A Assumptions of the Bell theorem

[Demystifier]

In Bohmian mechanics, macroscopic observables are supposed to be functions of the microscopic variables: $A = f(x_1, \ldots, x_n)$. You then get $A(t) = U^\dagger A U = f(U^\dagger x_1 U, \ldots, U^\dagger x_n U) = f(x_1(t), \ldots, x_n(t))$. So the non-commutativity $[x_i, x_i(t)]$ directly translates to the non-commutativity $[A,A(t)]$, i.e. the macroscopic variables don't commute at different times. So you can construct a contradiction to Kolmogorov probability again as above.

I don't think that the quantity $U^\dagger x_1 U$ makes sense. In Bohmian mechanics, the quantity $x_1$ you are referring to is not an operator.

 

[Nullstein]

I don't think that the quantity $U^\dagger x_1 U$ makes sense. In Bohmian mechanics, the quantity $x_1$ you are referring to is not an operator.

But Bohmian mechanics has to agree with the predictions of QM. The claim of Bohmian mechanics is that its predictions agree with the QM predictions when restricted to macroscopic observables $A = f(x_1, \ldots, x_n)$ (and their time evolved versions). So we're allowed to compute $A(t)$ using QM. If there was no agreement then BM and QM would lead to different time evolutions of these macroscopic variables.

 

[atyy]

But Bohmian mechanics has to agree with the predictions of QM. The claim of Bohmian mechanics is that its predictions agree with the QM predictions when restricted to macroscopic observables $A = f(x_1, \ldots, x_n)$ (and their time evolved versions). So we're allowed to compute $A(t)$ using QM. If there was no agreement then BM and QM would lead to different time evolutions of these macroscopic variables.

But if you don't restrict to those macroscopic observables, presumably you can use Kolmogorov probability, since one can do that for Bohmian mechanics?

 

[Demystifier]

But Bohmian mechanics has to agree with the predictions of QM. The claim of Bohmian mechanics is that its predictions agree with the QM predictions when restricted to macroscopic observables $A = f(x_1, \ldots, x_n)$ (and their time evolved versions). So we're allowed to compute $A(t)$ using QM. If there was no agreement then BM and QM would lead to different time evolutions of these macroscopic variables.

I see, you speak of time evolution in the Heisenberg picture. Can you translate your claims to the Schrodinger picture? That would help because Bohmian mechanics is much better understood in Scrodinger picture.

 

[Nullstein]

But if you don't restrict to those macroscopic observables, presumably you can use Kolmogorov probability, since one can do that for Bohmian mechanics?

I think BM generally satisfies Kolmogorov probability if you look at just one instant of time, because they only consider commuting variables at each instant. However, unitary time evolution will evolve this set of observables into a new set, which won't commute with the old set. This happens independently of whether the observables are microscopic or macroscopic, so the situation won't improve if we include the microscopic variables. I was only referring to the macroscopic observables, because stevendaryl suggested that the problem might go away if we restrict ourselves to them.

I think there is no doubt about the paradoxes at the microscopic level. The argument then just goes like this: $A=f(x)$ is diagonal in the position eigenbases and $A(t)=f(U^\dagger x U)$ is diagonal in the time evolved position basis. Since the time evolved position eigenbasis is incompatible with the time zero position eigenbasis, $A(t)$ is incompatible with $A$. So there are paradoxes at the microscopic level if and only if there are paradoxes at the macroscopic level (under the assumption that quantum mechanics is correct).

 

[Nullstein]

I see, you speak of time evolution in the Heisenberg picture. Can you translate your claims to the Schrodinger picture? That would help because Bohmian mechanics is much better understood in Scrodinger picture.

Hmm, there is not much of a difference I would say, except that one would write $U^\dagger(t) A U(t)$ instead of $A(t)$, i.e. $[A, U^\dagger A U] \neq 0$. But the idea is anyway that we find a regime where everyone agrees that QM and BM (and other interpretations as well) should produce the same results, so we can do the computation in the QM formalism.

 

[Fra]

Both many worlds and Bohmian mechanics are counteraxamples to this claim. Fundamentally they are unitary in the full system, but at the effective emergent level they explain the illusion of non-unitary collapse.

The problem I have with trying to avoid the observer bias via is, who is making the "division"?

