# Bell vs Kolmogorov: Unravelling Probability Theory Limits

• I
• Killtech
In summary: QT guarantees that the only states that survive post measurement are the eigenstates of the observable operator. In the absence of QT, any process could reach detailed balance.
martinbn said:
It is not an objection but something that bothers me because it is not clear to me. Changing it to what Bob would say is not really an improvement. Between Bob's measurement and the meeting with Alice there is nothing strange or unusual about the reality of the measurement outcome. So the question really is what will Bob measure.

What I don't understand is the following. If you can predict with 100% certainty the value of a dynamical variable, then the measurment only reviels a preexisting value, and the theory should account for that. That is how I understand EPR's what is reasanable to consider as an element of reality. And they argue that this applies to Bob's measurment. But for me to predict the outcome of Bob's measurement means that given all the data in the past light-cone of that event you can calculate the value Bob will get. And that is not possible. What they do is to look at Alice's outcome and inffer Bob's. But that is not a prediction. It is almost like looking at the answer before saying what the answer is. Of course it is not quite like that, so I am not objecting to anything, I am only stating what bothers me. The two possible outcome for Alice and Bob are the pairs {1,-1} and {-1,1} . So if you see the first half of the pair you can "predict" the other, so!

If someone (Alice) hands the correctly predicted answer to what Bob will observe, that is an EPR "element of reality" - and you already acknowledge that. They *assume* (explicitly) that whoever/however (Alice) came up with that answer, did NOT affect what Bob will observe (of course that is subject to challenge). Thus the answer MUST have been predetermined.

So it seems to me your (non-)objection - that QM does not supply a way to predict that answer from the initial conditions - is not relevant to their particular argument. After all, perhaps those mechanics will perpetually be obscured to us. The important item is whether you agree that Bob's outcome was predetermined (using their argument*).*Which also has the explicit assumptions of a) locality and b) realism (being the simultaneous existence of all identifiable "elements of reality"). This is of course what Bell hopped on.

stevendaryl said:
Yes, but I don’t understand what point you are making.

Let me introduce a fictional situation and see what you think about it. Suppose that there are a pair of coins. Each coin can be flipped to give a result of “heads” or “tails”. Looking at either coin in isolation reveals no pattern to the results, other than 50/50 chance for each outcome. Yet comparing the two coins shows that the nth flip of one coin always gives the opposite result of the nth flip of the other coin.

I think that most people confronted with such a coin would assume that either the results are predetermined, or that there is some kind of long range interaction between the coins.
Yes, that is what most/all will assume that. So?

DrChinese said:
If someone (Alice) hands the correctly predicted answer to what Bob will observe, that is an EPR "element of reality" - and you already acknowledge that. They *assume* (explicitly) that whoever/however (Alice) came up with that answer, did NOT affect what Bob will observe (of course that is subject to challenge). Thus the answer MUST have been predetermined.

So it seems to me your (non-)objection - that QM does not supply a way to predict that answer from the initial conditions - is not relevant to their particular argument. After all, perhaps those mechanics will perpetually be obscured to us. The important item is whether you agree that Bob's outcome was predetermined (using their argument*).*Which also has the explicit assumptions of a) locality and b) realism (being the simultaneous existence of all identifiable "elements of reality"). This is of course what Bell hopped on.
No, not relevant to their argument. My non-objection is about their definition of elements of reality.

martinbn said:
Yes, that is what most/all will assume that. So?
So that’s the intuition behind Einstein’s argument.

What Einstein seemed to believe is that what future results of a measurement are possible is determined by the state of the universe prior to the measurement. Furthermore, if all causal influences are limited by lightspeed, then the possibilities are in fact determined by just facts about the universe that are available in the backwards light cone.

The EPR experiment seems to violate this intuition, because Alice getting spin-up (in the anti-correlated half-spin case) means that spin-up is not a possible result for Bob, even though no facts available in his backwards light cone could tell him this.

Last edited:
Killtech said:
So in the thinking EBR such a "current state" would only be allowed to contain data from within the specific light cone? I suppose that EBR had something like a particle picture in mind where they though of something like local hidden variables for each particle that had the required information.
I don’t see what it has to do with particles. The question is: Does Alice’s measurement affect the “current state”, or not? If so, then her actions affect Bob’s results, contrary to the limitations of relativity.

stevendaryl said:
I don’t see what it has to do with particles. The question is: Does Alice’s measurement affect the “current state”, or not? If so, then her actions affect Bob’s results, contrary to the limitations of relativity.
So Alice measurements affect the "current state". This isn't contrary to relativity, since this change of current state isn't locally measurable. Relativity makes no statements for such things. It restricts anything that can be used as a signal which are inherently locally accessible.

stevendaryl said:
So that’s the intuition behind Einstein’s argument.

