How do entanglement experiments benefit from QFT (over QM)?

[A. Neumaier]

I don't understand this claim that QM does not have a single sample space.
Sample space is just a definition for a set of outcomes with few restrictions (from wikipedia):
- the outcomes must be mutually exclusive;
- the outcomes must be collectively exhaustive;
- we must remove irrelevant information from the sample space.

I consider an experiment E. It consists of choosing subexperiment X or Y and performing one of them. Experiment X has outcomes A and B, but experiment Y - outcomes C and D.
Sample space consists of A, B, C and D. It satisfies all three restrictions.
Why set of A, B, C and D can't be considered single sample space?

Because this set is not mutually exclusive in the sense implied by (the precise sources of) wikipedia. The probabilities for the four cases sum to 2 rather than to 1.

 

[zonde]

Because this set is not mutually exclusive in the sense implied by (the precise sources of) wikipedia. The probabilities for the four cases sum to 2 rather than to 1.

By the description of experiment they don't. Probabilities of performing subexperiment X or subexperiment Y sum to 1. Then probability of performing subexperiment X is split between outcomes A and B and probability of performing subexperiment Y is split between outcomes C and D. Of course the sum is 1 not 2.

 

[Morbert]

By the description of experiment they don't. Probabilities of performing subexperiment X or subexperiment Y sum to 1. Then probability of performing subexperiment X is split between outcomes A and B and probability of performing subexperiment Y is split between outcomes C and D. Of course the sum is 1 not 2.

In classical physics, different samples spaces can be related to one another by coarsening or refining, and there is a single sample space that is a common refinement of all others. In quantum physics, there isn't a common refinement.

 

[DarMM]

I don't understand this claim that QM does not have a single sample space.
Sample space is just a definition for a set of outcomes with few restrictions (from wikipedia):
- the outcomes must be mutually exclusive;
- the outcomes must be collectively exhaustive;
- we must remove irrelevant information from the sample space.

I consider an experiment E. It consists of choosing subexperiment X or Y and performing one of them. Experiment X has outcomes A and B, but experiment Y - outcomes C and D.
Sample space consists of A, B, C and D. It satisfies all three restrictions.
Why set of A, B, C and D can't be considered single sample space?

Because the four pairwise probabilities for Bell alignments $A, B, C, D$:
$$P\left(A,B\right), P\left(B,C\right), P\left(C,D\right), P\left(A,D\right)$$
can be proven to not be marginals of a single distribution $P\left(A,B,C,D\right)$ per Fine's theorem, thus they are not defined over a common sample space.

 

[zonde]

In classical physics, different samples spaces can be related to one another by coarsening or refining, and there is a single sample space that is a common refinement of all others. In quantum physics, there isn't a common refinement.

Are you saying that for my described experiment there is single sample space but when considering other experiments there is no common sample space?

 

[zonde]

Because the four pairwise probabilities:
$$P\left(A,B\right), P\left(B,C\right), P\left(C,D\right), P\left(A,D\right)$$
can be proven to not be marginals of a single distribution $P\left(A,B,C,D\right)$ per Fine's theorem, thus they are not defined over a common sample space.

So which rule do you say it violates? Do you say that four outcomes are not mutually exclusive? Or that there is some fifth outcome possible?
Or there is some other rule not specified by wikipedia?

 

[Morbert]

Are you saying that for my described experiment there is single sample space

If the selection of a subexperiment can be modeled by some appropriate variable like the outcomes of a coin flip that perfectly correlates with the subexperiment, there would be a common sample space, yes. E.g. The outcomes (heads, A),(heads,B),(tails,C),(tails,D).

but when considering other experiments there is no common sample space?

If you can carry out the experiment, you can build an appropriate sample space. But if you try to combine two incompatible sample spaces into a single common sample space a priori, you'll fail, and you will not be able to carry out the implied experiment.

 

[A. Neumaier]

By the description of experiment they don't. Probabilities of performing subexperiment X or subexperiment Y sum to 1. Then probability of performing subexperiment X is split between outcomes A and B and probability of performing subexperiment Y is split between outcomes C and D. Of course the sum is 1 not 2.