"System A: the system of interest
System B: the measuring device
System C: the rest of the universe"

This is to me an external view, detached from the agent. The reason there is no collapse in the full system is beacuse one allows nothing to happen. The implicit superobserver never makes a "measurement" on the full system. So it is not longer a theory of inference or measurement - which IMO is the target I seek. It's more like a "superfantasy", trivial as there is no measurments, but yet incredible complex as it contains the whole universe?

Wouldnt a more nautral intrinsic ansatz informally be like

"System B: the agent/observer itself (and it's encoded information about its own interaction history)
System A: the unknown that is on the exterior side of the communication channel or "sample space"

Any intrinsic "observable" should be defiend only in terms of sampling on the communication channel. And the "state" of the agent, should follow from a natural inference from the sampling history?

To me this seems more "clear", but it's also equally clear that QM as we know it, does not fit into this scheme - so we need modification if QM. In this picture it seems also that in the general case ant finite time translation evolves the effective probability into a NEW space. So $A \wedge B$ is undefined if A belongs to an evolved agent. OTOH, if the agent never interacts, it's all about the agents own internal expectations evolvling, into the one and same probability space (or it's generalisation). But it we disregard the real information updates, we have removed teh non-trivial element that drives the evolution, and we have left only the "self evolution of expectations" that unitary evolution represents in regular QM.

I think there is no way to escape this by looking at the whole system. I think one can also realize this without working out all explicit details of various thoughr experiments with entangled systems?

/Fredrik

 

[atyy]

I think BM generally satisfies Kolmogorov probability if you look at just one instant of time, because they only consider commuting variables at each instant. However, unitary time evolution will evolve this set of observables into a new set, which won't commute with the old set. This happens independently of whether the observables are microscopic or macroscopic, so the situation won't improve if we include the microscopic variables. I was only referring to the macroscopic observables, because stevendaryl suggested that the problem might go away if we restrict ourselves to them.

I think there is no doubt about the paradoxes at the microscopic level. The argument then just goes like this: $A=f(x)$ is diagonal in the position eigenbases and $A(t)=f(U^\dagger x U)$ is diagonal in the time evolved position basis. Since the time evolved position eigenbasis is incompatible with the time zero position eigenbasis, $A(t)$ is incompatible with $A$. So there are paradoxes at the microscopic level if and only if there are paradoxes at the macroscopic level (under the assumption that quantum mechanics is correct).

Is it simply due to contextuality? Non-commuting observables are modeled by different measurement apparatuses in Bohmian Mechanics, and one cannot measure position and momentum at the same time simply because the measurement setups for each is completely different.

 

[Demystifier]

Hmm, there is not much of a difference I would say, except that one would write $U^\dagger(t) A U(t)$ instead of $A(t)$, i.e. $[A, U^\dagger A U] \neq 0$. But the idea is anyway that we find a regime where everyone agrees that QM and BM (and other interpretations as well) should produce the same results, so we can do the computation in the QM formalism.

I meant to formulate it such that all time dependence (in standard QM) is in the time dependence of the state $|\psi(t)\rangle$.

 

[Demystifier]

Non-commuting observables are modeled by different measurement apparatuses in Bohmian Mechanics, and one cannot measure position and momentum at the same time simply because the measurement setups for each is completely different.

Different apparatuses are modeled by different Hamiltonians. If one considers a measurement the full Hamiltonian of which is the sum of two Hamiltonians (Hamilonian for position measurement plus Hamiltonian for momentum measurement), it turns out that it is effectively a measurement in the coherent state basis, in which both position and momentum are measured simultaneously, but neither with perfect precision. You can find the details in the Holland's book (but the result does not depend on the Bohmian interpretation).

 

[Nullstein]

Is it simply due to contextuality? Non-commuting observables are modeled by different measurement apparatuses in Bohmian Mechanics, and one cannot measure position and momentum at the same time simply because the measurement setups for each is completely different.

It's contextuality, but at different times, so it is possible to obtain both $A$ and $A(t)$ despite their non-commutativity. As you say, it's impossible to measure non-commuting variables at the same time (independent of the interpretation), but this isn't the situation we're interested in here.

 

[Nullstein]

I meant to formulate it such that all time dependence (in standard QM) is in the time dependence of the state $|\psi(t)\rangle$.

I understand, but in order to compute a commutator, we need to formulate it in terms of operators. But as soon as we agree that BM and QM agree on the predictions of macroscopic measurements, we should be able to perform the computation in the Heisenberg picture anyway.

 

[Kolmo]

I think along the lines that @Demystifier was saying, in the GHZ experiment, the events are not

$A_x = +1$
$A_x = -1$
$B_x = +1$
etc.

