How "spooky action...." may work?

[Markus Hanke]

A statistical correlation which one cannot explain with a common cause in the past.

I don't really understand this statement - the "common cause in the past" would be the initial interaction between the two particles which created the entanglement in the first place. After that initial interaction, no further "remote action" is needed or implied to uphold the relationship. The Bell inequalities are violated in this context precisely because no real, local "hidden variables" are needed to explain entanglement. This is not just conjecture, but pretty much an empirical finding.

Maybe it's just me, but after some initial study of quantum theory, there is nothing about entanglement that seems in any way unexplainable or mysterious, unless of course one insists that the world must be both real and at the same time Einstein-local, in a classical sense. But we already know ( from empirical finding ) that this is not the case.

 

[Ilja]

I refer here to Reichenbach's common cause principle. Roughly, if we observe a correlation, it should have a causal explanation. This causal explanation may be a direct one, A->B or B->A, but it may be also explained by a common cause C, with C->A and C->B.

This common cause principle is the base for, say, conclusions about the idea that smoking causes cancer. Without Reichenbach's common cause principle, the tobacco industry could simply say "oh, this is only a correlation, this does not prove any causal relation, so what". With the common cause principle, they are forced to find some alternative common cause explanations, else, the straightforward smoking->cancer explanation would have to be accepted. They are unable to present such common cause explanations - that means, the smoking->cancer explanation is what we believe.

Fortunately, if something is a common cause explanation or not has a simple precise meaning in probabilities: P(AB|C) = P(A|C)P(B|C). So, we do not have to accept vague feelings that blablabla is sufficient as an explanation.

And now we have the 100% correlation if we measure at A and B the same direction. Once you want to explain it with a common cause at the time of the creation of the pair, you have no other choice as to postulate that this common cause C defines, predefines, the result of the measurement. And, once this common cause cannot know what direction is measured, it has to predefine them all. And, so, we obtain the functions A(a,l), B(b,l) which describe how this common cause l predefines the measurement results. And then you can prove Bell's theorem, which is empirically falsified.

 

[Markus Hanke]

I am not an expert on this matter, so someone please correct me if I am wrong, but it would seem to me that the common cause principle is not fundamental to the universe, in the sense that it is not implied by any law of nature that I am aware of. In fact, I can immediately think of a number of circumstances where it is explicitly violated - for example, if you have exactly two electrons on the same atomic shell, their spins must be opposite as per the Pauli exclusion principle. This is a statistical correlation, since, if you perform a measurement on one of them, the outcome ( spin-up or spin-down ) is probabilistic, and not predetermined; however, once the outcome has been obtained on one electron, it is necessarily determined for the other electron as well, so as to stay true to Pauli's principle. There is no causal explanation for this, only the correlation implied by the laws of quantum theory.

Please don't misunderstand me here - I don't attempt to declare the common cause principle null and void, as it is quite useful and valid under certain circumstances, as you have demonstrated with the smoker example. I merely urge caution, in that I don't think this principle is universally applicable. Specifically, I don't think it applies to quantum entanglement, nor do I think it is needed in that context.

 

[Ilja]

I do not claim it is a law of nature, I think it is a base for doing scientific research, a necessary part of the scientific method. If causal explanations for observable correlations do not exist, science is as useless as collecting stamps.

If it is not universally applicable, it means that science is not universally applicable as a method to study the world around us. So, we can do science if we study human behavior or so, but to do science in fundamental physics would be a meaningless loss of time, all we can do there is to observe correlations and to do astrology based on such observed correlations without explanation.

In some sense, there are interesting parts of human thought which are not open to the scientific method already today. Like ideas about an immortal soul and so on. Even whole philosophical concepts like fatalism. So we cannot exclude, out of principle, that the quantum domain is also inaccessible to science, and accessible only to the observation of unexplainable correlations and subsequent astrology. But at least up to now I see no evidence for this. There exist realistic and causal interpretations of quantum theory. The only reason to reject them is that they violate a nice and reasonable but nonetheless metaphysical idea that relativistic symmetry is fundamental, instead of being only a large distance approximation valid only for some quantum equilibrium states.

 

[Markus Hanke]

If causal explanations for observable correlations do not exist, science is as useless as collecting stamps.

I agree with you, but I am not sure how this contradicts how quantum entanglement is commonly understood. There is a causal explanation for entanglement, but that "cause" is not an instantaneous physical action across arbitrarily large spatial distances, but rather an interaction between the particles in the past. In other words, the particles' histories are not independent of one another, so neither are the outcomes of measurements performed on them. This doesn't necessarily imply that those measurements are deterministic ( they aren't ), nor does it necessarily imply any type of interaction between the particles at the time the measurement takes place. At the same time, there is still a common cause for the correlation between the measurement outcomes, and that is their initial interaction in the past. I do not really see a contradiction here.

 

[rubi]

There is good reason to reject Reichenbach's principle in quantum theory and it has nothing to do with locality. Reichenbach's principle is based on ordinary probability theory, which needs a simplicial state space for its formulation. However, we know that the laws of physics are governed by a theory with a non-simplicial state space (quantum theory), so it would be unreasonable to apply concepts that only make sense in the context of ordinary probability theory to it.

The unapplicability of Reichenbach's principle in the context of quantum theory doesn't preclude the existence of a common cause. It only means that a criterion that captures our classical intuition about common causes isn't adequate to also capture the notion of a common cause in the generalized framework of quantum theory. It means that we don't have probabilistic method to identity what qualifies as a common cause. Unfortunately, nature just didn't provide us with such a method. Nevertheless, a common cause can still exist.

 

[Ilja]

... but I am not sure how this contradicts how quantum entanglement is commonly understood. There is a causal explanation for entanglement, but that "cause" is not an instantaneous physical action across arbitrarily large spatial distances, but rather an interaction between the particles in the past.

