A How does the thermal interpretation explain Stern-Gerlach?

  • Thread starter PeterDonis
  • Start date
I don't see the problem
Ok, let me try a different route. Consider a basic SG experiment with an N=1 beam. You claim the TI is deterministic. Accordingly, to encode this hidden determinism we should be able to write the state of the experiment *prior* to the detector click as

(|UP>| + |DOWN> ) ⊗ {up}

where the Dirac notation is the normal quantum state and {n} is the state of the hidden variable which deterministically predicts the click. In the TI, I believe {up} and {down} would represent different fine grained distinctions in the configuration of the detector itself (as opposed to BM, where it represents different configurations of the beam).

Do you agree with this description being faithful to the TI so far?
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
Ok, let me try a different route. Consider a basic SG experiment with an N=1 beam. You claim the TI is deterministic. Accordingly, to encode this hidden determinism we should be able to write the state of the experiment *prior* to the detector click as

(|UP>| + |DOWN> ) ⊗ {up}

where the Dirac notation is the normal quantum state and {n} is the state of the hidden variable. In the TI, I believe {up} and {down} would represent different fine grained distinctions in the configuration of the detector itself (as opposed to BM, where it represents different configurations of the beam).

Do you agree with this description being faithful to the TI so far?
No. The beables (hidden variables) are the collection of all q-expectations of the universe. Given a single spin prepared in a pure state ##\psi## we know at preparation time that for any 3-vector ##p## that the quantity ##S(p):=p\cdot\sigma## of the spin satisfies ##\langle S(p)\otimes 1\rangle= \psi^*S(p)\psi##. In your case, this is the sum of the four entries of ##S(p)##. Your curly up and down correspond to pointer readings, i.e., functions of q-expectations (beables, hidden variables) of the detector, not to a state of the detector. Many states of the detector lead to identical pointer readings.

This is completely independent of the deterministic dynamics, which is the Ehrenfest dynamics of the universe.

In the most general case we know nothing more, unless we make assumptions a similar kind about the environment, i.e., the state and the dynamics of the remainder of the universe and its interactions with the spin. These assumptions define a model for what it means that this environment contains a detector with a pointer or screen, that responds to the prepared spin in the way required to count as a measurement.

Thus you need to specify a complete model for the measurement process (including a Hamiltonian for the dynamics of th model universe) to conclude something definite. This is the reason why the arguments for analyzing meaurement are either very lengthy (as in the AB&N paper) or only qualitative (as in my Part III).
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
As I said in my previous post, being deterministic does not imply being superdeterministic. Classical mechanics is not superdeterministic.
In essence a deterministic world where the initial conditions never evolve into states corresponding to observers obtaining an accurate determination of the physical laws.
Do you have different notions of the meaning of superdeterminisitc?
 
Your curly up and down correspond to pointer readings, i.e., functions of q-expectations (beables, hidden variables) of the detector, not to a state of the detector. Many states of the detector lead to identical pointer readings
Ok it is possible I just don't get it or you are talking about hidden variables in a way very different from what I am used to. But this may all be semantics around the use of the word "state" so I want to rephrase:

All I am trying to pin down is whether or not the hidden variable descriptions are such that, just before the measurement, all HV descriptions of the detector that will lead to an observable "up" reading (for a particular choice of axis) are completely disjoint/distinct from all the HV descriptions that will lead to an observable "down" reading?

In essence, would knowledge of the hidden variable description of the detector at t<1 allow me to perfectly predict the observed click at t=1?
 

DarMM

Science Advisor
Gold Member
1,343
557
Do you have different notions of the meaning of superdeterminisitc?
I'm saying:
A superdeterministic world is a deterministic world where the initial conditions never evolve into states corresponding to observers obtaining an accurate determination of the physical laws.

A world can be deterministic without being superdeterministic if the initial conditions permit the development of observers who obtain accurate enough measurements to determine the laws of the world.

