\rangle_{\rm alive}\equiv {\rm Tr}O\rho_{\rm alive}(t)$ nor $\langle O(t)\rangle_{\rm dead}\equiv {\rm Tr}O\rho_{\rm dead}(t)$. This proves that beables of the thermal interpretation cannot solve the Schrodinger cat paradox. By a straightforward generalization of this proof, one can see that thermal interpretation cannot resolve the measurement problem of quantum physics in general.

Note that the cat beable can also be written as
$$\langle O(t)\rangle = {\rm Tr}_{\rm cat}O\rho^{\rm (cat)}(t)$$
where $\rho^{\rm (cat)}(t)$ (given by the first equation in Introduction above) satisfies a nonlinear equation and ${\rm Tr}_{\rm cat}$ denotes tracing over cat degrees of freedom. The thermal interpretation conjectures that this nonlinearity can somehow cause the decay towards an either dead or alive cat. What our proof shows is that this conjecture is not true, which is a consequence of the fact that the Universe as a whole obeys a linear evolution. No matter how complicated and apparently stochastic behavior of a subsystem may be, the unitary evolution of the whole Universe implies that it cannot solve the measurement problem within the thermal interpretation.

Finally a note on the ignored interference terms. If the initial state of the Universe is a coherent superposition
$$\frac{ |{\rm alive}\rangle + |{\rm dead}\rangle }{\sqrt{2}}$$
then the initial $\rho(0)$ has the additional interference term
$$\rho_{\rm interf}=\frac{1}{2}|{\rm alive}\rangle\langle {\rm dead}| + \frac{1}{2}|{\rm dead}\rangle\langle {\rm alive}|$$
In principle this contributes to beables via terms of the form
$${\rm Tr}O|{\rm alive}\rangle\langle {\rm dead}|$$
However, if $O$ is an observable that distinguishes a dead cat from an alive one, then terms of the above form are negligible. For instance, if the deat cat is distingushed from an alive one by having a closed/open eye, then $O$ can be taken to be the position operator $x$ describing the position of the eyelid, while $|{\rm alive}\rangle$ and $|{\rm dead}\rangle$ are proportional to two different eigenstates of $x$, in which case it's easy to see that the term above vanishes.

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#### A. Neumaier

Finally a note on the ignored interference terms. If the initial state of the Universe is a coherent superposition
$$\frac{ |{\rm alive}\rangle + |{\rm dead}\rangle }{\sqrt{2}}$$
then the initial $\rho(0)$ has the additional interference term
$$\rho_{\rm interf}=\frac{1}{2}|{\rm alive}\rangle\langle {\rm dead}| + \frac{1}{2}|{\rm dead}\rangle\langle {\rm alive}|$$
In principle this contributes to beables via terms of the form
$${\rm Tr}O|{\rm alive}\rangle\langle {\rm dead}|$$
However, if $O$ is an observable that distinguishes a dead cat from an alive one, then terms of the above form are negligible. For instance, if the deat cat is distingushed from an alive one by having a closed/open eye, then $O$ can be taken to be the position operator $x$ describing the position of the eyelid, while $|{\rm alive}\rangle$ and $|{\rm dead}\rangle$ are proportional to two different eigenstates of $x$, in which case it's easy to see that the term above vanishes.
In your final note, which is the only part relevant to the problem, you assumed the link between measurement results of O and eigenstates, which is not valid in the thermal interpretation.

#### Demystifier

2018 Award
In your final note, which is the only part relevant to the problem, you assumed the link between measurement results of O and eigenstates, which is not valid in the thermal interpretation.
That's not essential at all. Alternatively, I can take $|{\rm dead}\rangle$ and $|{\rm alive}\rangle$ to be proportional to two macroscopically different coherent states $|p,x_1\rangle$ and $|p,x_2\rangle$, in which case my argument that the interfernce term is negligible applies without having position eigenstates.

#### Mentz114

Gold Member
But what if initially we have the superposition of a dead and an alive cat? It is certainly possible as an initial condition, but the question is what happens with such a superposition
You are assuming this ridiculous fallacy without any physical justification or evidence. There is no such thing.

#### Demystifier

2018 Award
You are assuming this ridiculous fallacy without any physical justification or evidence. There is no such thing.
If you are suspicious about cats (despite the fact that Schrodinger proved that it is possible if Schrodinger equation is always true) , consider spin in a superposition up and down. The proof doesn't change.

