What does it mean for the Leggett Inequality to falsify realism in general in Quantum Mechanics? http://en.wikipedia.org/wiki/Leggett_inequality
As you can see in the above link, it claims that Bell's inequality ruled out local realism in quantum mechanics, and the violation of Leggett's inequalities is considered to falsify realism in general in quantum mechanics.
This would mean there is no material reality or objective reality! This is being said by Bernard Haisch as well, if I'm not mistaken.
Is this true, or is this false?
 A: Bell's theorem is interesting because it makes few assumptions about the hidden-variable theory. Essentially, the measurement outcomes can depend in an arbitrary way on anything in the universe except for the setting of the other polarizer (and anything from which that could be deduced).
Leggett, in contrast, makes very specific assumptions. To be vulnerable to his argument, a model of a Bell pair must include as hidden variables two unit vectors $\def\a{\mathbf a} \def\b{\mathbf b} \def\u{\mathbf u} \def\v{\mathbf v} \u,\v$, and the independent measurement statistics of the particles in the Bell pair must match the QM prediction for unentangled states $|\u\rangle$ and $|\v\rangle$, for all polarizer settings $\a$ and $\b$. This is formalized in his equations (3.3a) and (3.3b):
$$\begin{align}
\int A(\a,\b,λ) \, g_{\u\v}(λ) dλ &= 2(\u\cdot\a)^2-1 \\
\int B(\a,\b,λ) \, g_{\u\v}(λ) dλ &= 2(\v\cdot\b)^2-1
\end{align}$$
where $A,B,λ$ have essentially the same meaning as in Bell, and $g_{\u\v}$ is a probability distribution like Bell's $ρ$, but it ranges only over the values of $λ$ that include a specific $\u$ and $\v$. These equalities hold for all $\u,\v,\a,\b$.
I don't see why a hidden variable theory should be expected to have this form. It makes some sense to require it in the special case $\u=\a$ (likewise $\v=\b$), where it expresses that $\u$ ($\v$) is the true polarization of the particle. To account for experimental reality, it would need to hold approximately for $\u\approx\a$. But when $\u$ and $\a$ are arbitrarily misaligned, you aren't measuring the true polarization, and I see no reason to expect the quantum relationship (for, again, a different system than the one being modeled here) to apply.
I think that Bohmian mechanics doesn't even satisfy the special case $\u=\a$, since it has definite real values only for position, not arbitrary observables. Yet most people think that Bohmian mechanics is a realistic model.