If this would be a valid causal explanation, in agreement with Reichenbach's common cause and Einstein causality, we could prove Bell's theorem. Which is falsified by observation. So that your causal explanation is falsified by observation.

This doesn't necessarily imply that those measurements are deterministic ( they aren't ), nor does it necessarily imply any type of interaction between the particles at the time the measurement takes place. At the same time, there is still a common cause for the correlation between the measurement outcomes, and that is their initial interaction in the past. I do not really see a contradiction here.

That's not the point. Bell's formula
$$P(a,b) = \int d\lambda \rho(\lambda) A(a,\lambda) B(B,\lambda)$$
is also only a statistical formula, which includes a probability distribution, and has excluded any type of interaction between the particles at the time of measurement.

You have to understand that this formula follows from assuming a common cause by the interaction between the particles in the past.

Let's try. Common cause explanation by C for a 100% correlation between X, Y means P(X and Y|C) = P(X|C) P(Y|C).
X: measurement at A of a gives 1. Y: measurement at B of a gives 1. Once X and Y would violate the conservation law, we have P(X and Y) = 0, thus, also P(X and Y|C)=0. Thus, or P(X|C) = 0 or P(Y|C) = 0. If P(Y|C) = 0, it means the result at B is predefined by C as -1, which we can name simply "not Y". This gives, for 0 = P(not X and not Y|C) = P(not X|C), so, X is predefined by C too. And this works for all directions a. The free will of the experimenters to choose a and b, which decision is not known at preparation time, proves that the common cause C, even if hidden, can depend only on a probability distribution which does not depend on a and b. This is all we need for the formula above. Which is sufficient to derive a false conclusion, Bell's inequality.

 

[Ilja]

There is good reason to reject Reichenbach's principle in quantum theory and it has nothing to do with locality. Reichenbach's principle is based on ordinary probability theory, which needs a simplicial state space for its formulation.

No. Ordinary probability theory needs nothing but elementary logic. Read Jaynes, Probability theory: The logic of science.

You can always transform the space of some probability theory into a simplicial one. It is a trivial exercise. Simply postulate that the wave functions (the "pure states" of the non-simplicial theory) are the "hidden variables". That means that you define your "hidden variable" theory has the simplex created by the original pure states.

Above theories are equivalent, because in above theories all states can be described by a linear combination of pure states. The difference is that in the non-simplicial theory you have linear combinations of pure states which are indistinguishable by all observations allowed by the theory, and, therefore, identified. In the straightforward "hidden variable" theory they are nonetheless considered to be different states, even if indistinguishable by observation.

However, we know that the laws of physics are governed by a theory with a non-simplicial state space (quantum theory), so it would be unreasonable to apply concepts that only make sense in the context of ordinary probability theory to it.

No. We have decided, based on positivist philosophy, that states which we cannot distinguish by observation in the actual theory (which may be incomplete, but this is nothing a positivist would allow to think about) have to be really identical, and, based on this ideology, rejected the trivial, straightforward extension toward a simplicial probability theory. Or, in other words, we have a philosophical prejudice against hidden variables, formulated in a nice mathematical language, which suggests some deep mathematical insight behind it which in reality does not exist.

And we use this nicely formulated prejudice to reject one of the foundations of the scientific method itself - Reichenbach's principle of common cause, which allows to make a difference between astrology and science.

Sorry if the formulation is a bit too polemically formulated, but I hope you understand the point.

 

[rubi]

No. Ordinary probability theory needs nothing but elementary logic. Read Jaynes, Probability theory: The logic of science.

You can always transform the space of some probability theory into a simplicial one. It is a trivial exercise. Simply postulate that the wave functions (the "pure states" of the non-simplicial theory) are the "hidden variables". That means that you define your "hidden variable" theory has the simplex created by the original pure states.

This is provably wrong. For instance, you cannot possibly model the spin observables $S_i$ as observables on a single probability space. This precludes the formulation of quantum theory on a simplicial state space. (Bohmian mechanics can't do it either.)

No. We have decided

No. You have decided. Scientists have decided differently. Scientists believe that philosophical prejudice has no place in science.

And we use this nicely formulated prejudice to reject one of the foundations of the scientific method itself - Reichenbach's principle of common cause, which allows to make a difference between astrology and science.

Reichenbach tried to formulate mathematically what is common sense about the idea of a common cause. Quantum theory just shows that he failed, since it definitely is common sense to assume that the initial preparation of the system is the common cause for the correlations. That means that the mathematical formulation of our common sense (Reichenbach's principle) needs to be adjusted and not our common sense.

Sorry if the formulation is a bit too polemically formulated, but I hope you understand the point.

Sorry if the formulation is a bit too polemically formulated, but I hope you understand the point.

 

[vanhees71]

However, you do give the false impression that the 100% anti-correlation results (e.g. H vs V) are sufficient to undermine local hidden variables.

That's right, to demonstrate the violation of Bell's inequality you need different angles between A's and B's polarizers than $0$ or $\pi/2$, but within the minimal interpretation that doesn't either need any "spooky action at a distance" to explain the results, because it just says that the state is prepared before any measurement is done and local microcausal QFT tells you that there are only local interactions between the photons and the polarizers at A's and B's position. So the correlations, leading to the violation of Bell's inequality are there from the very beginning when the two photons were prepared and are not caused by A's or B's measurement at the other far distant respective other place.

 

[vanhees71]

I am a bit confused about the mechanics of splitting a photon and the process of parametric down-conversion. My searches only show that the photon passes through a NLO crystal, which seems to be always graphically represented by a box. I am aware of Compton and Thomson scattering in which a photon interacts with a charged particle, but what is happening within the NLO crystal to cause the split? Is it possible that the EM wave is split into its electric and magnetic components and then reacquiring their complimentary components after the split to create two photons of lower energy? But my real interest is in the mechanics of the split in some graphic form similar to how the Compton scattering can be illustrated. Thank you.

