Can the same density matrix represent two (or more) different ensembles? Given an ensemble i.e, a collection of states and the respective probabilities $\{p_i,|i\rangle\}$, one can uniquely construct the density matrix using $\rho=\sum_ip_i|i\rangle\langle i|$. Is the converse also true? Given a density matrix can we uniquely say which ensemble does it refer to i.e., reconstruct the set $\{p_i,|i\rangle\}$?
When I mean ensembles are different, I mean, can the ensembles be distinguished on the basis of some expectation value of some observable?
 A: No, we can't. For example, the ensembles
$$\left\{ \left( \frac{1}{3}, |\uparrow\rangle \right), \left( \frac{2}{3}, \frac{1}{\sqrt{2}} (|\uparrow\rangle + |\downarrow\rangle) \right) \right\}$$
and
\begin{align*}
\left\{ \left( \frac{1}{2} + \frac{\sqrt{5}}{6}, \sqrt{\frac{1}{2} + \frac{1}{2\sqrt{5}}} |\uparrow\rangle + \sqrt{\frac{2}{\sqrt{5}+5}} |\downarrow\rangle \right),\\
\left( \frac{1}{2} - \frac{\sqrt{5}}{6}, \frac{1-\sqrt{5}}{\sqrt{10-2 \sqrt{5}}} |\uparrow\rangle + \sqrt{\frac{1}{2}+\frac{1}{2 \sqrt{5}}}|\downarrow\rangle \right) \right\}
\end{align*}
both correspond to the same non-degenerate density matrix
$$\rho = \left(\begin{array}{cc} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{array} \right).$$
They are completely statistically indistinguishable. (Sorry, there's probably a simpler counterexample.)
A: No. The density matrix
$$
\rho=\frac12\left(\begin{matrix}1&0\\0&1\end{matrix}\right)
$$
can be e.g. decomposed as
$$
\rho=\tfrac12 |0\rangle\langle0|+\tfrac12 |1\rangle\langle1|
$$
or
$$
\rho=\tfrac12 |+\rangle\langle+|+\tfrac12 |-\rangle\langle-|
$$
with $|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$, or an an equal weight mixture of any other two orthogonal states.
Generally, there are infinitely many such decompositions for any non-pure $\rho$, which are related by an isometry.
In particular, the ensemble interpretation is an interpretation.  You cannot distinguish different ensembles which are described by the same $\rho$ experimentally, so it does not make sense to say that you have a given ensemble.