What Einstein seemed to believe is that what future results of a measurement are possible is determined by the state of the universe prior to the measurement. Furthermore, if all causal influences are limited by lightspeed, then the possibilities are in fact determined by just facts about the universe that are available in the backwards light cone.
This is exactly what confuses me. Future and prior in the relativistic setting are not the same as in the nonrelativistic. The measurement of Alice is not prior to that that of Bob's, nor is his in the future of that of Alice.
stevendaryl said:
The EPR experiment seems to violate this intuition, because Alice getting spin-up (in the anti-correlated half-spin case) means that spin-up is not a possible result for Bob, even though no facts available in his backwards light cone could tell him this.
But this is not strange. Look at this: you flip a coin, and I take a look at it. If it is heads it is impossible for you to see tails and nothing in the past of the flip can tell you that. Of course it is not exactly the same, but it looks similar. Almost as if knowing the answer allows me to predict it with 100% certainty.

Killtech said:
So Alice measurements affect the "current state". This isn't contrary to relativity, since this change of current state isn't locally measurable. Relativity makes no statements for such things.
No, you are misstating this. What relativity says is that there can be no such thing as a "current state" that includes spacelike separated events. So the view of QM you are using is fundamentally incompatible with relativity, since it makes use of a "current state" that does include spacelike separated events. Since the view of QM you are using is based on non-relativistic QM, that is not surprising; but it doesn't mean you can just wave your hands and use it in a context where relativity is clearly relevant.

You need to use relativistic QM in this context, and the only relativistic QM we have is quantum field theory. Quantum field theory says that spacelike separated measurements must commute, i.e., their results cannot depend on the order in which they are made. And that means that any claims about Alice's measurement affecting Bob's or Bob's affecting Alice's cannot be right in the context of QFT, since the measurement results are independent of their order so neither one can possibly affect the other.

Motore
PeterDonis said:
No, you are misstating this. What relativity says is that there can be no such thing as a "current state" that includes spacelike separated events.
But no such thing is required. If you think of events, the "current state" partly contains the event of calculating the correlation between Alice and Bob - but partly. The current state changes in what we can say about the possible correlations but since Alice "current state" does not yet include the information for what measurement Bob will ultimately settle for - because that event is indeed spacelike separated. Partial information makes it not a full event until it is combined with supplementing partial information and only then it can be related to other events. But the combination to a full event cannot happen outside either Alice nor Bobs light cone so it never contains any spacelike separated information.

So you can interpret it as Alice knows how she impacted Bobs result, but without knowing what Bob's input will be, she cannot calculate the outcome. It also means that the emergent behavior becomes relative depending on the observer, which is fine for as long as no FTL signaling can happen.

Killtech said:
no such thing is required
Then you should not be using the term "current state" at all, since that term does require that spacelike separated events are part of the "current state".

Killtech said:
the "current state" partly contains the event of calculating the correlation between Alice and Bob - but partly
This is just another way of saying that the "current state" contains spacelike separated events. Even if it only "partly" does so, that's enough to contradict relativity. It's also not anywhere in QFT. If you disagree, please show me where in QFT this "current state" you speak of is.

Killtech said:
So you can interpret it as Alice knows how she impacted Bobs result
In QFT, Alice cannot "impact" Bob's result. Alice's and Bob's measurements are spacelike separated, which means they must commute, which means neither one can "impact" the other's result.

Again, if you disagree, please show me where in QFT this "impact" you speak of is. You can't just wave your hands here. We're talking about physics, and physics makes predictions with math, not vague handwaving. So you need to back up your claims with math.

Killtech said:
It also means that the emergent behavior becomes relative depending on the observer
What does this even mean? Alice's and Bob's measurement settings and measurement results are perfectly objective; they are the same for all observers (though not all observers gain information about them at the same time).

vanhees71
PeterDonis said:
Then you should not be using the term "current state" at all, since that term does require that spacelike separated events are part of the "current state".
i used the "" for a reason and its a word @stevendaryl used which i originally replied to.

PeterDonis said:
This is just another way of saying that the "current state" contains spacelike separated events. Even if it only "partly" does so, that's enough to contradict relativity. It's also not anywhere in QFT. If you disagree, please show me where in QFT this "current state" you speak of is.
Then you clearly misunderstand. The difference is that before the measurement Alice would have described the system to be in ##|HV\rangle+|VH\rangle## whereas afterwards she would says its ##|H\rangle|V\rangle## assuming she got an ##H## result. This information is not spacelike separated since the ##|V\rangle## she has for Bobs photon isn't a measured value at Bob's location but represents her view of the photon.

