Bell experiment would somehow prove non-locality and information FTL?

[NateTG]

That's not the cleanest notation. The way you wrote it, A could be some kind of uniary set operation. It would be better to do something like:
$$H \equiv \{+,-\}^{\{A,B,C\}}$$
(which I was tempted to put into the post above which discusses equivalent sets.)

 

[NateTG]

2. A minor quibble : observing 2 of 3 settings is not the "fair sampling" assumption.

Apparently the earlier post was unclear:
Let's assume (for the sake of discussion) that we have a local hidden variable theory - so at the point where our EPR particles split, they have hidden states $\vec{h}$ and $-\vec{h}$. Where $\vec{h} \in H$ and
$$H=\{<-,-,->,<-,-,+>,<-,+,->,<-,+,+>,<+,-,->,<+,-,+>,<+,+,->, <+,+,+>\}$$

Now, we know that $h \in H$ with probability $1$, and that $h \in \null$ with probability $0$. Stipulating the fair sampling assumption, we can experimentally determine that the probability that $h \in \{<+,-,->,<+,-,+>,<+,+,->,<+,++>\}$ some real number $p_{<+,*,*>} \in [0,1]$ by measuring particle $\alpha$ directly, and the probabilities for the other 5 symetric subsets can also be tested.
Similarly, it's possible, by measuring both particles to determine the probability $p_{<+,+,*>}$ that $h \in \{<+,+,->,<+,+,+>\}$ or any of the other 11 subsets symetric to this one.

However, it's impossible to measure $p_{<+,+,+>}$, that is, the probability that $h \in \{<+,+,+>\}$, experimentally, but Bell's theorem assumes that it is a well defined real value i.e. that $p_{<+,+,+>} \in [0,1]$.

In a classical regime, where it's possible to make non-perturbing measurements, it's trivially possible to measure the concurrence of two things, so coincidence probabilities are well-defined, which validates the assignment of probabilities to these singletons. However, in a quantum setting, it's necessary to assume, without experimental justification, that these singleton probabilities exist and are well-defined for Bell's Theorem to be valid.

 

[JesseM]

However, it's impossible to measure $p_{<+,+,+>}$, that is, the probability that $h \in \{<+,+,+>\}$, experimentally, but Bell's theorem assumes that it is a well defined real value i.e. that $p_{<+,+,+>} \in [0,1]$.

But in a classical world, can't you assume that everything really is in a single definite state at all times, even if you can't measure that state? It's not necessary for the proof that we actually know the probability of these states, just there is some definite probability, which could be known by an imaginary omniscient observer who knows the complete state of the universe at every moment.

This criticism would make more sense if you were directing it at some specific hidden-variables theory which claimed to be a valid theory of physics, but since Bell's theorem is just trying to show that local hidden variable theories can't work, I don't think it's a weakness of the argument that even if they did "work" in the sense of not being ruled out a priori by violations of Bell inequalities, they still wouldn't "work" as good theories of physics since they would involve quantities which could never be measured by experiment.

 

[NateTG]

But in a classical world, can't you assume that everything really is in a single definite state at all times, even if you can't measure that state? It's not necessary for the proof that we actually know the probability of these states, just there is some definite probability, which could be known by an imaginary omniscient observer who knows the complete state of the universe at every moment.

Although, for the usual notion of probability, the coincidence of two states with well-defined probabilities always has a well-defined probability; the assumption of a definite state does not infer a well-defined probability for that state.

For example, under the usual notions, the probability that a random real number chosen from the interval $[0,1]$ is less than $\frac{1}{2}$, provided that it is a rational number is undefined. (This is effectively a division by zero.) On the other hand, the probability that a random real number from $[0,1]$ is rational and less than $\frac{1}{2}$ is zero.

 

[heusdens]

Well, if you're trying to contradict Bell's theorem with a classical example, there has to be a spacelike separation between measurement-events, because that's one of the conditions the theorem requires in order to guarantee the Bell inequalities are not violated classically.