But these are not the probabilities defined by Born's rule.

 

[zonde]

If the selection of a subexperiment can be modeled by some appropriate mixture like the outcomes of a coin flip that perfectly correlates with the subexperiment, there would be a common sample space, yes. E.g. The outcomes (heads, A),(heads,B),(tails,C),(tails,D).

If you can carry out the experiment, you can build an appropriate sample space. But if you try to combine two incompatible sample spaces into a single common sample space a priori, you'll fail, and you will not be able to carry out the implied experiment.

So as long as I talk only about experiments that can be carried out, there is a single sample space for those experiments. That's fine by me.
This means that arguments about absence of single sample space are irrelevant for any actual experiment e.g. Bell inequality test.
Then if two subsets of outcomes from two remote subexperiments can't be combined into single sample space when assuming that they happen in isolation then it would mean that the assumption of isolation is incorrect.

 

[DarMM]

So as long as I talk only about experiments that can be carried out, there is a single sample space for those experiments

No. There is a single sample space for each pair of Bell alignments, but not all of them together. All can be carried out. @Morbert is referring to the fact that their refinement cannot be carried out.

This means that arguments about absence of single sample space are irrelevant for any actual experiment

It is directly relevant, per Fine's theorem it explains the violations of the Bell inequalities by QM.

 

[zonde]

But these are not the probabilities defined by Born's rule.

Of course choice between experiment X and Y is not described by QM. Only probabilities of subexperiments are described by QM.

 

[DarMM]

So which rule do you say it violates? Do you say that four outcomes are not mutually exclusive? Or that there is some fifth outcome possible?
Or there is some other rule not specified by wikipedia?

Your original example isn't really like a Bell test. There we have two choices of observables at each location and for each choice of pairs we have a total of four outcomes.

I've relabelled things to be more accurate with $A, B$ referring to the measurement axis for Alice and $C, D$ referring to those for Bob, rather than referring to outcomes.

If we label the outcomes with something like $A_{0}, A_{1}$ then we find that one cannot define a probability measure on the space of pairs from $\left\{A_{0}, A_{1}, B_{0}, B_{1}, C_{0}, C_{1}, D_{0}, D_{1}\right\}$ that replicates the quantum predictions.

 

[Elias1960]

If the selection of a subexperiment can be modeled by some appropriate variable like the outcomes of a coin flip that perfectly correlates with the subexperiment, there would be a common sample space, yes. E.g. The outcomes (heads, A),(heads,B),(tails,C),(tails,D).

If you can carry out the experiment, you can build an appropriate sample space. But if you try to combine two incompatible sample spaces into a single common sample space a priori, you'll fail, and you will not be able to carry out the implied experiment.

But the proposal was exactly a simplified version of this: The combination of two incompatible sample spaces {A,B} and {C,D} using an experiment which obviously can be done.

So, you contradict yourself.

Essentially it is also the basic idea behind the trivial construction of that sample space for quantum theory given by Kochen and Specker. The experiment which is done is part of the element of the sample space too.

Or see this variant:

You're original example isn't really like a Bell test. There we have two choices of observables at each location and for each choice of pairs we have a total of four outcomes.

I've relabelled things to be more accurate with $A, B$ referring to the measurement axis for Alice and $C, D$ referring to those for Bob, rather than referring to outcomes.

If we label the outcomes with something like $A_{0}, A_{1}$ then we find that one cannot define a probability measure on the space $\left\{A_{0}, A_{1}, B_{0}, B_{1}, C_{0}, C_{1}, D_{0}, D_{1}\right\}$ that replicates the quantum predictions.

So what? Try the same for the sample space
$\left\{(A_{0},C_{0}), (A_{0},C_{1}), (A_{1},C_{0}),(A_{1},C_{1}),(B_{0},C_{0}),\ldots (B_{1},D_{1})\right\}$
The construction of a sample space is a triviality.