Yes, that confuses me for years too. I've not been able to find a microscopic model explaining the parametric down conversion and the preparation of polarization-entangled photon states. What you can find are (pretty simple) phenomenological models assuming the fact that in this way such two-photon states can be prepared (see e.g., [1,2]), but as I said no microscopic models, just using "in-medium QED". Maybe one of the experts here has such a reference?

[1] C. K. Hong, L. Mandel, PRA 31, 2409 (1985)
[2] M. H. Rubin et al, PRA 50, 5122 (1994)

 

[Ilja]

This is provably wrong. For instance, you cannot possibly model the spin observables $S_i$ as observables on a single probability space. This precludes the formulation of quantum theory on a simplicial state space. (Bohmian mechanics can't do it either.)

You have obviously misunderstood the construction.

You have a pure state in quantum theory $|\psi\rangle\langle\psi|$. You take all these states as the pure states of my HVT, in a single probability space. The resulting simplicial space consists of arbitrary sums (or integrals) of type
$$\hat{\rho} = \sum_i p_i |\psi_i\rangle\langle\psi_i|$$
which are in no way different from the states you have in ordinary quantum theory. So the construction does needs nor states not known in standard QT nor something else which does not exist already.

The only difference is that linear combinations which are indistinguishable by observation, because the resulting density operator is the same if the + is interpretated as the sum of operators in Hilbert space, and not as a sum of probability distributions over the basic states $|\psi_i\rangle\langle\psi_i|$, are, instead, different states of this HVT.

So, nor my construction, nor dBB theory does even try.

No. You have decided. Scientists have decided differently. Scientists believe that philosophical prejudice has no place in science.

Learn to read. My use of "we" was, of course, rhetorical and did not describe my own choice. I thought this would be obvious, just to quote the context:

No. We have decided, based on positivist philosophy, that states which we cannot distinguish by observation in the actual theory (which may be incomplete, but this is nothing a positivist would allow to think about) have to be really identical, and, based on this ideology, rejected the trivial, straightforward extension toward a simplicial probability theory. Or, in other words, we have a philosophical prejudice against hidden variables, formulated in a nice mathematical language, which suggests some deep mathematical insight behind it which in reality does not exist.

But, ok, let's assume you want to tell me my polemics was misguided, and Scientists have decided that such positivistic philosophical prejudice has no place in science. Then your argument disappears into nothing, because if there is no prejudice against theories with hidden variables, there is no base for a prejudice against the construction above, which gives a simplicial probability theory.

Reichenbach tried to formulate mathematically what is common sense about the idea of a common cause. Quantum theory just shows that he failed, since it definitely is common sense to assume that the initial preparation of the system is the common cause for the correlations. That means that the mathematical formulation of our common sense (Reichenbach's principle) needs to be adjusted and not our common sense.

No. It is not common sense to think that violations of Bell's inequality do not prove that there is some hidden communication channel. I agree that it is not our common sense which has to be adjusted, see http://ilja-schmelzer.de/realism/game.php about what I think common sense can tell us about this. But Reichenbach's formulation is fine too.

 

[rubi]

You have obviously misunderstood the construction.

Then you have misunderstood the argument. I'm saying that it is a mathematical theorem that no probability theory on a simplicial state space can reproduce the statistics of three non-commuting spin observables $[S_i, S_j] = i\epsilon_{ijk} S_k$. However, the statistics is experimentally confirmed, so physical descriptions of the situation based on ordinary probability theory are experimentally excluded. Hence, using concepts like Reichenbachs principle, which need ordinary probability theory, are unreasonable in the context of statistics that is incompatible with ordinary probability theory.

No. It is not common sense to think that violations of Bell's inequality do not prove that there is some hidden communication channel. I agree that it is not our common sense which has to be adjusted, see http://ilja-schmelzer.de/realism/game.php about what I think common sense can tell us about this. But Reichenbach's formulation is fine too.

You have explained the common sense of playing cards, which are classical objects. However, photons aren't playing cards, so we can't expect our common sense to apply to them. However, it is common sense to take the preparation of the quantum state to be the cause for the observed correlations. It's your philosophical prejudice that our common sense about playing cards can be carried over to photons.

 

[Ilja]

Then you have misunderstood the argument. I'm saying that it is a mathematical theorem that no probability theory on a simplicial state space can reproduce the statistics of three non-commuting spin observables $[S_i, S_j] = i\epsilon_{ijk} S_k$. However, the statistics is experimentally confirmed, so physical descriptions of the situation based on ordinary probability theory are experimentally excluded.

And I have given a simple counterexample for your claimed mathematical theorem. Which works in a quite general way, for every theory which reproduces whatever statistics. So, I would guess, your "theorem", whatever it is, makes some additional assumptions which exclude this simple and essentially trivial construction.

Again, the trivial construction: Find all the pure states of your theory which reproduces your statistics. Then, define the simplex which has all these pure states as
pure states. Then check that the this simplex reproduces the same statistics. Essentially by construction, because the pure states have the same statistics, and their affine combinations therefore too.

You have explained the common sense of playing cards, which are classical objects. However, photons aren't playing cards, so we can't expect our common sense to apply to them. However, it is common sense to take the preparation of the quantum state to be the cause for the observed correlations. It's your philosophical prejudice that our common sense about playing cards can be carried over to photons.

As well as it is my prejudice that classical logic can be applied everywhere. I'm not ready to accept logical contradictions in a theory simply because the contradictory theory is about some quantum or so things. And probability theory as well as Reichenbach's principle of common cause are, I think, on the same level, see Jaynes. They are the common sense base of science. To modify Kant, science is man's emergence from his self-imposed immaturity. And I'm not ready to accept that somebody else imposes some new Holy or quantum borders behind those it is not allowed to apply common sense, where he has to self-impose his immaturity.

