# Binney's interpretation of Violation of Bell Inequalities

stevendaryl
Staff Emeritus
Would you say that a geometrical feature like say equality of right angles implying isotropy of the space and that figures may be moved to any location and still be congruent is a local or a nonlocal property? Does the fact that figures with arbitrary spacelike separation can be measured to be congruent if they are put together again mean that measuring one figure is affecting the measure of the other figure making it congruent with itself?
Now this particular geometrical property is commutative so it would not violate the inequalities, but a non-commutative geometrical property like euclidean rotations is not commutative and does violate the inequalities, the spatial separation(local-vs nonlocal) is not the important feature for such properties because they are pervasive(ubiquitous) and independent of spatial separation of the objects on which it is observed.
Is this more understandable?
It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of $\pm 1$. The device has (at least) two possible settings, $a$ or $a'$. Bob similarly has a device with two possible settings, $b$ or $b'$. For many rounds, you perform the following procedure and collect statistics:

On round number $n$,
1. Alice chooses a setting $a_n$.
2. She performs a measurement, and gets a result $A_n$
3. Bob chooses a setting $b_n$.
4. Bob gets result $B_n$
Then we compute: $E(a,b) =$ the average, over all rounds $n$ such that $a_n = a$ and $b_n = b$, of $A_n \cdot B_n$

That's the context for which Bell derived his inequality. If you allow for Alice's choice $a_n$ to affect Bob's result, $B_n$, and vice-verse, then there is no reason to expect the inequality to hold. That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:
1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$
Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether $a$ and $b$ are described by a commutative or noncommutative algebra.

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings $a$ and $a'$ are described by noncommuting operators, and the two measurements corresponding to Bob's settings $b$ and $b'$ are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.

TrickyDicky and Derek Potter
stevendaryl
Staff Emeritus
Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:
1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$
Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$
I should say that the above form for $F_A$ and $F_B$ are not the most general form compatible with local realism. More generally, the outcome $A$ might also depend on an additional hidden variable $\alpha_n$ that is local to Alice, and Bob's outcome $B$ might depend on a hidden variable $\beta_n$ that is local to Bob. But the fact that in an EPR-type experiment, there are perfect correlations at some settings implies that there can't be any dependency on additional local variables.

Derek Potter
It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of $\pm 1$. The device has (at least) two possible settings, $a$ or $a'$. Bob similarly has a device with two possible settings, $b$ or $b'$. For many rounds, you perform the following procedure and collect statistics:

On round number $n$,
1. Alice chooses a setting $a_n$.
2. She performs a measurement, and gets a result $A_n$
3. Bob chooses a setting $b_n$.
4. Bob gets result $B_n$
Then we compute: $E(a,b) =$ the average, over all rounds $n$ such that $a_n = a$ and $b_n = b$, of $A_n \cdot B_n$

That's the context for which Bell derived his inequality. If you allow for Alice's choice $a_n$ to affect Bob's result, $B_n$, and vice-verse, then there is no reason to expect the inequality to hold. That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:
1. Each round, there is a hidden variable $\lambda_n$ influencing the results. The variable may take on different values on different rounds (hence the subscript).
2. There are deterministic functions $F_A(a, b, \lambda)$ and $F_B(a, b, \lambda)$ such that $A_n = F_A(a_n, b_n, \lambda_n)$ and $B_n = F_B(a_n, b_n, \lambda_n)$
3. Each round, $\lambda_n$ is chosen randomly according to some probability distribution $P(\lambda)$
Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether $a$ and $b$ are described by a commutative or noncommutative algebra.

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings $a$ and $a'$ are described by noncommuting operators, and the two measurements corresponding to Bob's settings $b$ and $b'$ are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.
Is $F_A× F_B$ the same as $F_B× F_A$ for the inequalities ?

stevendaryl
Staff Emeritus
Is $F_A× F_B$ the same as $F_B× F_A$ for the inequalities ?
Yes, because they are real numbers, not operators.

TrickyDicky and Derek Potter
Yes, because they are real numbers, not operators.
Got it, thanks!

No, it counters Popper's original claim that QM's indeterminacy(HUP) was falsified by a certain outcome of his thought experiment. Both Shih's 1999 and more recent 2015(see http://phys.org/news/2015-01-popper-againbut.html) papers agreed on this too.
Interesting...observations suggesting nonlocal interference between randomly paired - i.e. not pre-entangled - photons??

