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

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#### Mentz114

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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:

And these:

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

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#### Jimster41

Gold Member
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.

• julcab12

#### DarMM

Gold Member
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.

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#### PeterDonis

Mentor
for choices given by their angle with the zz-axis
Should $B$ be just $\pi / 4$ (instead of $5 \pi / 4$)?

#### DarMM

Gold Member
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

Gold Member
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.

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• mattt and Morbert

#### 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:

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

Gold Member
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.

• mattt and Auto-Didact

#### 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.

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• mattt and DarMM

#### 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

Gold Member
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.

• mattt, weirdoguy and Auto-Didact

#### 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

Gold Member
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.

• mattt, weirdoguy and dextercioby

#### Jimster41

Gold Member
Does that mathematically consistent single sample space have to contain a cat that is both dead and alive.

Trying to understand if the argument here is about what is mathematically consistent vs. what is both mathematically consistent and observable... or something like that.

#### Tendex

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.

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.

If the C*-algebraic structure and the associated dual state space of QM is "my interpretation" I hope I get my Nobel Prize soon.
Then the BM interpretation of QM is QM itself according to you and it also has a single sample space as constructed by Kochen and Specher not restricted to the noncommutative algebra(do you understand that in this generalization of the probability the noncommutative algebra is a special case?) besides multiple sample spaces for predictions involving just the noncommutative algebra.

I won't continue this exchange, as it seems you only keep repeating the same mantra to save face.

• weirdoguy

#### DarMM

Gold Member
Then the BM interpretation of QM is QM itself according to you
No, Bohmian Mechanics has a completely different mathematical structure. A structure which as of yet has not been shown to give results compatible with QFT, and possibly (there isn't complete proof in this regard) replicates non-relativistic QM when in equilibrium if one ignores iterated Wigner's friend scenarios. It is a different theory that isn't fully developed as of 2019.

QM itself, the fully developed framework of non-commutative C* algebras and states and unitaries upon them is the theory which matches all experimental predictions and it constitutes a multi-sample space probability theory.

it also has a single sample space
Yes, as I have said from the beginning Bohmian Mechanics has a single sample space.

do you understand that in this generalization of the probability the noncommutative algebra is a special case?
It isn't. I would study how QM's algebra is embedded in Bohmian observable algebra, it's quite subtle.

I won't continue this exchange, as it seems you only keep repeating the same mantra to save face.
I'm just glad the whole field uses my terminology and classifications, at least we all get to save face together. • mattt and weirdoguy

#### atyy

If locality and no retrocausality are retained then yes.
BM rejects locality and thus retains a common sample space. The quantum formalism retains locality and rejects a common sample space.
I don't understand how the quantum formalism can retain locality by rejecting a single sample space. If the quantum state is not real, then the quantum formalism is not nonlocal, thus saving locality in a sense. However, saying that the quantum state is not real does not mean that the quantum formalism is local, because realism is a precondition for locality. If the quantum state is real, then the quantum formalism is nonlocal.

So I link the nonreality of the quantum state to saving locality in some sense, whereas you link not having a single sample space to saving locality. Is the nonreality of the quantum state related to not having a single sample space (I don't see how it is)?

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