Well, my "experiment" does not take place in a spatial organized way, it is just a very abstract experiment (Gedanken experiment). For that reason probably it could show to conflict the Bell inequality, but for that same reason does not qualify as a "real" experiment, that could somehow break the inequality with a classical set-up.

But your answer doesn't make much sense to me in any case--are you sure you understand the meaning of the term "spacelike separation" in physics? If two events have a spacelike separation, all that means is that it would be impossible for a signal have left from the place and time of one event and reached the place and time of the other event, assuming the signal could not travel faster than light. So, for example, if I make a measurement on earth, and 1 year later you make a measurement 1.5 light-years away from Earth (as measured in some inertial reference frame, like the Earth's rest frame), it would be impossible for news of the result of my measurement to have reached you by the time you made your measurement, so these two measurement-events have a spacelike separation. On the other hand, if you made a measurement 1 year later but only 0.8 light-years away, you could have already learned the result of my measurement by the time you made yours (so you might adjust your detector setting based on that result), so in this case the measurement-events do not have a spacelike separation, instead they have a timelike separation.
If the source sends the streams at the speed of light in opposite directions, and Alice and Bob both make their measurements at the same time but different locations, this is enough to guarantee the two measurement-events have a spacelike separation. Does this clarify things?

Yes, I'm aware that in any real setup (wether classical or quantum) this is of importance; I'm aware of the difference between space-like and time-like seperations in SR, which has to do with the cone of worldlines; if A and B are on points well within (and including) each other cones, they are time-like separated, and if not, they are space-like separated.

Sorry, I thought you were talking about streams of information, which can be represented as a string of digits.

The actual stream of "information", what is encoded in the elements of the stream, might be anything that can be expressed mathematically (if it can be shown that anything that is expressable mathematically can be represented in digits in a unique way, then that may well be the case).

OK, but the question is, given that Alice has just received a particular set of elements from the source, and that her detector is set to 3, is that enough to completely determine whether she gets a + or a -, or is there an additional random aspect, so that even if you had two trials on which the source sent Alice an identical set of elements, and Alice had her detector set to 3 on both trials, she might get + on one trial and - on the other? If there were any randomness, then it seems you can't guarantee that each time Bob and Alice have the same detector setting they'll get the same answer, even if the source sent an identical set of elements to each one.

There could not likely be two trials in which the source sends out the same stream, it would be very improbable (would depend on the sample size, of course). And further, the detector settings determine what part of the stream actually get detected.

I think you misunderstood what I was saying, I didn't say that the source *does* know the settings in advance, the only assumption I'm making above is that once the source has sent the streams to Alice and she has chosen her setting, there is no additional random aspect (see my earlier comments)--either the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a +, or the streams have a combination of elements that make it guaranteed that if she chooses setting 2 she'll get a -. There are no combinations of elements that the source sends out such that if she chooses setting 2, she has a 70% chance of getting + and a 30% chance of getting a -; knowing the combination and her choice of detector setting completely determines the results (but of course, Alice herself does not know the combination, and we don't need to either, all we need to know is that any given combination of elements would have only a single possible outcome for each possible detector setting). Do you disagree with this?

Yes, but the stream contains more information then the correlated results.
Only, depending on detector settings, not all that information shows up in the results when measured. This is the crucial point.

But to this point, I think it is not worth discussing any more since:
a. This abstract model misses an aspect to make it a sensible "classical" setup which is in accordance with the prescriptions (space-like separation of the detectors).
b. I have not come up with an actual "mechanism" for explaining the behaviour of the "experiment", but just assumed such a "mechanism" could in theory exist (even if we were just providing the adequate mathematical expressions)

So, until I have come up with something to fix these omissions, we are not discussing something concrete, and we end up mere speculating.
If Bell's prediction is right of course we could not do it. I have not shown this wrong, and perhaps I can't.

 

[DrChinese]

Yes, but the stream contains more information then the correlated results.

Actually, a stream of entangled photons does not contain any information at all. Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)

 

[JesseM]

Although, for the usual notion of probability, the coincidence of two states with well-defined probabilities always has a well-defined probability; the assumption of a definite state does not infer a well-defined probability for that state.