 

[zonde]

If we label the outcomes with something like $A_{0}, A_{1}$ then we find that one cannot define a probability measure on the space $\left\{A_{0}, A_{1}, B_{0}, B_{1}, C_{0}, C_{1}, D_{0}, D_{1}\right\}$ that replicates the quantum predictions.

Yes, that's right.
And if you assume that experiments performed by Alice and Bob are independent then it should be possible to combine them into single sample space. Eberhard in his proof takes this as the very definition of locality.
Think about it. Locality is not property of reality. It is property of hypothetical model of reality. How would you define that two remote experiments in the model are considered independent? Performing one experiment should not change sample space (possible outcomes) of the other remote experiment and vice versa if they are independent.

 

[DarMM]

And if you assume that experiments performed by Alice and Bob are independent then it should be possible to combine them into single sample space.

In QM the experiments of Alice and Bob do not affect each other as shown by their marginals being unaffected by the other's choice of experiment and yet the pairs do not have a common sample space. So this does not follow.

Performing one experiment should not change sample space (possible outcomes) of the other remote experiment

It doesn't change the outcomes of the other experiment or their probabilities.

The fact that QM has multiple sample spaces is a fact of the formalism that has been known since the 1980s. I find it bizarre that people are so motivated to reject a simple consequence of the formalism in addition to decades old facts like QM being a generalization of probability.

Since this is the opinion of every expert in the subject the whole discussion here is verging into crankdom where people are just denying decades old well known aspects of the formalism.

I suggest just reading Chapter 6 of Streater or Summers paper I gave prior.

 

[Mentz114]

No. There is a single sample space for each pair of Bell alignments, but not all of them together. All can be carried out. @Morbert is referring to the fact that their refinement cannot be carried out.It is directly relevant, per Fine's theorem it explains the violations of the Bell inequalities by QM.

It does not explain anything physical - or there would not be dozens of threads arguing this.
When the data from an epr experiment are divided into the four categories each contingency table has expected marginals of 1/2 which is experimentally verified.

 

[DarMM]

So what? Try the same for the sample space
$\left\{(A_{0},C_{0}), (A_{0},C_{1}), (A_{1},C_{0}),(A_{1},C_{1}),(B_{0},C_{0}),\ldots (B_{1},D_{1})\right\}$
The construction of a sample space is a triviality.

You can't put a probability measure on that space that matches quantum theory.

 

[zonde]

In QM the experiments of Alice and Bob are independent as shown by their marginals being unaffected by the other's choice of experiment and yet the pairs do not have a common sample space. So this does not follow.

The devil is in details. Marginals are unaffected but in experiments you have actual detections. QM looks only at statistics but detections themselves are physical facts ignored by QM. And that's where this independence assumption fails.

It doesn't change the outcomes of the other experiment or their probabilities.

It does not change probabilities but it changes outcomes. QM does not look at outcomes so you can't argue that QM is counterexample.

 

[DarMM]

It does not explain anything physical - or there would not be dozens of threads arguing this.
When the data from an epr experiment are divided into the four categories each contingency table has expected marginals of 1/2 which is experimentally verified.

Well it's a property of quantum theory. If it doesn't explain things to you then tough, there isn't anything I can do. That's the theory.

It does have a physical meaning. Only those variables you measure obtain a value.

 

[DarMM]

The devil is in details. Marginals are unaffected but in experiments you have actual detections. QM looks only at statistics but detections themselves are physical facts ignored by QM. And that's where this independence assumption fails.

It does not change probabilities but it changes outcomes. QM does not look at outcomes so you can't argue that QM is counterexample.

I can't use QM in a discussion about the predictions of quantum theory? This has transcended the farcical.

It doesn't even change the outcomes. The outcomes of the local experiments at Alice's location are not affected by the experimental choice at Bob's location.

Have you gone through Fine's theorem, Streater's monograph or any texts on quantum probability?

 

[zonde]

I can't use QM in a discussion about the predictions of quantum theory? This has transcended the farcical.

Of course you can use QM. But you can't use QM to argue about things on which QM is silent.