So for example in 't Hooft's model Quantum Mechanics is literally completely wrong. Not approximately right but inaccurate in some remote regimes like the Early Universe, but literally completely wrong even in its predictions of say the Stern Gerlach experiment. However the initial conditions of the world are such that experimental errors occur that make it look correct.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
would knowledge of the hidden variable description of the detector at t<1 allow me to perfectly predict the observed click at t=1?
I don' think so, because in reading a discrete pointer there is a fuzzy decision boundary. This is like race conditions in computer science which may delay decisions indefinitely. Thus there is a partition into 3 sets, one deciding for spin up, one deciding for spin down, and one for indecision; the third one having positive measure that goes to zero only as the duration of the measurement goes to infinity.

In experimental practice, this accounts for the limited efficiency of detectors.
 
Last edited:

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
As I said in my previous post, being deterministic does not imply being superdeterministic. Classical mechanics is not superdeterministic.
A world can be deterministic without being superdeterministic if the initial conditions permit the development of observers who obtain accurate enough measurements to determine the laws of the world.
In a classical Laplacian universe, a Laplacian detector of finite size perfectly knowing its own state can never get an arbitrarily accurate estimate of a single particle state external to it. Thus a classical Laplacian universe would be superdeterministic. Do you mean that, @DarMM, contradicting @stevendaryl?

If so, the thermal interpretation is also superdeterministic, for essentially the same reason.
 
I don' think so, because in reading a discrete pointer there is a fuzzy decision boundary. This is like race conditions in computer science which may delay decisions indefinitely. Thus there is a partition into 3 sets, one deciding for spin up, one deciding for spin down, and one for indecision; the third one having positive measure that goes to zero only as the duration of the measurement goes to infinity.

In experimental practice, this accounts for the limited efficiency of detectors.
I don't understand how this answer is consistent with what you wrote in III.4.2, specifically:

"These other variables therefore become hidden variables that would determine the stochastic elements in the reduced stochastic description, or the prediction errors in the reduced deterministic description. The hidden variables describe the unmodeled environment associated with the reduced description.6 Note that the same situation in the reduced description corresponds to a multitude of situations of the detailed description, hence each of its realizations belongs to different values of the hidden variables (the q-expectations in the environment), slightly causing the realizations to differ."

Would your answer be different had I phrased my question as?:

would knowledge of the hidden variable description of the detector plus its local environment (eg, the detector casing or surrounding air in the lab) at t<1 allow me to perfectly predict the observed click at t=1?
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
I don't understand how this answer is consistent with what you wrote in III.4.2, specifically:

"These other variables therefore become hidden variables that would determine the stochastic elements in the reduced stochastic description, or the prediction errors in the reduced deterministic description. The hidden variables describe the unmodeled environment associated with the reduced description. Note that the same situation in the reduced description corresponds to a multitude of situations of the detailed description, hence each of its realizations belongs to different values of the hidden variables (the q-expectations in the environment), slightly causing the realizations to differ."

Would your answer be different had I phrased my question as?:

would knowledge of the hidden variable description of the detector plus its local environment (eg, the detector casing or surrounding air in the lab) at t<1 allow me to perfectly predict the observed click at t=1?
No. You can take the detector to be the whole orthogonal complement of the measured system, and my answer is still the same. You can also take it to be just the pointer variable; all other beables of the universe are effectively hidden variables, no matter whether they are actually hidden. My first response was less focussed and ignored the race conditions since your question was less clear.

This is because of the nature of a real detection process (which is what is modeled in the thermal interpretation). There is a continuous pointer variable ##x## (a function of the beables = hidden variables = q-expectations, all of them continuous) of the detector that is initially at zero. Suppose that the pointer readings for a decision up are close to ##1##, that for down are close to ##-1##, and a reading counts as definite only if the sign and one bit of accuracy have persisted for more than a minimal duration ##\Delta t##. This defines the three response classes up, down, and undecided. At short times after the preparation, the detector didn't have sufficient time to respond, and the third (undecided) set of conditions has measure essentially 1; the up and down measures are essentially zero. These measures are a continuous function of the observation time and gradually move to ##0,p,1-p##, but achieve these values only in the limit of infinite time.
 
