I The thermal interpretation of quantum physics

[A. Neumaier]

Nowhere does one make any approximate classical description of the motion of the atom

But you make one of the detector, which is formally entangled but you treat it as being in one of two definite states, which are not continuously connected.

 

[vanhees71]

Well, the detector is simply some plate, where you catch the Ag atoms to measure there position using a microscope. Of course, one can make a problem out of this simple procedure too, but for which purpose?

 

[A. Neumaier]

Well, the detector is simply some plate, where you catch the Ag atoms to measure there position using a microscope. Of course, one can make a problem out of this simple procedure too, but for which purpose?

For the purpose of checking whether the formal Rules of Quantum Mechanics together with the formal postulates of an interpretation justify this simple heuristic (of treating the plate classically). This is the measurement problem - on a heuristic level there is none.

The devil is always in the details!

 

[stevendaryl]

There is no indeterministic transition in what I said.

Well, the minimal interpretation claims that a measurement always produces an eigenvalue of the corresponding operator, with a probability given by the Born rule. But when you analyze the measurement process quantum mechanically, that's not true.

That's why I say that the minimal interpretation is inconsistent.

 

[ftr]

But when you analyze the measurement process quantum mechanically, that's not true.

So what is true?do you mean something like what Arnold is saying.

 

[vanhees71]

Well, the minimal interpretation claims that a measurement always produces an eigenvalue of the corresponding operator, with a probability given by the Born rule. But when you analyze the measurement process quantum mechanically, that's not true.

That's why I say that the minimal interpretation is inconsistent.

Can you specify an experiment where this contradiction occurs? If this were true, I'm sure it would make a big sensation since it would prove QT wrong, making a "revolutionary" new theory necessary. So far, there seems not to be any such sensation.

 

[stevendaryl]

Can you specify an experiment where this contradiction occurs? If this were true, I'm sure it would make a big sensation since it would prove QT wrong, making a "revolutionary" new theory necessary. So far, there seems not to be any such sensation.

No, people have been living with this contradiction for nearly 100 years. It doesn't bother them any more.

Asking for an experiment in which a contradiction shows up doesn't make any sense. A contradiction is a property of a theory. The real world can't have any contradictions.

 

[stevendaryl]

So what is true?do you mean something like what Arnold is saying.

I don't. Quantum mechanics in its standard formulation is just inconsistent. As I have said before, it's a "soft" inconsistency.

 

[*now*]

These papers are about the notion of temperature in classical general relativity. What should this have to do with my papers?

I think the question was more curiosity on my part given how the effect has been considered in literature concerning the thermal time hypothesis than commentary on the TI papers, so could be removed along with follow ups like this reply. The 2nd paper (4.3) of TI mentions that gravity should be discussed in a more detailed account of the universe and so touch on the open question of quantum gravity, but that the papers offer a consistent framework for such discussions. Some thermal time hypothesis papers consider that temporal aspect thermodynamically and in non-general-relativistic physics, and the effect is compared in terms of Boltzmann Gibbs equilibrium state and general relativistic statistics, generalising to stationary spacetime. There may be some parallels with perspectival accounts, like allowing for different pictures and views like the statistical minimal one, or in comparisons of the effect with the TI. It might be relevant or useful, or not.

Quantum mechanics predicts the right correlations, but does not explain so far how it is possible that these are actually obtained!

What of attempts so far?

 

[lukesfn]

Sorry late reply...

In relativistic quantum field theory, q-expectations of fields refer to single measurements, all fields are local and objective, and single measurements are fully explained through the effects of coarse-graining and the resulting dissipation.

That was non sequitur. I prefer to think of expectations values applying to individual measurements myself.

However, maybe I was looking at TI too much through the lenses of the Ensemble interpretation. I can see how TI could apply to single measurement results, where on my terms it would be compatible with unique objective measurement results, but non-local. (I’d have to call extended objects explicitly non local)

Trying to understand TI is an interesting thought exercise, however I can’t help but feel it has a lot of unnecessary baggage, obfuscation, and new definitions, where it could have been more simple and more general, although I suppose some people will find it all well motivated, after all, it could be argued that all interpretations introduce unnecessary baggage.

It all feels a bit backwards from how I like to think about things, even as I try to understand it, I end up changing it to fit better with my thinking, so it probably isn't the most helpful way of looking at things for me. Anyway, I doubt my contributions are particularly helpful here.

 

[ftr]

Can you specify an experiment where this contradiction occurs? If this were true, I'm sure it would make a big sensation since it would prove QT wrong, making a "revolutionary" new theory necessary. So far, there seems not to be any such sensation.

I think it all boils down to what experiment proves that the position of a particle is an eigenvalue of the corresponding operator. can you name one?

 

[A. Neumaier]

the position of a particle is an eigenvalue of the corresponding operator.

The position operator has a purely continuous spectrum, hence no eigenvalues.

 

[ftr]

So what experiment determines this specrum?

 

[A. Neumaier]

So what experiment determines this spectrum?

Every real number is in the spectrum of the position operator, so the part of Born's rule you refer to is here trivially valid.

 

[ftr]

Do you mean no experiment is needed to validate?

