1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 '18 at 21:20
  • $\begingroup$ @NorbertSchuch Emilio had not yet posted his answer when SRS asked this question. $\endgroup$ – tparker May 10 '18 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 '18 at 6:49
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ That's really a complicated example! ;-) $\endgroup$ – Norbert Schuch May 10 '18 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 '18 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 '18 at 6:49
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.