For example, under the usual notions, the probability that a random real number chosen from the interval $[0,1]$ is less than $\frac{1}{2}$, provided that it is a rational number is undefined.

Isn't this a consequence of having an infinite number of possible outcomes, with "rationality/nonrationality" dependent on knowing an infinite number of digits for every outcome? In this situation it's not clear that the idea of picking a "random real" between 1 and 0, with each outcome being either definitely rational or definitely non-rational, makes sense in the first place...it would depend on your views on some tricky philosophy-of-math issues, I think. On the other hand, in the proof of Bell's theorem it can be assumed that every state emitted by the source must either fall into the category of "would yield a + if measured on detector setting 2" or "would yield a - if measured on detector setting 2", since we know that whenever both experimenters do use detector setting 2 they always get definite opposite answers, and the source has no way of knowing which trials the experimenters will use which detector settings in advance.

So if we assume that on each trial the hidden variables were definitely either of type +2 or -2, then there must be some definite ratio (number of trials where state was of type +2)/(total number of trials), and presumably the ratio would have to approach a single value in the limit as the number of trials goes to infinity. Even if there was a physical process which as an outcome generated an object with an infinite number of properties which could be encoded as a string of 1's and 0's (and with each individual property being equally like to be 1 or 0), and if the experimenter would get a + on setting 2 whenever this string was rational and a - on setting 2 whenever it was irrational, the ratio of one type of outcome to another would have to be well-defined for any given series of trials...again, the problem is that we can't be sure if it even makes sense to imagine an observable measurement whose outcome is dependent on whether an infinite random string is rational or irrational, it's a bit like imagining an observable measurement whose outcome is dependent on whether the continuum hypothesis is true or false (a question which may not even have an objective answer depending on your views of the philosophy of mathematics).

edit: Also, since the set of rational numbers between 0 and 1 is countable, the notion of randomly picking a rational number between 0 and 1 should be equivalent to the notion of picking a random integer from 0 to infinity, with each one equally likely. This clearly leads to nonsensical conclusions; for example, if you pick two such numbers randomly and put each one in an envelope, then open one and look at it, you should conclude that there is a 100% probability the second envelope will contain a larger number, regardless of which envelope you opened first, since there are an infinite number of integers larger than the first one you looked at and only a finite number smaller. Indeed, in probability it is forbidden to have a uniform distribution on all the numbers from 0 to infinity.

 

[MeJennifer]

Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

Actually we cannot claim that at all, that is completely wrong.

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

 

[DrChinese]

Actually we cannot claim that at all, that is completely wrong.

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

Huh? (Maybe we are talking about 2 different things...)

Of course the polarization of Bell test photon pairs is demonstrably random. Let's consider Type II PDC, for example. One member of the pair is vertically polarized, the other member is hortizonally polarized. You simply never know which is which. Because of that, it is not possible to encode any information in the stream.

Now you might believe there is a "prior" cause that we simply do not know about, and that might seem a reasonable hypothesis at first glance. Except for one minor detail. The correlated polarization statistics for photon pairs that are not entangled are different that ones that are entangled. IF there were a prior "cause" that we could potentially tap into - and therefore use as a mechanism for encoding a bit of information - then the pair would no longer be represented by a single wave function. It would now be represented by 2 separated wave functions. Doesn't matter if we actually look at the prior cause or not. IF we could have, then that would be enough to eliminate the entangled state.

If a pair of photons is polarization entangled, then the polarization is fully random (to whatever degree of randomness you choose to measure) and NO information can be encoded in the stream.

QED.

 

[MeJennifer]

Huh?

Of course the polarization of Bell test photon pairs is demonstrably random. Let's consider Type II PDC, for example. One member of the pair is vertically polarized, the other member is hortizonally polarized. You simply never know which is which. Because of that, it is not possible to encode any information in the stream.