#### A. Neumaier

That's not essential at all. Alternatively, I can take $|{\rm dead}\rangle$ and $|{\rm alive}\rangle$ to be proportional to two macroscopically different coherent states $|p,x_1\rangle$ and $|p,x_2\rangle$, in which case my argument that the interfernce term is negligible applies without having position eigenstates.
But then your argument about properties is no longer valid. O is not a quantity you can freely choose in your argument.

#### stevendaryl

Staff Emeritus
You are assuming this ridiculous fallacy without any physical justification or evidence. There is no such thing.
That's harsh, and also unhelpful (I think). It's true that there aren't actual states $|alive\rangle$ and $|dead\rangle$, but is this oversimplification important for the point @Demystifier is making? If so, can you show how a more careful treatment of cats would lead to a different conclusion? If not, then your remark is unhelpful.

What I think a more careful treatment would like is something like this:

Presumably, a macroscopic configuration (a description at the level of cats and cyanide canisters) corresponds to some equivalence class of microscopic states. Some microscopic states are incompatible with there being a live cat. So I assume that for a macroscopic configuration $c$ (a description of the locations, types, shapes, and health of cats and so forth) there is a corresponding projection operator $\Pi_c$ such that if microstate $|\psi\rangle$ is compatible with configuration $c$, then $\Pi_c |\psi\rangle = |\psi\rangle$, and if $|\psi\rangle$ is incompatible with $c$, then $\Pi_c |\psi\rangle = 0$.

Then instead of talking about the states $|alive\rangle$ and $|dead\rangle$, we can talk about the projection operators.

#### stevendaryl

Staff Emeritus
But what if initially we have the superposition of a dead and an alive cat? It is certainly possible as an initial condition, but the question is what happens with such a superposition later? Is it stable or unstable? To simplify the analysis we shall assume that the initial superposition is incoherent, i.e. that
$$\rho(0)=\frac{1}{2}\rho_{\rm alive}+\frac{1}{2}\rho_{\rm dead}$$
without the interference term. (We shall show later that inclusion of the interference terms does not change the final results.) Hence the linearity of evolution for the whole Universe implies
$$U(t)\rho(0)U^{\dagger}(t)=\frac{1}{2}\rho_{\rm alive}(t)+\frac{1}{2}\rho_{\rm dead}(t)$$
This proves that the superposition is stable, i.e. that there is no decay to $\rho_{\rm alive}(t)$ or $\rho_{\rm dead}(t)$.
I don't know about the thermal interpretation, but in some interpretations of quantum mechanics, the density matrix is interpreted to include subjective uncertainty. So being a mix of "alive" and "dead" is compatible with the cat being alive or dead, and you just don't know which (until you peek, to resolve the subjective uncertainty).

If you consider the "density matrix of the universe" to be the most complete information there can be about the state of the universe, then I guess that out isn't possible.

#### Demystifier

2018 Award
I don't know about the thermal interpretation, but in some interpretations of quantum mechanics, the density matrix is interpreted to include subjective uncertainty.
That's not the case with thermal interpretation.

#### Demystifier

2018 Award
But then your argument about properties is no longer valid. O is not a quantity you can freely choose in your argument.
All my argument requires is that O is a quantity that distinguishes a dead cat from an alive one. This requirement is indeed necessary if one wants the corresponding beable to determine whether the cat is dead or alive.

#### A. Neumaier

I don't know about the thermal interpretation, but in some interpretations of quantum mechanics, the density matrix is interpreted to include subjective uncertainty.
In the thermal interpretation, the density operator is fully objective. It encodes true properties, not what we know about them. The latter would be additional uncertainty about the true density operator.

#### A. Neumaier

All my argument requires is that O is a quantity that distinguishes a dead cat from an alive one. This requirement is indeed necessary if one wants the corresponding beable to determine whether the cat is dead or alive.
But O is a macroscopic pointer reading, not an operator acting on the cat state as you pretend!

You also ignore what happens to the additional interference terms during the evolution of the state of the universe. They become very complicated stuff about which you habe no information, but just postulate that they should remain zero. This is wishful thinking.

#### Demystifier

2018 Award
But O is a macroscopic pointer reading, not an operator acting on the cat state as you pretend!
No. $O$ is an operator, $\langle O\rangle$ is a pointer reading.