PeterDonis said:
What does this even mean? Alice's and Bob's measurement settings and measurement results are perfectly objective; they are the same for all observers (though not all observers gain information about them at the same time).
It means that the change of state Alice registers is not purely an update of knowledge about the unknows of the system and does not transform as such information would otherwise do. It instead additionally includes some change of the system, rendering some previous knows unknow, specifically she impacted/changed the axis at which Bobs results will be deterministic/maximally anticorrelated.

vanhees71
Another point of view is much more natural, and it's best "visualized" using the Heisenberg picture of time evolution: The state describes the preparation procedure, i.e., in this case the two-photon polarization state as the Bell state ##\hat{\rho}=|\Psi \rangle \langle \Psi|##, whose state ket is
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle).$$
In the Heisenberg picture this state and the state ket are time-independent. This implies that, if A measures the polarization of her photon to be ##H## then she immediately knows that B will find ##V##, provided he also measures the polarization in the same direction as she did with her photon, i.e., the 100% correlation for this particular case of joint measurements of the polarization of both photons is already imprinted by the state preparation procedure, leading to the said Bell state for the two photons. This view is confirmed in experiments, where the registration events of the photons and thus the measurements of their polarization are space-like separated, such that there cannot be any influence of one measurement on the other (according to local, microcausal QFT which describes photons very well since QED is the most accurate theory ever), i.e., although the single-photon polarizations are utmost indetermined (they are exactly unpolarized photons) before A's or B's measurement, the 100% correlation is still implied by the description of the entangled Bell quantum state.

Killtech said:
i used the "" for a reason and its a word @stevendaryl used which i originally replied to.
Unlike you, @stevendaryl correctly stated that if Alice can affect the "current state", that is contrary to relativity. You claimed it isn't. Your claim is wrong. My statements about the term "current state" were part of explaining why your claim is wrong. @stevendaryl didn't make your claim, you did.

Killtech said:
Then you clearly misunderstand.
No, you misunderstand. The whole notion of a "state" that includes both Alice's and Bob's qubit, since they are spatially separated, is not consistent with relativity. If you are going to try to describe what happens in a way that is consistent with relativity, you cannot use these kinds of states at all. You need to use a correct QFT description of what is going on.

Killtech said:
It means that the change of state Alice registers is not purely an update of knowledge about the unknows of the system and does not transform as such information would otherwise do. It instead additionally includes some change of the system, rendering some previous knows unknow, specifically she impacted/changed the axis at which Bobs results will be deterministic/maximally anticorrelated.
I don't know where you're getting all this from, but it's not from any correct understanding of QM and relativity. It's not even consistent with what you said earlier in the same post.

vanhees71 said:
The state describes the preparation procedure
Which took place at a single, localized spacetime event, so it does not raise the difficulties that a "state" that purports to describe both qubits after they are spatially separated would.

vanhees71
Exactly! In this case it's not so clear what "spatially separated" really means since you cannot even localize photons theoretically, because they do not have a position observable to begin with. All you can say is that a detector at some place registered a photon at some time (that's what I mean when I write "detection event"). Together with some optical elements you can measure its polarization (e.g., using a polaroid filter or a polarizing beam splitter).

vanhees71 said:
In this case it's not so clear what "spatially separated" really means
Alice's and Bob's measurement events are spatially separated, so the two qubits they measured were spatially separated, at the very least, when they were measured. I agree that localizing photons is problematic, but photons are not the only possible realizations of qubits; we could use electron spins, for example, where the localization issues don't arise and we can unproblematically talk about the electrons being spatially separated after they are prepared and while they are en route to being measured by Alice and Bob.

vanhees71
Yes, but that doesn't even matter, because the detection events are local and since they are spatially separated one cannot influence causally the other, and that's true also within relativistic local QFT (in this case QED) due to the microcausality property used in its very conceptual foundation.

PeterDonis said:
Alice's and Bob's measurement events are spatially separated, so the two qubits they measured were spatially separated, at the very least, when they were measured. I agree that localizing photons is problematic, but photons are not the only possible realizations of qubits; we could use electron spins, for example, where the localization issues don't arise and we can unproblematically talk about the electrons being spatially separated after they are prepared and while they are en route to being measured by Alice and Bob.
However, the measurement of the correlation between Bob and Alice is another matter. This event differs drastically from other events as it does not form a spacetime point but rather a hypersurface where it can obtained initially. This is because it cannot be measured at a single spacetime point i.e. locally but requires two separate measurements to obtain it. The operator that represents such an observable is therefore strictly non-separable, i.e. non local. The correlation of two spatially separated entities itself isn't a local quantity yet it is an "element of reality".

Killtech said:
the measurement of the correlation between Bob and Alice is another matter. This event differs drastically from other events as it does not form a spacetime point but rather a hypersurface where it can obtained initially.
No, it doesn't. The measurement of the correlation requires the information about the two measurement results to be brought to the same spacetime event and compared there.

Killtech said:
This is because it cannot be measured at a single spacetime point i.e. locally but requires two separate measurements to obtain it.
Wrong. The measurement of the correlation is a single measurement that compares results from Alice and Bob. It is not the same as Alice's or Bob's measurements themselves.