 

[rubi]

And I have given a simple counterexample for your claimed mathematical theorem. Which works in a quite general way, for every theory which reproduces whatever statistics. So, I would guess, your "theorem", whatever it is, makes some additional assumptions which exclude this simple and essentially trivial construction.

Again, the trivial construction: Find all the pure states of your theory which reproduces your statistics. Then, define the simplex which has all these pure states as
pure states. Then check that the this simplex reproduces the same statistics. Essentially by construction, because the pure states have the same statistics, and their affine combinations therefore too.

No, your construction won't recover the statistics of three non-commuting spin observables. It doesn't even work out mathematically, since there are uncountably many pure states, so your sum will diverge. Furthermore, a density matrix doesn't constitute a simplicial state space. The theorem doesn't make any additional assumptions. The only thing you need is the probability distributions that are predicted by QT for non-commuting spin observables. No ordinary probability theory can recover them. It is a fact, which you will have to accept.

As well as it is my prejudice that classical logic can be applied everywhere. I'm not ready to accept logical contradictions in a theory simply because the contradictory theory is about some quantum or so things. And probability theory as well as Reichenbach's principle of common cause are, I think, on the same level, see Jaynes. They are the common sense base of science. To modify Kant, science is man's emergence from his self-imposed immaturity. And I'm not ready to accept that somebody else imposes some new Holy or quantum borders behind those it is not allowed to apply common sense, where he has to self-impose his immaturity.

Quantum theory is logically consistent and the quantum borders are imposed upon us by nature itself, not by any human. Self-imposed immaturity would be to reject mathematical theorems based on ones philosophical prejudices.

 

[Ilja]

No, your construction won't recover the statistics of three non-commuting spin observables. It doesn't even work out mathematically, since there are uncountably many pure states, so your sum will diverge.

Clearly wrong. Learn how to take integrals over probability distributions, say $\int \rho(x) dx$, over uncountably many real numbers (hint: the result is, in this case, 1).

Furthermore, a density matrix doesn't constitute a simplicial state space.

As if I would have claimed this. It is an element of the quantum state space. If it is a pure density matrix $|\psi\rangle\langle\psi|$, then we take it and use it as a vertex of some other, simplicial state space. Ok, let's use a different denotation for this element, $S_{|\psi\rangle\langle\psi|}$. Better?
Then you construct the simplicial state space created by these density matrices as basic states. Each element is a formal linear combination $\sum_i p_i S_{|\psi\rangle\langle\psi|}$. Feel free to replace the sum (with $\sum_i p_i = 1, p_i\ge 0$) by an integral (with $\int_\lambda \rho(\lambda) d\lambda = 1, \rho(\lambda)\ge 0$).

Each element of this simplex then defines a density matrix by a straightforward projection operator $\sum_i p_i S_{|\psi\rangle\langle\psi|} \to \sum_i p_i |\psi\rangle\langle\psi|$. So for every element of this simplex there exists a corresponding density matrix.

The theorem doesn't make any additional assumptions. The only thing you need is the probability distributions that are predicted by QT for non-commuting spin observables. No ordinary probability theory can recover them. It is a fact, which you will have to accept.

No, I don't have to accept claims which I can easily prove to be wrong, and exist only because you claim it. Give a link to a paper in a peer-reviewed journal where this proof has been made, then we will see what are the additional assumptions.

Self-imposed immaturity would be to believe your claims of existence of some Theorem.

 

[stevendaryl]

I thought it was always due to a conservation law, not just often, are there exceptions?

Well, entanglement doesn't have to involve conservation laws. Any time you have a pair of systems that are in a superposition of states of the form:

$|\Psi\rangle = \alpha |\phi_A\rangle |\chi_A\rangle + \beta |\phi_B\rangle |\chi_B\rangle$

you have entanglement, and measurement of the first system instantly tells you about the second system, no matter how far away. So the idea of entanglement doesn't have anything to do with conservation laws. However, it might be that in practice, the only way to get such an entangled state is by through conservation laws: Either the total energy, or the total momentum, or the total angular momentum, etc., is fixed, but the partitioning of the quantity between the two systems is different in the two possible elements of the superposition.

 

[rubi]

Clearly wrong. Learn how to take integrals over probability distributions, say $\int \rho(x) dx$, over uncountably many real numbers (hint: the result is, in this case, 1).

No such integral is defined on the space of operators that you want to sum up.

No, I don't have to accept claims which I can easily prove to be wrong, and exist only because you claim it. Give a link to a paper in a peer-reviewed journal where this proof has been made, then we will see what are the additional assumptions.

Self-imposed immaturity would be to believe your claims of existence of some Theorem.

The proof is really trivial high school mathematics and well-known to every researcher in quantum theory, so there is really no point to doubt it. But here you go:
We try to reproduce the statistics of a quantum state $\left|\Psi\right>=\left|z+\right>$. As everybody knows, the statistics is given by $P(z+)=1$, $P(z-)=0$, $P(x+)=\frac{1}{2}$ and $P(x-)=\frac{1}{2}$.

We want to know whether there is a joint probability distribution $P(s_x,s_z)$ such that $P(x_\pm)=P(\pm,+)+P(\pm,-)$ and $P(z_\pm) = P(+,\pm)+P(-,\pm)$. In particular, we would have $0=P(z-)=P(+,-)+P(-,-)$, i.e. $P(+,-) = - P(-,-)$. Hence, either $P(+,+)$ is negative or $P(-,-)$ is negative or $P(+,+)=P(-,-)=0$.

