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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?

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    $\begingroup$ Had you read Emilio Pisanty's answer to your post physics.stackexchange.com/questions/404093, you would know the answer. $\endgroup$
    – Norbert Schuch
    May 10, 2018 at 21:20
  • $\begingroup$ @NorbertSchuch Emilio had not yet posted his answer when SRS asked this question. $\endgroup$
    – tparker
    May 10, 2018 at 22:24
  • $\begingroup$ @tparker I realized after posting the comment. Yet, in the light of the discussion which already had happened there, the answer should have been clear. $\endgroup$
    – Norbert Schuch
    May 11, 2018 at 6:49

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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.)

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  • $\begingroup$ That's really a complicated example! ;-) $\endgroup$
    – Norbert Schuch
    May 10, 2018 at 21:22
  • $\begingroup$ @NorbertSchuch I wanted to give an example for which the Schmidt weights are non-degenerate, in order to show why ACuriousMind's (deleted) answer is incorrect. The first ensemble was the simplest one I could think of where the states weren't orthogonal, and the second ensemble is just the corresponding density matrix's eigendecomposition. $\endgroup$
    – tparker
    May 10, 2018 at 22:29
  • $\begingroup$ Oh, I see. In that case, one has to work more indeed. Still, I imagine the use of pythagorean triples might help. $\endgroup$
    – Norbert Schuch
    May 11, 2018 at 6:49
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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.

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