Is Bell's Theorem a Valid Solution to the Locality Versus Nonlocality Issue?

[billschnieder]

I object to the language - "Then whatever Alice does (or fails to do)" implies she has a choice, while the idea that CFD denies free will contradicts this.

I think "free will" may be a classical concept, with more and more limited applicability as you go to the quantum realm. I say "may be" because I cannot prove it. Thus, I think accepting CFD may be a classical prejudice. When you say "Hint: admit the validity of CFD"... why?

The point is that the argument appears to suggest that everything that will happen in the future is already settled and nothing anyone does or does not do can change it. Do you agree or do you disagree with this and if so why? There is a modal error in such an argument, which I will present in moment after others have had a chance to think about the issue. It will reveal a subtle error being made by Bell proponents over and over.

To give a further hint, consider the following:
a) Statement made on Monday: "Bill will eat do-nuts for breakfast on Tuesday"
b) Statement made on Wednesday: "Bill ate do-nuts for breakfast on Tuesday"
Bill did in fact eat do-nuts for breakfast on Tuesday. So let us rewind time back to Monday, is statement (a) True? Of course both statements are true. Does it mean Bill had no choice or control over what Bill did on Tuesday simply because on Monday it was already true that he will eat do-nuts on Tuesday?

 

[Rap]

The point is that the argument appears to suggest that everything that will happen in the future is already settled and nothing anyone does or does not do can change it. Do you agree or do you disagree with this and if so why? There is a modal error in such an argument, which I will present in moment after others have had a chance to think about the issue. It will reveal a subtle error being made by Bell proponents over and over.

To give a further hint, consider the following:
a) Statement made on Monday: "Bill will eat do-nuts for breakfast on Tuesday"
b) Statement made on Wednesday: "Bill ate do-nuts for breakfast on Tuesday"
Bill did in fact eat do-nuts for breakfast on Tuesday. So let us rewind time back to Monday, is statement (a) True? Of course both statements are true. Does it mean Bill had no choice or control over what Bill did on Tuesday simply because on Monday it was already true that he will eat do-nuts on Tuesday?

I had hoped that I had made it clear that I am not an advocate for rejecting CFD, nor do I advocate against it. I just want to understand the implications of rejecting it, which I do not, at present.

Regarding Bill and the do-nuts. Of course both statements are true, but this is not an illustration of CFD. An illustration would be "If Bill buys do-nuts monday night, he will eat do-nuts for breakfast on Tuesday and if he does not, he will not" According to our model of this phenomenon, this will always happen. We observe a thousand times, Bill buys do-nuts monday night, and every time, he eats do-nuts for breakfast on Tuesday. A thousand times, Bill does not and he does not. Our model has been correct. Now suppose one monday night, Bill does not buy do-nuts, and does not eat do-nuts for breakfast on Tuesday. What is the truth value of the statement "if he had bought do-nuts, he would have eaten do-nuts for breakfast on Tuesday"? This is not the same as saying that if he does, he will. It is saying something definite (he would have eaten do-nuts) about something which is counter-factual (he in fact did not buy do-nuts). On the other hand, to say that if he does, he will, is not counterfactual.

If we reject CFD, then it is improper to even speak about what would have happened. Rejecting CFD, quantum mechanics then only presumes to speak about the future and the factual past, not about a hypothetical past. I don't so much think that rejection of CFD denies free will as it renders the concept improper, not a proper subject for scientific inquiry, an untestable concept, much like the concept of simultaneous position and momentum. And, much like simultaneous position and momentum, in the limit of classical physics, CFD gains meaning. I think that rejecting CFD may be the solution to Bell's paradox, but I don't know. I can't wrap my mind around the concept yet, just like once upon a time I couldn't wrap my mind around not knowing position and momentum simultaneously. Until I learned to identify and reject what amounts to a classical prejudice.

I mean, consider this counter-factual situation: suppose I prepare a system and measure the position of a particle and after many repeated preparations and measurements on many particles, I get the same answer. If, after one of those preparations, I had instead measured the momentum, would the position of the particle at that time been the same as what I have in fact measured in the past? The answer is to reject CFD in this case - the answer to the question is not yes or no, the answer is that the question is improper. If I accept CFD and say yes, it will have the same position, then, knowing the position and momentum will allow me to calculate exactly where it will be measured to be at some time in the future, and when I look for it, chances are it will not be there. A paradox - which may be resolved by rejecting CFD in this case (or assuming hidden variables).

I'm not saying this is a perfect analogy to the Bell paradox, it just ... resonates in my mind.

 

[SpectraCat]

The point is that the argument appears to suggest that everything that will happen in the future is already settled and nothing anyone does or does not do can change it. Do you agree or do you disagree with this and if so why? There is a modal error in such an argument, which I will present in moment after others have had a chance to think about the issue. It will reveal a subtle error being made by Bell proponents over and over.

To give a further hint, consider the following:
a) Statement made on Monday: "Bill will eat do-nuts for breakfast on Tuesday"
b) Statement made on Wednesday: "Bill ate do-nuts for breakfast on Tuesday"
Bill did in fact eat do-nuts for breakfast on Tuesday. So let us rewind time back to Monday, is statement (a) True? Of course both statements are true. Does it mean Bill had no choice or control over what Bill did on Tuesday simply because on Monday it was already true that he will eat do-nuts on Tuesday?

First of all, it seems like statement a) is only "true" if you assume complete determinism of the universe. I don't think everyone is willing to grant that ...

What if I add the following statement on Monday:

a') Courtney will assassinate Bill on Monday evening?

Now it sure seems like only one statement, a or a', can be "true", and we can't know which one until after the events have passed. Therefore it seems that one cannot make judgments about the truth of a (or a') until the event has already occurred. In a deterministic universe, I suppose one of the statements could be said to have been "true all along" ... but you cannot know that until after the fact, so it just seems like interprative post-rationalization to me. There is no way to use such scenarios to empirically deduce whether or not the universe is deterministic.

 

[Rap]

Now it sure seems like only one statement, a or a', can be "true", and we can't know which one until after the events have passed. Therefore it seems that one cannot make judgments about the truth of a (or a') until the event has already occurred. In a deterministic universe, I suppose one of the statements could be said to have been "true all along" ... but you cannot know that until after the fact, so it just seems like interprative post-rationalization to me. There is no way to use such scenarios to empirically deduce whether or not the universe is deterministic.

I'm taking the statement "Bill will eat do-nuts on Tuesday for breakfast" in the same sense that if Alice and Bob align their detectors and make measurements, "Alice and Bob will measure equal and opposite spins".

 

[SpectraCat]

I'm taking the statement "Bill will eat do-nuts on Tuesday for breakfast" in the same sense that if Alice and Bob align their detectors and make measurements, "Alice and Bob will measure equal and opposite spins".

But it's not the same. The corresponding statement about Bill that could (perhaps) be correlated the the Alic and Bob example is, if Bill is alive on Tuesday, and if he is awake in the morning, and if he is hungry, and if donuts are the only available food, (there are lots more qualifiers needed, but I guess you get the idea) then Bill will eat donuts for breakfast on Tuesday.

The Alice & Bob statement is based on a theory of physics that has been extensively tested and has never been found to be false. Making that statement is equivalent to saying, "Quantum mechanics is expected to still be valid when Alice and Bob make their measurements".

 

[billschnieder]

Most scientists do not accept that there is a value to unmeasured particle observables. They reject CFD. That is mainline QM. There are the various interpretations such as MWI, BM, Copenhagen, etc. which all make the same predictions.

Again you demonstrate that you do not understand what CFD means.


a) If I look at the moon, I will see it.
b) Had I looked at the moon, I would have seen it.

(a) is a True statement. In this case, it is implicit that the possibility of either looking at the moon or not looking at the moon still exists. (b) is a counter-factually definite statement. Statement (b) will be valid even if it is impossible for me to look at the moon now. Accepting (b) as a valid statement does not mean:

c) "Seeing the moon" exists prior to me looking at the moon.

Your description above of CFD is similar to statement (c) which demonstrates a lack of basic understanding of philosophy and logic.

The result which Alice will get when she tilts her device to angle b, could not possibly exist before Alice actually makes a measurement! To suggest that any realist has ever made such a claim is naive at best. I have pointed out this error to you multiple times but you continue to insist on making it so in case it is still not clear let me elaborate:


Let us denote observable "what Alice observes when she tilts her device to angle a" as A, and "what Alice observes when she tilts her device to angle b" as B and "what Alice observes when she tilts her device to angle c" as C.