In the first two cases, we wouldn't have a probability distribution, since probabilities must be positive. In the third case, the system of equations reduces to $1=P(-,+)$, $0=P(+,-)$, $\frac{1}{2}=P(+,-)$ and $\frac{1}{2}=P(-,+)$. Obviously, this system has no solutions ($\frac{1}{2} \neq 1$!), hence no such joint probability distribution exists and hence no observables modeled as random variables on a probability space can reproduce the statistics of the quantum state $\left|\Psi\right>$, since if there were such observables, they would have a well-defined joint probability distribution.

 

[Markus Hanke]

If this would be a valid causal explanation, in agreement with Reichenbach's common cause and Einstein causality, we could prove Bell's theorem.

I disagree. You could prove Bell's theorem only if both realism and Einstein locality hold simultaneously at the time of measurement ( or at least in the period between first interaction and measurement ), implying that there are some form of local hidden variables. Clearly, this is not the case, as evidenced by experiment and observation. The point is that this does not make any reference to, nor does it require, Reichenbach's principle.

On the other hand, the interaction between the two particles in the past demonstrably is the causal explanation of entanglement, because after the interaction has taken place the total information contained in the composite system is greater than the the sum of the information carried by the two subsystems considered in isolation ( I believe the term for this is "subadditivity" ). This would not be the case without the interaction in the past, so causality seems quite obvious to me. The "extra" information contained in the composite system is precisely the correlation implied by entanglement. All of this is of course purely statistical.

Making this mathematically precise isn't so easy for me ( I'm not a scientist ), but my first impulse here would be to write down the reduced density matrices for the two subsystems, and then calculate the entropy from it. The total entropy of the composite system should be less than the sum of the entropies for each particle in isolation, which is a direct causal consequence of their past interaction ( without which we would get an equation, instead of an inequality ).

 

[stevendaryl]

On the other hand, the interaction between the two particles in the past demonstrably is the causal explanation of entanglement, because after the interaction has taken place the total information contained in the composite system is greater than the the sum of the information carried by the two subsystems considered in isolation ( I believe the term for this is "subadditivity" ). This would not be the case without the interaction in the past, so causality seems quite obvious to me. The "extra" information contained in the composite system is precisely the correlation implied by entanglement. All of this is of course purely statistical.

I'm not sure that entanglement necessarily requires interaction in the past, but I don't know of a counter-example.

But the issue with causality is not really about the creation of the entanglement, but about it's consequences.

Alice measures an electron to have spin-up in the z-direction. Because she knows that their particles are entangled, she immediately knows something about Bob's situation--that Bob will not measure spin-up in the z-direction. So think about that statement: "Bob will not measure spin-up in the z-direction". It seems that if the statement is true (in Many-Worlds, it's not true, or no more true than its negation), then we have two possibilities:

1. It became true at some moment.
2. It has always been true.

It's hard to make sense of the first statement without nonlocal effects, and it's hard to make sense of the second statement (in light of Bell's inequality) without superdeterminism.

 

[Ilja]

No such integral is defined on the space of operators that you want to sum up.

You want to tell me that such a space of operators does not allow the definition of a probability measure? Really?

The proof is really trivial high school mathematics and well-known to every researcher in quantum theory, so there is really no point to doubt it. But here you go:
We try to reproduce the statistics of a quantum state $\left|\Psi\right>=\left|z+\right>$. As everybody knows, the statistics is given by $P(z+)=1$, $P(z-)=0$, $P(x+)=\frac{1}{2}$ and $P(x-)=\frac{1}{2}$.

Fine. That means, this state defines a pure state $\left|z+\right>\left<z+\right|$, thus, a vertex of the construction of the simplex.

We want to know whether there is a joint probability distribution $P(s_x,s_z)$ such that ...

And where do you think my construction needs any such "joint probability distribution"? You obviously have not understood the construction at all, invent some meaning of it which requires some "joint probability distribution". Sorry, no. Read again the construction. And recognize that for every state, I repeat every state of the construction there exists an image defined by the map $\sum_i p_i S_{|\psi\rangle\langle\psi|} \to \sum_i p_i |\psi\rangle\langle\psi|$ which is a well-defined density operator of standard QM. I do not have to construct some "joint probability distributions".

Recommended reading: Holevo, Probabilistic and statistical aspects of quantum theory, North Holland, Amsterdam 1982

Theorem 7.1. Any separated statistical model ... is a reduction of a classical model with a restricted class of measurements.

 

[rubi]

And where do you think my construction needs any such "joint probability distribution"?

No, your construction would have to imply the existence of a joint probability distribution, since this existence is automatically guaranteed by probability theory. If I can prove that this joint probability distribution cannot exist, I have also proven that no underlying probability space can exist. You have obviously not understood basic probability theory.

Theorem 7.1. Any separated statistical model ... is a reduction of a classical model with a restricted class of measurements.

This has exactly nothing to do with it.

 

[Ilja]

I disagree. You could prove Bell's theorem only if both realism and Einstein locality hold simultaneously at the time of measurement ( or at least in the period between first interaction and measurement ), implying that there are some form of local hidden variables. Clearly, this is not the case, as evidenced by experiment and observation. The point is that this does not make any reference to, nor does it require, Reichenbach's principle.

May be you know about a variant which needs what you tell us, but does not need Reichenbach's common cause. So what? I have described here already a variant which uses, instead, Reichenbach's common cause together with Einstein causality.

And once there exists such a variant of the proof, that means that once the result - Bell's inequality - is empirically false, or Reichenbach's common cause or Einstein causality is false.

On the other hand, the interaction between the two particles in the past demonstrably is the causal explanation of entanglement, because after the interaction has taken place the total information contained in the composite system is greater than the the sum of the information carried by the two subsystems considered in isolation ( I believe the term for this is "subadditivity" ). This would not be the case without the interaction in the past, so causality seems quite obvious to me. The "extra" information contained in the composite system is precisely the correlation implied by entanglement. All of this is of course purely statistical.

