The ontology problem of thermal interpretation

[Demystifier]

TL;DR Summary

The thermal interpretation of QM contains both local and nonlocal beables. Taking the local ones for granted, I argue that nonlocal beables are either inconsistent with the local ones or consistent but superfluous.

I have already discussed the ontology problem of thermal interpretation (TI) of quantum mechanics (QM) several times in the main thread on TI. The following is supposed to be the final refined version of my argument, so I don't want it to be lost among other posts in the main TI thread. Therefore I open a new thread about it.

Suppose that Alice measures the field $\phi(x)$ at the point $x_A$ and Bob measures the field at the point $x_B$. The values of the field obtained by Alice and Bob are denoted by $\varphi_A$ and $\varphi_B$, respectively. After that Alice and Bob meet and tell their results of measurements to each other. Then they compute the product of those results, the numerical value of which is
$$\varphi_{AB}=\varphi_A\varphi_B$$
But according to TI we have
$$\varphi_A=\langle\phi(x_A)\rangle, \;\;\; \varphi_B=\langle\phi(x_B)\rangle$$
$$\varphi_{AB}=\langle\phi(x_A)\phi(x_B)\rangle$$
which implies
$$\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle = \langle\phi(x_A)\phi(x_B)\rangle \;\;\;\;\; (1)$$
On the other hand, in QM (or QFT, to be more precise) in general we have
$$\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle \neq \langle\phi(x_A)\phi(x_B)\rangle$$
so it seems that QM is in contradiction with TI. In other words, the nonlocal TI beable $\langle\phi(x_A)\phi(x_B)\rangle$ seems inconsistent with the local TI beables $\langle\phi(x_A)\rangle$ and $\langle\phi(x_B)\rangle$.

A possible way out of this conundrum is to take into account quantum contextuality. Perhaps the equality (1) is not always satisfied, but only at the time of measurement. And perhaps at that time the state $\rho$ is something close to $|\psi\rangle\langle\psi|$ where $|\psi\rangle$ is an eigenstate of both $\phi(x_A)$ and $\phi(x_B)$. If so, then (1) would be at least approximately valid, which might resolve the problem in the FAPP sense.

But there are problems. First, in my opinion, it is not clear how can $\rho$ in TI evolve in such a way to provide the FAPP equality (1). I have discussed this in a somewhat different context in https://www.physicsforums.com/threads/proof-that-the-thermal-interpretation-of-qm-is-wrong.970038/ . Second, even if I ignore this problem, or assume that it is somehow solved in a way I don't understand, there still remains another problem. If FAPP equalities of the form (1) are valid contextually, i.e. whenever the corresponding experiments are performed, then, in the FAPP sense, all measurements can be reduced to measurements of local beables such as $\langle\phi(x_A)\rangle$ and $\langle\phi(x_B)\rangle$. The nonlocal beables such as $\langle\phi(x_A)\phi(x_B)\rangle$ become superfluous.

Finally a comment on nonlocal action at a distance. If only local beables exist, then the Bell theorem implies that there must be a nonlocal action at a distance between them. If, on the other hand, a measurement of correlation is really a measurement of a nonlocal beable such as $\langle\phi(x_A)\phi(x_B)\rangle$, one might think that it can avoid a nonlocal action at a distance. But it is true only if the problem of evolution of $\rho$ discussed in the paragraph above is solved without introducing a nonlocal mechanism such as an objective collapse. But, as I argued in the link above, I think the problem of evolution of $\rho$ cannot be solved in that way.

 

[PeterDonis]

it seems that QM is in contradiction with TI

You've just restated Bell's Theorem; basically you're saying that TI must satisfy Bell's definition of a "local" theory (i.e., that joint probability distributions must factorize). But I don't think that's the case, since TI just uses the standard math of QM like every other interpretation, so its experimental predictions are the same as those of any other interpretation, i.e., it predicts that the Bell inequalities will be violated in appropriate experimental setups.

In the specific case you mention, while the quantity $\langle \phi(x_a) \phi(x_B) \rangle$ is indeed a nonlocal beable in TI (since every q-expectation is a beable and this is a nonlocal q-expectation), I don't see why TI must assert your equality (1), since TI just uses the standard math of QM to compute q-expectations and (1) contradicts the standard math of QM.

according to TI we have
$$
\varphi_A=\langle\phi(x_A)\rangle, \;\;\; \varphi_B=\langle\phi(x_B)\rangle
$$

No, we don't, because the measurement results Alice and Bob obtain do not have to be equal to the q-expectations.

