Where is the flaw with predetermined entanglement state?

[carllooper]

Bell doesn't make an assumption as such. He is introducing the hypothesis in question: the EPR hypothesis.

He will go on to test that hypothesis, rather than simply assume it.

Assumptions are like an hypothesis, but assumptions ask to be treated as if they were not in question.

C

 

[LsT]

@carlhooper
As I understand it, Bell clearly makes the assumption (not hypothesis) that in a local world the result must be predetermined to agree with QM. Upon this he continues, and in the end undoubtly "confines" this predetermined-result locality (aka local hidden variables) within testable limits. And local hidden variables is proved experimentally wrong by Bell's theorem, I have no doubts about this. But a locality without predetermined results is addressed in Bell's paper only through this assumption. (maybe just I am unaware of a proof of this assumption?)

 

[atyy]

As I understand it, Bell clearly makes the assumption (not hypothesis) that in a local world the result must be predetermined to agree with QM. Upon this he continues, and in the end undoubtly "confines" this predetermined-result locality (aka local hidden variables) within testable limits. And local hidden variables is proved experimentally wrong by Bell's theorem, I have no doubts about this. But a locality without predetermined results is addressed in Bell's paper only through this assumption. (maybe just I am unaware of a proof of this assumption?)

No, Bell's does not assume that a local hidden variables theory must agree with QM. Bell defines what he means by a local hidden variables theory, and shows that this entire class of theories cannot match the predictions of QM.

 

[Nugatory]

These are my problems with it -
1] To say it "has to be" is based on the same line of thought that says "it has to be" that velocity additions may exceed c.
...

Suppose I write down a long random sequence of the letters U and D (standing for "up" and "down"). I make two copies of this sequence on two separate sheets of paper, give one to you and keep one for myself. You change three randomly chosen letters on your sheet of paper and I change three randomly letters on mine. We then get back together, compare our lists.

You're not seriously suggesting that we might find more than six disagreements between our two lists, I hope? All we're doing is counting the number of places where your list doesn't match mine, and noting that everywhere that happens then one or both of our lists must also not match the original list.

 

[stevendaryl]

So, Bell does make such an assumption.

It's not an assumption, since it follows from the fact of perfect correlations. If something is provable from other assumptions, then it isn't an additional assumption.

 

[stevendaryl]

Ok that is what I am saying also. Probability $P_A$ does not depent on $\beta$ and vice versa, because $P_A = P_B = \frac{1}{2}$.

No, you don't get to assume that P_A(\alpha, \lambda) = P_B(\alpha, \lambda) = \frac{1}{2}. We don't measure the particle state \lambda. So the probabilities that we measure are averaged over all possible values of \lambda:

\int P(\lambda) P_A(\alpha, \lambda) d\lambda = \int P(\lambda) P_B(\beta, \lambda) d\lambda = \frac{1}{2}

If it's always the case that P_A = P_B = \frac{1}{2}, then you would always have that the probability of both Alice and Bob measuring spin-up would be:

P_A \cdot P_B = \frac{1}{4}

That isn't what we observe. Instead, we observe:
P_A \cdot P_B = \frac{1}{2} cos^2(\theta)

where \theta is the angle between Alice's setting and Bob's setting. (In the spin-1/2 EPR experiment, the prediction is \frac{\theta}{2}

 

[stevendaryl]

@carlhooper
As I understand it, Bell clearly makes the assumption (not hypothesis) that in a local world the result must be predetermined to agree with QM. Upon this he continues, and in the end undoubtly "confines" this predetermined-result locality (aka local hidden variables) within testable limits. And local hidden variables is proved experimentally wrong by Bell's theorem, I have no doubts about this. But a locality without predetermined results is addressed in Bell's paper only through this assumption. (maybe just I am unaware of a proof of this assumption?)

I sketched the proof. If the result is not predetermined, but is instead probabilistic, then you could not have perfect correlation in the case where Alice and Bob choose the same setting.