I do not doubt that it is possible to redefine "explanation" so that it becomes possible to interpret this text as an explanation. It is not a common cause explanation in agreement with Reichenbach's common cause principle, because I cannot see any evidence of type P(AB|C) = P(A|C) P(B|C).

 

[Ilja]

No, your construction would have to imply the existence of a joint probability distribution, since this existence is automatically guaranteed by probability theory. If I can prove that this joint probability distribution cannot exist, I have also proven that no underlying probability space can exist. You have obviously not understood basic probability theory.

Given that my construction is a trivial construction equivalent to quantum theory, you have proven that quantum theory does not exist. Congratulations.

This has exactly nothing to do with it.

It has nothing to do with your "no joint probability distribution" argument, indeed. My construction is his, I simply take all the extremal points of the statistical model, in the case of quantum theory the pure states $|\psi\rangle\langle\psi|$, and use them as points in the classical phase space. And it has, in the same way as Holevo's construction, nothing to do with your "no joint probability distribution" argument.

 

[rubi]

Given that my construction is a trivial construction equivalent to quantum theory, you have proven that quantum theory does not exist. Congratulations.

Wrong. I have proven that the statistics of the quantum observables $S_x$ and $S_z$ in the state $\left|\Psi\right>$ cannot be modeled by random variables on a probability space. The argument is watertight, otherwise, you would be able to point out an error, rather than refer to your construction, of which you haven't even attempted to prove that it reproduces said statistics.

It has nothing to do with your "no joint probability distribution" argument, indeed. My construction is his, I simply take all the extremal points of the statistical model, in the case of quantum theory the pure states $|\psi\rangle\langle\psi|$, and use them as points in the classical phase space. And it has, in the same way as Holevo's construction, nothing to do with your "no joint probability distribution" argument.

If you believe that you can reproduce the statistics of the quantum system from my post using a classical probability space, then please just define the objects that are needed, i.e. a probability space $(\Omega, \Sigma)$ with a probability measure $\mu:\Sigma\rightarrow\mathbb R_+$ and the random variables $S_x, S_z :\Omega\rightarrow\left\{-1,1\right\}$ and prove that their probability distributions reproduce the statistics. So far you haven't done this. Just carry out the calculations, please, instead of telling us that it will obviously work out.

Hint: You will not succeed, since I have proven it to be impossible. (If it were possible, I would be able to compute the joint probability distribution of $S_x$ and $S_z$, which I have shown to not exist.)

 

[Markus Hanke]

Because she knows that their particles are entangled, she immediately knows something about Bob's situation

Ok, I think I understand your point ( at least I hope so ). However, it seems to me that the knowledge of the particles being entangled is something that has been added into the mix from the "outside". If we assume that the Alice-Bob system ( with their respective particles ) is isolated in space and time, how would Alice by herself know by performing a measurement on her particle whether it is entangled with a distant particle or not ? Only by either having been present during the initial interaction between them ( classical exchange of information across time ), or by subsequently comparing her results with those of Bob - which is a classical information exchange across space. Without either information exchange or prior interaction ( at some point along Alice's world line ), the outcome of both measurements would appear completely random to both Alice and Bob in isolation. In that sense, it is either the initial interaction that caused the correlation, or the act of comparing the measurement outcomes ( which is always a classical channel ). Without either, the concept of entanglement becomes meaningless. Both cases involve some form of non-locality - either non-locality in space, or non-locality in time, so either way Bell's inequalities will be violated, just as we empirically observe.

Or am I seeing this wrong / missing something ? I am still actively learning about this whole subject matter.

 

[Ilja]

If you believe that you can reproduce the statistics of the quantum system from my post using a classical probability space, then please just define the objects that are needed, i.e. a probability space $(\Omega, \Sigma)$ with a probability measure $\mu:\Sigma\rightarrow\mathbb R_+$ and the random variables $S_x, S_z :\Omega\rightarrow\left\{-1,1\right\}$ and prove that their probability distributions reproduce the statistics. So far you haven't done this. Just carry out the calculations, please, instead of telling us that it will obviously work out.

Ok, let's define the space $\Omega$ as the space consisting of all states $\omega = |\psi\rangle\langle\psi|$. Then, for a probability measure $\mu$ I defined on this space the expectation value of your $S_x$ or whatever will be $\int d \mu(|\psi\rangle\langle\psi|) \langle\psi| S_x |\psi\rangle$. As I said, a triviality.

But, I see, you want something more than the space of states being a simplex. You want some $S_x, S_z :\Omega\rightarrow\left\{-1,1\right\}$. But what does it have to do with the probability space being or not being a simplex?

 

[rubi]

Ok, let's define the space $\Omega$ as the space consisting of all states $\omega = |\psi\rangle\langle\psi|$. Then, for a probability measure $\mu$ I defined on this space the expectation value of your $S_x$ or whatever will be $\int d \mu(|\psi\rangle\langle\psi|) \langle\psi| S_x |\psi\rangle$. As I said, a triviality.

In order to have a probability space, you needn't give me an expectation value functional, but the measure $\mu$ itself. You need to tell me how to assign a number $0\leq\mu(A)\leq 1$ to any (measurable) subset $A\subseteq\Omega$ (in such a way that the axioms of probability theory are satisfied). What you have given me is just the expectation value of a mixed state $Tr(\rho S_x)$.

Given a probability space $(\Omega,\Sigma,\mu)$ and a set of random variables, I can define an expectation value functional on the set of random variables $E(S)=\int_\Omega S(x)\mathrm d\mu(x)$. However, given an expectation value functional on some set of observables, there isn't in general a probability space, such that the expectation value functional is given by taking the expectation values with respect to the probability space. In particular, in the case of quantum spin 1/2 particles, no expectation value functional can possibly be the expectation value functional of a classical probability theory.

