Heinera wrote:Nope, that won't work. Your hidden variable lambda = (a,b) actually depends on the first argument to A(a, lambda).
Sorry, that works just fine. If lambda is (a,b) then A(a, lambda) is A(a, b), and integral A(a, lambda) B(b, lambda) rho(lambda) d lambda, is just integral of A(a, b) B(a,b) rho(a,b) d (a,b), you have a problem with that?
So your definition of A and B only applies to a very restricted subset of the arguments. .. And the subset is too restricted to make sense:
So what? It is a non-local model, it is already nonsensical. The outcome at Alice is A(a, b) (
well defined), the outcome at Bob is B(a,b) (
well defined), the product of the paired outcomes is
A(a, b)B(a,b),
well definedAll well defined, if you repeatedly measure two particle pairs at Alice and Bob, the average of the product is
well defined, the expectation value of that product will indeed be well defined and will be the integral, contrary to your earlier outburst. There is nothing about the expectation value of a paired product that is specific to LHV. What exactly are you arguing against here?
If we go back to the original integral A(a, lambda) B(b, lambda) rho(lambda) d lambda, you will see that the integral is over all possible values of lambda, while still keeping a and b fixed. This is not well defined with your definition of lambda as equal to (a,b).
Wrong. lambda is (a,b), you can argue that there is only one value of lambda if a and b are fixed and lambda = (a,b) but what relevance has that got to do with the fact that the expression is exactly the same well defined integral? There is nothing in your argument. Zilch.