 

[stevendaryl]

These are my problems with it -
1] To say it "has to be" is based on the same line of thought that says "it has to be" that velocity additions may exceed c.
2] To say "That's just counting" is similar to insisting that velocity addition is an arithmetic sum and not bounded by c.
3] Ever since I first learned that velocity addition is not arithmetic I have kept a suspecting watch for similar assumptions.
4] To say "it has to be" does not agree with the experimental facts of the cosine squared curve; how can an argument presented for logical evaluation and hypothesis testing be justified when it is known up front from experiment to be incorrect?
5] All the above suggests to me that the logical math accounting for the number of mismatches is flawed, or incomplete, or missing something critical.

It seems to me that you are arguing at cross-purposes to Bell. He's not saying that there is no way to reproduce the EPR correlations. There certainly is: QM. What he's saying is that there is no way to reproduce the correlations using a classical notion of local state information propagating no faster than the speed of light.

The comparison with SR would be: There is no way for the speed of light to be the same in every reference frame if we are to use the classical notion of velocity addition. That's true. And so is what Bell claimed. He's saying that there is no classical, local, hidden-variables explanation. He's not saying that there is no explanation.

 

[LsT]

No, Bell's does not assume that a local hidden variables theory must agree with QM.

We agree, because I have never said that.

It's not an assumption, since it follows from the fact of perfect correlations. If something is provable from other assumptions, then it isn't an additional assumption.

You said that, I am saying otherwise, the question is I am not getting a proof of this (continue reading to see why I insist to this)

No, you don't get to assume that PA(α,λ)=PB(α,λ)=12P_A(\alpha, \lambda) = P_B(\alpha, \lambda) = \frac{1}{2}. We don't measure the particle state λ\lambda. So the probabilities that we measure are averaged over all possible values of λ\lambda:
?P(λ)PA(α,λ)dλ=?P(λ)PB(β,λ)dλ=12\int P(\lambda) P_A(\alpha, \lambda) d\lambda = \int P(\lambda) P_B(\beta, \lambda) d\lambda = \frac{1}{2}

You are probably right, but I am not testing a general model here, I just test a specific hypothesis, so I can definitely say that $P_A = P_B = \frac{1}{2}$ just because it is experimental fact, or this is not the case? If it is, how you claim to disprove it?. (btw, how you quote latex correctly ??)

If it's always the case that PA=PB=12P_A = P_B = \frac{1}{2}, then you would always have that the probability of both Alice and Bob measuring spin-up would be:
PA?PB=14P_A \cdot P_B = \frac{1}{4}

To derive to this result, you have to consider the probabilities independent. Otherwise you cannot do that.
(for example, you cannot do this if Alice and Bob are playing by standing at 180 deg. in a wheel of fortune)

I will show again, in my hypothesis, what happens at zero detectors angle difference:

A -> 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0
B -> 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0
C -> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Here probability of Alice's Detection is 1/2 as is Bob's (as is always). And the perfect correlation is a fact. The probabilities here are related of course.

But for the probabilities to be related in a local world, I am saying, it is not necessary to assume predetermined results for each detector.

I don't claim to have a formal proof for this, but I will try demonstrating it by the following thought experiment:

This thought experiment is not identical to particle entanglement of course, but it serves the purpose of demonstrating a perfect correlation between results of two distant, non-local detectors each having a probability outcome of 1/2, without any violation of locality, and without predetermined results per each detector (aka hidden variables). The setup is like this:

Alice and Bob are in separated sites. In each site there is a camera for taking photos. Each camera is looking horizontally. Each camera is mounted in a vertical axis rotating mechanism, such as that by the rotation of the mechanism, the camera stands in the perimeter of a circle and keeps pointing at the center of it, for the purpose of taking photos of around an object, that is placed in the center. Now make the rotation of this mechanism/camera independent in such a way that the two sites are considered non-local. Alice and Bob measure the angle of their mechanism, the time and keep the photo for each object that is going through the center of that circle.

