Calculating the reduced density matrix of two separate experiments Consider setting up two experiments $\alpha $ and $\beta$.
In experimental setup $\alpha$ we can prepare $n$ possible pure states $\{| \psi_1 \rangle, | \psi_2 \rangle, ... | \psi_n \rangle\}$ with associated probabilities $\{p_1,p_2,...p_n\}$.
In experimental setup $\beta$ we prepare $m$ non-interacting systems. Each system  prepared in its corresponding lower energy states $\{| \phi_1 \rangle, | \phi_2 \rangle, ... | \phi_n \rangle\}$.
My question is how would one deduce the density matrix operators for the quantum states prepared in each experiment.
Is it simply a matter of noting that one can denote the density operator as  a sum of a set of projectors  as follows: $\rho = \sum_j p_j |j\rangle \langle j|.$
So then I may kind of assume that the density matrix operator for states prepared in $\alpha $ will be $\rho_{\alpha} = \sum_j p_i |\psi_i\rangle \langle \psi_i|.$
But I am unsure if this is correct I am also unsure how to construct the equivalent for experiment $\beta$.
 A: With some more thought you arrive at the answer for the second part of your question, since you have the correct understanding of the first part.
Setup $\alpha$: we know one system has been prepared in one of the $n$ possible pure states $\{\lvert\psi_1\rangle,\dotsc,\lvert\psi_n\rangle\}$. We don't know which, but we assign probabilities $p_1,\dotsc,p_n$ to the $n$ possibilities. Then the density matrix representing our knowledge of this setup is, as you correctly wrote, $\sum_i p_i \lvert\psi_i\rangle\langle\psi_i\rvert$.
Setup $\beta$: we know that $m$ systems have been prepared in $m$ pure states (possibly identical for some of them), each from among the set $\{\lvert\phi_1\rangle,\dotsc,\lvert\phi_n\rangle\}$. We also know the preparation state of each system, say $\lvert\phi_{i_k}\rangle$ for the $k$th system.
This means that their joint system has been prepared in the pure state $\lvert \phi_{i_1},\dotsc,\phi_{i_m}\rangle$, which is equivalent to the density matrix $\lvert \phi_{i_1},\dotsc,\phi_{i_m}\rangle\langle \phi_{i_1},\dotsc,\phi_{i_m}\rvert \equiv \lvert \phi_{i_1}\rangle\langle \phi_{i_1}\rvert \otimes \dotsb \otimes \lvert \phi_{i_m}\rangle\langle \phi_{i_m}\rvert$. (We're assuming that there are no bosonic or fermionic symmetrization complications.)
Saying that they're non-interacting is unimportant for the specification of the state (kinematics): that qualification refers to the dynamics of the joint system, meaning that it has the form $U_1 \otimes \dotsb \otimes U_m$, where $U_k$ is the evolution operator acting on the $k$th system.
Just to add a reference, my favourite book about these topics: Bengtsson, Życzkowski: Geometry of Quantum States: An Introduction to Quantum Entanglement (2nd ed., Cambridge 2017).