Ok I appreciate the details, but I don't think this is necessary for the heart of my question. Some finite time after the N=1 beam has become incident on the detector, the pointer is going to visibly have pointed towards 1 or -1. I am not concerned with how quickly this happens.

All I want to know is: would a full hidden variable/beable description of the detector/environment at some time before the beam is incident be sufficient to predict whether the detector eventually reads 1 or -1 (for any given beam).

I take this to be the minimal definition of hidden variable determinism in quantum foundations, so if you say no to this, I don't understand how you claim the TI is deterministic (except in the classical limit where all interpretations are effectively deterministic) or has meaningful hidden variables. Hidden variables that don't make this sort of prediction are not really fulfilling their defined purpose.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
the pointer is going to visibly have pointed towards 1 or -1. I am not concerned with how quickly this happens.
or continues to oscillate, or is stuck near zero, due to race conditions. If you ignore this, you ignore a loophole that makes a practical difference - real efficiency is never 100%, and a good model of a deterministic universe must predict this reduced efficiency!

would a full hidden variable/beable description of the detector/environment at some time before the beam is incident be sufficient to predict whether the detector eventually reads 1 or -1 (for any given beam).
It would, in all cases where a definite decision is reached, and it would predict when this is the case.
 
Last edited:
It would, in all cases where a definite decision is reached, and it would predict when this is the case.
Ok perfect. So then something you need to explain is how, in an EPR experiment, the hidden variables describing the configuration of detector 1, and (when applicable) predicting the outcome of its measurement, are able to coordinate with the hidden variables doing the same for detector 2, such that Bell violations become possible.

The only known hidden variable solutions to this problem are

A) add a non-local pilot wave that can surgically adjust the local HVs as needed to create Bell violations, while using an absolute definition of simultaneity

B) superdeterminism, where the Bell violations are ultimately just the result of dumb luck in the initial conditions, and entanglement itself is just an illusion of this coincidence.

C) Retrocausality is an option too, but that's not quite a hidden variable approach per se.

But I also get the sense from earlier in the discussion you think its okay to stop short of making this choice and therefore don't need to engage with their perceived downsides in the existing foundations literature. I don't agree, and I expect you'll have a hard time getting folks to adopt (or even know if they'd want to adopt) the TI while not clearly biting one of these bullets. So that's my main point.

(One extra thing I hope for clarity: the idea that Bohmian mechanics has non-local hidden variables is not really accurate. What it has are local hidden variables that receive the benefit of non-local corrections via the pilot wave in order to permit Bell violations. And I don't see how a "non-local" hidden variables interpretation could be anything other than this.)
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
The only known hidden variable solutions to this problem are
Nothing in your arguments forbids that the thermal interpretation provides an additional, previously
unknown way to achieve that. @DarMM gave in post #268 of the main thread on the thermal interpretation a nice summary of the thermal interpretation, where he addresses this in his point 4.
 
Nothing in your arguments forbids that the thermal interpretation provides an additional, previously
unknown way to achieve that. DarMM gave in post #268 of the main thread on the thermal interpretation a nice summary of the thermal interpretation, where he addresses this in his point 4.
I don't agree that what he calls correlator properties in that post can be consistent hidden variable determinism.

A correlator is a conditional of the form: "when subsystem A takes value x, B takes y; when A takes y, B is x". By construction, it requires that there is some uncertainty in the local description of each subsystem.

If such a correlator description is complete, the hidden variable descriptions of each local detector will not be definite and so will not satisfy the determinism/predictability condition we just established.

In hidden variable determinism, a non-local HV description is only of the form: "detector A will measure UP and B will measure DOWN" which is of course consistent with the truth of "detector A will measure UP" on its own. There are no conditionals. And just like local HVs, these trivial non-local HVs will not violate Bell ineqs without adopting one of the previously discussed options.
 