For the purpose of simplicity, a hypothetical object is created (this does not affect the perfectness of the correlations):
This is a coin, with a zero width, such as when you take a photo of it, you have to see one (or the other) side.

In the center of the distance between Alice and Bob, we have another mechanism:

This mechanism, is throwing a pair of coins, each one to the direction of each rotating mechanism/camera center, at the speed of light (the speed of light it is not crucial, just a hypothesis to preserve the non-locality between the detectors and the coin), such as each coin is rotating in the vertical axis with the same speed, in the same direction, and with a very specific, constant angle difference that is preserved between them. Now, the angle that each coin has when it leaves the detector, is random, because let's say the mechanism is problematic. But their relative angle is constant and the speed of rotation and direction is the same

When each coin reaches the center of the photographic rotation mechanism, the camera takes a photo, in a random angle. Alice and Bob record all the results for X coin throws, and they compare their results.

The outcome is obvious: The probability of each is 1/2 and for any given angle difference between the mechanisms, the result cannot be but perfectly correlated.

 

[atyy]

But for the probabilities to be related in a local world, I am saying, it is not necessary to assume predetermined results for each detector.

I don't quite understand your question, but maybe it's related to this? http://arxiv.org/abs/1303.3081 (p10)

Proposition 2.1. A family of probability distributions P_X,Y can be explained with pre- established agreement if and only if it can be explained with deterministic local variables.

In other words, although local probabilistic hidden variables may seem be a more general concept than local deterministic hidden variables, for the purposes of deriving Bell's theorem for local probabilistic hidden variables, it is sufficient to derive it for local deterministic hidden variables.

 

[stevendaryl]

You said that, I am saying otherwise

Okay, but you're wrong.

 

[stevendaryl]

You are probably right, but I am not testing a general model here, I just test a specific hypothesis, so I can definitely say that $P_A = P_B = \frac{1}{2}$

But that's not true. If Alice has already observed a polarized photon at angle 0^0 then the probability of Bob measuring a photon polarized at angle 90^o is ZERO, not \frac{1}{2}. Probability \frac{1}{2} is the average over all possible runs, but each individual measurement doesn't have probability \frac{1}{2}

 

[stevendaryl]

To derive to this result, you have to consider the probabilities independent.

Okay, Bell elaborates in his "Theory of Local Beables" that his notion of locality assumes that if two probabilities for events are not independent, then it implies there is a common causal influence on both of them. If all causal influences propagate at lightspeed or slower, then that common causal influence is in their common backward lightcone. You can certainly reject this notion of causality, but that's really the whole point of Bell's theorem, is to show that the predictions of QM are inconsistent with a certain common-sensical notion of causality.

It doesn't do any good for you to say: Well, there might be some other alternative that he didn't think of. He wasn't trying to be exhaustive. He was laying out a particular type of model, and showing that QM can't possibly be that type of model. Once again, you're arguing at cross-purposes to Bell.

 

[stevendaryl]

(for example, you cannot do this if Alice and Bob are playing by standing at 180 deg. in a wheel of fortune).

That doesn't contradict Bell's assumptions---that's an EXAMPLE of the sort of hidden-variables theory that Bell was talking about. Alice and Bob's results in that case become independent once you take into account the position of the wheel.

 

[bahamagreen]

I was looking at various Bell presentations and in my searching I came across a 2008 paper by Andrei Khrennikov "Bell-Boole Inequality: Nonlocality or Probabilistic Incompatibility of Random Variables?" reviewing the history of mathematical probability assumptions of Bell type inequalities going all the way back to Boole's original development, showing that mathematicians have taken a somewhat different view of the probability assumptions in these inequalities than subsequent physicists, and suggesting a different interpretation of the experimental results with respect to locality.

From the Conclusion section
"...nonexistence of a single probability space does not imply that the realistic description (a map lamda ->a(lamda) is impossible to construct. Bell’s type inequalities were considered as signs (sufficient conditions) of impossibility to perform simultaneous measurement (of/on) all random variables from a family under consideration." (italics in original).