The question is: Do quantum expectation values arise as expectation values of random variables on a classical probability space, i.e. is there a $\mu_\rho$ for each $\rho$ such that $Tr(\rho X) =\int_\Omega \hat X \mathrm d\mu_\rho$, where $\hat X$ is the corresponding random variable? And the answer is negative. Quantum theory is more general than classical probability theory. It can't be reformulated as a classical probability theory. Hence, concepts from classical probability theory don't generally apply to it.

But, I see, you want something more than the space of states being a simplex. You want some $S_x, S_z :\Omega\rightarrow\left\{-1,1\right\}$. But what does it have to do with the probability space being or not being a simplex?

You don't have a probability space in the first place! You have a state space together with an expectation value functional. A (classical) probability theory is by definition a probability space $(\Omega,\Sigma,\mu)$ together with random variables defined on it. You want to define something, which isn't a classical probability theory. If you do this, the usual laws of probabilities won't hold anymore, since they are derived from the concept of a probability space (Kolmogorov's axioms). The very formulation of Reichenbach's principle (and also Bell's theorem) crucially depends on all these concepts from probability theory, such as conditional probabilities and the rules how to combine them. And what I'm saying is that no classical probability theory can reproduce the statistics of quantum spin 1/2 particles, hence it is unreasonable to expect concepts from classical probability theory to apply to it.

Since the statistics of quantum theory (which is consistent with experiments) isn't compatible with classical probabiliy theory, classical probability as a foundational concept for physics is dead forever. It is a theorem that it can't be saved. Thus all physical concepts that depend on classical probability theory need to be modified.

 

[Ilja]

You don't have a probability space in the first place! You have a state space together with an expectation value functional.

Indeed. So let's look what we are talking about, what I have objected to, your quote from #41:

Reichenbach's principle is based on ordinary probability theory, which needs a simplicial state space for its formulation. However, we know that the laws of physics are governed by a theory with a non-simplicial state space (quantum theory), so it would be unreasonable to apply concepts that only make sense in the context of ordinary probability theory to it.

My point is that the non-simplicial state space is not a problem at all. Because of the quite trivial construction from Holevo.

And then I have objected that you started to argue about something completely different.

If you accept that the remark at #41 was misguided, fine. If not, let's forget about the probability spaces and talk about the state space.

Except you start with a completely different argument about the applicability of probability theory. In this case, I would not use the Holevo construction, but, instead, an established hidden variable theory like de Broglie-Bohm, which nicely recovers quantum probabilities and has no problem with Reichenbach's common cause principle. Feel free to tell me that dBB theory uses something different than usual probability theory. Alternatively, we can start to read together Bohm 1952 to see that the physical predictions are equivalent to those of quantum theory, despite its use of classical probability theory.

 

[rubi]

Indeed. So let's look what we are talking about, what I have objected to, your quote from #41:

My point is that the non-simplicial state space is not a problem at all. Because of the quite trivial construction from Holevo.

First of all, the space of density matrices is not at all a simplicial state space, but rather more like a sphere. But the point is that you need a space of states of classical probability distributions, which constitutes a simplex. Do you agree that Reichenbach's principle depends crucially on the concept of conditional probability? The concept of conditional probability is only well-defined in the context of classical probability theory.

Except you start with a completely different argument about the applicability of probability theory. In this case, I would not use the Holevo construction, but, instead, an established hidden variable theory like de Broglie-Bohm, which nicely recovers quantum probabilities and has no problem with Reichenbach's common cause principle. Feel free to tell me that dBB theory uses something different than usual probability theory. Alternatively, we can start to read together Bohm 1952 to see that the physical predictions are equivalent.

No, Bohmian mechanics cannot formulate the spin observables of a particle as random variables on a classical probability space in such a way that it is consistent with the predictions of quantum mechanics. This is a mathematical theorem. If you object to this, then find the error in my argument in post #53. No theory of classical probabilities can reproduce the statistics of quantum spin 1/2 particles. Reichenbach's principle depends on concepts from classical probability theory, hence it depends on concepts that are not generally valid in any theory that describes observed phenomena of nature.

 

[Jilang]

Rubi, you say that no classical theory can reproduce the statistics of spin. Would that also apply to a four dimensional model of spin projected onto a three dimensional space?

 

[Ilja]

First of all, the space of density matrices is not at all a simplicial state space, but rather more like a sphere.

Of course, I have never questioned this, the point was that this can be easily modified with Holevo's construction.

But the point is that you need a space of states of classical probability distributions, which constitutes a simplex. Do you agree that Reichenbach's principle depends crucially on the concept of conditional probability? The concept of conditional probability is only well-defined in the context of classical probability theory.

I do not see any problem with conditional probabilities. I see probability as the logic of plausible reasoning, which I can always apply. See Jaynes. And conditional probability is part of this logic. Simplices are quite irrelevant for this.

No, Bohmian mechanics cannot formulate the spin observables of a particle as random variables on a classical probability space in such a way that it is consistent with the predictions of quantum mechanics. This is a mathematical theorem. If you object to this, then find the error in my argument in post #53. No theory of classical probabilities can reproduce the statistics of quantum spin 1/2 particles. Reichenbach's principle depends on concepts from classical probability theory, hence it depends on concepts that are not generally valid in any theory that describes observed phenomena of nature.

It can and does. Not in a non-contextual way, of course. But this is not obligatory.

In #53 you want some joint probability distribution - for things where we have no joint experiments. What is this? Some metaphysical idea how these evil "hidden variables" have to look like? A hidden variable theory has to recover only the results of quantum experiments, not the ideas of opponents of hidden variable theories.

So, simply an explanation how dBB works: If you have a "measurement", you have to describe the interaction between the system and the "measurement instrument". This interaction depends, in dBB theory, in general also on the hidden variables of the measurement device. So, there is no "measurement" of some inherent "property" of the system, but a result of an interaction. And this result has nothing to do with another result of a completely different interaction where something different is "measured".

