# A How does the thermal interpretation explain Stern-Gerlach?

#### A. Neumaier

How do "multilocal hidden variables" explain the EPR results?
They don't explain them by itself, they are just outside the scope of Bell's arguments since his assumptions are incompatible with multilocal hidden variables.

The explanation of the nonclassical long-distance correlations is the standard quantum dynamics, which is assumed exact in the thermal interpretation and predicts the standard correlations. These are bilocal beables, approximately measurable (by the weak law of large numbers as any q-expectation; see Section 3 of Part II of my series of papers) through averaging over many independent realizations of discrete tests. The discrete response is explained by environment-induced randomness and environment-induced dissipation, as discussed in Subsections 4.3 and 5.1 of Part III.

The thermal interpretation does not give an explanation, though, of how Nature is able to figure out how to behave according to the quantum laws. It just follows them.

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#### DarMM

Gold Member
If there is a fixed deterministic value, it does not mean anything to say it is bilocal
I think a deterministic variable can be bilocal is a well defined concept, I don't understand how it is meaningless. It might be wrong, but to have no meaning seems unlikely to me.

A fixed, deterministic bilocal value is of the form "particle A is spin up and particle B is spin down". All you really have here are two separate, local claims, namely: "particle A is spin up"; "particle B is spin down." Assigning bilocal HVs like this is isomorphic to assigning local HVs and cannot by themselves violate Bell ineqs.
That's not what is happening in the Thermal Interpretation, the fixed deterministic value is not "A is up and B is down" or anything like that. First before I explain further, have you read @A. Neumaier 's papers in detail? In order to know what level of detail to go into.

#### Demystifier

2018 Award
That's not what is happening in the Thermal Interpretation, the fixed deterministic value is not "A is up and B is down" or anything like that.
Then what is it? Perhaps something like $(A,B)=(up,down)$?

#### DarMM

Gold Member
Then what is it? Perhaps something like $(A,B)=(up,down)$?
One has the quantity $\langle AB\rangle$, not understood in the typical probabilistic manner as a correlation between $A$ and $B$ but as a quantity in and of itself. This can take a continuous range of values.

There are then the local variables $\langle A\rangle$ and $\langle B\rangle$. The quantum mechanical state does imply that there is a constraint between their values, but fundamentally they are separate quantities.

Local observations of $\langle A\rangle$ and $\langle B\rangle$ appear discrete due to how the device's slow modes evolve.

However repeated observation of $\langle A\rangle$ and $\langle B\rangle$ allows one to obtain statistical estimates on their values.

Then by comparing the joint statistics of $\langle A\rangle$ and $\langle B\rangle$ you can also obtain an estimate of the value of $\langle AB\rangle$.

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#### A. Neumaier

Then by comparing the statistics of $\langle A\rangle$ and $\langle B\rangle$ you can also obtain an estimate of the value of $\langle AB\rangle$.
by comparing their joint statistics

#### charters

That's not what is happening in the Thermal Interpretation, the fixed deterministic value is not "A is up and B is down" or anything like that. First before I explain further, have you read @A. Neumaier 's papers in detail? In order to know what level of detail to go into.
I have read the papers, but I think what will be more helpful is for you to simply state a fixed, deterministic HV that is a counterexample. This disagreement is not really about the TI, it is a much more general question of what it means for anything to be a hidden variable.

My argument is very simple: in a deterministic hidden variables interpretation, you must assign hidden variable which uniquely predicts the outcome of each local measurement. Otherwise, local measurements cannot be deterministic. If local measurements are not deterministic, the interpretation is not deterministic. The Bell's theorem non-locality only enters as a means of correcting the HVs when necessary, reacting to other local measurements performed elsewhere. This will be a de facto pilot wave.

If, as you say, "the fixed deterministic value is not "A is up and B is down" or anything like that" then I do not see what it can even mean to call these values fixed or deterministic.

Arnold has suggested:

There are more general deterministic bilocal properties, those of the kind ''The bilocal variable $C(x,y)$ has a given value at a pair of spacetime positions $x,y$''.
But I can just turn this around into: "the value at spacetime point x is C(x)" and "the value at spacetime point y is C(y)". As long as C(x,y) maps to a pair of unique values (not a statistical mixture at each local site), this is clearly isomorphic to the above. If C(x,y) maps to statistical mixtures at each local site, this bilocal variable is not a deterministic HV for local measurements.

Dar, you said:

However repeated observation of ⟨A⟩⟨A⟩\langle A\rangle and ⟨B⟩⟨B⟩\langle B\rangle allows one to obtain statistical estimates on their values.

Then by comparing the joint statistics of ⟨A⟩⟨A⟩\langle A\rangle and ⟨B⟩⟨B⟩\langle B\rangle you can also obtain an estimate of the value of ⟨AB⟩⟨AB⟩\langle AB\rangle.
If each individual observation of ⟨A⟩ and ⟨B⟩ reveals a preexisting hidden variable of ⟨A⟩ or ⟨B⟩, the joint stats will not violate Bell ineqs without assuming something more (pilot wave, superdeterminism).

If there are no preexisting hidden variables for ⟨A⟩ and ⟨B⟩ individually, local measurements are not deterministic, and this is not a deterministic HV interpretation at all.

#### PeterDonis

Mentor
As long as C(x,y) maps to a pair of unique values
Which it might not; that's exactly the point. Not every function of two variables can be decomposed into two functions, each of one variable.

#### charters

Which it might not; that's exactly the point. Not every function of two variables can be decomposed into two functions, each of one variable.
I agree. But if it doesn't do this, it is not going to make deterministic predictions for local measurements of an individual subsystem.