For a single photon, All three observables A, B, C are possible, however if Alice never measures anything, none of them exist as actual observables. It is easy for Bellists to state without substantiation that realism implies the three observables must exist prior to measurement but I'm holding your feet to the fire to not only use terms A, B, C but spell out descriptively what A, B, and C mean. Let us take A as defined above,

A: "what Alice observes when she tilts her device to angle a"

Why would anyone with more than a single brain cell expect Alice to observe anything without performing a measurement. In other words, why would you expect me to see the moon without looking at it. Therefore although A,B,C are all "possible", only the one which Alice actually performs becomes actual. The others remain counter-factual definite.

By mixing "possibilities" with "actualities", you obtain paradoxes. So before you jump to deny CFD make sure you understand what it means. QM can also make predictions about experiments that are never performed and in some cases can no longer be performed. Yet the QM prediction tells us what we would have actually obtained had we performed the measurements. Why is that not CFD? You can not deny CFD without denying logic.

On the the other hand if the definition of CFD used by Bell proponents, is the idea that A actually exists prior to Alice making her measurement, or rather that "what Alice observes when she tilts her device to angle b" actually exists before Alice actually tilts her device and makes the measurement, such a ridiculous idea is non-classical, illogical and nonsensical. So attributing this ridiculous idea to "realists" or "classical systems" is naive at best.

 

[SpectraCat]

Again you demonstrate that you do not understand what CFD means.


a) If I look at the moon, I will see it.
b) Had I looked at the moon, I would have seen it.

(a) is a True statement. In this case, it is implicit that the possibility of either looking at the moon or not looking at the moon still exists. (b) is a counter-factually definite statement. Statement (b) will be valid even if it is impossible for me to look at the moon now. Accepting (b) as a valid statement does not mean:

c) "Seeing the moon" exists prior to me looking at the moon.

Your description above of CFD is similar to statement (c) which demonstrates a lack of basic understanding of philosophy and logic.

The result which Alice will get when she tilts her device to angle b, could not possibly exist before Alice actually makes a measurement! To suggest that any realist has ever made such a claim is naive at best. I have pointed out this error to you multiple times but you continue to insist on making it so in case it is still not clear let me elaborate:


Let us denote observable "what Alice observes when she tilts her device to angle a" as A, and "what Alice observes when she tilts her device to angle b" as B and "what Alice observes when she tilts her device to angle c" as C.

For a single photon, All three observables A, B, C are possible, however if Alice never measures anything, none of them exist as actual observables. It is easy for Bellists to state without substantiation that realism implies the three observables must exist prior to measurement but I'm holding your feet to the fire to not only use terms A, B, C but spell out descriptively what A, B, and C mean. Let us take A as defined above,

A: "what Alice observes when she tilts her device to angle a"

Why would anyone with more than a single brain cell expect Alice to observe anything without performing a measurement. In other words, why would you expect me to see the moon without looking at it. Therefore although A,B,C are all "possible", only the one which Alice actually performs becomes actual. The others remain counter-factual definite.

By mixing "possibilities" with "actualities", you obtain paradoxes. So before you jump to deny CFD make sure you understand what it means. QM can also make predictions about experiments that are never performed and in some cases can no longer be performed. Yet the QM prediction tells us what we would have actually obtained had we performed the measurements. Why is that not CFD? You can not deny CFD without denying logic.

On the the other hand if the definition of CFD used by Bell proponents, is the idea that A actually exists prior to Alice making her measurement, or rather that "what Alice observes when she tilts her device to angle b" actually exists before Alice actually tilts her device and makes the measurement, such a ridiculous idea is non-classical, illogical and nonsensical. So attributing this ridiculous idea to "realists" or "classical systems" is naive at best.

Thank you for that description .. I think I finally understand at least one of the points you are trying to make. I am not sure whether or not I agree yet, but that post was quite helpful. I also think that what I laid out in my previous reply to you is basically saying the same thing .. namely that events (i.e. measurements) obtain objective reality only after they occurred. Similarly, the CFD-based statements about the other possibilities which were NOT observed, only become valid after the measurement as well .. is that what you are saying?

 

[DrChinese]

Again you demonstrate that you do not understand what CFD means.


a) If I look at the moon, I will see it.
b) Had I looked at the moon, I would have seen it.

(a) is a True statement. In this case, it is implicit that the possibility of either looking at the moon or not looking at the moon still exists. (b) is a counter-factually definite statement. Statement (b) will be valid even if it is impossible for me to look at the moon now. Accepting (b) as a valid statement does not mean:

c) "Seeing the moon" exists prior to me looking at the moon.

Your description above of CFD is similar to statement (c) which demonstrates a lack of basic understanding of philosophy and logic.

The result which Alice will get when she tilts her device to angle b, could not possibly exist before Alice actually makes a measurement! To suggest that any realist has ever made such a claim is naive at best. I have pointed out this error to you multiple times but you continue to insist on making it so in case it is still not clear let me elaborate:


Let us denote observable "what Alice observes when she tilts her device to angle a" as A, and "what Alice observes when she tilts her device to angle b" as B and "what Alice observes when she tilts her device to angle c" as C.

For a single photon, All three observables A, B, C are possible, however if Alice never measures anything, none of them exist as actual observables. It is easy for Bellists to state without substantiation that realism implies the three observables must exist prior to measurement but I'm holding your feet to the fire to not only use terms A, B, C but spell out descriptively what A, B, and C mean. Let us take A as defined above,

A: "what Alice observes when she tilts her device to angle a"

Why would anyone with more than a single brain cell expect Alice to observe anything without performing a measurement. In other words, why would you expect me to see the moon without looking at it. Therefore although A,B,C are all "possible", only the one which Alice actually performs becomes actual. The others remain counter-factual definite.

By mixing "possibilities" with "actualities", you obtain paradoxes. So before you jump to deny CFD make sure you understand what it means. QM can also make predictions about experiments that are never performed and in some cases can no longer be performed. Yet the QM prediction tells us what we would have actually obtained had we performed the measurements. Why is that not CFD? You can not deny CFD without denying logic.

On the the other hand if the definition of CFD used by Bell proponents, is the idea that A actually exists prior to Alice making her measurement, or rather that "what Alice observes when she tilts her device to angle b" actually exists before Alice actually tilts her device and makes the measurement, such a ridiculous idea is non-classical, illogical and nonsensical. So attributing this ridiculous idea to "realists" or "classical systems" is naive at best.

I repeat: Most scientists do not accept that there is a value to unmeasured particle observables.

Not sure what the rest of this is all about. I am not debating the definition of CD, nor am I asserting that this is important to Bell. Bell says that using the EPR definition of realism, there can be no local realism. Not that hard really.

There are those who accept that local realism is ruled out and still believe in CD. Guess what, that is not inconsistent (because of nonlocality). But I would still say most scientists accept my statement above.

And by the way, QM does NOT make predictions about counterfactual setups. But if you try to (assuming some form of counterfactual reasoning), you can get a paradox. So I wouldn't recommend that.

 

[DrChinese]

Let us denote observable "what Alice observes when she tilts her device to angle a" as A, and "what Alice observes when she tilts her device to angle b" as B and "what Alice observes when she tilts her device to angle c" as C.

For a single photon, All three observables A, B, C are possible, however if Alice never measures anything, none of them exist as actual observables. ... Why would anyone with more than a single brain cell expect Alice to observe anything without performing a measurement.

Answer: because I can predict the results with certainty to anyone of those. That is what those single brain cell authors EPR believed anyway.

 

[DrChinese]

For a single photon, All three observables A, B, C are possible, however if Alice never measures anything, none of them exist as actual observables. It is easy for Bellists to state without substantiation that realism implies the three observables must exist prior to measurement but I'm holding your feet to the fire to not only use terms A, B, C but spell out descriptively what A, B, and C mean.

As I have told you any number of times previously, it is up to the realist to define realism! I am using the one from EPR, as Bell did. But if you want to give some improved definition that is useful, go for it! The only reason us poor Bellists use this definition is to hold the realist to something more firm than jelly.

So using your A, B and C: I would say that realism is having a definite value for these simultaneously.

 

[SpectraCat]

QM can also make predictions about experiments that are never performed and in some cases can no longer be performed. Yet the QM prediction tells us what we would have actually obtained had we performed the measurements. Why is that not CFD?