 

[LsT]

atyy that's interesting, maybe it answers my questions but first I have to read it (lol), thanx

But that's not true. If Alice has already observed a polarized photon at angle 000^0 then the probability of Bob measuring a photon polarized at angle 90o90^o is ZERO, not 12\frac{1}{2}. Probability 12\frac{1}{2} is the average over all possible runs, but each individual measurement doesn't have probability 12

$P_A$ is the probability that "Alice detects a photon"
The probability that "Alice detects a photon if somenthing", is only the same with $P_A$ if you consider that this "something" is independent of Alice's result. I am not doing that if "something" is the result of Bob.

Okay, Bell elaborates in his "Theory of Local Beables" that his notion of locality assumes that if two probabilities for events are not independent, then it implies there is a common causal influence on both of them. If all causal influences propagate at lightspeed or slower, then that common causal influence is in their common backward lightcone.

I am in agreement with this notion. In my hypothesis, each photon caries information from their past, but this information is a real state and only in relation to each other, not in relation to any detector.

It doesn't do any good for you to say: Well, there might be some other alternative that he didn't think of. He wasn't trying to be exhaustive. He was laying out a particular type of model, and showing that QM can't possibly be that type of model. Once again, you're arguing at cross-purposes to Bell.

No, this is my first time, last time was bahamagreen (not the same person) :)

And I am not arguing with anyone. In fact Bell seems very clear to me (emphasis mine):

VI. Conclusion
In a theory in which parameters are added to quantum mechanics to determine the results of individual measurements, without changing statistical predictions, there must be a mechanism whereby the setting of one measuring device can influence the reading of another instrument, however remote.(...)

That doesn't contradict Bell's assumptions---that's an EXAMPLE of the sort of hidden-variables theory that Bell was talking about. Alice and Bob's results in that case become independent once you take into account the position of the wheel.

(you mean dependent). Ok. I am not saying that the wheel of fortune is my hypothesis. That was just to demonstrate the previous sentence.

 

[stevendaryl]

$P_A$ is the probability that "Alice detects a photon"
The probability that "Alice detects a photon if somenthing", is only the same with $P_A$ if you consider that this "something" is independent of Alice's result.

That's a very silly distinction. Every probability is a conditional probability. To say that Bob has a probability of \frac{1}{2} is conditional on his having an appropriate photon detector, and there being an appropriate source of photons, and so forth. So you seem to be saying that conditional probability is not a probability? That's absurd.

 

[stevendaryl]

(you mean dependent).

No, I mean INDEPENDENT. That's why I said "independent". Alice's result is independent of Bob's result, once you've taken into account the "hidden variable", which is the position of the wheel.

Let A be Alice's result (either 1 if she detects a photon, or 1 if she doesn't). Let B be Bob's result. Let \alpha be Alice's setting. Let \beta be Bob's setting. Let \lambda be the value of the hidden variable (the position of the wheel, for instance).

So let P(A, B, \alpha, \beta, \lambda) be the probability that Alice will get result A and Bob will get result B, given her setting being \alpha and Bob's setting being \lambda and the hidden variable having value \lambda.

To say that Alice's results and Bob's results are statistically independent means that there are two functions P_A(A, \alpha, \lambda) and P_B(B, \beta,\lambda) such that P(A, B, \alpha, \beta, \lambda) = P_A(A, \alpha, \lambda) \cdot P_B(B, \beta,\lambda). That definitely is the case with the wheel of fortune example.

Now, if you average over \lambda, you will find that the averaged probabilities are NOT independent:

Let P_{av}(A, B, \alpha, \beta) = average over \lambda of P(A, B, \alpha, \beta, \lambda)

In general, the average probability P_{av} will not be factorable into P_{av, A}(A, \alpha) \cdot P_{av, B}(B, \beta). So if you ignore hidden variables, then the probabilities seem dependent, but that's because you threw away information.