Do you understand the experiment, Pat? I can't make sense of the article.

morrobay
Gold Member
It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of $\pm 1$. The device has (at least) two possible settings, $a$ or $a'$. Bob similarly has a device with two possible settings, $b$ or $b'$. For many rounds, you perform the following procedure and collect statistics:

On round number $n$,
1. Alice chooses a setting $a_n$.
2. She performs a measurement, and gets a result $A_n$
3. Bob chooses a setting $b_n$.
4. Bob gets result $B_n$
Then we compute: $E(a,b) =$ the average, over all rounds $n$ such that $a_n = a$ and $b_n = b$, of $A_n \cdot B_n$

That's the context for which Bell derived his inequality.

Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality.
In summery then the above applies to this inequality : (AB) + (AB') + (A'B) - (A'B') ≤ 2
where A,A',B,B' = ± 1
With assumptions that p (a,b) depend only on past variable λ and local measurements at x and y
p(ab|xy,λ) = p(a|x,λ) p(b|y,λ)

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Do you understand the experiment, Pat? I can't make sense of the article.
If you can't, Derek, I don't think there's much chance I can!

It does say:

"The randomly paired photons do not have any pre-prepared entanglement, which means they are considered to be a classical system. This raises the question, how could a classical system produce the same result as a quantum system?"

"...The observations suggest that the photon pair is interfering with itself instantaneously in a phenomenon called nonlocal interference. In the new experiment, the randomly paired photons have two different yet indistinguishable properties to be simultaneously annihilated at two distant photodetectors. The observations are the result of the superposition of these two probability amplitudes."

Quote from the abstract of the paper itself (I can't access the full text without buying it):

"Although the observation cannot be considered as a violation of the uncertainty relation as Popper believed, this experiment reveals a concern about nonlocal interference of a random photon pair, which involves the superposition of multi-photon amplitudes, and multi-photon detection events at a distance."

That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Those three define what Bell means by a "hidden variables theory". The special case of a "local" hidden variables theory makes the additional assumption that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$. That is,

$A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether $a$ and $b$ are described by a commutative or noncommutative algebra.
I still see something confusing about what it implies to assume locality, if we agree on defining locality for two spacelike separated subsystems as the absence of influence of measurement of one subsystem on the other, I'm not sure how the expression above exhausts this definition.
It is obvious that assuming that $F_A$ does not depend on $b$ and $F_B$ does not depend on $a$ assures us locality, but that is not the same as saying that assumptions that violate it are automatically not obeying the locality principle, because there might be ways to assure locality independent of that assumption that we are not aware of.

In other words to say that a theory is local only if and only if it assumes the inequalities is a very strong assertion and assumes that any correlated measurement of spatially separated systems implies instantaneous influence of measurements.

Is there any logical obstruction to the possibility that there are conditions independent from $A_n = F_A(a_n, \lambda_n)$ and $B_n = F_B(b_n, \lambda_n)$
that are compatible with locality? The fact that noone has come up with them shouldn't be a definitive impediment. In general in science statistical correlation does not necessarily imply direct influence, why should it be so in physics?

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings $a$ and $a'$ are described by noncommuting operators, and the two measurements corresponding to Bob's settings $b$ and $b'$ are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.
The experiments usually have the measurements being simultaneous so Alice's measurements versus Bob's commutativity is not what matters here. The commutativity (or lack of) is indeed about the choice(the angles) that Alice and Bob make, i.e, the measurements each one of them performs, in the inequalities the choices must commute . Why should all theories obeying locality (absence of influence at a distance) have this feature about the choice $a$ and $a'$? Are all possible theories describing the measurements settings with operators automatically incompatible with absence of instantaneous influences?

stevendaryl
Staff Emeritus
In other words to say that a theory is local only if and only if it assumes the inequalities is a very strong assertion and assumes that any correlated measurement of spatially separated systems implies instantaneous influence of measurements.
No, it absolutely does not assume that. That was the whole point of the inequality, was to distinguish nonlocality from mere correlation.

For example, suppose that you have a process that takes a pair of shoes and randomly selects one shoe out of a pair to send to Alice and the other one to send to Bob. When Alice sees that she has a left shoe, she immediately knows that Bob received a right shoe, and vice-verse. So the observations made by Alice and Bob are perfectly correlated. But that correlation does not imply that FTL signals go from Alice to Bob or vice-verse.

The whole point of Bell's inequality was to be able to distinguish between correlations that (classically) require distant influences from those that do not.

stevendaryl
Staff Emeritus
No, it absolutely does not assume that. That was the whole point of the inequality, was to distinguish nonlocality from mere correlation.

