In Bell are pairs independent?

[jk22]

let A B B' be Bell usual matrices for spin 1/2.

Are the results of $$A\otimes B$$ and $$A\otimes B'$$ independent ?

I cannot imagine the contrary since on what could they depend all angles are fixed.

 

[Strilanc]

You need to be clearer; give more context.

What exactly are the matrices $B$ and $B'$? Googling "Bell matrix" brings up nothing obviously applicable. And what is $A$? And what results are you talking about? There's fixed angles somewhere?

You're probably asking something about tests of Bell inequalities, but I'm not sure what exactly.

 

[jk22]

A and B are the projection of the spin operator along direction of measurement : $$A=\vec{\sigma}\cdot\vec{n}_A$$

All matrices are defined the same way, we can take an angle in the x z plane for exsmple.

Now the result of measurement of $$C=A\otimes B$$ is either 1 or -1. My question is, if we define $$C'=A\otimes B'$$ : do we have p(C=1,C'=1)=p(C=1)p(C'=1) ? Or are those dependent ?

 

[Strilanc]

I don't think the probability p(C=1, C'=1) is well defined in this case, because the measurements may not commute.

For example, for the bell pair $\left| 00 \right\rangle + \left| 11 \right\rangle$, the result of $X \otimes X$ is always the same so you don't affect the system whereas the result of $X \otimes Z$ is 50/50ish and does mess up the system. So if you measure $X \otimes X$ then $X \otimes Z$ you'll get both =1 about 50% of the time, but if you measure $X \otimes Z$ then $X \otimes X$ it will drop to 25%.

So at the very least you must be explicit about the measurement order, I think.

 

[jk22]

In the notation p(a=1,b=1) the order is important since the function is not forcedly symmetrical. However, The measurement are supposed to be simultaneous or better said with the same initial state so that in that case i suppose the order is not relevant

 

[Avodyne]

You cannot measure B and B' simultaneously. So, in QM, p(B,B') is not defined.

In hidden variable theory, p(B,B') is defined. That's essentially what it means to have a hidden variable theory.

 

[jk22]

So in qm we could define measuring B before B' ?

What about if we pass the photon to a half-silvered mirror and send half of it to B and the other to B' ?

 

[mfb]

So in qm we could define measuring B before B' ?

You could, but then P(B') is fully determined by the measurement result of B and has nothing to do with the entanglement any more.

What about if we pass the photon to a half-silvered mirror and send half of it to B and the other to B' ?

Then you'll detect a photon "at B" or "at B'" - half of the time you perform one experiment, half of the time another, but you never get two measurements of the same photon then.

 

[jk22]

So passing through a beam splitter does not make the photon go in superposition of both path ?

 

[mfb]

It does, but measuring it later means you only observe one path.

 

[gill1109]

A and B are the projection of the spin operator along direction of measurement : $$A=\vec{\sigma}\cdot\vec{n}_A$$

All matrices are defined the same way, we can take an angle in the x z plane for exsmple.

Now the result of measurement of $$C=A\otimes B$$ is either 1 or -1. My question is, if we define $$C'=A\otimes B'$$ : do we have p(C=1,C'=1)=p(C=1)p(C'=1) ? Or are those dependent ?

Your observables C and C' do not commute so they cannot be measured at the same time.