For classical systems, it's always the case that probabilities becomes independent once you take into account common causal influences.

 

[stevendaryl]

I guess there is a question about whether Bell's factorizability condition on probabilities is an assumption, or a definition, or is provable from some more basic assumptions.

People are always reminded, whenever statistics show a correlation between A and B that "correlation doesn't prove causality". However, it is usually assumed that correlations are either spurious (that is, they are just artifacts of having insufficient data) or are due to an unknown causal influence affecting both A and B. If there is a correlation between, say, black hair and being lactose-intolerant, then people of course don't assume that black hair causes lactose-intolerance, but they assume that something caused both of them. Maybe having a non-European ancestry makes both black hair and lactose-intolerance more likely.

The usual assumption is that once you've controlled for all the variables that might affect both A and B, either the correlation will go away, or you will have discovered a real causal influence.

I think that's what Bell's factorizability assumption formalizes: If A and B are spacelike separated (so neither can cause the other, if relativity is correct), then either they are statistically independent, or there is a common cause for both of them.

 

[LsT]

I 'll try to say it in another way:

But that's not true. If Alice has already observed a polarized photon at angle 000^0 then the probability of Bob measuring a photon polarized at angle 90o90^o is ZERO, not 12\frac{1}{2}

That is circular, because I allready had made the hypothesis (which is consistent with experiments) that $P_A=\frac{1}{2}$ so even if "the probability that B measures a photon if A measures a photon at 90" (lets name this $P_C$) is indeed zero, that does not make $P_B$ zero. As I have already said, $P_B$ and $P_C$ are different probabilities per definition because you change the conditions of the later. Of course I understand now (with your help), that I cannot consider in this hypothesis $P_A$ as $(\alpha, \lambda)$ and $P_B$ as $(\beta, \lambda)$.

As per your example with the wheel I get your point now and this makes perfect sense to me. If you consider $\lambda$ as the state of the wheel, probabilities obviously become independent. But the wheel was not supposed to be a representative example of my hypothesis. Such an example (in the respect of perfect corelations only) is my thought experiment in the post no. #39 There I demonstrate that you can have perfect correlations without a complete specification of the state $\lambda$ that gives a probability other than 1/2 per detector. In fact this is the crucial point in my hypothesis, the state $\lambda$ has to be defined in such a way that it does NOT influence the probabilities of each detector.

I am thinking something in the lines of:

$P_A(\alpha, \theta)$
$P_B(\beta, \kappa)$

where $\theta$ and $\kappa$ is the polarization angle of each photon such as
$\theta - \kappa = 0$
but
$(\theta+\kappa) \sim U([0,720])$ (aka random).

Because of the condition $\theta - \kappa = 0$ I think this does not violate locality

 

[stevendaryl]

I 'll try to say it in another way:
That is circular, because I allready had made the hypothesis (which is consistent with experiments) that $P_A=\frac{1}{2}$ so even if "the probability that B measures a photon if A measures a photon at 90" (lets name this $P_C$) is indeed zero, that does not make $P_B$ zero

What is your definition of P_B? I'm using P_B to mean the probability that Bob will detect a photon. That probability is not a fixed number, it depends on what other facts you know (such as whether Alice measured a photon).

 

[stevendaryl]

I am thinking something in the lines of:

$P_A(\alpha, \theta)$
$P_B(\beta, \kappa)$

where $\theta$ and $\kappa$ is the polarization angle of each photon such as
$\theta - \kappa = 0$
but
$(\theta+\kappa) \sim U([0,720])$ (aka random).

Because of the condition $\theta - \kappa = 0$ I think this does not violate locality

No, of course not. That's a hidden-variables theory of the form that Bell proved cannot explain the EPR correlations.

 

[LsT]

My definition of $P_B$ is "Probability that Bob detects a photon per photon "emited" (creation of entangled pair)

No, of course not. That's a hidden-variables theory of the form that Bell proved cannot explain the EPR correlations.