The events are:

• Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$
• Someone measures the spin particle $A$ in the x-direction, and the result is $-1$
• etc

These events have well-defined probabilities that obey the Kolmogorov axioms.

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

If one just tries to directly take the QM predications it also won't make sense. Consider $X_{+1}$, $X_{-1}$ the obvious $X$-axis events (i.e. somebody measures $X$ and gets $+1$) and $Z_{+1}$, $Z_{-1}$ similar for the $Z$-axis. Then the event:

• $X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}$

would have a probability exceeding unity.

 

[Nullstein]

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

I don't think one can exclude the possibility of formulating this within a more detailed description completely, but as I said: If your most detailed description is still a quantum theory, the paradoxes won't go away and we would gain no fundamental insight. Maybe we can shift the problem to a realm where it doesn't bother us anymore, but people who care about these things in the first place can hardly be satisfied with such a non-solution.

 

[Demystifier]

I understand, but in order to compute a commutator, we need to formulate it in terms of operators. But as soon as we agree that BM and QM agree on the predictions of macroscopic measurements, we should be able to perform the computation in the Heisenberg picture anyway.

In principle, yes. But when we compare measurements at different times, then we must take into account the effect of "wave function collapse" (or state update, or whatever one likes to call it). This effect is awkward to take into account in the Heisenberg picture.

EDIT: Let me briefly explain an example how collapse can be accounted for in the Heisenberg picture, by first describing it in the Schrodinger picture. Suppose that the state is first prepared in the initial state $|\psi_0\rangle$ at time $t_0$ and then measured at time $t_{\rm m}>t_0$. This means that the state first evolves unitary with the unitary evolution operator $U(t-t_0)$, then at time $t_{\rm m}$ the state gets projected by some randomly chosen projection operator $\pi$. (For POVM measurements it does not need to be a projection operator, but the point is that at $t_{\rm m}$ the unitary evolution is interrupted by a random non-unitary operator.) After measurement we again have unitary evolution described by $U(t-t_{\rm m})$. For $t>t_{\rm m}$ this evolution is described by the non-unitary operator
$$V(t)=U(t-t_{\rm m}) \pi U(t_{\rm m}-t_0)$$
More precisely, the state can be written as
$$|\psi(t)\rangle= \frac{V(t)|\psi_0\rangle}{|V(t)|\psi_0\rangle|}$$
where the denominator accounts for the correct normalization of the state, needed because $V(t)$ is not unitary. Then for an arbitrary observable $A$ we can write
$$\langle\psi(t)|A|\psi(t)\rangle = \langle\psi_0|A(t)|\psi_0\rangle$$
where
$$A(t) \equiv \frac{V^{\dagger}(t)AV(t)}{\langle\psi_0|V^{\dagger}(t)V(t)|\psi_0\rangle}$$
The last equation is naturally interpreted as evolution in the Heisenberg picture, with the effect of measurement taken into account. Due to the measurement, the evolution of the observable is non-unitary (because $V(t)$ is non-unitary), random (because $\pi$ is random) and non-linear (because of the denominator).

 

[Kolmo]

I don't think one can exclude the possibility of formulating this within a more detailed description completely

It can't be eliminated by a more detailed quantum description, you mean some kind of new theory I assume?

My point is more to do with Quantum Theory not being a Kolmogorov probability theory. That would be the standard academic view, i.e. that it breaks Kolmogorov's axioms. "Lost Causes in and Beyond Physics" by R.F. Streater explains this quite well in Chapter 8. This paper by Summers and Rédei explains it well:

https://arxiv.org/abs/quant-ph/0601158

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

 

[stevendaryl]

I don't think they do as for instance if we take:

• $E =$ "Someone measures the spin of particle $A$ in the x-direction, and the result is $+1$"

can one actually work out $P(E)$?
I don't think one can since one would have to work out the probability of selecting a device measuring a specific spin angle which isn't really something covered by the observable algebra for the particle.

What's computed from quantum mechanics is the conditional probability:

"The probability of measuring +1 given that the experimenter chose angle $\alpha$"

The event "Someone measures the spin and the result is +1" can certainly be given an empirical (statistical) probability, but you're right, quantum mechanics by itself doesn't allow us to compute the probability (at least not in practice).

If one just tries to directly take the QM predications it also won't make sense. Consider $X_{+1}$, $X_{-1}$ the obvious $X$-axis events (i.e. somebody measures $X$ and gets $+1$) and $Z_{+1}$, $Z_{-1}$ similar for the $Z$-axis. Then the event:

• $X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}$

would have a probability exceeding unity.