 

[rubi]

Rubi, you say that no classical theory can reproduce the statistics of spin. Would that also apply to a four dimensional model of spin projected onto a three dimensional space?

Yes, that's right. It's completely independent of how the statistics came about. It needn't even be derived from quantum theory. Whenever a model predicts the probabilities $0$ and $1$ for one two-valued observable and $\frac{1}{2}$ and $\frac{1}{2}$ for the other, no classical probability theory is compatible with this prediction. (Of course, this also applies to other numerical values for the probabilities. I just chose one specific example in order to turn the proof into high school mathematics, which I hope is accessible to anyone.)

I do not see any problem with conditional probabilities. I see probability as the logic of plausible reasoning, which I can always apply. See Jaynes. And conditional probability is part of this logic. Simplices are quite irrelevant for this.

The problem is that you can't find a concept of conditional probabilities in a non-simplicial state space. You will always violate some basic axiom of classical probability theory, like probabilities adding up to $1$. For instance in quantum theory, the concept only makes sense for commuting observables, and this is exactly the case, where quantum probabilities are consistent with classical probabilities. If you include non-commuting observables, the concept ceases to make sense.

It can and does. Not in a non-contextual way, of course. But this is not obligatory.

It is obligatory if you want to be consistent with classical probability theory.

In #53 you want some joint probability distribution - for things where we have no joint experiments. What is this? Some metaphysical idea how these evil "hidden variables" have to look like? A hidden variable theory has to recover only the results of quantum experiments, not the ideas of opponents of hidden variable theories.

You still misunderstand the proof. I don't want joint probabilities. I get them for free by classical probability theory. You cannot possibly have a classical probability theory without joint probabilities. Hence, if I can show that no joint probability distribution can exist, I have automatically proven that the statistics is incompatible with classical probability theory. It's just basic logic. If $A\Rightarrow B$ and I can prove $\neg B$, then I also know $\neg A$. I don't assume anything about hidden variables except that they can be modeled on a classical probability space. The joint probabilities are just an intermediate tool, which I can assume, since their existence is guaranteed by probabilty theory.

So, simply an explanation how dBB works: If you have a "measurement", you have to describe the interaction between the system and the "measurement instrument". This interaction depends, in dBB theory, in general also on the hidden variables of the measurement device. So, there is no "measurement" of some inherent "property" of the system, but a result of an interaction. And this result has nothing to do with another result of a completely different interaction where something different is "measured".

It doesn't matter whether BM can recover the predictions of QM. The thing that matters is whether they are compatible with classical probability theory. We don't even need quantum theory at all. It can already be proven from the observed statistics. No theory that attempts to predict the observed statistics, be it ordinary QM, Bohmian mechanics, or something completely different, can predict the probabilities in such a way that they are compatible with classical probability theory. Hence, it is an experimental fact that classical probability cannot serve as the basis for the foundations of physics. Thus, all concepts that require it for their formulation must be modified.

 

[wle]

Yes, that's right. It's completely independent of how the statistics came about. It needn't even be derived from quantum theory. Whenever a model predicts the probabilities $0$ and $1$ for one two-valued observable and $\frac{1}{2}$ and $\frac{1}{2}$ for the other, no classical probability theory is compatible with this prediction. (Of course, this also applies to other numerical values for the probabilities. I just chose one specific example in order to turn the proof into high school mathematics, which I hope is accessible to anyone.)

Huh? You can set $P(s_{\mathrm{x}}, s_{\mathrm{z}}) = P(s_{\mathrm{x}}) P(s_{\mathrm{z}})$ to trivially construct the sort of joint probability distribution you describe. For the example from your post #53 this would get you $$\begin{eqnarray}
P(+_{\mathrm{x}}, +_{\mathrm{z}}) &=& 1/2 \,, \qquad P(+_{\mathrm{x}}, -_{\mathrm{z}}) &=& 0 \,, \\
P(-_{\mathrm{x}}, +_{\mathrm{z}}) &=& 1/2 \,, \qquad P(-_{\mathrm{x}}, -_{\mathrm{z}}) &=& 0 \,.
\end{eqnarray}$$ You can easily check that this reproduces the marginals $P(+_{\mathrm{z}}) = 1$, $P(-_{\mathrm{z}}) = 0$, and $P(+_{\mathrm{x}}) = P(-_{\mathrm{x}}) = 1/2$ from your post #53.

The problem in your proof seems to be here:

In particular, we would have $0=P(z-)=P(+,-)+P(-,-)$, i.e. $P(+,-) = - P(-,-)$. Hence, either $P(+,+)$ is negative or $P(-,-)$ is negative or $P(+,+)=P(-,-)=0$.

It looks like $P(+, -)$ accidentally got changed to $P(+, +)$ in the second sentence.

 

[Zafa Pi]

That's right, to demonstrate the violation of Bell's inequality you need different angles between A's and B's polarizers than $0$ or $\pi/2$, but within the minimal interpretation that doesn't either need any "spooky action at a distance" to explain the results, because it just says that the state is prepared before any measurement is done and local microcausal QFT tells you that there are only local interactions between the photons and the polarizers at A's and B's position. So the correlations, leading to the violation of Bell's inequality are there from the very beginning when the two photons were prepared and are not caused by A's or B's measurement at the other far distant respective other place.

I'm not sure I get it. To make sure we are on the same page let's refer to my post #39 on the thread https://www.physicsforums.com/account/posts/5494960/ .
It is true the QM predicts the correlations for the entangled state when it's created, but to check the validity we must step into reality and perform experiments. The results (denying local realism, i.e. the Inequality of #39) would cause people to refer to "spooky action at a distance". In spite of being used to it I still find it mysterious/spooky.

What am I missing? And BTW, what is a reference to Einstein's regret about the EPR paper?