Care to make it more clear to me why this is the case?

 

[stevendaryl]

Care to make it more clear to me why this is the case?

That's exactly what Bell proved.

The claim is this: Let P(A,B, \alpha,\beta) be the probability that Alice will get result A and Bob will get result B, given that Alice's detector has setting \alpha and Bob's detector has setting \beta. For an EPR-type experiment, A = 0 or A=1, depending on whether Alice detects a photon, or not. Similarly for B. The prediction of QM for the two-photon case is:

P(A,B,\alpha,\beta) = \frac{1}{2} cos^2(\alpha - \beta) (if A = B) and P(A,B,\alpha,\beta) = \frac{1}{2} sin^2(\alpha - \beta) (if A \neq B)
(That's assuming perfect detection, and no spurious photons. I don't know how the formula should be adjusted to allow for detector mistakes)

We can say that the correlation P(A,B,\alpha,\beta) has a "local causal explanation" if there is some property, or variable \lambda characterizing the photon creation event such that the probabilities can be written as follows:

P(A, B, \alpha, \beta) = average over all \lambda of P_A(A, \alpha, \lambda) \cdot P_B(B, \alpha, \lambda)

Your hypothesis about \theta, \kappa is exactly of this form.

 

[LsT]

P(A,B,α,β)=P(A, B, \alpha, \beta) = average of λ\lambda of PA(A,α,λ)⋅PB(B,α,λ)P_A(A, \alpha, \lambda) \cdot P_B(B, \alpha, \lambda)

Your hypothesis about θ,κ\theta, \kappa is exactly of this form.

In my hypothesis, even if you consider the state $\lambda$, the probabilities $P_A$ and $P_B$ remain dependent. So you cannot do this.

 

[stevendaryl]

In my hypothesis, even if you consider the state $\lambda$, the probabilities $P_A$ and $P_B$ remain dependent. So you cannot do this.

Give me an actual probability function for P_A and P_B, and I will show you how it works.

You said that P_A depends on \alpha and \theta and P_B depends on \beta and \kappa. So isn't it the case that:

P(A \&B) = average over \theta of P_A(\alpha, \theta) \cdot P_B(\beta, \theta)

 

[LsT]

I post this just for completeness: ***

Definitions:
1. $P_A$ probability to detect a photon per photon emitted for Alice
2. $P_B$ " " for Bob
3. $H_1$ photon heading to Alice
4. $H_2$ photon heading to Bob
5. $\lambda =$ polarization angle of photon $H_1 =$ polarization angle of photon $H_2$
6. $\lambda \sim \cup([0, 360])$
7.$\alpha$ polarization angle of Alice's detector
8.$\beta$ polarization angle of Bob's detector

Assumptions:
1. The photons are supposed to have a definite and real polarization that is defined the moment of pair creation
2. The probability $P_\delta$ that a single photon passes through a polarization filter is dependent only in their relative angle $\delta$, such as $P_\delta = 1$ IF $\delta < 45$ and $P_\delta = 0$ IF $\delta > 45$

$P_A = \cos^2(\alpha - \lambda)$

$P_B = \cos^2(\beta - \lambda)$

***
I just realized that the crucial assumtion 2 (I mention it for the first time...) that I have made from the beggining of this hypothesis in my mind, is wrong because its not in accordance with malus Law, and I can realize that without this, my hypothesis either falls under Bell's umbrella of local hidden variables or you cannot have perfect correlations. So I can consider my hypothesis wrong. Despite this I enjoyed this conversation, I have learned something and I hope that I was not too irritating. Thanks to everyone that aswered me.

 

[DrChinese]

I will show again, in my hypothesis, what happens at zero detectors angle difference:

A -> 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0
B -> 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0
C -> 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Here probability of Alice's Detection is 1/2 as is Bob's (as is always). And the perfect correlation is a fact. The probabilities here are related of course.

Yes, perfect correlations for 0 angle difference is explained by your local hypothesis. Unfortunately, that is one of the few settings that are.