It doesn't even change the outcomes. The outcomes of the local experiments at Alice's location are not affected by the experimental choice at Bob's location.

Outcomes = sample space. You yourself say that there is no single sample space in QM. And yet you say QM does not change outcomes. It's different words but the same meaning.

 

[DarMM]

And yet you say QM does not change outcomes

"QM changing outcomes" is a meaningless phrase. I am saying that Alice's outcomes only depend on her choice of measurement.

At this point it's just people axe grinding.

For everybody else, check out Chapter 6 of Streater's text or Summers paper, they're a really nice run down. There is also these lecture notes:
http://info.phys.unm.edu/~crosson/Phys572/Physics572Index.html
And these:
https://www.math.tamu.edu/~jml/trentosurvey.pdf
I'll leave the choice of whether to believe Streater, Summers and other experts or @zonde and @Elias1960 up to you.

 

[Jimster41]

If you can carry out the experiment, you can build an appropriate sample space. But if you try to combine two incompatible sample spaces into a single common sample space a priori, you'll fail, and you will not be able to carry out the implied experiment.

To me the key thing is “a-priori”. The role of location in time order gets lost here a lot I think.

Not saying there is a thing that is really Time. I’m just saying Time order is a critical component of the description of nature being debated here.

The cat is alive and dead a-priori. Well ontologically - that is one odd cat.

I But cripes it is subtle, so I’m still probably not getting it.

 

[DarMM]

So what? Try the same for the sample space
$\left\{(A_{0},C_{0}), (A_{0},C_{1}), (A_{1},C_{0}),(A_{1},C_{1}),(B_{0},C_{0}),\ldots (B_{1},D_{1})\right\}$
The construction of a sample space is a triviality.

Just quickly for others, for choices given by their angle with the $z$-axis:
$$
A = S_{0}\\
B = S_{\pi/4}\\
C = S_{\pi/2}\\
D = S_{3\pi/4}
$$
The probabilities QM gives for each outcome pair $\left\{00,01,10,11\right\}$ are given by:

	0	1
0	$\frac{1}{4}\left(1 - \cos(\theta)\right)$	$\frac{1}{4}\left(1 + \cos(\theta)\right)$
1	$\frac{1}{4}\left(1 + \cos(\theta)\right)$	$\frac{1}{4}\left(1 - \cos(\theta)\right)$

Where $\theta$ is the difference in the angles for the two observables.

You'll see that if applied to @Elias1960 's set this gives results going over unity. Thus it isn't a sample space.

 

[PeterDonis]

for choices given by their angle with the zz-axis

Should $B$ be just $\pi / 4$ (instead of $5 \pi / 4$)?

 

[DarMM]

Should $B$ be just $\pi / 4$ (instead of $5 \pi / 4$)?

Sorry corrected now. I've put in the conventional choices with $B = \frac{\pi}{4}$ and $C = \frac{\pi}{2}$

 

[Elias1960]

You can't put a probability measure on that space that matches quantum theory.

I can. However you construct your experiment, you will with some quite classical probability make a choice what to measure, AC, BC, AD or BD. Let's name these classical probabilities $P(AC), P(BC),P(AD),P(BD)$. They sum up to 1 as classical probabilities. Then, for each of the four choices, you have the quantum probabilities for that particular experiment. Let's name them $Q_{AC}(0,0), Q_{AC}(1,0), Q_{AC}(0,1), Q_{AC}(1,1)$ and so on. They sum up to 1 for each of the quantum experiments considered separately.

So, the obvious rule is $P(A_0,C_0) = P(AC)Q_{AC}(0,0), P(A_0,C_1) = P(AC)Q_{AC}(0,1)$ and so on in the straightforward way.

 

[DarMM]

Things like $P\left(AC\right)$ refer to probabilities for choosing the equipment set up, they're nothing to do with the electron.