Now you might believe there is a "prior" cause that we simply do not know about, and that might seem a reasonable hypothesis at first glance. Except for one minor detail. The correlated polarization statistics for photon pairs that are not entangled are different that ones that are entangled. IF there were a prior "cause" that we could potentially tap into - and therefore use as a mechanism for encoding a bit of information - then the pair would no longer be represented by a single wave function. It would now be represented by 2 separated wave functions. Doesn't matter if we actually look at the prior cause or not. IF we could have, then that would be enough to eliminate the entangled state.

If a pair of photons is polarization entangled, then the polarization is fully random (to whatever degree of randomness you choose to measure) and NO information can be encoded in the stream.

You seem to mixup the question if nature is fundamentally random with the ability for us to determine that.

QED.

The only QED we have here is that you cannot admit you made a mistake.

 

[NateTG]

...So if we assume that on each trial the hidden variables were definitely either of type +2 or -2, then there must be some definite ratio (number of trials where state was of type +2)/(total number of trials), and presumably the ratio would have to approach a single value in the limit as the number of trials goes to infinity.

That what you'd think, but why? And, how does that compare with the other possibilities?

 

[wm]

Actually, a stream of entangled photons does not contain any information at all. Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information.

(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)

DrC, I'm wondering if the wording here is confusing?

1. Is it not the case that each PAIR is in an identical (random) state? Such that a test on one twin reveals information about a property of the other (correlated) twin; that is, information about how the untested twin will respond to a specific test?

So the stream does encode information; though that information is not encoded by us?

2. Also: It seems to me that you learn about the detector settings later on; by directly comparing detector settings. Not by comparing test results?

So other information that you learn later on (when the test-results are correlated) is about each twinned-pair's response to each specific pair of detector settings?

wm

 

[DrChinese]

You seem to mixup the question if nature is fundamentally random with the ability for us to determine that.


The only QED we have here is that you cannot admit you made a mistake.

LOL.

I didn't say nature was fundamentally random, because no one knows this for a fact. I said that all of the following are true:

1. It is not possible to encode any information in a polarization entangled pair of photons. This statement was a response to a comment by heusdens. If you know a way to do this, please let me know.

2. It is impossible to distinguish the polarization values of a set of such photons from that of a truly random source. This makes it clear that your statement is merely conjecture, as the evidence points the other way. IE if it walks like a duck, and quacks like a duck, it is truly random.

3. It is possible to distinguish a stream of polarization entangled pairs of photons from a stream which is not polarization entangled. This is a simple consequence of experiment, and goes to show that in fact there is no information (i.e. cause) that we are simply "missing" (because if there were, the entanglement would immediately disappear).

You are welcome to dismiss the science of this, if you prefer to operate at a semantical level.

 

[DrChinese]

DrC, I'm wondering if the wording here is confusing?

1. Is it not the case that each PAIR is in an identical (random) state? Such that a test on one twin reveals information about a property of the other (correlated) twin; that is, information about how the untested twin will respond to a specific test?

So the stream does encode information; though that information is not encoded by us?

2. Also: It seems to me that you learn about the detector settings later on; by directly comparing detector settings. Not by comparing test results?

So other information that you learn later on (when the test-results are correlated) is about each twinned-pair's response to each specific pair of detector settings?

wm

You cannot transmit information in such a stream (at least using an entangled attribute).

1. If the polarization values are NOT assigned by us, how would we encode anything? heusdens was commenting - really a side comment as I read it - that there was information in the stream but all of it was not usable. I was merely making a point that such a stream consists of random values. To encode information, you would choose to polarize a photon as vertical or horizontal and then establish bit values (0 or 1) to correspond with what the receiver detects. If you try to do this, you will NOT have polariztion entanglement. This is both per theory and per experiment.

2. Sure, you learn about the detector settings when you compare the results later. In fact, I specifically said that... "(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)" So no disagreement there.

Hopefully, we are on the same page.

 

[MeJennifer]

It is impossible to distinguish the polarization values of a set of such photons from that of a truly random source. This makes it clear that your statement is merely conjecture, as the evidence points the other way. IE if it walks like a duck, and quacks like a duck, it is truly random.

