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Define product states as simple tensors $\rho_1 \otimes \rho_2$ and separable states to be (trace) norm limits of convex combinations of product states. We say that a state $\rho$ admits a hidden variable model if there is a probability space $(X,\Sigma,\mu)$ such that given two observables with spectral decomposition $A=\sum_n \alpha_n P_n, B=\sum_m \beta_m Q_m$ one can find the so called response functions $x \mapsto F_A(n|x),x \mapsto F_B(m|x)$ such that
$$\operatorname{tr}((P_n \otimes Q_m)\rho)=\int_X F_A(n|x)F_B(m|x)d \mu(x) $$ (one assumes that these functions are measurable, positive and sum up to $1$). It is easy to see that if $\rho=\sum_{r=1}^N p_r S_r \otimes T_r$ where $p_r \geq 0, \sum_{r=1}^N p_r=1$ then one can construct hidden variable model: just take $X:=\{1,2,...,N\}$ with $\mu(\{r\})=p_r$ and $F_A(n|r)=\operatorname{tr}(P_n S_r)$ and $F_B(m|r)=\operatorname{tr}(Q_m T_r)$.

How to prove that if $\rho$ is a (trace norm) limit of convex combinations of product states then $\rho$ also admits hidden variable model?

In particular I'm not sure about a general form of a separable state: I believe that it is not the case that any such state is of the form $\sum_{r=1}^{\infty} p_r S_r \otimes T_r$ where $\sum_{r=1}^{\infty}p_r=1$ and one shuld not confuse a decomposition into product states with the decomposition into pure states.

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    $\begingroup$ what's the point of bringing the observables $A,B$ into it? Their eigenvalues don't matter, so in practice the question is only a function of the pair of projective measurements $(P_n)_n$ and $(Q_m)_m$. That aside, I'm not sure I understand the final question. Separable states are by definition convex combinations of (not necessarily pure) product states. What other definition are you going by? $\endgroup$
    – glS
    May 16 at 9:15
  • $\begingroup$ Thanks, only the spectral decomposition enter into the formula so the observables are not entirely relevant. Regarding separability: I have defined separable states to be limits of convex combinations of product states: I don't see why any such limit should again be convex combination of product states. $\endgroup$
    – truebaran
    May 16 at 14:09
  • $\begingroup$ ah, I see. In which case, mathoverflow.net/q/332478/84108 is probably relevant $\endgroup$
    – glS
    May 16 at 16:30
  • $\begingroup$ Thank you, this discussion is very interesting and nontrivial: I suspect that the proof that any separable state admits hidden variable model should be easier but I don't see how to construct the apriopriate probabilistic space. $\endgroup$
    – truebaran
    May 16 at 17:48
  • $\begingroup$ One straighforward way to proceed is to show that the set of states admitting hidden variable model is closed-but I don't whether this statement is true... $\endgroup$
    – truebaran
    May 16 at 19:29

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