Why would that be?

 

[stevendaryl]

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

Okay.

 

[Kolmo]

What's computed from quantum mechanics is the conditional probability:

"The probability of measuring +1 given that the experimenter chose angle α"

Yeah so that's certainly true. The probabilities are intrinsically conditional, Kolmogorov however formulated probability theory with unconditional probabilities as the base and conditionals as derived. So if you fundamentally only have conditionals it's not a Kolmogorov theory.

The event "Someone measures the spin and the result is +1" can certainly be given an empirical (statistical) probability, but you're right, quantum mechanics by itself doesn't allow us to compute the probability (at least not in practice

I would agree that one could give a probability to the event, but this would be a purely subjective credence/guess not something one could work out from quantum theory. Even in principle, I wouldn't say it is an in practice thing.

Why would that be?

The events have no intersection, i.e. $X_{+1} \land Z_{-1}$ has zero probability, so

$P(X_{+1} \lor X_{-1} \lor Z_{+1} \lor Z_{-1}) = P(X_{+1}) + P(X_{-1}) + P(Z_{+1}) + P(Z_{-1})$

and if you put in the predictions a spin-z $+1$ eigenstate for each of these cases then it exceeds $1$.

 

[Demystifier]

So if you fundamentally only have conditionals it's not a Kolmogorov theory.

I disagree. Conditional probabilities satisfy all axioms of Kolmogorov probability, so conditional probabilities are just a special case. Just because Kolmogorov haven't explicitly studied this case does not mean that it's not Kolmogorov probability.

 

[Kolmo]

I disagree. Conditional probabilities satisfy all axioms of Kolmogorov probability, so conditional probabilities are just a special case. Just because Kolmogorov haven't explicitly studied this case does not mean that it's not Kolmogorov probability.

I don't really think there is mathematical scope for disagreement. Kolmogorov's axioms formulate probability theory as a measure function of total weight one assigned to a sigma algebra. By the Gelfand representation such a set up is always equivalent to a commutative C*-algebra, so quantum theory cannot be equivalent to such a system.

The point is that individual conditionals might satisfy Kolmogorov's axioms, but the totality of conditionals does not. So the conditional probabilities for a z-axis spin measurement obey Kolmogorov theory when considered alone, as do x-axis measurements, however both of them together do not. There's no Kolmogorov theory containing the four CHSH measurement contexts for example.

The breaking of Kolmogorov's axioms is the condition in Bell's theorem that quantum theory breaks, as explained in Streater's monograph I mentioned above.

Bell’s inequalities, which hold in any Kolmogorovian theory, do not hold in quantum mechanics

 

[Demystifier]

Kolmogorov formulates probability theory as a measure on the sigma algebra of some space. It's Gelfand's famous representation theorem that shows a Kolmogorov probability space is always the representation of some commutative C*-algebra. Thus quantum theory, which uses a non-commutative algebra, isn't a Kolmogorov theory.

I will not pretend that I fully understand this. However, with my boldings of "some" and "a", it should be clear that there is no contradiction. Just because quantum theory uses a non-commutative algebra does not imply that there is no some other commutative algebra that represents probability. For instance, the probability in the momentum space can be represented with one commutative algebra and the probability in the momentum space can be represented with another one commutative algebra. Of course, those two commutative algebras are different from the non-commutative algebra of quantum observables, but so what? I don't see any contradiction with statements of the theorems.

A real issue is the fact that there is no Kolmogorov probability $P(x,p)$ (with $x$ and $p$ being the position and momentum, respectively) such that the marginals
$$P_1(p)=\int dx \, P(x,p)$$
$$P_2(x)=\int dp \, P(x,p)$$
coincide with quantum probabilities, but again I don't see a contradiction with any Kolmogorov axiom.

 

[Kolmo]

I don't see a contradiction with any Kolmogorov axiom

Each individual context obeys Kolmogorov axioms, but there is no Kolmogorov model of all the variables at once, thus a quantum system as a whole (not focusing on an individual context) does not obey Kolmogorov's axioms.
The fact that the probabilities for each pair of variables in a CHSH test are not marginals of some common Kolmogorov model for all four variables is what allows violation of the Bell inequalities.

The fact that the theory does obey Kolmogorov's axioms within each context is what allows one to rewrite it as a Topos in more advanced treatments.

