# Measurement operator for Chsh experiment

1. Jul 27, 2015

### jk22

The experiment is described at p26 of http://arxiv.org/abs/quant-ph/0402001

In this experiment we see that we sum measurement results and not measure the sum.

Is then the quantum measurement operator not :

$$S=A\otimes B\otimes\mathbb{1_{64}}-\mathbb{1_{4}}\otimes A\otimes B'\otimes\mathbb{1_{16}}+\mathbb{1_{16}}\otimes A'\otimes B\otimes\mathbb{1_{4}}+\mathbb{1_{64}}\otimes A'\otimes B'$$

Since we have the eigenvalues of S $$eig(S)\in\{-4,-2,0,2,4\}$$ ?

2. Aug 1, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Aug 2, 2015

### jk22

In other words it seems that in this experiment we do the sum of measurement results $$v(AB)-v(AB')+v(A'B)+v(A'B')$$

We are searching for the operation $$\diamond$$ such that we have $$v(A)+v(B)=v(A\diamond B)$$

This cannot be the usual sum of operators but $$A\diamond B=A\otimes 1+1\otimes B$$ satisfies the condition above.

So that for four terms it is easily generalized and we get that in fact the measurement operator corresponding to a Bell Chsh 2 channel experiment is a huge 256x256 matrix.

Does this makes any sense ?

Last edited: Aug 2, 2015