atyy said:
How about if I have a pure state, say two unentangled photons of the same definite vertical polarization (0##^{\circ}##)? If both polarizers are set vertical (0##^{\circ}##) or horizontal (45##^{\circ}##), then each photon will definitely pass or not pass. But in a Bell test, the polarizer angles used may be 0##^{\circ}##, -45##^{\circ}##, and 22.5##^{\circ}##, so not all angles have results that are predicted with certainty. Would you consider this to be a state that is excluded by a Bell inequality violation?
Separable states don't lead to a Bell violation. The most general situation in quantum mechanics is that the two parties in a Bell-type experiment (Alice and Bob) can perform POVM measurements on a shared mixed state. Unentangled mixed states are generally defined as those that can be decomposed in the form $$\rho_{\mathrm{AB}} = \sum_{\lambda} p_{\lambda} \rho_{\mathrm{A}}^{(\lambda)} \otimes \rho_{\mathrm{B}}^{(\lambda)} \,,$$
in which ##p_{\lambda}## are a set of probability coefficients and ##\rho_{\mathrm{A}}^{(\lambda)}## and ##\rho_{\mathrm{B}}^{(\lambda)}## are density operators defined on Alice's and Bob's Hilbert spaces respectively. If Alice has a set ##\{M_{a}^{(x)}\}## of POVM measurements she can perform (indicated by an index ##x## denoting the choice of measurement, with the index ##a## indicating the result) and Bob similarly can perform the set ##\{N_{b}^{(y)}\}## of POVMs, then the joint probabilities predicted by quantum mechanics just reduce to the definition of a local model: $$\begin{eqnarray}
P(ab \mid xy) &=& \mathrm{Tr} \bigl[ M_{a}^{(x)} \otimes N_{b}^{(y)} \rho_{\mathrm{AB}} \bigr] \\
&=& \sum_{\lambda} p_{\lambda} \, \mathrm{Tr} \bigl[\bigl( M_{a}^{(x)} \otimes N_{b}^{(y)} \bigr) \, \bigl( \rho_{\mathrm{A}}^{(\lambda)} \otimes \rho_{\mathrm{B}}^{(\lambda)} \bigr) \bigr] \\
&=& \sum_{\lambda} p_{\lambda} \, \mathrm{Tr}_{\mathrm{A}} \bigl[ M_{a}^{(x)} \rho_{\mathrm{A}}^{(\lambda)} \bigr] \, \mathrm{Tr}_{\mathrm{B}} \bigl[ N_{b}^{(y)} \rho_{\mathrm{B}}^{(\lambda)} \bigr] \\
&=& \sum_{\lambda} p_{\lambda} \, P_{\mathrm{A}}(a \mid x; \lambda) \, P_{\mathrm{B}}(b \mid y; \lambda) \,,
\end{eqnarray}$$ with ##P_{\mathrm{A}}(a \mid x; \lambda) = \mathrm{Tr}_{\mathrm{A}} \bigl[ M_{a}^{(x)} \rho_{\mathrm{A}}^{(\lambda)} \bigr]## and ##P_{\mathrm{B}}(b \mid y; \lambda) = \mathrm{Tr}_{\mathrm{B}} \bigl[ N_{b}^{(y)} \rho_{\mathrm{B}}^{(\lambda)} \bigr]## according to the Born rule. So for nonentangled states, you always trivially have a local model that makes the same predictions as quantum mechanics which, of course, won't violate any Bell inequality.
An unentangled pure state is just the special case of a density operator of the form ##\rho_{\mathrm{AB}} = \lvert \psi \rangle \langle \psi \rvert_{\mathrm{A}} \otimes \lvert \phi \rangle \langle \phi \rvert_{\mathrm{B}}##. In that case, the quantum predictions factorise completely: $$\begin{eqnarray}
P(ab \mid xy) &=& \langle \psi \rvert M_{a}^{(x)} \lvert \psi \rangle_{\mathrm{A}} \, \langle \phi \rvert N_{b}^{(y)} \lvert \phi \rangle_{\mathrm{B}} \\
&=& P_{\mathrm{A}}(a \mid x) \, P_{\mathrm{B}}(b \mid y) \,.
\end{eqnarray}$$
As far as I know, the converse isn't so clear. Specifically, I think it's known that all entangled pure states can predict a Bell violation, but I don't think it's known for arbitrary entangled mixed states (though this isn't a topic I know much about, so don't quote me on this).
(EDIT: Section III of the review I linked to covers all of this.)