For example, suppose that you have a process that takes a pair of shoes and randomly selects one shoe out of a pair to send to Alice and the other one to send to Bob. When Alice sees that she has a left shoe, she immediately knows that Bob received a right shoe, and vice-verse. So the observations made by Alice and Bob are perfectly correlated. But that correlation does not imply that FTL signals go from Alice to Bob or vice-verse.

The whole point of Bell's inequality was to be able to distinguish between correlations that (classically) require distant influences from those that do not.
Bell absolutely does not assume that correlation implies a causal influence. What he does assume is that correlation between two measurements implies something about their common past. Actually, it's not so much that he assumes that, but that it's part of what he means by a "hidden variable theory". The whole point of a hidden variable theory is to explain correlations by invoking a shared variable whose value was determined in their shared past (the overlap of their past light-cones, according to SR causality).

Let me try another way to describe it, in terms of flow of information. Roughly speaking, information is knowing the answer (or probabilities for possible answers) to some question about the history of the universe. Bell's concept of "local realism" basically amounts to the assumption that information about the history of the universe propagates at the speed of light (or slower). If information is available in some localized region of space, then that information must either have been created there, or it must have propagated there from some source in the backwards lightcone. If Alice at time $t+\delta$ knows something (say, about Bob's measurement results), then that information in principle follows from
1. Information that was already available to Alice at time $t$
2. Information that came into existence in the region near Alice (a distance of less than or equal to $c \delta$ from her)
3. Information that flowed into that region from elsewhere.
When I say information that was already available at time $t$, I mean information that could be deduced, in principle, from detailed knowledge of the little region near Alice. Classically, that would mean that the information is deducible from the positions and momenta of the particles near Alice, as well as the values of fields in that region. Similarly, "information that flowed into that region from elsewhere" classically would mean either particles that flowed into the region, or energy/momentum that flowed into the region, or waves that propagated into that region. When I say "information that came into existence", that is allowing for intrinsic nondeterminism. If I flip a coin and the coin is intrinsically nondeterministic, then the result is new information that didn't exist prior to flipping the coin.

Quantum mechanics seems to violate this concept of information as being created locally and flowing at the speed of light or slower.

No, it absolutely does not assume that. That was the whole point of the inequality, was to distinguish nonlocality from mere correlation.

For example, suppose that you have a process that takes a pair of shoes and randomly selects one shoe out of a pair to send to Alice and the other one to send to Bob. When Alice sees that she has a left shoe, she immediately knows that Bob received a right shoe, and vice-verse. So the observations made by Alice and Bob are perfectly correlated. But that correlation does not imply that FTL signals go from Alice to Bob or vice-verse.

The whole point of Bell's inequality was to be able to distinguish between correlations that (classically) require distant influences from those that do not.
But your example doesn't violate the inequalities, and for that case I said they are fine, they discern correlation from action at a distance, the problem is in assuming that there is not locality from violation of the inequalities.

Bell absolutely does not assume that correlation implies a causal influence. What he does assume is that correlation between two measurements implies something about their common past. Actually, it's not so much that he assumes that, but that it's part of what he means by a "hidden variable theory". The whole point of a hidden variable theory is to explain correlations by invoking a shared variable whose value was determined in their shared past (the overlap of their past light-cones, according to SR causality).

Let me try another way to describe it, in terms of flow of information. Roughly speaking, information is knowing the answer (or probabilities for possible answers) to some question about the history of the universe. Bell's concept of "local realism" basically amounts to the assumption that information about the history of the universe propagates at the speed of light (or slower). If information is available in some localized region of space, then that information must either have been created there, or it must have propagated there from some source in the backwards lightcone. If Alice at time $t+\delta$ knows
results), then that information in principlefollows from Information that was already available to Alice at time $t$
1. Information that came into existence in the region near Alice (a distance of less than or equal to $c \delta$ from her)
2. Information that flowed into that region from elsewhere.
When I say information that was already available at time $t$, I mean information that could be deduced, in principle, from detailed knowledge of the little region near Alice. Classically, that would mean that the information is deducible from the positions and momenta of the particles near Alice, as well as the values of fields in that region. Similarly, "information that flowed into that region from elsewhere" classically would mean either particles that flowed into the region, or energy/momentum that flowed into the region, or waves that propagated into that region. When I say "information that came into existence", that is allowing for intrinsic nondeterminism. If I flip a coin and the coin is intrinsically nondeterministic, then the result is new information that didn't exist prior to flipping the coin.

