Measurement problem and computer-like functions

[jk22]

Neverthless I computed the probabilities with hidden variable and i got p(-4)=(3/4)^4 aso

They differ from qm and give the average S=2

The problem i see is numerically and experimentally : those are always finite number of trials and the statistics can vary.

If -4 arrives as the sum of a single trial then we could imagine we could select a sample where we get a violation ?

 

[stevendaryl]

Neverthless I computed the probabilities with hidden variable and i got p(-4)=(3/4)^4 aso

They differ from qm and give the average S=2

The problem i see is numerically and experimentally : those are always finite number of trials and the statistics can vary.

If -4 arrives as the sum of a single trial then we could imagine we could select a sample where we get a violation ?

Yes, I think that a local hidden variables model can give a violation for a small sample size. the assumption is that

The average of A(\alpha) B(\beta) over the sample is approximately equal to \int_\lambda P(\lambda) d\lambda A(\alpha, \lambda) B(\beta, \lambda). If you have a violation of CHSH that approximate equality can't hold.

 

[gill1109]

Yes, I think that a local hidden variables model can give a violation for a small sample size. the assumption is that

The average of A(\alpha) B(\beta) over the sample is approximately equal to \int_\lambda P(\lambda) d\lambda A(\alpha, \lambda) B(\beta, \lambda). If you have a violation of CHSH that approximate equality can't hold.

In fact, if local hidden variables are true, and you do the experiment, you will probably violate the CHSH bound with probability 50%. Nowadays we have exact finite N probability bounds: assuming LHV, the chance to violate CHSH: "S < = 2" in an experiment with N trials by more than epsilon, is less than ... (something like A exp( - B N eps^2).)

See e.g. http://arxiv.org/abs/1207.5103 Statistical Science 2014, Vol. 29, No. 4, 512-528 Theorem 1 (assuming no memory). Not the best result at all, but as simple as possible and with a relatively simple proof (elementary discrete probability ... at least, elementary for mathematicians. A first year undergraduate course is enough). A = 8 and B = 1 / 256

 

[jk22]

For the results I obtained in qm

$$p(-4)=(1/2(1+1/\sqrt{2}))^4$$
$$p(-2)=4(1/2(1+1/\sqrt{2}))^3(1/2(1-1/\sqrt{2}))$$

And lhv

$$p(-4)=(3/4)^4$$
$$p(-2)=4(3/4)^3(1/4)$$

Hence in qm -4 appears more frequently than -2 whereas for hidden variables it is the opposite.

 

[gill1109]

For the results I obtained in qm

$$p(-4)=(1/2(1+1/\sqrt{2}))^4$$
$$p(-2)=4(1/2(1+1/\sqrt{2}))^3(1/2(1-1/\sqrt{2}))$$

And lhv

$$p(-4)=(3/4)^4$$
$$p(-2)=4(3/4)^3(1/4)$$

Hence in qm -4 appears more frequently than -2 whereas for hidden variables it is the opposite.

I have no idea what these calculations are supposed to refer to.

 

[jk22]

These should be the probabilities for the measurement results of AB-AB'+A'B+A'B' for the angles of measurement 0,Pi/4,Pi/2,3Pi/4 for A B A' B' respectively.

The lhv model considered was given in a previous post, it's the signum of the projection of the hidden vector on the direction of measurement.

 

[jk22]

By the way shouldn't the measurement operator for CHSH not be $$A\otimes B\ominus A\otimes B '\oplus A'\otimes B\oplus A'\otimes B'$$

where $$\oplus$$ is the Kronecker sum ?

I think of that because in a CHSH experiment we sum eigenvalues of measurement.

 

[gill1109]

By the way shouldn't the measurement operator for CHSH not be $$A\otimes B\ominus A\otimes B '\oplus A'\otimes B\oplus A'\otimes B'$$

where $$\oplus$$ is the Kronecker sum ?

I think of that because in a CHSH experiment we sum eigenvalues of measurement.