My statement merely conjecture?
It seems I have to remind you who actually made what statement.

Your statement:

"Each member of each pair is in a purely random state of polarization, and so by definition it cannot contain any information. "

I answered:

We simply do not know how those states of polarization are determined, it could be random but that is certainly not the only option.

We cannot determine how they get their values, but what we can do is repeat the same experiment and make predictions based on the statistics. But in no way does that prove that each experiment is random.

And now my statement is supposedly "merely conjecture"
Anyway it demonstrates my prior point, you can't admit you made a mistake and furthermore when caught with one you attempt to wiggle yourself out of it.

Anyway there seems to be no point in cluttering this topic any further. Feel free to PM me.

 

[wm]

You cannot transmit information in such a stream (at least using an entangled attribute).

1. If the polarization values are NOT assigned by us, how would we encode anything? heusdens was commenting - really a side comment as I read it - that there was information in the stream but all of it was not usable. I was merely making a point that such a stream consists of random values. To encode information, you would choose to polarize a photon as vertical or horizontal and then establish bit values (0 or 1) to correspond with what the receiver detects. If you try to do this, you will NOT have polariztion entanglement. This is both per theory and per experiment.

2. Sure, you learn about the detector settings when you compare the results later. In fact, I specifically said that... "(The pairs themselves of course have an interesting property because they are entangled, which does allow you to learn about the detector settings when the results are correlated later on.)" So no disagreement there.

Hopefully, we are on the same page.

DrC, it seems that we are not yet on the same page re this small matter.

1. I did NOT say that we can transmit information is such a stream.

Rather: Accepting the existence of the twin-streams; one stream correlated with the other ... then, from our interrogation of one stream, we learn something [ = information] about the other stream.

This correlative information was encoded in the streams by virtue of the creation of each pair (each set of twins) in the (highly-correlated) singlet-state.

2. Now BIG disagreement/misunderstanding here: It is by NO property of the PAIRS that we learn of the detector settings. We learn that by direct observation of each setting; or by direct communication; IRRESPECTIVE of the PAIRS.

I'm a bit surprised at our entanglement here, wm

 

[wm]

No, because the stellar light is still a collection of photons. You don't know the polarization of any particular one, just as you don't know the polarization of individual photons from a collection of entangled particles in a Bell test.

Clarification of terminology, please:

Question: Is there such a thing as an unpolarised photon?

OR: When we see the phrase (in physics papers) ''unpolarised photon''; this is always to be interpreted as ''photon with polarisation unknown''?

Thanks, wm

 

[DrChinese]

Anyway there seems to be no point in cluttering this topic any further.

The door is that way, for those that want to make an unwarranted assertion that they prefer not to defend...

 

[JesseM]

That what you'd think, but why? And, how does that compare with the other possibilities?

Well, you'd agree that for any finite number of trials, there must be a definite ratio, right? It seems to me that the only way that the ratio would not converge to a single value as the number of trials approached infinity is if there were some time-dependence in the hidden variables, like if the first 10 were 2+, the next 100 were 2-, the next 1000 were 2+, the next 10000 were 2-, and so forth. But if there were any time-dependent pattern, then when using the same measurement setting on trial after trial you would notice some sort of pattern in the +'s vs. the -'s, it wouldn't look like a completely random string.

Anyway, it seems to me that the proof of Bell's theorem does not actually depend on saying anything about the probabilities of different hidden states, just that there is a definite hidden state on each trial, and that there is no statistical correlation between the choice of detector settings and the hidden state. Suppose each hidden state must definitely either be of type 1+ (meaning if you use detector setting 1 you're guaranteed to get a + with that hidden state) or 1-, and likewise each hidden state must definitely be of type 2+ or 2-, and must definitely either be of type 3+ or 3-. In this case we can make the following statement about the number of trials on which the hidden state is of one type or another, a statement which has nothing to do with probabilities:

Number(state of type 1+ and 2-) + Number (state of type 2+ and 3-) greater than or equal to Number(state of type 1+ and 3-)

This has to be true, because on every trial where the state was of type 1+ and 3-, it must have been either of type 2+ or 2-, and if it was of 2- this would also add one to Number(state of type 1+ and 2-), while if it was of type 2+ it would add one to Number(state of type 2+ and 3-). So, every trial which adds one to the number on the right side of the inequality must add one to the number on the left side to, showing the left side must be greater than or equal to the right side.

