Let's consider a system composed of the tensor product of two qubits in the following states:
$$|\Psi\rangle = \alpha |00\rangle - \beta |11\rangle$$
$$\rho_{AB} = \rho_1|\Psi^{-}\rangle \langle \Psi^{-}| + \rho_2|\Psi^{+}\rangle \langle \Psi^{+}| + \rho_{3}|\psi^{-}\rangle \langle \psi^{-}| + \rho_{4}|\psi^{+}\rangle \langle \psi^{+}|$$
$$\rho_{AB} = \rho|\psi^{-}\rangle\langle \psi^{-}| + (1-\rho)|00\rangle\langle 00|$$
Where
$$|\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|00\rangle\pm\langle 11|\biggr)$$
$$|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|01\rangle\pm\langle 10|\biggr)$$
are different entangled states. If the two qubits are separate, as in Bell inequality experiment, how could we find the state of Alice's subsystem in the three states?
If I am not mistaken, $|\Psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|00\rangle\pm\langle 11|\biggr)$, $|\psi^{\pm}\rangle = \frac{1}{\sqrt{2}}\biggr(|01\rangle\pm\langle 10|\biggr)$ form a Bell basis for the quantum system $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. Then, could we approach this problem by representing $\rho_{AB}$ as a density matrix (by acting on Bell basis), and then by taking the partial trace $Tr_{A}(\rho_{AB})$?
Your exposition of the problem isn’t terribly clear. I gather that the first index in the states and the subscript $A$ refer to Alice’s subsystem and the second index and the subscript $B$ presumably to Bob’s. The state of Alice’s subsystem is obtained by tracing over Bob’s state, not over Alice’s. I’m not sure why you want to do this in a Bell basis – that doesn’t seem particularly practical, as you already have the states expressed in an unentangled basis for the individual subsystems.
As an example, for the third state we have
\begin{eqnarray} \operatorname{Tr}_B(\rho_{AB}) &=& \operatorname{Tr}_B(\rho|\psi^-\rangle\langle\psi^-|+(1-\rho)|00\rangle\langle00|) \\ &=& \operatorname{Tr}_B\left(\frac\rho2(|01\rangle-|10\rangle)(\langle01|-\langle10|)+(1-\rho)|00\rangle\langle00|)\right) \\ &=& \operatorname{Tr}_B\left(\frac\rho2(|01\rangle\langle01|-|10\rangle\langle01|-|01\rangle\langle10|+|10\rangle\langle10|)+(1-\rho)(|00\rangle\langle00|\right) \\ &=& \frac12\rho(|0\rangle_A\langle0|_A+|1\rangle_A\langle1|_A)+(1-\rho)(|0\rangle_A\langle0|_A \\ &=& \left(1-\frac\rho2\right)|0\rangle_A\langle0|_A+\frac\rho2|1\rangle_A\langle1|_A\;, \end{eqnarray}
where the subscript $A$ indicates the state of Alice’s subsystem. Tracing over Bob’s state eliminates the off-diagonal terms $|10\rangle\langle01|$ and $|01\rangle\langle10|$.