Your sample space there is the Cartesian product of $\Sigma_{1} \times \Sigma_{2}$ where $\Sigma_{1}$ is the space of my choices and $\Sigma_{2}$ is the space of electron pair observable outcomes. It's not a single sample space for the actual observables of the electron and things like $P\left(AC\right)$ are not electron observable probabilities and not part of the predictions of quantum theory.

You're avoiding the fact that the probabilities for the electron's outcomes exceed unity by weighting them by my set up choices. That's like saying that if classical statistical mechanics predicts a mercury fluid has a 40% chance of being 300K, you say "No, it's only 20% because I'm only going to choose a thermometer 50% of the time"

Also done in full generality your "construction" is going to be infinitely larger than the sample spaces in QM.

I'm not sure what you're even arguing for.

 

[atyy]

That Bell's theorem is related to the number of sample spaces is a result known as Fine's theorem. It shows that assuming a single space (and locality and no retrocauslity, etc) gives the Bell inequalities. Two proofs are here:
https://arxiv.org/pdf/1403.7136.pdf
This is important to know because the lack of a single sample space is how QM itself manages to violate Bell's theorem. Violating it via nonlocality, retrocausality, etc is the approach of alternate theories.

Is it correct to say that Fine's theorem says that if the CHSH equality is violated, there cannot be a single sample space in the measurement outcomes?

If so, then wouldn't a BM-like theory that reproduces the violation of the CHSH inequality also not have a single sample space in that sense?

 

[DarMM]

Is it correct to say that Fine's theorem says that if the CHSH equality is violated, there cannot be a single sample space in the measurement outcomes?

If locality and no retrocausality are retained then yes.

If so, then wouldn't a BM-like theory that reproduces the violation of the CHSH inequality also not have a single sample space in that sense?

BM rejects locality and thus retains a common sample space. The quantum formalism retains locality and rejects a common sample space.

 

[Morbert]

But the proposal was exactly a simplified version of this: The combination of two incompatible sample spaces {A,B} and {C,D} using an experiment which obviously can be done.

I don't think it was a simplified version of this. But let my try to address possible ambiguities.

If we have a Hilbert space of the X apparatus $\mathcal{H}_X$, Y apparatus $\mathcal{H}_Y$, the microscopic system $\mathcal{H}_s$, and a coin $\mathcal{H}_c$, and if we model the subexperimental outcomes as $I_{\mathcal{H}_X\otimes\mathcal{H}_s} = P_A+P_B$ and $I_{\mathcal{H}_Y\otimes\mathcal{H}_s} = P_C+P_D$, as well as a coin flip $I_{\mathcal{H}_c} = P_{\rm heads} + P_{\rm tails}$ then we can use the projective decomposition$$I_{\mathcal{H}_X\otimes\mathcal{H}_Y\otimes\mathcal{H}_s\otimes\mathcal{H}_c}=P_{\rm heads}P_AI_{\mathcal{H}_Y} + P_{\rm heads}P_BI_{\mathcal{H}_Y} + P_{\rm tails}P_CI_{\mathcal{H}_X}+P_{\rm tails}P_DI_{\mathcal{H}_X}$$and build a sample space from these four mutually exclusive outcomes.However, if the samples spaces $\{A,B\}$ and $\{C,D\}$ are instead alternative results of measuring the microscopic system such that $I_{\mathcal{H}_s} = P_A+P_B = P_C+P_D$, then neither $I_{\mathcal{H}_s} = P_A+P_B+P_C+P_D$ nor $P_iP_j = \delta_{ij}P_i$ hold, and $\Omega = \{A,B,C,D\}$ is not a valid sample space of mutually exclusive alternatives, resolved by any experiment.

 

[Tendex]

DarMM said:
"I'll leave the choice of whether to believe Streater, Summers and other experts or @zonde and @Elias1960 up to you."

Would Kochen and Specker in the article communicated by Gleason that Elias1960 has referenced several times in this thread count as experts to you? Maybe experts enough to be believed?

They say referring to formula (1) in that article that defines the probability measure of the single sample space of a classical system in terms of a measurable susbset of the reals: "We may always introduce, at least mathematically, a phase space $\Omega$ into a theory so that (1) is satisfied."