From this non-probabilistic statement, it shouldn't be too hard to show that if the choice of detector settings is random on each trial and is uncorrelated with the hidden state, and Alice and Bob always get the same result when they choose the same detector setting, this leads naturally to the conclusion that

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

 

[DrChinese]

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

 

[NateTG]

Well, you'd agree that for any finite number of trials, there must be a definite ratio, right? It seems to me that the only way that the ratio would not converge to a single value as the number of trials approached infinity is if there were some time-dependence in the hidden variables, like if the first 10 were 2+, the next 100 were 2-, the next 1000 were 2+, the next 10000 were 2-, and so forth. But if there were any time-dependent pattern, then when using the same measurement setting on trial after trial you would notice some sort of pattern in the +'s vs. the -'s, it wouldn't look like a completely random string.

Assuming local realism, any finite number of trials should, indeed, produce a ratio. However, for a non-stochastic process, the central limit theorem does not apply, and, as a consequence, there is no confidence interval.

Now, you'd think leads to a problem because you could have one person observing at one frequency (say, every trial) and another person observing at a different frequency (like pairs of trials), leading to a contradiction, but the quantity we're talking about is unobservable.

Anyway, it seems to me that the proof of Bell's theorem does not actually depend on saying anything about the probabilities of different hidden states, just that there is a definite hidden state on each trial, and that there is no statistical correlation between the choice of detector settings and the hidden state.

The process for deriving Bell's Inequality involves adding or subtracting the probabilities for hidden states which is only sensible if those probabilities are well defined. I would be very interested and surprised to see any derivation of Bell's Inequality that only involves experimentally measurable probabilities.

Number(state of type 1+ and 2-) + Number (state of type 2+ and 3-) greater than or equal to Number(state of type 1+ and 3-)

Sure.

From this non-probabilistic statement, it shouldn't be too hard to show that if the choice of detector settings is random on each trial and is uncorrelated with the hidden state, and Alice and Bob always get the same result when they choose the same detector setting, this leads naturally to the conclusion that

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

With the usual assumptions about experimental bias and detector choices, doesn't that work out to:
$$\frac{1}{16}+\frac{1}{16} \geq \frac{1}{16}$$

Which hardly seems contradictory to me.

 

[JesseM]

Assuming local realism, any finite number of trials should, indeed, produce a ratio. However, for a non-stochastic process, the central limit theorem does not apply, and, as a consequence, there is no confidence interval.

Now, you'd think leads to a problem because you could have one person observing at one frequency (say, every trial) and another person observing at a different frequency (like pairs of trials), leading to a contradiction, but the quantity we're talking about is unobservable.

Can you give an example of a non-stochastic process or algorithm that would generate a series of + and - results, where the pattern of results shows no time-dependence, and yet the ratio of +'s to -'s would not approach some fixed ratio as the number of trials goes to infinity? I'm having trouble imagining how this would work.

The process for deriving Bell's Inequality involves adding or subtracting the probabilities for hidden states which is only sensible if those probabilities are well defined.

I don't think it necessarily does--as I suggested, you can just talk about the number of different hidden states in a large set of trials, and then the only probabilistic assumption you need is the assumption that the choice of detector settings on each trial is random and there is no correlation between the detector setting and the hidden state emitted by the source.

Probability(Alice chooses setting 1, gets + and Bob chooses setting 2, gets -) + Probability(Alice chooses setting 2, gets + and Bob chooses setting 3, gets -) greater than or equal to Probability(Alice chooses setting 1, gets + and Bob chooses setting 3, gets -)

With the usual assumptions about experimental bias and detector choices, doesn't that work out to:
$$\frac{1}{16}+\frac{1}{16} \geq \frac{1}{16}$$

Which hardly seems contradictory to me.