DarMM

Science Advisor
Gold Member
1,343
557
In a classical Laplacian universe, a Laplacian detector of finite size perfectly knowing its own state can never get an arbitrarily accurate estimate of a single particle state external to it. Thus a classical Laplacian universe would be superdeterministic. Do you mean that, @DarMM, contradicting @stevendaryl?

If so, the thermal interpretation is also superdeterministic, for essentially the same reason.
No it's not just about not being able to obtain total precision. Its more that the initial state is conspiratorial. Let me take a real world example.

There was a recent test of CHSH violations that used light from distant quasars to select the spin orientations.

In a superdeterministic world quantum mechanics is actually false, but the light from the quasars happens to always select the correct orientation to incorrectly give the impression the CHSH inequalities are violated.

So it's not just a lack of arbitrary accuracy it's that the observers are determined to come to false conclusions about the physical laws that apply to their world.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
will not violate Bell ineqs without adopting one of the previously discussed options.
Where is the theorem you refer to? I don't know of any theorem that has as one of its necessary alternatives a pilot wave statement as your post #62 states. There is a big difference between
known hidden variable solutions to this problem
and necessary properties.

In any case, all interpretation have open research questions, and the thermal interpretation has these, too; some of these are discussed in post #293 of the main thread. No interpretation must indicate how it falls into a particular classification, though those interested in classifying interpretations may want to investigate these issues. Those interested in understanding quantum mechnaics only need one plausible interpretation they can make sense of.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
No it's not just about not being able to obtain total precision. Its more that the initial state is conspiratorial. Let me take a real world example.

There was a recent test of CHSH violations that used light from distant quasars to select the spin orientations.

In a superdeterministic world quantum mechanics is actually false, but the light from the quasars happens to always select the correct orientation to incorrectly give the impression the CHSH inequalities are violated.

So it's not just a lack of arbitrary accuracy it's that the observers are determined to come to false conclusions about the physical laws that apply to their world.
In this sense, the thermal interpretation is definitely not superdeterministic. Very coarse knowledge of the state of the universe at preparation time, together with some more details about the detector and how it works, are sufficient to predict with the traditional approximations everything known.
 
Where is the theorem you refer to?
I don't think I need a theorem. I'm only listing the solutions I am aware of and agree are viable, but I don't mean to be closed off to alternatives I've never contemplated.

However, I would say the burden of proof is on the proponent of a new interpretation to convince readers they've indeed found such a viable alternative to the accepted approaches to HVs that works in light of Bell's theorem. It is not enough just to say you have non-local, determinist HVs and be done with it. You need to elaborate on what exactly this means, especially when you claim its emphatically not a pilot wave or superdeterministic.

In any case, all interpretation have open research questions, and the thermal interpretation has these, too; some of these are discussed in post #293 of the main thread. No interpretation must indicate how it falls into a particular classification
Sure, and I would expect this will be one of the particular open questions that folks who think a lot about foundations will want to see tackled in the TI context. This is more than just a sociological classification exercise. It speaks to what the ontology of the TI is, what it claims the universe is like.
 

DarMM

Science Advisor
Gold Member
1,343
557
In this sense, the thermal interpretation is definitely not superdeterministic
Yes, I would have said the thermal interpretation is deterministic, but not superdeterministic.

A correlator is a conditional of the form: "when subsystem A takes value x, B takes y; when A takes y, B is x". By construction, it requires that there is some uncertainty in the local description of each subsystem.

If such a correlator description is complete, the hidden variable descriptions of each local detector will not be definite and so will not satisfy the determinism/predictability condition we just established
Remember one of the main differences between the thermal interpretation and other views is really that probability theory itself is given a different interpretation.

In the Thermal Interpretation a correlator does not have the meaning you give. Rather it is a bilocal property, that is a nonlocal property that requires measurements at two locations to ascertain. It has a fixed deterministic value.