 

[ftr]

Also do you agree that say if we do measurement for a particle in a box we can confirm the density function?

 

[A. Neumaier]

I think it all boils down to what experiment proves that the position of a particle is an eigenvalue of the corresponding operator. can you name one?

Do you mean no experiment is needed to validate?

Also do you agree that say if we do measurement for a particle in a box we can confirm the density function?

Please ask about such elementary things in a different thread. They have nothing to do with the thermal interpretation.

 

[ftr]

Ok, I will.

 

[DrDu]

The official description of the thermal interpretation of quantum physics can be found in my just finished papers

Foundations of quantum physics I. A critique of the tradition,
Foundations of quantum physics II. The thermal interpretation,
Foundations of quantum physics III. Measurement.

They are also accessible through the arXiv at
arXiv:1902.10778 (Part I), arXiv:1902.10779 (Part II), arXiv:1902.10782 (Part III).

If you are short of time, start reading at post #260. DarMM gave in post #268 a nice summary of the thermal interpretation. The simplest quantum system, a qubit, was already described by Stokes 1852, in terms essentially equivalent to the thermal interpretation.

Three reviews (Part I, Part II, Part III) are on PhysicsOverflow, together with some comments by me.

The articles are very interesting. Nevertheless I do not like the style, especially of paper I.
It insinuates that there is only one universal "traditional" way of learning QM which follows the steps listed in 5.2.
Although it is more than 30 years now that I learned QM, this was certainly not the way I learned it.

 

[stevendaryl]

I think it all boils down to what experiment proves that the position of a particle is an eigenvalue of the corresponding operator. can you name one?

It's more interesting with an operator that has discrete eigenvalues, such as angular momentum. Experiments can confirm that a measurement of any component of angular momentum always yields an integer or half-integer multiple of $\hbar$.

The point I have made is that this prediction of quantum mechanics mixes up some things that are true by definition and some things that are empirically verified.

By definition, a device can only be said to measure an observable $A$ if distinct values of $A$ lead to macroscopically distinguishable states of the measuring device. If $A$ has eigenvalues $a_1, a_2, ...$, then that means that for each $j$, there is a corresponding distinct macroscopic state of the device $S_j$ such that if the system being measured is initial in an eigenstate of $A$ with value $a_j$, then the device will make a transition into macroscopic state $S_j$. That's the definition of what it means to measure $A$.

So the rule saying that measurements always produce eigenvalues of the observable being measured follows from the definition of "measure", plus the linearity of the evolution of quantum states, plus the macroscopic rule: If a macroscopic system would, by Schrodinger's equation be in a superposition of macroscopically distinguishable states, then it will be found to be in one or the other of those states, with probabilities given by the square of the corresponding amplitude of the wave function of the macroscopic system.

In other words, the empirical content of the Born rule is implied by the claim that every system evolves unitarily according to Schrodinger's equation, except that macroscopic systems are always observed to be in states with definite values for all macroscopic quantities (with probabilities given by the square of the amplitude corresponding to each value). There is no need for the Born rule to apply to all observables; it is enough to apply it to macroscopic observables.

 

[ftr]

I don't. Quantum mechanics in its standard formulation is just inconsistent. As I have said before, it's a "soft" inconsistency.

Why do you say that, in what sense it is soft.
Just saw your post thanks. Ignore.

 

[ftr]

There is no need for the Born rule to apply to all observables; it is enough to apply it to macroscopic observables.

Now I remember your other post that you elaborated on this point. But isn't still you need to explain the wavefunction since it is a pre-measurement entity. I think this problem is related to physics where usually we define variable as related to some physical character, but psi is an after thought ( maybe it is correct we are forced into it, maybe the model is simply awkward).

 

[stevendaryl]

Why do you say that, in what sense it is soft.

Let's suppose that we have a measuring device that starts off in some start state, $S_0$. It measures an electron's spin along the z-axis, and if it measures spin-up, it makes a transition to the state $S_{up}$, and if it measures spin-down, it makes a transition to the state $S_{down}$. Let $S_{other}$ be some third state of the measuring device.

Now, suppose we prepare the electron so that it's in a superposition of spin-up (with amplitude $\alpha$) and spin-down (with amplitude $\beta$). We want to compute the probability that the measuring device winds up in state $S_{other}$.

We can first compute amplitudes, and then square them to get probabilities. Let $\phi_{up, other}$ be the amplitude for the device to transition from state $S_{up}$ to state $S_{other}$. let $\phi_{down, other}$ be the amplitude to transition from $S_{down}$ to $S_{other}$. Then the amplitude for the device to end up in state $S_{other}$ is given by:

$\chi_{other} = \alpha \phi_{up, other} + \beta \phi_{\beta, other}$

The probability is the absolute square of the amplitude, so the probability of ending up in state $S_{other}$ is:

$|\chi_{other}|^2 = |\alpha|^2 |\phi_{up, other}|^2 + |\beta|^2 |\phi_{down, other}|^2 + \alpha^* \phi_{up, other}^* \beta \phi_{down, other} + \alpha \phi_{up, other} \beta^* \phi_{down, other}^*$

If we define $P_{other} = |\chi_{other}|^2$, $P_{up, other} = |\alpha|^2 |\phi_{up, other}|^2$, $P_{down, other} = |\beta|^2 |\phi_{down, other}|^2$ and $I = \alpha^* \phi_{up, other}^* \beta \phi_{down, other} + \alpha \phi_{up, other} \beta^* \phi_{down, other}^*$, then this becomes:

$P_{other} = P_{up, other} + P_{down, other} + I$

where $I$ is an interference term between the two intermediate possibilities.