It is not CFD because CFD requires that ALL properties of a system simultaneously possesses reality and thus have values that can be predicted with certainty prior to measurement. QM specifically tells us that not all properties of a system can be simultaneously known to arbitrary accuracy. It tells us that you can construct a sequence of measurements that is fundamentally unpredictable .. for example the triple Stern-Gerlach experiment. It is not unpredictable because of random collisions, or chaos theory .. it is unpredictable because you can only know one vector component of the spin at a time .. the others exist in superposition states with no definite value. Thus standard QM does not satisfy CFD.

You can not deny CFD without denying logic.

Who said QM was supposed to be logical?

 

[DrChinese]

Oh, and Bill: if you don't want to give a useful definition of realism that has some bite to it... perhaps you should come over to the other side and visit us Bellists.

 

[Rap]

You can not deny CFD without denying logic.

I question that statement. In #102 of this thread, I think I stated clearly what CFD is. Maybe #102 was too long to read, if so I apologize.

Can you give an example of a logical contradiction that occurs when denying CFD, as I defined it in #102?

 

[harrylin]

No, Bell did not try to prove the inverse, i.e. he never tried to prove that NOT (values predictable in advance)=>NOT (predetermined values prior to measurement)

As I understood it, they believed that processes are "local" and they tried to prove that if something can be predicted, there must be an element of reality that corresponds to that prediction. Note that "roughly" is - again roughly speaking! - the inverse of a mathematical statement.

 

[billschnieder]

Similarly, the CFD-based statements about the other possibilities which were NOT observed, only become valid after the measurement as well .. is that what you are saying?

NO! Let us be very careful with our choice of words here.

The statements are valid and True. But they are not actual. If you insist that they are not valid, then the predictions of QM are not valid either, and no prediction can ever be valid until an experiment is performed. You have to carefully distinguish in your mind between validity and actuality.

QM can make simultaneous predictions about position and monentum:

"If the position is measured x will be obtained"
"If the momentum is measured p will be obtained"

Will you say both predictions are valid? Sure. But once you have measured the position, it becomes impossible to measure the momentum. Does that mean the momentum prediction was not valid? Of course not. It becomes a counterfactual statement. Had the momentum been measured instead, we would have obtained p.

 

[billschnieder]

Not sure what the rest of this is all about. I am not debating the definition of CD, nor am I asserting that this is important to Bell. Bell says that using the EPR definition of realism, there can be no local realism. Not that hard really.

But you and other Bellists continue to perform smoke and mirrors by suggesting that "realism" means observables must exist prior to measurement. And I am pointing out to you that such a ridiculous idea not only does not make any sense, but no realist has ever claimed as such ever.

And before you respond, affirming such a definition, remember what you would be saying:

that "I can see the moon without looking at it", "The result which Alice obtains by measuring a photon at angle a must exist as an actual result prior to Alice actually making the measurement". And you want to attribute such a ridiculous idea to realists? That is not being serious.

 

[billschnieder]

Oh, and Bill: if you don't want to give a useful definition of realism that has some bite to it... perhaps you should come over to the other side and visit us Bellists.

Realism means particles have objective properties at all times, whether or not they are measured or not, whether they are measurable or not.

According to realists, an observable is an outcome of a measurement and can either be a direct revealing of the underlying property, or a result of multiple influences such as all the objective properties of all the entities and variables involved at the moment of measurement.

All of the above is consistent with realism. Realists make no claim whether direct observation of particle properties in quantum scale particles are possible.

 

[yoda jedi]

it appears that they meant with "reality" quite the same as you and me.

not for me.

it is not necessary to have a defined value (CFD) to be real..

 

[Rap]

QM can make simultaneous predictions about position and monentum:

"If the position is measured x will be obtained"
"If the momentum is measured p will be obtained"

QM would never make such a statement, at least for the same point in time. It would imply that the wavefunction was simultaneously in an eigenstate of both position and momentum, which is impossible, since the two operators do not commute.

 

[SpectraCat]

QM would never make such a statement, at least for the same point in time. It would imply that the wavefunction was simultaneously in an eigenstate of both position and momentum, which is impossible, since the two operators do not commute.

I tried to point that out already .. maybe your explanation will make more sense to him.

 

[harrylin]

After I wrote that it appears that EPR meant with "reality" quite the same as you and me:

not for me.
it is not necessary to have a defined value (CFD) to be real.
.

Please tell us where you think EPR made such a claim

 

[harrylin]

I repeat: Most scientists do not accept that there is a value to unmeasured particle observables.

Not sure what the rest of this is all about. I am not debating the definition of CD, nor am I asserting that this is important to Bell. Bell says that using the EPR definition of realism, there can be no local realism. Not that hard really.
[..]

Billschnieder explained in a neat way why most realist don't think that there should be a value to unmeasured particle observables.

However, your next claim brings us back to my little question of before - for the only EPR definition of realism that I know of, is an incomplete one - merely a "criterion", as they put it. You did not answer my question if that is relevant - but here you seem to suggest that it is very relevant.
Using again my apple example: EPR say that if they see an apple, this implies the existence of a fruit. And now you apparently claim that using this criterion of fruit as a definition of fruit, Bell says that there can be no square fruit.
I find it hard to believe that; surely such a stupid misunderstanding can't be the basis of a famous theorem?!

 

[billschnieder]

QM would never make such a statement, at least for the same point in time. It would imply that the wavefunction was simultaneously in an eigenstate of both position and momentum, which is impossible, since the two operators do not commute.

You are thinking about one single wavefunction for both measurements. But think about the questions:

"Given particle Z, if you would measure it's position at point t, what would you obtain", QM can give you a prediction for that.

"Given particle Z, if would measure it's momentum at point t, what would you obtain", QM can give you a prediction for that.

Are you suggesting that QM is unable to answer such questions?

 

[yoda jedi]

After I wrote that it appears that EPR meant with "reality" quite the same as you and me:

Please tell us where you think EPR made such a claim

my claim.
not EPR claim.

it is not necessary to have a defined value (CFD) to be real.

.

 

[SpectraCat]

You are thinking about one single wavefunction for both measurements.

Of course we are talking about a wavefunction .. what else would you use to describe the system in terms of standard quantum mechanics (which is what we are discussing here)? The "particle Z" system that you mention below *is* a wavefunction in terms of standard QM.

But think about the questions:

"Given particle Z, if you would measure it's position at point t, what would you obtain", QM can give you a prediction for that.

"Given particle Z, if would measure it's momentum at point t, what would you obtain", QM can give you a prediction for that.

Are you suggesting that QM is unable to answer such questions?

Yes, that is what we are saying. In standard QM, "real particles" are described by wavepackets, which have non-zero finite widths in both position and momentum space, and thus can only provide probabilities of observing particular eigenvalues of position and momentum. A measurement samples the probability distribution defined by the square modulus of the wavepacket, and thus prior to the measurement, it is fundamentally impossible to predict the definite values of observables that are implied by the context of your question. In other words, the proper way to frame the question in QM is, "Given particle Z, what is the probability that a measurement of position (or momentum) at time t, would have the value x (or p)?"

 

[billschnieder]

Of course we are talking about a wavefunction .. what else would you use to describe the system in terms of standard quantum mechanics (which is what we are discussing here)? The "particle Z" system that you mention below *is* a wavefunction in terms of standard QM.



Yes, that is what we are saying. In standard QM, "real particles" are described by wavepackets, which have non-zero finite widths in both position and momentum space, and thus can only provide probabilities of observing particular eigenvalues of position and momentum. A measurement samples the probability distribution defined by the square modulus of the wavepacket, and thus prior to the measurement, it is fundamentally impossible to predict the definite values of observables that are implied by the context of your question. In other words, the proper way to frame the question in QM is, "Given particle Z, what is the probability that a measurement of position (or momentum) at time t, would have the value x (or p)?"

My argument does not depend on the type of prediction made by QM, but the fact that a prediction is being made. So let us try again using your terminology:

Prediction 1: "A position measurement on particle Z at time t has a probability distribution ρ(x)"
Prediction 2: "A momentum measurement on particle Z at time t has a probability distribution ρ(p)"

Surely you admit that such predictions can be, and are made in QM. If you disagree, say so.

Both predictions are valid! Do you disagree? Now suppose, I decide to perform the position measurement at time t. I will in fact find that the observed position is in agreement with the QM prediction, confirming that QM made a valid prediction. However, it is no longer possible to do a momentum measurement on the same particle, but had I performed the momentum measurement instead, I would have definitely obtained a result consistent with QM.