Can you explain how you got the 1/16 figure? What are the "usual assumptions" about detector settings? It may be that for the detector settings used in some other common version of the experiment, this inequality would not be violated, but you can violate it with the right combination of detector settings--see the section on "Applying Bell's inequality to electron spin" in this article on Bell's theorem.

 

[NateTG]

Can you give an example of a non-stochastic process or algorithm that would generate a series of + and - results, where the pattern of results shows no time-dependence, and yet the ratio of +'s to -'s would not approach some fixed ratio as the number of trials goes to infinity? I'm having trouble imagining how this would work.

It's somewhat ugly.
Consider the interval $[0,1]$.
For example, let's say we have an evenly distributed random sequence $x_n$ on $[0,1]$. Now, for $A$, some non-measurable subset of $[0,1]$, define $y_n$ as follows:
$$y_n=+ \rm{if} x_n \in A$$
$$y_n=- \rm{otherwise}$$
Now, while a particular $x_n$ and $A$ might lead to a $y_n$ that has a limit frequency, that's not provably the case.

Can you explain how you got the 1/16 figure? What are the "usual assumptions" about detector settings?

I had a brain fart - I was thinking the settings were orthogonal.

 

[wm]

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

Regards, wm

 

[DrChinese]

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

Regards, wm

You'll need to provide some context. How does that quote relate to your original question? You say it is a definition, what is it a definition of? What is the thing you are really asking?

Polarized light has a simple property: it will pass through a polarizer set at an appropriate angle with nearly 100% intensity. Unpolarized light has a 50% chance of passing through that same polarizer.

 

[wm]

You'll need to provide some context. How does that quote relate to your original question? You say it is a definition, what is it a definition of? What is the thing you are really asking?

Polarized light has a simple property: it will pass through a polarizer set at an appropriate angle with nearly 100% intensity. Unpolarized light has a 50% chance of passing through that same polarizer.

Doc, I was seeking to understand the nature of single photon polarisation. You supplied some definitions and I then asked how this piece from a text-book fitted with your definitions.

So my question was:

How does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

PS: My own understanding is that paired photons in the singlet state are not polarised and that is why the singlet-state has spherical symmetry. (Which is a far greater symmetry than symmetry about the line-of-flight axis.)

Would ''unpolarised'' photons per your definition yield this spherical symmetry?

Thanks, wm

 

[DrChinese]

1. Doc, I was seeking to understand the nature of single photon polarisation. You supplied some definitions and I then asked how this piece from a text-book fitted with your definitions.

2. PS: My own understanding is that paired photons in the singlet state are not polarised and that is why the singlet-state has spherical symmetry. (Which is a far greater symmetry than symmetry about the line-of-flight axis.)

Would ''unpolarised'' photons per your definition yield this spherical symmetry?

Thanks, wm

wm,

1. Your question probably makes perfect sense to you. But I have no idea what the quote is supposed to represent as an idea. You were asking about polarized vs. unpolarized, and you introduced this quote into the picture. I have no idea how it relates. Is it supposed to be some kind of definition? You know what you have in your mind, but I don't and so you should explain this more clearly.

2. Entangled photon pairs - the kind used for Bell tests - are unpolarized. I am not sure what you mean when you say "spherical symmetry". I guess they are, seems reasonable, and the individual members certainly exhibit a strong symmetry in every respect that I am aware of - when considered as a pair. I have never heard this term used with entanglement though, so I am trying to understand how it became a part of the discussion.

 

[wm]

wm,

1. Your question probably makes perfect sense to you. But I have no idea what the quote is supposed to represent as an idea. You were asking about polarized vs. unpolarized, and you introduced this quote into the picture. I have no idea how it relates. Is it supposed to be some kind of definition? You know what you have in your mind, but I don't and so you should explain this more clearly.
<SNIP>

Doc, I was seeking to understand the nature of photon polarisation; in essence, to understand your view of Bell's theorem. For we read (in the QM literature) of polarised and unpolarised photons.