 

[Demystifier]

Each individual context obeys Kolmogorov axioms, but there is no Kolmogorov model of all the variables at once, thus a quantum system as a whole (not focusing on an individual context) does not obey Kolmogorov's axioms.

OK, but what's wrong with using many Kolmogorov models, one for each context?

Besides, if you add the measuring apparatus into the system as a whole, then all predictions of QM can be reduced to measurement of only one commuting set of observables - the position observables. In this way the quantum system as a whole can be reinterpreted as one Kolmogorov model, which indeed is a basis for the Bohmian interpretation.

Perhaps we can formulate all this as follows. Bell's theorem assumes that there is only one Kolmogorov model and from this one derives nonlocality. Indeed, Bohmian mechanics is an example of this general result. If, on the other hand, we use another Kolmogorov model for each context, then we arrive at consistent histories (which obey locality but non-classical logic when different contexts are discussed at once).

 

[Kolmo]

OK, but what's wrong with using two Kolmogorov models, one for each context?

There's nothing wrong with it. It just means a quantum system as a whole isn't something which obeys the Kolmogorov axioms and that's why we get CHSH violations.

That'd only be a problem if we didn't have some other probability theory to use, but in fact we do as we have quantum theory itself.

 

[Demystifier]

There's nothing wrong with it. It just means a quantum system as a whole isn't something which obeys the Kolmogorov axioms and that's why we get CHSH violations.

That'd only be a problem if we didn't have some other probability theory to use, but in fact we do as we have quantum theory itself.

OK, let me then look at it from a slightly different perspective. I think the Bell's theorem can now be concisely stated as follows: There is no single local Kolmogorov model compatible with all probabilistic predictions of QM.

So it seems that we have 3 options:
(i) abandoning locality (Bohmian mechanics, GRW, ...)
(ii) using many models of Kolmogorov axioms, one for each context (Bohr's complementarity, consistent histories, ...)
(iii) abandoning Kolmogorov axioms at some deeper level (??)

My question for you is this: Is there an explicit proposal of an interpretation in the category (iii)? (Perhaps many worlds?)

 

[Kolmo]

My question for you is this: Is there an explicit proposal of an interpretation in the category (iii)?

Well a quantum system, despite not obeying the Kolmogorov axioms, still obeys certain probabilistic constraints. We could consider giving these up, I'll give an example.

Say the variables $A,B$ have a common Kolmogorov model¹. Then imagine the same is true also for $B, C$ and $A, C$. Quantum Theory implies that since there is a common Kolmogorov model for each pair drawn from $A,B,C$ then for the triplet $A,B,C$ as a whole there is a single Kolmogorov model². In fact you can derive quantum theory as the most general probability theory which permits this structure.

You could imagine probability theories more general than this, i.e. each pair having a Kolmogorov model does not imply the triplet has one. However such theories would violate the Tsirelson bound and thus don't have empirical support.

¹ Conveyed in quantum theory by $[A,B] = 0$
² In QM itself the mathematical expression of this is if each pair has a common eigenbasis, then all three do

 

[Demystifier]

Say the variables $A,B$ have a common Kolmogorov model¹. Then imagine the same is true also for $B, C$ and $A, C$. Quantum Theory implies that since there is a common Kolmogorov model for each pair drawn from $A,B,C$ then for the triplet $A,B,C$ as a whole there is a single Kolmogorov model². In fact you can derive quantum theory as the most general probability theory which permits this structure.

That's axiomatization (which is interesting), but I asked for interpretation. Interpretation is something that is supposed to give an intuitive conceptual picture explaining why is that so. But if thinking in terms of interpretations is not your style, that's OK too.

 

[Kolmo]

That's axiomatization (which is interesting), but I asked for interpretation. Interpretation is something that is supposed to give an intuitive conceptual picture explaining why is that so

I'm not sure I understand what you mean, but the conceptual picture is fairly clear here I would say. In our world being able to perform simultaneous measurements on pairs drawn from some set of variables, implies you can design a single measurement which measures all variables in the set. Quantum Theory is then a probability theory obeying that concept with no further assumptions¹.

¹ No further assumptions because it is the most general probability theory compatible with that assumption.

 

[Demystifier]

I'm not sure I understand what you mean, ... Quantum Theory is then a probability theory ...

I guess I was right that interpretations is not your style. Roughly speaking, in an interpretation quantum theory is supposed to be a physical theory, rather than a probability theory. But of course, it's hard to define precisely what "physical" means, so if you don't get it, forget it!