In other words, QM predictions about previously possible experiments which are no longer possible are conterfactual definite statements just as well. If you deny this, the only alternative is to conclude that once the position measurement is performed, the momentum prediction becomes invalid and you run into a situation in which QM can not make a valid prediction at all.

Now in case the above point is still not clear, let me give another example closer to Bell. Bell's inequality is the following:

|P(a,b) - P(a,c)| <= P(b,c) + 1

Let us write down the meaning of the terms in plain English:

A stream of particle pairs are heading towards Alice and Bob each of whom will set their detectors to a chosen angle and measure an outcome"
1) What is the expectation value of the paired-product of the outcomes if Alice and Bob choose settings (a,b) respectively? --> P(a,b)
2) What is the expectation value of the paired-product of the outcomes if Alice and Bob choose settings (a,c) respectively? --> P(a,c)
3) What is the expectation value of the paired-product of the outcomes if Alice and Bob choose settings (b,c) respectively? --> P(b,c)


As you can see, before any experiment is performed, all three terms are possible and valid terms to be calculated. But clearly, if Alice and Bob have already set their devices to (a,b), in one experiment, the other two terms, are counterfactual definite (ie, "had Alice and bob set their devices to (b,c), they would have obtained the expectation value P(b,c)", same for (a,c))

It is already clear from this, why it is impossible to perform an experiment to verify Bell's inequality. Because it is impossible to recover the particles and measure them at a different pair of settings. If you measure three different streams of particles, the terms you get from those streams can not just be mixed in the inequality willy-nilly without paradoxes. It will be similar to mixing and matching terms from the following three inequallities, and expecting to get the same results:

|P(a1,b1) - P(a1,c1)| <= P(b1,c1) + 1
|P(a2,b2) - P(a2,c2)| <= P(b2,c2) + 1
|P(a3,b3) - P(a3,c3)| <= P(b3,c3) + 1

Although each one is a valid Bell's inequality which can never be violated, there is no mathematical or logical justification for expecting the inequality:

|P(a1,b1) - P(a2,c2)| <= P(b3,c3) + 1

to be valid. This is why you get a violation from experiments. If anybody thinks the above inequality is valid, they should be able to derive it easily.

What about QM. Same thing, the three terms calculated from QM are also related in a similar way. If one is actualized in a real experiment, the other two are counterfactually definite and therefore violate the inequality. The predictions from QM for P(a,b), P(a,c), P(b,c), although all valid, are not all actual. In fact only one can ever be actualized in a given experiment. So if you are suggesting that in my previous position/momentum example, the momentum prediction becomes invalid after the position measurement, you must also conclude in this case that P(a,c) and P(b,c) become invalid after the P(a,b) measurement, in which case using them in the same expression as Bell did will be a mathematical error! But if you admit as I do that they are all valid, but not actual, then you must admit also that, an experiment which can produce the three terms for Bell's inequality is impossible to perform.

As I hope is clear now, the violation of Bell's inequality by QM and experiment, has nothing to do with whether particles have objective properties at all times, or whether locality or non-locality is involved. It has simply to do with a misunderstanding of the difference between possibilities and actualities.

Everything that is actually true is possibly true; but not everything that is possibly true is actually true.
Everything that is necessarily true is actually true; but not everything that is actually true is necessarily true.

 

[SpectraCat]

Bill, I don't have time to work through your Bell example right now, I will respond when I have more time, but I want to address the first part of your post.

My argument does not depend on the type of prediction made by QM, but the fact that a prediction is being made. So let us try again using your terminology:

Prediction 1: "A position measurement on particle Z at time t has a probability distribution ρ(x)"
Prediction 2: "A momentum measurement on particle Z at time t has a probability distribution ρ(p)"

Surely you admit that such predictions can be, and are made in QM. If you disagree, say so.

I agree that such statements are consistent with QM, but I am not sure that I agree that they qualify as predictions in the current context. This is because a single measurement (position or momentum) on your "particle Z" does not in fact reveal anything about the underlying probability distribution. So, I really need to modify the wording in my original modification of your predictive statements:

I believe the *only* way to phrase such statements so that they are consistent with standard QM would be:

"If an infinite ensemble of identical particles Z were prepared, and then subjected to position (momentum) measurements at some time t that is defined identically for each measurement, then the statistics of the set of measurements would correspond to the probability distribution predicted from the square-modulus of the position-(momentum-) space wavefunction for the identical particles Z."

As you can see, that is a special kind of prediction. It does not say anything about discrete events, but only describes what ought to happen for an unrealizable theoretical case. The only prediction standard QM makes for single measurements is:

"If you measure a property of a quantum system, you will observe *one* of the possible eigenvalues for that property."

So, since the above is an example of the only sort of predictions that can be made about individual measurements (without additional restrictions, such as knowing the system starts out in an eigenstate of whatever property is being measured prior to the actual measurement), and such a statement is completely useless from a CFD sense (at least I think it is), I must conclude that QM is incompatible with CFD.

Both predictions are valid! Do you disagree? Now suppose, I decide to perform the position measurement at time t. I will in fact find that the observed position is in agreement with the QM prediction, confirming that QM made a valid prediction. However, it is no longer possible to do a momentum measurement on the same particle, but had I performed the momentum measurement instead, I would have definitely obtained a result consistent with QM.

I disagree, for the reasons I laid out above.

In other words, QM predictions about previously possible experiments which are no longer possible are conterfactual definite statements just as well. If you deny this, the only alternative is to conclude that once the position measurement is performed, the momentum prediction becomes invalid and you run into a situation in which QM can not make a valid prediction at all.

I think that perhaps it is the "definite" part of counterfactual defniteness that is the issue in our discussion. I take it to mean that a theory consistent with CFD allows statements to be made about *specific* values of measurements on the properties of the system. Such statements would be equally valid both before and after a measurement has been performed.

I am not saying that QM is incompatible with CFD because statements about *definite values* of properties that were valid prior to the measurement lose their meaning after it is conducted. Rather, I am saying that such statements are NEVER valid in the context of QM, either before or after the measurement has been conducted.

 

[billschnieder]

I believe the *only* way to phrase such statements so that they are consistent with standard QM would be:

"If an infinite ensemble of identical particles Z were prepared, and then subjected to position (momentum) measurements at some time t that is defined identically for each measurement, then the statistics of the set of measurements would correspond to the probability distribution predicted from the square-modulus of the position-(momentum-) space wavefunction for the identical particles Z."

As you can see, that is a special kind of prediction. It does not say anything about discrete events, but only describes what ought to happen for an unrealizable theoretical case. The only prediction standard QM makes for single measurements is:

"If you measure a property of a quantum system, you will observe *one* of the possible eigenvalues for that property."

In other words, you are saying when Bell writes P(a,b) = - a.b in his equation (3) as the quantum mechanical expectation value, it is not valid? It looks to me like a definite prediction. Even according to your definition, - a.b is a *specific value* for the system of particles measured by Alice and Bob with their devices oriented along the vectors (a, b) respectively. Do you disagree?

What you are not getting is that my argument has nothing to do with individual events, it has to do with definite predictions, whether those predictions have to do with individual events or expectation values for a large number of events does not matter. All that matters is that there is an unambiguous prediction.

So again let me ask. Are you suggesting that QM does not make unambiguous predictions about expectation values? Or are you suggesting that QM makes such predictions but those predictions are never valid.

Note also that in my Bell explanation in the previous post, I was talking specifically about multiple events and expectation values, so if you could respond to that part it will be clear to me what part of my argument you still disagree with.

 

[harrylin]

my claim.
not EPR claim.
.

What does your claim have to do with EPR or with what I wrote?

 

[billschnieder]

Why would anyone with more than a single brain cell expect Alice to observe anything without performing a measurement. In other words, why would you expect me to see the moon without looking at it.

Answer: because I can predict the results with certainty to anyone of those. That is what those single brain cell authors EPR believed anyway.

You have misunderstood their paper. They made no such claim that observables exist independent of actual measurement. Could you provide the quote from the EPR paper which supports your claim.

Apparently, you are confused between possiblities and actualities. A prediction is a possiblity. You can predict an event with certainty and yet the event will not occur. That is why I mentioned the paradox in my post #89.

I can predict with certainty that "If Joe's head is cut off and burned, he will die". Yet, just because I have made such a prediction does not bind Joe's fate to my prediction. Your misunderstanding comes from separating the outcome from the precondition embedded in the prediction.