YOU then supplied the following definitions (emphasis added):

wm,

There are the following definitions:

Polarized photon: polarization known (or could have been known).

Unpolarized photon: polarization not known (and could not have been known). In the terminology of QM, and following the Heisenberg Unvertainty Principle, the polarization is not definite until observed. You do NOT need to accept this explanation to use the term "unpolarized" as the term merely means that no polarization observation has been performed yet.

Polarization entangled photon: an unpolarized photon that shares a wave function with one or more particles.

Then I replied:

DrC, thanks for this. BUT I still find it confusing.

For example, how does the following fit with your definitions, please?

Gottfried and Yan (Quantum Mechanics: Fundamentals 2nd Edition, 2003; page 447): 'Hence the polarization of an arbitrary one-photon state of a given momentum is specified by a real three ''vector'' E ... and it has no polarization whatsoever when E = 0.' (Emphasis added.)

THAT IS: Your definition indicates that an unpolarised photon is such that ''unpolarised'' may be taken to mean that no polarization observation has been performed yet.

The textbook appears to be discussing a photon state with no polarisation whatsoever.

SO I ask:

1. How does the textbook sit with your definition?

That is: How does the textbook's ''no polarisation whatsoever'' sit with YOUR definition that an unpolarised photon is one on which ''no polarisation observation has yet been made''?

2. How does your definition sit with common-sense?

ESPECIALLY given that I can send you photons in a definite state of polarisation; photons on which ''no polarisation observation has yet been made'' by anyone. That is: Definitely polarised photons are, by your definition, unpolarised because neither YOU (nor anyone else) have yet observed them??

THE LAST SENTENCE (appearing to be directly derived from you definition) APPEARS TO ME to be confusing; even crazy!?

Hope this helps, wm

 

[DrChinese]

Doc, I was seeking to understand the nature of photon polarisation; in essence, to understand your view of Bell's theorem. For we read (in the QM literature) of polarised and unpolarised photons.

...

ESPECIALLY given that I can send you photons in a definite state of polarisation; photons on which ''no polarisation observation has yet been made'' by anyone. That is: Definitely polarised photons are, by your definition, unpolarised because neither YOU (nor anyone else) have yet observed them??

THE LAST SENTENCE (appearing to be directly derived from you definition) APPEARS TO ME to be confusing; even crazy!?

Hope this helps, wm

I don't really follow what you are saying, because you seem to be mixing and matching words. The result doesn't match too well with the common way of expressing polarization.

I DO NOT claim that: an unpolarized photon has a polarization, but that we don't know what it is.

1. An unpolarized photon is in a mixed state (where H is horizontal and V is vertical):

H> + V>

2. A polarized photon is in a pure state:

H> (in whatever basis you choose to observe it)

3. A pair of entangled photons are also in a mixed state:

H>V> + V>H> (assuming Type II PDC)

As far as I know, it takes an observation to cause the mixed state to collapse to a pure state.

 

[calamero]

hello drchinese ,anonyn,everyone
think I've had an Einstein moment were everything becomes clear .but I am shaking and scared to say what i think BUT this idea I've just had when looking at why the wave collapses at dual slit I've hit upon a different way at looking at it and guess what ..here just now a minute ago i applied the same idea to that expirement wer how does the quanta know to show both red or both green etc when the switches at the collector points are set the same but random at other times its so simple an idea i don't know if anyones thought what I am thinking before but i havnt seen it so far and it all makes sense now i can explain why that happens or I've an idea why that happens and it makes sense to me ..im new here but I've wondered about that dual slit for ages now and past year i lie awake at night thinking about it so i have gave it a lot of thought ..ill sleep on this and think about it more in other situations before asking what you all think i just wanted to come here and say something..."heres hoping NOONE has had the same idea " because if they had and you lot arnt talking about it then it must have been crap :p

calamero