In an ideal CHSH experiment, we many times either simultaneously measure A on subsystem 1 and B on subsystem 2, or A on subsystem 1 and B' on subsystem 2, or A' on subsystem 1 and B on subsystem 2, or A' on subsystem 1 and B' on subsystem 2. Each time, the two subsystems have been yet again prepared in the same joint state.

 

[jk22]

Indeed i saw a paper that shows the whole cannot be measured simultaneously : http://arxiv.org/abs/quant-ph/0206076

However If we use beam splitter instead of fast changing switcher could we say this were experimentally simultaneous ?

 

[jk22]

The idea is that for every pair, the quantity A(a,\lambda) B(b,\lambda) + A(a&#039;,\lambda) B(b,\lambda) + A(a,\lambda) B(b&#039;,\lambda) - A(a&#039;,\lambda) B(b&#039;,\lambda) has to be less than 2.

What i meant is that we have A(a,\lambda_1) B(b,\lambda_1) + A(a&#039;,\lambda_2) B(b,\lambda_2)+ A(a,\lambda_3) B(b&#039;,\lambda_3)- A(a&#039;,\lambda_4) B(b&#039;,\lambda_4) has to be less than 4 so that a violation is possible. we don't measure all 4 terms for each pair, we can only measure one term. However, if we average that quantity over lambda_i we get:

\langle A(a) B(b) \rangle + \langle A(a) B(b&#039;) \rangle + \langle A(a&#039;) B(b) \rangle - \langle A(a&#039;) B(b&#039;) \rangle \leq 2 [/QUOTE]

 

[Simon Phoenix]

Suppose we define the measurement of an observable A by v(A) v being an 'algorithm giving out one of the eigenvalues each time it is called' (we accept the axiom of choice)

Sorry, I'm a bit late to this thread and there have been many good answers, but I was struck by your initial question. Actually I think it's a good question because it really brings out an essential difference between classical and quantum thinking.

There is a difference between a measurement that 'chooses' one out of a set of pre-existing values, and a measurement that 'generates' a value that is a member of a set.

In standard QM there's no pre-existing value to 'choose' from.

In the usual Bell set up we have Alice and Bob, and at least conceptually we can imagine Alice to be on Earth and Bob to be on Pluto. One of the particles is winging its way to Bob who has set up his apparatus to measure some property. Now we could suppose that the particle is somehow carrying the set of possible values with it and all the measurement is going to do is to pick one of them. But what if Bob changes his mind about what to measure at the very last moment? Is the particle also carrying the new set of possible values with it in some pre-existing sense?

It's this kind of question that the Bell set up really tackles very beautifully. It asks what are the limitations on what we measure if we do assume that in some appropriate sense these properties have some kind of 'pre-existence'.

One of the things that took me a little while to appreciate when I first tried to understand Bell's arguments was the assumption that the result of Alice's measurement cannot depend on the setting of Bob's measurement device (and vice versa). It's so obvious - and it's also true in QM too. The only way we could have a dependence (assuming Alice and Bob are actually free to choose the settings) is if some information about Bob's setting reached Alice and affected the result she obtained.

Put this together with the assumption that there are some real properties that are orchestrating things (our so-called hidden variables) and one consequence of this is the Bell inequality.

OK - some of that is a little vague and imprecise, but I'm trying to highlight the essential components in an intuitive way. I don't think anyone would question too much the assumption that local results can't depend on distant settings but this question of whether physical properties pre-exist before measurement in some appropriate sense is really the mind-bender, for me at least.

 

[jk22]

Yess^the locality should hold, but it is maybe the question if it is in principle possible to determine completely with lambda the result, or if we should let the door open for an indeterminacy that would be determined afterwards.

However stebendaryl showed that if we can predict with certainty in some cases, then there is no place for indeterminacy, so that the parameters : angles and lambda should determine the result. Then the correlation is classical, it can have no bumps, or else it is a saw curve.