This amounts to applying the Born rule only at the end. On the other hand, if we apply the Born rule at the point of measurement, then we get:

$P_{other} = P_{up, other} + P_{down, other}$

(without the $I$). So saying that the Born rule applies to every measurement gives a different answer for probabilities than if you only apply the Born rule at the end. That's a contradiction, it seems to me.

On the other hand, the difference between the two predictions is the interference term $I$. Interference terms between macroscopically distinct configurations are practically impossible to measure. It's impossible to calculate, in the first place, and is likely to be completely negligible.

So the contradiction is that there are two different ways to compute a probability, and they give different values. But it's a soft contradiction in the sense that those two different values are neglibly different.

 

[A. Neumaier]

The articles are very interesting. Nevertheless I do not like the style, especially of paper I.
It insinuates that there is only one universal "traditional" way of learning QM which follows the steps listed in 5.2.
Although it is more than 30 years now that I learned QM, this was certainly not the way I learned it.

Thanks for your critique. This section was kind of a caricature. Of course, not everyone is exposed to everything mentioned, also the order may be quite different. Nevertheless, one comes across all this stuff sooner or later when one is doing enough quantum mechanics.

I learned quantum mechanics not in a course but by self-teaching from books and later, articles. Certainly all these things puzzled me when I encountered them, and it took me a long time to figure out how to think of all this in a coherent way.

 

[vanhees71]

No, people have been living with this contradiction for nearly 100 years. It doesn't bother them any more.

Asking for an experiment in which a contradiction shows up doesn't make any sense. A contradiction is a property of a theory. The real world can't have any contradictions.

Ok, then tell me what you consider a contradiction. As far as I know there's no intrinsic contradiction in QT at all.

 

[vanhees71]

I think it all boils down to what experiment proves that the position of a particle is an eigenvalue of the corresponding operator. can you name one?

Well, the spectrum of any position operator is $\mathbb{R}$. I don't think it makes sense to ask this very question in this case since by definition we measure positions with real numbers. It was not me who made the bold claim that there's a contradiction within QT! I don't think that there is one!

 

[stevendaryl]

Ok, then tell me what you consider a contradiction. As far as I know there's no intrinsic contradiction in QT at all.

I already told you the contradiction.

1. On the one hand, the minimal interpretation claims that a measurement of an observable produces a result that is one of the eigenvalues of that observable.
2. On the other hand, if the system being measured is in a superposition of eigenstates, and we treat the measuring device quantum-mechanically, then the device itself ends up in a superposition of different results.

That's a contradiction. According to 1, the device will end up in one of a number of possible macroscopic states, with probability given by the Born rule. According to 2, the device will definitely end up in a superposition state that is none of those possibilities.

 

[DarMM]

I already told you the contradiction.

1. On the one hand, the minimal interpretation claims that a measurement of an observable produces a result that is one of the eigenvalues of that observable.
2. On the other hand, if the system being measured is in a superposition of eigenstates, and we treat the measuring device quantum-mechanically, then the device itself ends up in a superposition of different results.

That's a contradiction. According to 1, the device will end up in one of a number of possible macroscopic states, with probability given by the Born rule. According to 2, the device will definitely end up in a superposition state that is none of those possibilities.

That's not a contradiction in my opinion, more a sign of possible incompleteness. So if we have the system:
$$\mathcal{H}_t = \mathcal{H}_{s}\otimes\mathcal{H}_{d}\otimes\mathcal{H}_{e}$$
where $s$ is the atomic system, $d$ is the device and $e$ is the entire rest of the lab environment.

A quantum mechanical model will say that the state of the whole system $|\psi\rangle \in \mathcal{H}_t$ is in a superposition after measurement and a superobserver will use such a superposed state.

However to date nobody has produced an actual proof that this is in contradiction with the subsystem $s + d$ being in a definite state. Frauchiger-Renner and Brukner's objectivity theorems are attempts at this, but the consensus by now is that they don't succeed.

 

[A. Neumaier]

would you say that because a system is in a state for which
$$P\left(S_z = \frac{1}{2}\right) = 1$$
that means the particle in fact already has the spin value of $S_z = \frac{1}{2}$ or does it only mean if you set up an $S_z$ measurement it's guaranteed to produce a specific result?

The answer to this would allow me to know if you're closer to Brukner-Zellinger or others like Haag.

Guaranteed? With probability zero (but still in finitely many of infinitely many cases) it could also have another value, form a purist point of view...

 

[DarMM]

Guaranteed? With probability zero (but still in finitely many of infinitely many cases) it could also have another value, form a purist point of view...

It is a discrete outcome space though.