 

[Rap]

My argument does not depend on the type of prediction made by QM, but the fact that a prediction is being made. So let us try again using your terminology:

Prediction 1: "A position measurement on particle Z at time t has a probability distribution ρ(x)"
Prediction 2: "A momentum measurement on particle Z at time t has a probability distribution ρ(p)"

Surely you admit that such predictions can be, and are made in QM. If you disagree, say so.

Both predictions are valid! Do you disagree? Now suppose, I decide to perform the position measurement at time t. I will in fact find that the observed position is in agreement with the QM prediction, confirming that QM made a valid prediction. However, it is no longer possible to do a momentum measurement on the same particle, but had I performed the momentum measurement instead, I would have definitely obtained a result consistent with QM.

In other words, QM predictions about previously possible experiments which are no longer possible are conterfactual definite statements just as well. If you deny this, the only alternative is to conclude that once the position measurement is performed, the momentum prediction becomes invalid and you run into a situation in which QM can not make a valid prediction at all.

I agree with the above, both are valid, and the last paragraph is true. But denying CFD does not mean that prediction 2 is false. Once a measurement is made at time t, prediction 2 is no longer a prediction, since time t is past. Prediction 2 is now stated in the conditional perfect sense - a counterfactually definite statement:

"If a momentum measurement on particle Z had been made at time t, it would have had a probability distribution ρ(p)"

Denying CFD does not mean it is false, only that its truth value cannot be a subject of scientific inquiry. In the conditional perfect tense, is an untestable statement. In this particular case, this causes no problem, it is never used anyway.

The case for Bell is different. The predictions are the usual, if Alice and Bobs detectors are aligned they will record 0 coincidence (complementary spins), if they are at 90 degrees, they will have 50% coincidence, etc. The counterfactual statements relevant to Bell are not simply turning these statements into conditional perfect statements. Suppose Alice and Bob aligned their detectors at 45 degrees and made measuremnts. The counter factually definite statement is not simply "If Alice had aligned her detector with Bobs, they would have measured complementary spins", but rather "If Alice had aligned her detector with Bob's, she would have measured the complement of what Bob has in fact measured". Now rejecting CFD and stating that the truth value of this statement is not a subject of scientific inquiry is important.

 

[billschnieder]

"If a momentum measurement on particle Z had been made at time t, it would have had a probability distribution ρ(p)"

You can not deny CFD without at the same time denying the validity of the above statement. Period!

Denying CFD does not mean it is false, only that its truth value cannot be a subject of scientific inquiry. In the conditional perfect tense, is an untestable statement. In this particular case, this causes no problem, it is never used anyway.

Then you are confused about what it means to deny CFD. Denying CFD means you are claiming CFD statements are invalid.

The case for Bell is different. The predictions are the usual, if Alice and Bobs detectors are aligned they will record 0 coincidence (complementary spins), if they are at 90 degrees, they will have 50% coincidence, etc. The counterfactual statements relevant to Bell are not simply turning these statements into conditional perfect statements.

Please read the Bell portion of my argument again carefully, it does not depend on any complementary spins or angles. My argument depends only on the fact that the three terms involved, although each individually "possible", only one of them can be simultaneously "actual". That is the crux of the argument. Trying to focus on specific angles and complementary spins, misses the point completely.

The counter factually definite statement is not simply "If Alice had aligned her detector with Bobs, they would have measured complementary spins", but rather "If Alice had aligned her detector with Bob's, she would have measured the complement of what Bob has in fact measured".

Again this is missing the point. So let me ask you specifically, about the three terms in Bell's inequality P(a,b), P(b,c), P(a,c). Do you or do you not agree that until the measurement on the stream of particles is made, all three are "possible", but once the measurement (a,b) is made only the one actually measured is "actual", the rest are then counterfactually definite statements about what would have been observed, had the settings been (b,c) or (a,c). Please think carefully about this and note that each term involves measurements from both Alice and Bob and that is why I say you are missing the point.

Ultimately then, it doesn't matter whether you accept or reject CFD, you still end up with the same conundrum:

If you deny CFD, then you are effectively saying no experiment to test Bell's inequality with those three terms included can ever be performed. It doesn't matter whether you try to get the terms from QM or experiment, they will not be valid terms for use in Bell's inequality (in your own words: "rejecting CFD and stating that the truth value of this statement is not a subject of scientific inquiry".)

If you accept CFD, then you also recognize that CFD statements although true, are not actual so, no experiment which can test a CFD statement can ever be performed.

Hopefully you see that maligning CFD as an escape from Bell's paradox does not help. In the worst case, denying the validity of CFD statements, forces you to deny the validity of the terms P(b,c), P(a,c) from QM as well which is not desirable. In the case in which (a,b) was actually measured, if you say the terms P(b,c), P(a,c) in Bell's inequality are valid CFD statements which but which are not subject to scientific inquiry, then you still reach the conclusion that Bell's inequalities can not be tested experimentally.

 

[SpectraCat]

Bill,

I have read your Bell example carefully, and come to the conclusion that you are just using a lot of words to say, "you can never cross the same river twice". In other words, you are denying that experiments carried out on ensembles of identically prepared particles can give predictable results. You are absolutely right that the 3 Bell distributions P(a,b), P(b,c) and P(a,c) will always be measured with different ensembles of particles ... but so what? The QM predictions about the probability distributions are determined by the settings a,b or c, not the particular ensemble of *identical particles* that is used to generate the distribution. So I don't see how your criticism pertains to the Bell experiments.

Also, I notice that you never addressed my points about the probabilistic predictions made by QM being incompatible with CFD ... after all, you can never say anything about a probability distribution based on the results of a single measurement. As I understand CFD, it requires that you be able to make definite predictions about individual measurements. Is that not correct? You mentioned a couple of times about predictions of expectation values, but that is exactly what I am saying .. expectation values are averages ... they cannot be determined by single experiments, but only by (large) sets of repeated experiments on the same system.

So to summarize, I agree that the following statements are valid based on QM, both before and after any particular ensemble of measurements has been used to generate an expectation value.

Given an ensemble of identically prepared entangled particles, and a pair of detectors (Alice and Bob) with 3 settings {a,b,c} then sufficiently large sets of measurements with identical settings will yield the following results:

1) if Alice and Bob set their detectors to a & b, respectively, they will measure the expectation value P(a,b)
2) if Alice and Bob set their detectors to a & c, respectively, they will measure the expectation value P(a,c)
3) if Alice and Bob set their detectors to b & c, respectively, they will measure the expectation value P(b,c)

I do not think those statements are consistent with CFD, because I have laid out above (and previously) they are statements about ensembles, rather than discrete events. However, if I am wrong, and CFD predictions can be stated probabilistically, as opposed to definitely (which seems inconsistent with its name), then I suppose those statements would be consistent with CFD.

Furthermore, I agree with Rap's argument that the following is also true based on QM:

Suppose we had done the P(a,b) experiment and obtained the expectation value. We can say that if Alice and Bob had instead chosen b and c as the settings, they would have measured P(b,c), and likewise if they had chosen a & c they would have measured P(a,c).

Those statements seems consistent with a putative probabilistic form of CFD as well.

So I guess I don't understand what your point is, beyond the (sorry to say) sohpists argument that "you can never measure the same entangled pair twice".

 

[harrylin]

Originally Posted by rlduncan
"Theoretically, Bell’s theorem can be not be violated by any experiment when applied to a two-valued variables, such as S(T,F), S(H,T), or S(+,-). Whether the measured values are true/false, heads/tails, or up spin/down spin, etc. Bell’s theorem is a mathematical truth, a tautology. If misapplied by not meeting the conditions of the theorem, then violations may occur. Two examples will demonstrate using a coin tossing experiment where the upper most face is observed and the sequence of heads and tails is recorded. Three coins are tossed simultaneously by three individuals. For simplicity, let's them be a, b, and c and each coin is tossed eight times."
...

Despite some of the comments above, your argument is valid. You can find more rigorous variations of it in the following articles:

* Hess, K. and Michielsen, K. and De Raedt, H. Possible experience: From Boole to Bell. 2009. EPL (Europhysics Letters), 87:60007. http://arxiv.org/pdf/0907.0767
[..] [/url]

A comprehensive overview and follow-up, based on those earlier papers by different authors, has just been published (it was available on Arxiv for some time, but that doesn't count for physicsforums).
As this thread seems to wander off in different directions with personal arguments by the participants, I now started a thread to focus on the arguments as presented in that new paper by De Raedt et al:

https://www.physicsforums.com/showthread.php?t=499002

Some of the arguments of this thread are likely pertinent for that discussion, but it's not clear to me which. For example, I searched that paper for "counterfactual" but got no hit; so I wonder if their arguments are related to the discussion here. At first sight, billschnieder's argument looks similar to that of De Raedt et al. It would be helpful to clarify that there.