You also ignore what happens to the additional interference terms during the evolution of the state of the universe. They become very complicated stuff about which you habe no information, but just postulate that they should remain zero. This is wishful thinking.
No. I have shown that their contribution is negligible in situations of interest. Note that from
$$|{\rm dead}(t)\rangle =U(t) |{\rm dead}(0)\rangle$$
$$|{\rm alive}(t)\rangle =U(t) |{\rm alive}(0)\rangle$$
$$\langle {\rm alive} (0)|{\rm dead} (0)\rangle \ll 1$$
if follows that
$$\langle {\rm alive} (t)|{\rm dead} (t) \rangle \ll 1$$
for all $t$. The interference terms give rise to contributions of the form
$${\rm Tr}O|{\rm alive}(t)\rangle\langle {\rm dead}(t)|=\langle {\rm dead}(t)|O|{\rm alive}(t)\rangle$$
which are small for observables $O$ which distinguish dead from alive.

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#### A. Neumaier

If you are suspicious about cats (despite the fact that Schrodinger proved that it is possible if Schrodinger equation is always true) , consider spin in a superposition up and down. The proof doesn't change.
I prefer to discuss this version, since one sees better your hidden assumptions.
No. $O$ is an operator, $\langle O\rangle$ is a pointer reading.
I meant that you took for O an operator on the system (the spin state), not one on the detector (which you don't model at all)! Thus O is not a pointer. Your spin measures itself!

No. The interference terms give rise to contributions of the form
$${\rm Tr}O|{\rm alive}(t)\rangle\langle {\rm dead}(t)|=\langle {\rm dead}(t)|O|{\rm alive}(t)\rangle$$
which are small for observables $O$ which distinguish dead from alive.
This is only true because you don't consider the detector at all. Nothing in your setting measures anything.

#### Demystifier

2018 Award
This is only true because you don't consider the detector at all. Nothing in your setting measures anything.
That is simply not true. My $\rho$ describes the whole universe, which includes the detector. My $O$ is any operator that distinguishes one state from the other. If those states are states of the detector (instead of eyelids of the cat), then your objection does not apply.

#### A. Neumaier

My $\rho$ describes the whole universe, which includes the detector.
No. The devil is in the details!

You assumed in your final note that the whole state of the universe is a pure state:
If the initial state of the Universe is a coherent superposition
But the state of a system containing a detector (which has positive temperature) is never pure.
Thus your assumption excludes a detector.

Your ''proof'' would invalidate the analysis of every bistable quantum system, of which many are realizable experimentally: There are quite a number of papers on optical bistability that show how
coarse-grained bistability arises from a quantum model by projecting out irrelevant degrees of freedom. See, e.g., the time-honored papers
• P.D. Drummond and D.F. Walls, Quantum theory of optical bistability I. Nonlinear polarisability model, J. Physics A Math. Gen. 13 (1980), 725--741.
• M.L. Steyn-Ross and C.W. Gardiner, Quantum theory of excitonic optical bistability, Phys. Rev. A 27 (1983), 310--325.

#### DarMM

Gold Member
In the thermal interpretation, the density operator is fully objective. It encodes true properties, not what we know about them. The latter would be additional uncertainty about the true density operator.
Not only in the thermal interpretation, but I think the idea that density matrices reflect uncertainties about the "true" pure state are basically untenable.

If they really were uncertainty about the pure state you'd expect $\mathcal{L}^{1}\left(\mathcal{H}\right)$ as the space of mixed states rather than $Tr\left(\mathcal{H}\right)$. Not to mention the absence of pure states in QFT.

#### Demystifier

2018 Award
You assumed in your final note that the whole state of the universe is a pure state:
But the state of a system containing a detector (which has positive temperature) is never pure.
Thus your assumption excludes a detector.
My final note is not at all essential for the proof. The proof itself is presented in the subsection called "The proof", where no assumption that the state is pure is used. In fact, in this section I explicitly assume that the state is not pure (which simplifies the calculation), while the only purpose of the final note is to demonstrate that the result is essentially the same even when the state is pure. You are barking up the wrong tree.

And by the way, even though it is not essential for the argument, note that even a pure state can have a temperature. To see that, consider the state of the form
$$\sum_n e^{i\varphi_n}e^{-\beta E_n/2}|n\rangle$$
Thus it is not true that a system with a detector cannot be pure.