However it is the metastability of the slow modes of the devices at each location that cause them to develop discrete inaccurate readings of this quantity. This inaccuracy requires one to use several measurements to determine the correlator.

So in this view it's not fundamentally a conditional and it doesn't require that there is (fundamental) uncertainty in each local device. That just arises as it normal does in the Thermal Interpretation.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
. It is not enough just to say you have non-local, determinist HVs and be done with it. You need to elaborate on what exactly this means
It is enough to explain how this is compatible with long-distance entanglement experiments, and I did this in Part II of my series of papers.
 
Rather it is a bilocal property, that is a nonlocal property that requires measurements at two locations to ascertain. It has a fixed deterministic value.
If there is a fixed deterministic value, it does not mean anything to say it is bilocal, and it would be trivial to describe it this way. 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.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
A fixed, deterministic bilocal value is of the form "particle A is spin up and particle B is spin down".
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##''.
 
Last edited:

stevendaryl

Staff Emeritus
Science Advisor
Insights Author
8,338
2,513
In a classical Laplacian universe, a Laplacian detector of finite size perfectly knowing its own state can never get an arbitrarily accurate estimate of a single particle state external to it. Thus a classical Laplacian universe would be superdeterministic.
Why does that make it superdeterministic? The reason superdeterminism is relevant to interpretations of quantum mechanics, as I said, is because superdeterminism is a loophole in Bell's argument against local hidden variables theories. In an attempt to reproduce the statistics of spin-1/2 EPR, you would have five "players":
  1. Alice, who chooses a detector orientation ##\overrightarrow{\alpha}##
  2. Bob, who chooses a detector orientation ##\overrightarrow{\beta}##
  3. Charlie, who chooses a value for ##\lambda## according to some probability distribution ##P(\lambda)##
  4. Alice's detector, that computes a result ##A(\overrightarrow{\alpha}, \lambda) = \pm 1## based on ##\overrightarrow{\alpha}## and ##\lambda##
  5. Bob's detector, that picks a result ##B(\overrightarrow{\beta}, \lambda) = \pm 1## based on ##\overrightarrow{\beta}## and ##\lambda##
Bell shows that there is no probability distribution ##P(\lambda)## that can reproduce the correlations predicted by quantum mechanics. However, there are a number of loopholes in the argument:

  1. if Bob's detector is allowed to depend on ##\overrightarrow{\alpha}##, or if Alice's detector is allowed to depend on ##\overrightarrow{\beta}## (nonlocality) then there is no problem in reproducing the predictions of quantum mechanics.
  2. If Charlie's choice of ##\lambda## is allowed to depend on ##\overrightarrow{\alpha}## and ##\overrightarrow{\beta}## (superdeterminism), then there is no problem in reproducing the predictions of quantum mechanics.
Alice being a deterministic machine doesn't make her choice predictable, because as I argued earlier, Alice can consult the whole rest of the universe in order to make her choice. Simple determinism would allow us to predict Alice's choice based on her state plus the inputs she receives from the rest of the universe. Superdeterminism would require that not only Alice but the whole rest of the universe be known and predictable.
 

A. Neumaier

Science Advisor
Insights Author
6,043
2,197
Why does that make it superdeterministic?
It was, in my tentative understanding of DarMM's definition of superdeterminism. In the mean time, he clarified his definition, and in my resulting understanding, my old comment makes no longer sense.
superdeterminism is a loophole in Bell's argument against local hidden variables theories.
Since the thermal interpretation has multilocal hidden variables, Bell's argument doesn't apply anyway.
 

stevendaryl

Staff Emeritus
Science Advisor
Insights Author
8,338
2,513
Since the thermal interpretation has multilocal hidden variables, Bell's argument doesn't apply anyway.
How does "multilocal hidden variables" explain the EPR results?
 

Want to reply to this thread?

"How does the thermal interpretation explain Stern-Gerlach?" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top