Thanks,
Harald

 

[DrChinese]

But you and other Bellists continue to perform smoke and mirrors by suggesting that "realism" means observables must exist prior to measurement. And I am pointing out to you that such a ridiculous idea not only does not make any sense, but no realist has ever claimed as such ever.

And before you respond, affirming such a definition, remember what you would be saying:

that "I can see the moon without looking at it", "The result which Alice obtains by measuring a photon at angle a must exist as an actual result prior to Alice actually making the measurement". And you want to attribute such a ridiculous idea to realists? That is not being serious.

The outcome VALUE of an observable must pre-exist. I think it is clear that is what EPR thought. Else what are you asserting as being realism? Gimme a useful definition that others might be able to use.

 

[DrChinese]

You have misunderstood their paper. They made no such claim that observables exist independent of actual measurement. Could you provide the quote from the EPR paper which supports your claim.

Apparently, you are confused between possiblities and actualities. A prediction is a possiblity. You can predict an event with certainty and yet the event will not occur. That is why I mentioned the paradox in my post #89.

I can predict with certainty that "If Joe's head is cut off and burned, he will die". Yet, just because I have made such a prediction does not bind Joe's fate to my prediction. Your misunderstanding comes from separating the outcome from the precondition embedded in the prediction.

If I can predict the outcome of A in advance every time, and the outcome of B every time, and of C every time, according to EPR these constitute elements of reality. This is by their definition. Now, the only thing left in doubt here is whether these are simultaneously elements of reality. In my opinion (and that of virtually all), EPR intended this to be YES, they are simultaneous elements of reality.

If you disagree with this, which seems apparent now, then you are a realist in name only. You must be a closet Bellist.

 

[yoda jedi]

What does your claim have to do with EPR or with what I wrote?

EPR requires counterfactual definiteness (CFD).

.

 

[billschnieder]

The outcome VALUE of an observable must pre-exist. I think it is clear that is what EPR thought.

Wrong. Provide the EPR quote which informs such a ridiculous and nonsensical claim. Although I notice you are now saying this is what EPR "thought" rather than what they "said". But if you truly believe they "thought" this, you should be able to provide quotes from them justifying this, which you still haven't done.

Else what are you asserting as being realism? Gimme a useful definition that others might be able to use.

I already gave you a definition of realism which you ignored. Here it is again:

Realism means particles have objective properties at all times, whether or not they are measured or not, whether they are measurable or not.

According to realists, an observable is an outcome of a measurement and can either be a direct revealing of the underlying property, or a result of multiple influences such as all the objective properties of all the entities and variables involved at the moment of measurement.

All of the above is consistent with realism. Realists make no claim whether direct observation of particle properties in quantum scale particles are possible.

 

[billschnieder]

If I can predict the outcome of A in advance every time, and the outcome of B every time, and of C every time, according to EPR these constitute elements of reality.

Again please provide the exact quote which claims that. This is what EPR said as a sufficient criterion for reality

"If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to that quantity."

Note it does not say the quantity itself must be the element of reality only that it corresponds to one.

This is by their definition. Now, the only thing left in doubt here is whether these are simultaneously elements of reality. In my opinion (and that of virtually all), EPR intended this to be YES, they are simultaneous elements of reality.

Please provide the quote where they say these must be simultaneous elements of reality.

 

[DrChinese]

Again please provide the exact quote which claims that. This is what EPR said as a sufficient criterion for reality

"If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to that quantity."

Note it does not say the quantity itself must be the element of reality only that it corresponds to one.


Please provide the quote where they say these must be simultaneous elements of reality.

I fully agree with your first point above ("corresponding"), although I fail to see its significance. I sometimes use shorthand lingo when I don't think it matters, but I will do my best to honor your wishes on this.

Second point: I believe I have said repeatedly that they DIDN"T say the word "simultaneous", they assumed the informed scientific reader would pick up this rather straightforward point. As far as I recall, Bell didn't use the word either, although his logic doesn't go very far if you don't assume this is the intended context. Nonetheless, EPR has little meaning if their assertion is that reality depends on the measurement performed by a distant observer (i.e. an actual experiment). In fact they specifically denied that.

So my question to you remains: if reality does not simultaneously and collectively correspond to the individually observable elements a, b, c ... and so on, then what does it mean to be a realist, Bill? Does an individual photon have a polarization "answer" for observations at all angle settings (simultaneously) independent of an observer's choice of same? Or does the observer shape reality (contrary to the EPR assertion that this is unreasonable)?

 

[Gordon Watson]

The outcome VALUE of an observable must pre-exist. I think it is clear that is what EPR thought. Else what are you asserting as being realism? Gimme a useful definition that others might be able to use.

1. What do you mean, please, by the outcome VALUE.

2. What is the value associated with a spin-half particle, "spin-up at 45 degrees"?

3. What is the value associated with a pristine spin-half particle, "entangled"?

4. Is it not the case that Bell [1964: equation (1)] assigns the VALUES ± 1 to outcomes?

5. Surely ± 1-s don't pre-exist?

6. AS FOR FOR ME, a dedicated local realist: In Bell (1964), pristine lambda represents the INITIAL VALUE of the following REAL hidden-variable: the orientation of each pristine (and entangled) particle's principal axis associated with total spin.

7. After "measurement", lambda remains a principal axis: BUT it is now that associated (in QM) with intrinsic spin.

8. So we have the transformation of pristine lambda (perturbed by "measurement") to a new variable [lambda --> +a, say] as a result of the "measurement" interaction.

I'd welcome clarification, or your opinion, on each of these points.

 

[DrChinese]

1. What do you mean, please, by the outcome VALUE.

2. What is the value associated with a spin-half particle, "spin-up at 45 degrees"?

3. What is the value associated with a pristine spin-half particle, "entangled"?

4. Is it not the case that Bell [1964: equation (1)] assigns the VALUES ± 1 to outcomes?

5. Surely ± 1-s don't pre-exist?

6. AS FOR FOR ME, a dedicated local realist: In Bell (1964), pristine lambda represents the INITIAL VALUE of the following REAL hidden-variable: the orientation of each pristine (and entangled) particle's principal axis associated with total spin.

...

I'd welcome clarification, or your opinion, on each of these points.

1. 2. 4. You can label those outcome/result values (or whatever your choose to call them) however you like, +1 or -1 (following Bell), H or T, up or down, I fail to see what difference it makes. It is the local realist who is asserting these exist, not me. I tend to see the result of an observation as being "real".

3. 5. I don't think there is such. But I am not a realist anyway. (Hey, all my life I've been told my ideas are unrealistic so I guess that fits.)

6. This is obviously false and I am shocked you would state this as being your position. Even to a local realist it should be obvious that you don't get perfect EPR correlations from a (single/dual/triple) principal axis. You need a lot more encoded in those babies than a few bits to get the ability to predict spin with certainty.

However, I give you 10 points for at least owning up to some kind of a definition. That is more than more local realists will do.

 

[Gordon Watson]

1. 2. 4. You can label those outcome/result values (or whatever your choose to call them) however you like, +1 or -1 (following Bell), H or T, up or down, I fail to see what difference it makes. It is the local realist who is asserting these exist, not me. I tend to see the result of an observation as being "real".

3. 5. I don't think there is such. But I am not a realist anyway. (Hey, all my life I've been told my ideas are unrealistic so I guess that fits.)

6. This is obviously false and I am shocked you would state this as being your position. Even to a local realist it should be obvious that you don't get perfect EPR correlations from a (single/dual/triple) principal axis. You need a lot more encoded in those babies than a few bits to get the ability to predict spin with certainty.

However, I give you 10 points for at least owning up to some kind of a definition. That is more than more local realists will do.

Added emphasis by GW.

Thanks Doc. SOS!

Why not view this [ ] before elaborating on point #6? I'm not much into believing in the "obviously false" -- nor overlooking the "should be obvious".

