I Bell vs Kolmogorov: Unravelling Probability Theory Limits

[PeterDonis]

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.

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).

 

[PeterDonis]

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.

 

[Killtech]

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.

 

[jedishrfu]

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