#### Demystifier

2018 Award
Your ''proof'' would invalidate the analysis of every bistable quantum system, of which many are realizable experimentally: There are quite a number of papers on optical bistability that show how
coarse-grained bistability arises from a quantum model by projecting out irrelevant degrees of freedom. See, e.g., the time-honored papers
• P.D. Drummond and D.F. Walls, Quantum theory of optical bistability I. Nonlinear polarisability model, J. Physics A Math. Gen. 13 (1980), 725--741.
• M.L. Steyn-Ross and C.W. Gardiner, Quantum theory of excitonic optical bistability, Phys. Rev. A 27 (1983), 310--325.
Those papers assume that semi-classical approximation can be used. An interpretation of QM must explain why the semi-classical approximation can be used, without assuming it. My proof shows that the thermal interpretation cannot explain that.

In other words, I am not saying that those bistable systems don't exist. I am saying that the thermal interpretation cannot explain it.

#### A. Neumaier

My final note is not at all essential for the proof. The proof itself is presented in the subsection called "The proof", where no assumption that the state is pure is used.
Instead you make the ridiculous assumption that the state of the universe is the uniform mixture of the states of two universes, one with a dead cat and one with an alive cat. Let $t_A$ and $t_D$ be the times where the universe is in the alive-cat and dead-cat state, respectively, then at no time the universe is in the state you analyze, since this state develops deterministically according to the unitary dynamics.

Thus "The proof" is completely bogus.

while the only purpose of the final note is to demonstrate that the result is essentially the same even when the state is pure.
At the time of preparation, a Schrödinger cat is in a pure state while the remaining universe is always in a state approximately described by a local equilibrium state (at each time different). This enters the standard analysis of open systems resulting without any interpretation (pure formal manipulation) in a disspative subsystem behavior.

You, however, replace these physical (preparable!) states by weird unphysical states, from which one of course gets only nonsense, ''proving'' that all statistical mechanics is bogus.
And by the way, even though it is not essential for the argument, note that even a pure state can have a temperature. To see that, consider the state of the form
$$\sum_n e^{i\varphi_n}e^{-\beta E_n/2}|n\rangle$$
Thus it is not true that a system with a detector cannot be pure.
You thoroughly misunderstand the meaning of temperature in statistical mechanics. It is a Lagrange multiplier in the expression for the density operator, not just a variable multiplying energies. That your pure state depends on such a variable $\beta$ doesn't endow this variable with the physical (measurable) meaning of temperatue.

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#### A. Neumaier

An interpretation of QM must explain why the semi-classical approximation can be used, without assuming it.
Oh, really? 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!

Instead I have seen the standard textbook expositions of traditional interpretations make liberal use of semiclassical approximations without any explanation other than that it makes sense to use it. The Stern-Gerlach experiment only treats the spin by quantum mechanics, everything else is treated semiclassically. Long distance photon experiments use semiclassical paths to analyze the setups.

Without assuming semiclassical approximations for all but the relevant core of a quantum system, nothing at all can be done with traditional interpretations.

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#### A. Neumaier

The proof itself is presented in the subsection called "The proof"
Thus "The proof" is completely bogus.
For a true proof you'd have to assume the following:

The universe is at each time in a state $\rho(t)$ following a unitary dynamics . At some preparation time $t_0$, the reduced density matrix of some spin to be measured is $\rho^S$.

The basic claim of the thermal interpretation is that, if conditions characterizing a measurement situation hold, there is a pointer variable $X$ on the remaining universe such that at some reasonable time $t>t_0$, the q-expectation of $X$ allows one to read off spin up or down with the Born probabilities determined by $\rho^S$.

To prove the thermal interpretation wrong you need to show that this is impossible, i.e., that such a pointer variable never exists.

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#### vanhees71

Gold Member
Well, it's hard to prove something which is not clearly formulated. So I can't say whether #1 disproves anything or not, because it's still not clarified what the interpretation of the "thermal interpretation" really is (you only told us what it is not ;-)).

The one thing I can say is that of course the SG experiment can be completely described quantum mechanically without recourse to semiclassical approximations for the motion of the atoms. This quantum mechanical description, however, shows of course that the semiclassical approximation is justified very well. Anyway, here's a paper using numerics to solve the Schrödinger equation; see, e.g.,

https://arxiv.org/abs/quant-ph/0409206

#### A. Neumaier

the SG experiment can be completely described quantum mechanically without recourse to semiclassical approximations for the motion of the atoms.
It still treats the electromagnetic field and the detector semiclassically.

Of course, all this can be justified, without recourse to interpretation, in contrast to what Demystifier assumes. Unless one wants to derive everything from the unitary dynamics of the universe, semiclassical approaches are necessary and usually justified.

#### charters

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

"Proof that the thermal interpretation of QM is wrong"

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