 

[JesseM]

1. What do you mean, please, by the outcome VALUE.

2. What is the value associated with a spin-half particle, "spin-up at 45 degrees"?

3. What is the value associated with a pristine spin-half particle, "entangled"?

4. Is it not the case that Bell [1964: equation (1)] assigns the VALUES ± 1 to outcomes?

5. Surely ± 1-s don't pre-exist?

I think Gordon may still be unclear on the basic notion of the EPR/Bell argument involving predetermined values, so I want to elaborate a bit on DrChinese's answer here. ± 1 are indeed assigned to outcomes, but EPR/Bell assume that if by measuring particle #1 we can perfectly predict what the outcome will be if a given type of measurement is made on particle #2 far away (and at a spacelike separation), then in a local realistic theory there must have been some local "elements of reality" associated with particle #2 beforehand that predetermined what the outcome of such a measurement would be. It's not that the particle already has a property equal to that outcome, since the act of measurement could very well alter the particle's properties: for example, even if a particle was predetermined to show momentum p if its momentum was measured (and we can predict this by measuring the momentum of its entangled twin and invoking the rule that momentum is always measured to be conserved), its "hidden" momentum prior to measurement could have been some different value p', but it must in that case have had some local hidden variables that ensured that if a momentum measurement were performed on it, its momentum would change to the value p and that would be the observed outcome of the measurement.

Einstein uses the analogy of a pair of boxes, such that whenever we open one, if we see a ball inside then it's guaranteed no ball will be found when the other is opened, and vice versa. Under local realism, one way to explain this is just to say the boxes each already had the property of "ball inside" or "no ball inside" beforehand. But as I discussed in [post=3270631]this post[/post], you could also come up with more complicated explanations where the "ball inside" property did not itself exist prior to measurement, but they boxes had other properties which predetermined whether you'd see a ball inside when opened or not:

In terms of the box analogy, one might imagine that instead of one box containing a ball before being opened, they both contain computers connected to holographic projectors, and the computers can sense when the lid is being opened and depending on their programming they will either respond by projecting an image of a ball, or projecting the image of an empty box. In this case the local variables associated with each box would not consist of "ball" or "no ball", but rather would be a detailed specification of the programming of each computer. But it would still be true based on the separation principle and the perfect correlation between results that if one was programmed to project a ball when the box was opened, that must mean the other was programmed to project an empty box, so the local variables (the program of each computer) would still predetermine the fact that one would give the measurement result "saw a ball" and the other would give the result "didn't see a ball".

Gordon, if you don't think that particles have local hidden variables that predetermine the results they will give when measured in this scenario (which I guess in terms of your #6 would mean that even given advance knowledge of "the orientation of each pristine (and entangled) particle's principal axis associated with total spin" for each particle, we couldn't say in advance with perfect certainty whether it would give +1 or -1 if the spin was measured at a given angle), does that mean you think there is a random element in what result they give? If so, as a local realist how can you explain the fact (in this scenario) that whenever both experimenters choose the same property to measure, they are guaranteed to get the same (or opposite) results? If there was a random element to the outcomes, and the random events in the neighborhood of one particle couldn't have a nonlocal influence on random events in the neighborhood of the other in the case of measurements at a spacelike separation, wouldn't that mean there would be some nonzero probability they would fail to give the same (or opposite) results? Think of the box scenario with the holographic projectors, but suppose each box also has something like a true random number generator that determines whether it will project an image of a ball or not--since the two random numbers picked are statistically independent, even if the odds are stacked so it's unlikely they would both project holographic balls (say the first box is programmed to randomly pick a number from 1-100 and project a ball if the number is anywhere from 1 to 99, while the second box also picks a number from 1-100 and only projects a ball if the result is 1), there's always going to be a nonzero probability they both will (in this case the probability both will project images of balls is (99/100)*(1/100)).

 

[Gordon Watson]

I think Gordon may still be unclear on the basic notion of the EPR/Bell argument involving predetermined values, so I want to elaborate a bit on DrChinese's answer here. ± 1 are indeed assigned to outcomes, but EPR/Bell assume that if by measuring particle #1 we can perfectly predict what the outcome will be if a given type of measurement is made on particle #2 far away (and at a spacelike separation), then in a local realistic theory there must have been some local "elements of reality" associated with particle #2 beforehand that predetermined what the outcome of such a measurement would be. It's not that the particle already has a property equal to that outcome, since the act of measurement could very well alter the particle's properties: for example, even if a particle was predetermined to show momentum p if its momentum was measured (and we can predict this by measuring the momentum of its entangled twin and invoking the rule that momentum is always measured to be conserved), its "hidden" momentum prior to measurement could have been some different value p', but it must in that case have had some local hidden variables that ensured that if a momentum measurement were performed on it, its momentum would change to the value p and that would be the observed outcome of the measurement.

Einstein uses the analogy of a pair of boxes, such that whenever we open one, if we see a ball inside then it's guaranteed no ball will be found when the other is opened, and vice versa. Under local realism, one way to explain this is just to say the boxes each already had the property of "ball inside" or "no ball inside" beforehand. But as I discussed in [post=3270631]this post[/post], you could also come up with more complicated explanations where the "ball inside" property did not itself exist prior to measurement, but they boxes had other properties which predetermined whether you'd see a ball inside when opened or not:

Gordon, if you don't think that particles have local hidden variables that predetermine the results they will give when measured in this scenario (which I guess in terms of your #6 would mean that even given advance knowledge of "the orientation of each pristine (and entangled) particle's principal axis associated with total spin" for each particle, we couldn't say in advance with perfect certainty whether it would give +1 or -1 if the spin was measured at a given angle), does that mean you think there is a random element in what result they give? If so, as a local realist how can you explain the fact (in this scenario) that whenever both experimenters choose the same property to measure, they are guaranteed to get the same (or opposite) results? If there was a random element to the outcomes, and the random events in the neighborhood of one particle couldn't have a nonlocal influence on random events in the neighborhood of the other in the case of measurements at a spacelike separation, wouldn't that mean there would be some nonzero probability they would fail to give the same (or opposite) results? Think of the box scenario with the holographic projectors, but suppose each box also has something like a true random number generator that determines whether it will project an image of a ball or not--since the two random numbers picked are statistically independent, even if the odds are stacked so it's unlikely they would both project holographic balls (say the first box is programmed to randomly pick a number from 1-100 and project a ball if the number is anywhere from 1 to 99, while the second box also picks a number from 1-100 and only projects a ball if the result is 1), there's always going to be a nonzero probability they both will (in this case the probability both will project images of balls is (99/100)*(1/100)).

Thanks Jesse,

To put your mind at ease:

1. I believe that we live in a quantum world and that classical analogies are (consequently) often misleading. Bell's reference (even deference) to a classical analogy by d'Espagnat was, imho at the time, laughable. [Edit: See Bell, in Bertlmann's socks: "To explain this denouement without maths I cannot do better than follow d'Espagnat." Still laughable; even desperate, imho.]

2. There is no random element in the outcome you refer to.

3. Determinism is not a rude word with me.

 

[JesseM]

Thanks Jesse,

To put your mind at ease:

1. I believe that we live in a quantum world and that classical analogies are (consequently) often misleading. Bell's reference (even deference) to a classical analogy by d'Espagnat was, imho at the time, laughable. [Edit: See Bell, in Bertlmann's socks: "To explain this denouement without maths I cannot do better than follow d'Espagnat." Still laughable; even desperate, imho.]

2. There is no random element in the outcome you refer to.

3. Determinism is not a rude word with me.

So, do you think a hypothetical omniscient observer with complete knowledge of one particle's local properties (which might include the particle's own "orientation" in your model, but wouldn't include any information about the other particle) at some time prior to measurement (but after the past light cones of the two measurement regions have ceased to overlap, as in "region 3" of fig. 2 at the top of p. 3 in this paper) would be able to predict in advance with total certainty what outcome would be seen if the particle were measured at any of three detector settings? So that the observer could say something like "this particle's local properties ensure it is predetermined to give +1 if measured at angle a, -1 if measured at angle b, and -1 if measured at angle c"? And if both particles are always found to give identical observed results when measured at the same angle, would you agree this implies (under local realism) that for each pair emitted by the source, their local properties must be correlated in such a way as to ensure that one particle must have the same three predetermined results as the other one?

 

[billschnieder]

Bill,

I have read your Bell example carefully, and come to the conclusion that you are just using a lot of words to say, "you can never cross the same river twice". In other words, you are denying that experiments carried out on ensembles of identically prepared particles can give predictable results.