Quantum mechanics seems to violate this concept of information as being created locally and flowing at the speed of light or slower.
Does a non-deterministic theory(I don't mean non-determinism only in the stochastic or probabilistic sense that you use in your post, I mean for instance in the nonlinear sense) by definition violate the inequalities?

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stevendaryl
Staff Emeritus
No, I'm not. Determinism, in the context of information, would mean that information never increases. So the answers to any question is already present in the initial conditions of the universe. But if there is nondeterminism, information can be created. For example, if flipping a coin is nondeterministic, then the information about what the result of a coin flip will be doesn't exist until the moment the coin is flipped.

Does a non-deterministic theory(I don't mean non-determinism in the stochastic or probabilistic sense that you use in your post, I mean for instance in the nonlinear sense) by definition violate the inequalities?
I don't know what you mean by "in the nonlinear sense". I use nondeterminism to mean that the state of the universe at a future time is not uniquely determined by the state at earlier times.

No, I'm not. Determinism, in the context of information, would mean that information never increases. So the answers to any question is already present in the initial conditions of the universe. But if there is nondeterminism, information can be created. For example, if flipping a coin is nondeterministic, then the information about what the result of a coin flip will be doesn't exist until the moment the coin is flipped.

I don't know what you mean by "in the nonlinear sense". I use nondeterminism to mean that the state of the universe at a future time is not uniquely determined by the state at earlier times.
But you surely muat have heard about nonlinear equations not being deterministic? There is more to non-determinism than the coin flip probabilistic concept.

stevendaryl
Staff Emeritus
But you surely muat have heard about nonlinear equations not being deterministic? There is more to non-determinism than the coin flip probabilistic concept.
No, I don't know what you mean by that. Why would a nonlinear equation be nondeterministic? Are you talking about chaos?

But you surely muat have heard about nonlinear equations not being deterministic? There is more to non-determinism than the coin flip probabilistic concept.
I'm sure stevendaryl will have heard of it, but others here, like me for instance, may not. In fact I wouldn't be at all surprised if some people are wondering whether you're confusing non-computability with indeterminacy.

No, I don't know what you mean by that. Why would a nonlinear equation be nondeterministic? Are you talking about chaos?
Well, for example, I know they are considered deterministic traditionally but they aren't precisely in the increase of information sense you mentioned above.

stevendaryl
Staff Emeritus
Well, for example, I know they are considered deterministic traditionally but they aren't precisely in the increase of information sense you mentioned above.
If you are talking about chaotic systems, then they are still deterministic. The information about future conditions is present in the initial conditions, it's just computationally infeasible to extract it.

Have you considered multivalued branching functions? There is, for example, The Free-Will Function. Nobody knows how it arrives at a value and it changes each time you use it.

stevendaryl
Staff Emeritus
Have you considered multivalued branching functions? There is, for example, The Free-Will Function. Nobody knows how it arrives at its value.
Well, this is a little off the track. The fact is that whether things are deterministic or not doesn't really affect Bell's inequality, if we assume that any nondeterminism is localized. A coin flip, or a choice made by a person using free will are localized. It doesn't make any difference for Bell's theorem whether the results are nondeterministic, or are just very difficult to predict.

Well, this is a little off the track. The fact is that whether things are deterministic or not doesn't really affect Bell's inequality, if we assume that any nondeterminism is localized. A coin flip, or a choice made by a person using free will are localized. It doesn't make any difference for Bell's theorem whether the results are nondeterministic, or are just very difficult to predict.
It is off the track, wich wasn't really my intention, So back to Bell and determinism, you are saying that violating the inequalities doesn't imply non-determinism?

Even if they are easy to predict the theorem doesn't say anything about the correlations unless the system fulfills the theorem's conditions. But to revert to the previous point, the mathematical representation of locality is not obvious. TrickyDicky seems to have sidetracked himself into determinism but I'm still not sure whether we got to the bottom of his previous question, how BI violation implies that F(A) must depend on b. Or conversely if F(A) does not depend on b then the BI must hold. It's probably elementary but I seem to have a blind spot to it. Or perhaps the way the Wikipedia[/PLAIN] [Broken] article suddenly plunges into rather formal language like this:
"Implicit in assumption 1) above, the hidden parameter space Λ has a probability measure ρ and the expectation of a random variable X on Λ with respect to ρ is written

where for accessibility of notation we assume that the probability measure has a density ρ that therefore is nonnegative and integrates to 1
"
... is a clue that it's not trivial (and perhaps that I should chew on it a bit more).

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TrickyDicky
stevendaryl
Staff Emeritus