Best to take A=0, B=120 and C=240 degrees as I have mentioned before. According to QM: P(A=B) = P(A=C) = P(B=C) = .25. Attempt to complete the example below where this is true and you will soon see the futility of the exercise.

A (000) -> 0 1 0 1 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0
B (120) ->
C (240) ->

 

[LsT]

DrChinese I consider the matter settled, I have no reason to continue this, It was all about a wrong assumption of mine. See previous post. Thanks anyway.

 

[Carl B]

The spin states of the entangled photons could have been pre-determined from the beginning of the experiment, it requires simply that certain assumptions on which quantum physics is based are either untrue or misleading and you may not wish to consider this!
The first assumption I discuss is one of Dirac's. In his introduction to his theory of the Principles of Quantum Mechanics (Fourth Edition revised) he states that “Only questions about the results of experiments have real significance for the physicist.” and then “ The foregoing discussion about the result of an experiment with a single obliquely polarized photon incident on a crystal of tourmaline answers all that can be legitimately be asked about what happens to an obliquely polarised photon when it reaches the tourmaline. Questions about what decides whether the photon is to go through or not and how it changes its direction of polarisation when it goes through cannot be investigated by experiment and should be regarded as outside the domain of science... Nevertheless some further description is necessary in order to correlate the results of this experiment with the results of other experiments that might be performed with photons and to fit them into a general scheme. Such further description should be regarded, not as an attempt to answer questions outside the domain of science, but as an aid to the formulation of rules for expressing concisely the results of large numbers of experiments.”
Obviously when Dirac was formulating these statistical rules he was deeply concerned about the practicalities of making precise measurements on individual particles although he explicitly accepted that there may be deterministic rules guiding things.
All of quantum mechanics is now based on Dirac's derivation of statistical rules based on the idea that determinism is impossible to prove so we will adopt a statistical approach. When we interpret the sayings of physicists then, we should bear in mind that QM is only statistical because we do not have the fineness and gentleness of experiment to be able to conduct the experiment deterministically.

To then come to the issue of Bell's inequality which has been mentioned. Bell himself made the statement in a 1985 radio interview (from wiki) “There is a way to escape the inference of superluminal speeds and spooky action at a distance. But it involves absolute determinism in the universe, the complete absence of free will. Suppose the world is super-deterministic, with not just inanimate nature running on behind-the-scenes clockwork, but with our behavior, including our belief that we are free to choose to do one experiment rather than another, absolutely predetermined, including the ‘decision’ by the experimenter to carry out one set of measurements rather than another, the difficulty disappears. There is no need for a faster-than-light signal to tell particle A what measurement has been carried out on particle B, because the universe, including particle A, already “knows” what that measurement, and its outcome, will be.”. I suggest that this super-determinism is excessive and that only individual photon exchanges need to be pre-determined, which would be the strict answer to what you ask, but it then begs follow on questions such as how such a 'local' (individual photon exchange) determinism maybe achieved in isolation from a more general non-deterministic environment and this is answered as follows:

There is no example of a photon existing independently and outside of an exchange between two atoms. The only way we can detect a photon is through its acceptance and incorporation into the structure of a receiving atom. Therefore we may consider a photon to be a singular exchange of an indivisible quantum of energy between two atoms. Over a large number of photons the pattern of distribution of those exchanges is then determined to fit a statistical pattern as described by what we know as electromagnetic waves. An exchange is a transaction or an instantaneous moment in time occurring between a giver and a receiver. I would suggest that at the moment of exchange of any photon its destination atom is known and 'agreed' with that destination.

But then if this transaction is instantaneous the question then arises as to how the time-delay appears which gives rise to a speed associated with the photon's time of flight from source atom to destination atom. This question also has a complete answer but which would be a separate topic.

In conclusion I would say that it is possible for the spin of entangled photons to be pre-determined,there is no experiment or theory which precludes this possibility. In Dirac's words what you ask is “outside the domain of science”!