(1)$$P_{A\psi}(U)=\mu_\psi (f_A^{-1}(U))$$ with $P_{A\psi}$ the probability measure assigned to an observable A and mixed state $\psi$

They then describe exactly how to obtain such single sample space in QM, and go on to explain the reasons to introduce what they call "this somewhat trivial construction" [ I have in previous posts commented about this physical triviality but the existence of this mathematical construction follows from consistency of any mathematical physics theory based on standard mathematical logic.] and one of the reasons is that it "indicates the direction in which the condition (1) is inadequate[for constructing a classical hidden variables theory given it applies also to QM]. For each state $\psi$ as interpreted in the space $\Omega$, the functions $f_A$ are easily seen to be measurable functions with respect to the probability measure $\mu_\psi$. In the language of probability theory the observables are thus interpreted as random variables for each state $\psi$. It is not hard to show furthermore that in this representation the observables appear as independent random variables"

(My own aclaratory comments are bewteen brackets)

So I'm still pondering why would anyone deny these facts with such zeal with counterarguments that are totally orthogonal to this construction that is not related to the Gelfand algebra that restricts to each specific quantum experiment and the predictions on them and is included in the generalization by definition. Is it too much to ask to discern between the phenomenology and the math restricted to it from the global consistence of a mathematical theory?

 

[DarMM]

Would Kochen and Specker in the article communicated by Gleason that Elias1960 has referenced several times in this thread count as experts to you?

Nothing in that article contradicts what I'm saying. It's just the ontological models framework as most generally formulated by Spekkens in an earlier form. Which is essentially the most general framework for non-retrocausal hidden variable theories. Indeed there one has that the random variable for a POVM outcome $\Gamma_{E}\left(\lambda\right)$ is required to be generalised to $\Gamma_{E,M}\left(\lambda\right)$ as I have already stated, with $M$ any partition of the identity containing $E$. In fact I've already dealt with this.

My point is that QM itself does not do this. QM itself has multiple sample spaces, as it must because it has a non-commutative C*-algebra.

So I'm still pondering why would anyone deny these facts with such zeal

Because they're not facts.

 

[Tendex]

Nothing in that article contradicts what I'm saying. It's just the ontological models framework as most generally formulated by Spekkens in an earlier form. Which is essentially the most general framework for non-retrocausal hidden variable theories. Indeed there one has that the random variable for a POVM outcome $\Gamma_{E}\left(\lambda\right)$ is required to be generalised to $\Gamma_{E,M}\left(\lambda\right)$ as I have already stated, with $M$ any partition of the identity containing $E$. In fact I've already dealt with this.

My point is that QM itself does not do this. QM itself has multiple sample spaces, as it must because it has a non-commutative C*-algebra.Because they're not facts.

Of course QM phenomenology has multiple sample spaces depending on the experiment, nobody denies this, but the generalization of the quantum mathematical theory also has Kochen and Specker construction.
Also what are you calling "QM itself"? That is non-standard. Having QM phenomenology and making the same predictions? BM, an interpretation of QM, so with the same empirical physics, can have a single sample space(as you have finally admitted to atyy), how is a QM interpretation with QM predictions and experiments not QM itself?
Is your own interpretation only QM itself?

 

[DarMM]

Also what hat are you calling "QM itself"? That is non-standard

A non-commutative C*-algebra with normed states on it? That is utterly standard. That is QM. A non-commutative C*-algebra has multiple sample spaces. The proof being there is no Gelfand homomorphism that covers the whole algebra. The End.

Having QM phenomenology and making the same predictions? BM, an interpretation of QM, so with the same empirical physics, can have a single sample space(as you have finally admitted to atyy)

I've been saying Bohmian Mechanics has a single sample space from the beginning, before you got involved. I've never denied this.

Look just look at post #376, I think it is clear you are not actually understanding the papers and material being referenced here.

Is your own interpretation only QM itself?

If the C*-algebraic structure and the associated dual state space of QM is "my interpretation" I hope I get my Nobel Prize soon.