No. I'm not saying experiments carried out on ensembles can not give predictable results. Rather, I'm using a lot of words to say, "the average height of 100 people means something completely different from the average of a single person's height measured 100 times", both can give predictable results but are not necessarily the same. In more relevant terms, I am saying the three terms from an experiment or from QM can not be used simultaneously in a single inequality, since they represent alternate possibilities only one of which can ever be actual, whereas Bell's inequality is dealing with an abstract thought experiment in which all three are simultaneously available.

You are absolutely right that the 3 Bell distributions P(a,b), P(b,c) and P(a,c) will always be measured with different ensembles of particles ... but so what?

Bell's inequality derivation relies on the fact that they originate from THE SAME ensemble, therefore you cannot use three different ensembles and expect them to just work. If you think three different ensembles should work, you should start out from that assumption and derive the inequalities and show that you can still obtain them. However, many authors have done that, and obtained different inequalities which no experiment or QM has ever violated. See the articles I mentioned earlier.

Also, I notice that you never addressed my points about the probabilistic predictions made by QM being incompatible with CFD ... after all, you can never say anything about a probability distribution based on the results of a single measurement. As I understand CFD, it requires that you be able to make definite predictions about individual measurements. Is that not correct?

A definite prediction is one that is unambiguous. I guess if the experiment one single event then the result will be ambiguous, but the experiment in this case is not one event, it is a series, and the prediction from QM is not about a single event but an unambiguous expectation value. It does not matter anyhow because rejection of CFD is just a red-herring and does not address the main issues.

You mentioned a couple of times about predictions of expectation values, but that is exactly what I am saying .. expectation values are averages ... they cannot be determined by single experiments, but only by (large) sets of repeated experiments on the same system.

It will be interesting to know what you mean by system here. Since clearly each event is a different system.

Suggesting that the averages on the first 100 photons must be the same as the averages on the next 100 is similar to saying the average stock price of a stock for the first 100 days of the year must be the same as the average for the next 100 days of the year. It is easy to make that mistake if you are really thinking that you are measuring the same photon everytime, (or the same stock tick every time), which is impossible to do, so you just naively measure a different photon and hope that the averages are the same.

Given an ensemble of identically prepared entangled particles, and a pair of detectors (Alice and Bob) with 3 settings {a,b,c} then sufficiently large sets of measurements with [/b]identical[/b] settings will yield the following results:

1) if Alice and Bob set their detectors to a & b, respectively, they will measure the expectation value P(a,b)
2) if Alice and Bob set their detectors to a & c, respectively, they will measure the expectation value P(a,c)
3) if Alice and Bob set their detectors to b & c, respectively, they will measure the expectation value P(b,c)

Note that you are relying on the idea that everything is identical, presumably because you are hoping that the results will be equivalent from one photon to the next...

I do not think those statements are consistent with CFD, because I have laid out above (and previously) they are statements about ensembles, rather than discrete events. However, if I am wrong, and CFD predictions can be stated probabilistically, as opposed to definitely (which seems inconsistent with its name), then I suppose those statements would be consistent with CFD.

In my previous posts I have explained why your definition of CFD will not eliminate the conundrum so I will say here it doesn't matter, as the real issue is elsewhere.

I suppose you know about the triangle inequality which says for any triangle with sides labeled x, y, z where x, y, z represents the lengths of the sides

z <= x + y

Note that this inequality applies to a single triangle. What if you could only measure one side at a time. Assume that for each measurement you set the label of the side your instrument should measure and it measured the length destroying the triangle in the process. So you performed a large number of measurements on different triangles. Measuring <z> for the first run, <x> for the next and <y> for the next.

Do you believe the inequality
<z> <= <x> + <y>

Is valid? In other words, you believe it is legitimate to use those averages in your inequality to verify its validity?

 

[Gordon Watson]

So, do you think a hypothetical omniscient observer with complete knowledge of one particle's local properties (which might include the particle's own "orientation" in your model, but wouldn't include any information about the other particle) at some time prior to measurement (but after the past light cones of the two measurement regions have ceased to overlap, as in "region 3" of fig. 2 at the top of p. 3 in this paper) would be able to predict in advance with total certainty what outcome would be seen if the particle were measured at any of three detector settings? So that the observer could say something like "this particle's local properties ensure it is predetermined to give +1 if measured at angle a, -1 if measured at angle b, and -1 if measured at angle c"? And if both particles are always found to give identical observed results when measured at the same angle, would you agree this implies (under local realism) that for each pair emitted by the source, their local properties must be correlated in such a way as to ensure that one particle must have the same three predetermined results as the other one?

Sure; why not?

 

[JesseM]

Sure; why not?

Because this line of argument leads inevitably to Bell inequalities, as I and others have been trying to explain to you since you started posting here. Suppose we have some large number of particle pairs, from the above you should agree that in each pair, the two particles should have some definite set of predetermined results like [+ on a, - on b, + on c] or [- on a, +on b, + on c] etc.? And for any collection of things (like particle pairs) where each member of the collection either does or doesn't have each of three possible properties A, B, and C (say A=+ on angle a, B=+ on angle b, C=+ on angle c, so "not A"=- on angle a, "not B"=- on angle b, and "not C"=- on angle c), simple arithmetic shows the whole collection must satisfy this inequality:

Number(A, not B) + Number(B, not C) ≥ Number(A, not C)

There's a proof on this page, but I think their proof is not as simple as it could be, the simplest way of seeing it is this:

Number(A, not B) = Number(A, not B, C) + Number(A, not B, not C) [since any member of the group satisfying A, not B must either have or not have property C]
Number(B, not C) = Number(A, B, not C) + Number(not A, B, not C)
Number(A, not C) = Number(A, B, not C) + Number(A, not B, not C)

And plugging this into the above inequality and cancelling like terms from both sides gives:

Number(A, not B, C) + Number(not A, B, not C) ≥ 0

Which obviously must be true since the number with any given set of properties must be ≥ 0!

Anyway, whether you like my proof or the one on the page I linked to better, hopefully you agree that if we knew the complete set of three predetermined properties for a collection of particle pairs, the inequality Number(A, not B) + Number(B, not C) ≥ Number(A, not C) would be satisfied? If so, it's a short step from there to the statement that if you measure two properties for a large number of particle pairs, P(A, not B|measured a and b) + P(B, not C|measured b and c) ≥ P(A, not C|measured a and c) (basically the only extra assumption needed is that the probability the experimenters will pick a given pair of axes to measure is uncorrelated with the triplet of predetermined results prior to measurement). I discuss this more on post #11 here, but we can also discuss it here if you agree with the inequality Number(A, not B) + Number(B, not C) ≥ Number(A, not C) for all particle pairs but don't agree that for measurements this implies P(A, not B|measured a and b) + P(B, not C|measured b and c) ≥ P(A, not C|measured a and c).

 

[JesseM]

Bell's inequality derivation relies on the fact that they originate from THE SAME ensemble, therefore you cannot use three different ensembles and expect them to just work.

Well, you can if you have a single ensemble of particle pairs, and then for each pair you choose which combination of properties to measure using a rule that is statistically uncorrelated with the hidden properties of each pair--the "no-conspiracy assumption" which you seem to have forgotten about. Read my post #11 on this thread for more on this point, and consider the following part in particular:

If you disagree, think of it this way. Suppose we generate a hypothetical list of the predetermined values for each in a series of N trials, where N is fairly large, say N=100, like this:

trial #1: [A, B, C]
trial #2: [A, not-B, not-C]
trial #3: [not-A, B, not-C]
trial #4: [A, B, not-C]
...
trial #100: [A, not-B, not-C]

You can use any algorithm you want to generate this list, including one where you pick the values for each trial based on a probability distribution for all 8 possible combinations, and the probability distribution itself changes depending on the number of the trial (equivalent to De Raedt's notion that the probability distribution for lambda might be time-dependent). Anyway, once you have the list, then select which two the imaginary experimenters are going to sample using a rule that is random with respect to the actual set of predetermined values on that trial--for example, you could use this random number generator with Min=1 and Max=3, and then on each trial if it gives "1" you say that the measurement was a,b, if it gives "2" you say the measurement was b,c, and if it gives "3" you say the measurement was a,c. I would say that regardless of what algorithm you chose to generate the original list of predetermined values, the fact that the choice of which values were sampled on each trial was random ensures that if the number of entries N on the list is large, the probability is very small that you'll get a violation of the inequality above involving measured subsets. Would you disagree with that?