Killtech said:
The operator that represents such an observable is therefore strictly non-separable, i.e. non local.
If you think there is such a "non-separable", "non-local" operator for "measurement of the correlation", please show your math and give a reference to back it up.

Motore and vanhees71
PeterDonis said:
No, it doesn't. The measurement of the correlation requires the information about the two measurement results to be brought to the same spacetime event and compared there.
And there you have it: the measurement has a new "bringing" part with isn't part of any local measurement. You forget that if no one tells Alice and Bob to signal their results to a specified location... then the the correlation becomes unmeasurable everywhere. Hence, the signaling part is non-omittable and needs to be factored in. The correlation is therefore technically a function to everything the signals encountered on their way to the point where they were combined.

PeterDonis said:
If you think there is such a "non-separable", "non-local" operator for "measurement of the correlation", please show your math and give a reference to back it up.
It can be directly derived from expectation value of the correlation: it's the sum of the probability of each possible state times the value it yields for the correlation. So its operator is the sum of the projection operators onto those states time the correlations they will yield. Since the possible states are continous, this is an integral.

weirdoguy and Motore
Killtech said:
the measurement has a new "bringing" part with isn't part of any local measurement.
The "bringing" part isn't new; the original Alice & Bob measurements had it too. The two qubits had to be prepared in the entangled state at a single spacetime event, and then each qubit had to be brought to a different spacetime event to be measured.

Killtech said:
The correlation is therefore technically a function to everything the signals encountered on their way to the point where they were combined.
And by the same logic, the results of Alice's and Bob's measurements are a function of everything the qubits encountered on their way to the point where they were measured. Which in practice, in both cases, means nothing, since by hypothesis the qubits encountered no significant interactions between preparation and measurement, and equally, the signals carrying the information about Alice's and Bob's results (and these "signals" could just be Alice and Bob themselves traveling to meet each other) encountered no significant interactions between their souce (Alice's and Bob's measurements) and the measurement of the correlation. If they did, the information would be garbled, and we're assuming it's not (since our purpose is to discuss the meaning of the correlations, not to discuss possible sources of noise that could garble our measurements of them).

vanhees71
Killtech said:
It can be directly derived from expectation value of the correlation: it's the sum of the probability of each possible state times the value it yields for the correlation. So its operator is the sum of the projection operators onto those states time the correlations they will yield. Since the possible states are continous, this is an integral.
Do you have a reference to back this up? Otherwise it's your personal theory, which is off limits here.

Lynch101, Motore and vanhees71
PeterDonis said:
Do you have a reference to back this up? Otherwise it's your personal theory, which is off limits here.
Hmm, I would have though that is pretty basic QM stuff directly based on how the theory treats observables. On the one hand correlations are values which can be measured, hence they are observables in term of the theory and in a experimental sense, no? On the other hand QM definition of an observable is very general, so there is one for anything we need, though QM is a little vague on how identifying them.

For the correlation we have: for each two particle quantum state ##|\phi\rangle## the correlation ##c_{\phi}## (i leave out indices specifying the measurement axes of Alice and Bob) can be calculated from the theory or taken from experiments and yields just a simple value. Each projection operator ##\langle \phi|## is an observable on its own according to QM, since it is a linear real valued operator. Scaling it with some real constant doesn't change that, so ##c_{\phi}\langle \phi|## is still an observable. And finally the sum ##C = \sum_{i} c_{\phi_i}\langle \phi_i|## where the sum goes over a basis of states is just another observable.

The expectation of such an observable always gives the expectation of correlation in question, hence it it represents that correlation as an observable, no?

I was merely discussing the consequences of such measurements in QM yielding such an observable structure. They are quite more general then just summarizing the measurements of Alice and Bob, hence i don't see how one could simply interpret them as just a local event. Instead it does rather represent a property of the quantum sate (technically with its own quantum numbers, as it is an operator) and not a specific experiment since there are other ways to measure it. You could alternatively take the beams at Alice and Bob and interfere them, measure the outcome and deduct a value for ##C## from there. The operator leaves it open how the correlation is actually measured taking into account all possibilities, however regardless how you do it, you have to bring two spatially separated information together resulting in the operator having a non-separable structure in general as you can easily check yourself.

weirdoguy
this thread has run its course and it is now a good time to close it and say thanks to all who contributed here.

weirdoguy

• Quantum Physics
Replies
0
Views
770
• Quantum Physics
Replies
1
Views
860
• Quantum Physics
Replies
72
Views
4K
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
4
Views
755
• Quantum Physics
Replies
50
Views
4K
• Quantum Physics
Replies
5
Views
1K
• Quantum Physics
Replies
80
Views
4K
• Quantum Physics
Replies
16
Views
2K
• Quantum Physics
Replies
1
Views
1K