#### ftr

I think a deterministic variable can be bilocal is a well defined concept, I don't understand how it is meaningless. It might be wrong, but to have no meaning seems unlikely to me.
I have already given such example in a post two years ago, which is nothing but a generalization of the concept I gave in this post, the lines connecting two points can represent energy(or momentum times speed of light).

#### A. Neumaier

I have read the papers, but I think what will be more helpful is for you to simply state a fixed, deterministic HV that is a counterexample. This disagreement is not really about the TI, it is a much more general question of what it means for anything to be a hidden variable.

My argument is very simple: in a deterministic hidden variables interpretation, you must assign hidden variable which uniquely predicts the outcome of each local measurement. Otherwise, local measurements cannot be deterministic. If local measurements are not deterministic, the interpretation is not deterministic. The Bell's theorem non-locality only enters as a means of correcting the HVs when necessary, reacting to other local measurements performed elsewhere. This will be a de facto pilot wave.

If, as you say, "the fixed deterministic value is not "A is up and B is down" or anything like that" then I do not see what it can even mean to call these values fixed or deterministic.

Arnold has suggested:

But I can just turn this around into: "the value at spacetime point x is C(x)" and "the value at spacetime point y is C(y)". As long as C(x,y) maps to a pair of unique values (not a statistical mixture at each local site), this is clearly isomorphic to the above. If C(x,y) maps to statistical mixtures at each local site, this bilocal variable is not a deterministic HV for local measurements.

Dar, you said:

If each individual observation of ⟨A⟩ and ⟨B⟩ reveals a preexisting hidden variable of ⟨A⟩ or ⟨B⟩, the joint stats will not violate Bell ineqs without assuming something more (pilot wave, superdeterminism).

If there are no preexisting hidden variables for ⟨A⟩ and ⟨B⟩ individually, local measurements are not deterministic, and this is not a deterministic HV interpretation at all.
There are local pointer variables a(x) and b(y) but their deterministic dynamics depends on bilocal and mulilocal variables.

#### charters

There are local pointer variables a(x) and b(y) but their deterministic dynamics depends on bilocal and mulilocal variables.
Ok this I can agree is a workable and well understood solution to Bell's theorem. Basically, instead of saying a "pilot wave" is steering the deterministic time evolution of the local pointer variables, the TI says it is "multilocal variables" doing so.

#### A. Neumaier

Ok this I can agree is a workable and well understood solution to Bell's theorem. Basically, instead of saying a "pilot wave" is steering the deterministic time evolution of the local pointer variables, the TI says it is "multilocal variables" doing so.
The nonlocal dynamics is the Ehrenfest dynamics introduced in Section 2 of my Part II paper.

#### Demystifier

2018 Award
There are local pointer variables a(x) and b(y) but their deterministic dynamics depends on bilocal and mulilocal variables.
I don't think that their dynamics depends on bilocal and multilocal variables. Their dynamics is given by equations of the form
$$\langle a(x,t)\rangle={\rm Tr}\rho(t)a(x)$$
which depends only on the state of the Universe $\rho(t)$.

#### A. Neumaier

$$\def\<{\langle} \def\>{\rangle}$$
I don't think that their dynamics depends on bilocal and multilocal variables. Their dynamics is given by equations of the form
$$\langle a(x,t)\rangle={\rm Tr}~\rho(t)a(x)$$
which depends only on the state of the Universe $\rho(t)$.
Only in the noncovariant Schrödinger picture, where the essence of the thermal interpretation is hidden: Your $a(x)$ is a space-dependent operator depending on spatial coordinates $x$. On the other hand,
The nonlocal dynamics is the Ehrenfest dynamics introduced in Section 2 of my Part II paper.
I was using covariant beables $a(x) =\langle A(x)\rangle={\rm Tr}~\rho A(x)$ in the covariant Heisenberg picture dependent on a vector of spacetime coordinates $x$, with a fixed state of the universe. Their dynamics is given by the covariant Ehrenfest equation (Section 4.2 of Part II)
$\frac{d}{dx_\nu} \<A\>_x=\<p_\nu~\angle~ A(x)\>$, specialized to $t=x_0/c$, giving
$$\frac{d}{dt} \<A\>_x=\<H~\angle~ A(x)\>,$$
where $H=cp_0$ is the (frame-dependent) Hamiltonian of the quantum field theory of the universe and $A ~\angle~ B:=\frac{i}{\hbar}[A,B]$; see Section 2 of Part II.

Since $H$ is a sum of integrals over multilocal operators, the right hand side is a sum of integrals over multilocal q-expectations.

#### Demystifier

2018 Award
$H$ is a sum of integrals over multilocal operators
Can you better explain what do you mean by that? (Pinpoiting to the right part of your paper would be OK.)

#### A. Neumaier

Can you better explain what do you mean by that? (Pinpointing to the right part of your paper would be OK.)
In my paper I didn't discuss the detailed form of the Hamiltonian of the universe. It depends on stuff yet to be discovered about how to represent quantum gravity. But the general form of the Hamiltonian is already visible from simpler quantum field theories such as QED, where it is derived as usual from the action, and later modified through renormalization.

Already a free Hamiltonian contains a term with a spatial integral over quadratic expressions, and interactions plus renormalization at all orders add terms of all higher degrees, which become multilocal when inserted into the Ehrenfest dynamics. (Search for Hamiltonian in this Wikipedia article to find the explicit unrenormalized expression for scalar field theory; the integration variable runs over points distinct from the $x$ in the Ehrenfest equation.)

"How does the thermal interpretation explain Stern-Gerlach?"

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