 

[A. Neumaier]

Suppose that Alice measures the field $\phi(x)$ at the point $x_A$ and Bob measures the field at the point $x_B$. The values of the field obtained by Alice and Bob are denoted by $\varphi_A$ and $\varphi_B$, respectively. After that Alice and Bob meet and tell their results of measurements to each other. Then they compute the product of those results, the numerical value of which is
$$\varphi_{AB}=\varphi_A\varphi_B$$
But according to TI we have
$$\varphi_A=\langle\phi(x_A)\rangle, \;\;\; \varphi_B=\langle\phi(x_B)\rangle$$
$$\varphi_{AB}=\langle\phi(x_A)\phi(x_B)\rangle$$

No. TI claims instead only
$$\varphi\approx\langle\phi(x_A)\rangle, \;\;\; \varphi_B\approx\langle\phi(x_B)\rangle$$
$$\varphi_{AB}\approx\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle$$
which is obviously true.

In this sense Bohmian mechanics is a theory of fundamental local beables (particles have well defined positions in space) with nonlocal interactions. Many-world interpretation, on the other hand, is a theory of nonlocal fundamental beables (the state in the Hilbert space is not defined at a space point). Thermal interpretation is somewhere in between, because it contains both local fundamental beables (e.g. $\langle\phi(x)\rangle$) and nonlocal fundamental beables (e.g. $\langle\phi(x)\phi(y)\rangle$).

This mean that
$$X=\langle\phi(x_A)\rangle, \;\;\; Y=\langle\phi(x_B)\rangle, \;\;\;Z=\langle\phi(x_A)\phi(x_B)\rangle$$
are three independent fundamental beables. There is no reason at all to suppose that $Z=XY$, which is almost never the case. If $Z=XY$ were generlly true then $Z$ would not be fundamental.

 

[Demystifier]

No. TI claims instead only
$$\varphi\approx\langle\phi(x_A)\rangle, \;\;\; \varphi_B\approx\langle\phi(x_B)\rangle$$
$$\varphi_{AB}\approx\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle$$
which is obviously true.

This mean that
$$X=\langle\phi(x_A)\rangle, \;\;\; Y=\langle\phi(x_B)\rangle, \;\;\;Z=\langle\phi(x_A)\phi(x_B)\rangle$$
are three independent fundamental beables. There is no reason at all to suppose that $Z=XY$, which is almost never the case. If $Z=XY$ were generlly true then $Z$ would not be fundamental.

So basically you are saying that $\varphi_{AB}\approx\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle$ and that $\varphi_{AB}$ is not even approximately equal to $\varphi_{AB}'\equiv\langle\phi(x_A)\phi(x_B)\rangle$. Am I right?

But in the Bell-correlations experiments the measured quantities are of the $\varphi_{AB}$ type, and not of the $\varphi_{AB}'$ type. Hence $\varphi_{AB}'$ plays no role in experimental violation of Bell inequalities. Hence only the local Bell-type beables are relevant in such experiments, which makes the nonlocal beables superfluous. How then TI avoids the Bell-type action at a distance?

 

[A. Neumaier]

So basically you are saying that $\varphi_{AB}\approx\langle\phi(x_A)\rangle\langle\phi(x_B)\rangle$

For example, if the field is Hermitian then $X,Y,XY$ are real but $Z$ is generally complex.

and that $\varphi_{AB}$ is not even approximately equal to $\varphi_{AB}'\equiv\langle\phi(x_A)\phi(x_B)\rangle$.

Your $\varphi_{AB}$ equals $XY$ and may or may not be close to $Z$, depending on the nature of the state. $|XY-Z|$ is bounded by the product of the uncertainties of $X$ and $Y$. If these are tiny (as when $X,Y$ are macroscopic quantities), the approximation is good.

But in the Bell-correlations experiments the measured quantities are of the $\varphi_{AB}$ type. [...] Hence only the local Bell-type beables are relevant in such experiments, which makes the nonlocal beables superfluous. How then TI avoids the Bell-type action at a distance?

The transmission of the signal involves time, and hence the dynamics. The time derivative of a local beable involves nonlocal stuff. The situation is analogous to that in Bohmian mechanics, where the particles are also local.

 

[Demystifier]

The transmission of the signal involves time, and hence the dynamics. The time derivative of a local beable involves nonlocal stuff. The situation is analogous to that in Bohmian mechanics, where the particles are also local.

Given the analogy with Bohmian mechanics (BM), what would be the main conceptual difference between BM and TI? In simple terms, what makes TI better than BM?

 

[A. Neumaier]

Given the analogy with Bohmian mechanics (BM), what would be the main conceptual difference between BM and TI? In simple terms, what makes TI better than BM?

In contrast to the beables of Bohmian mechanics, the beables of the TI are standard objects in the quantum formalism, and their dynamics follows from standard quantum theory.