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I have 2 qubits which are in an unknown pure state i.e. their density matrix $\rho$ can be expressed as $|\psi\rangle\langle\psi|$.

Let the initial state be $|\psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle$. These coefficients $c_{i}$ are unknown.

I wish to find out if they are entangled or seperable? Can this be done by constructing a Quantum Oracle or some other circuit?

The Rules are:

  • We do not have access to the circuit which created this state $|\psi\rangle$.
  • The initial state of the qubits can be destroyed at the end of the measurement.
  • Any method which has $>0.5$ chance of success is acceptable.
  • Free to use ancillary qubits.

Edit: I have tried to rewrite the question in order to clarify it as per suggestions in the comments.

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  • $\begingroup$ Can you clarify the question please? Do you mean you have 2 two-qubit states or do you mean you have 2 one-qubit states that you call $q_1$ and $q_2$? Because if you have 2 one-qubit states given as $q_1$ and $q_2$ and they are known to be pure, then the total state must be a product state. $\endgroup$ – keisuke.akira Jul 21 at 20:53
  • $\begingroup$ 1. Do you know exactly the way (e.g. a certain circuit) which these qubits are entangled (in case they were entangled) and their initial values ​​(before possible entanglement)? 2. Can you prepare this state many times? $\endgroup$ – Aleksey Zhuravlev Jul 23 at 3:31
  • $\begingroup$ No state cannot be prepared more than once. $\endgroup$ – vasjain Jul 23 at 3:53
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No.

For instance, if I either give you $|00\rangle$ or $|11\rangle$ with 50% probability each, or $|00\rangle\pm|11\rangle$ with 50% probability each, there is no way to distinguish these two cases - not even with any whatsoever small probability. The mathematical reason is that those are described by the same density matrix - but you always get some pure state!

However, one of those sets consists of entangled states, while the other one doesn't.

The situation is different if you get the same state several times and can do a number of tests on each of those copies.

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  • $\begingroup$ I appreciate the quick answer. However my question actually assumed pure states and not mixed states. I was absent minded to not mention this specifically. $\endgroup$ – vasjain Jul 21 at 20:25
  • $\begingroup$ These are pure states. My point is: You either get one of two pure entangled states or one of two pure non-entangled states. There is no way to tell the difference. And you always get pure states. All I did was to pick an example where the math is simple. $\endgroup$ – Norbert Schuch Jul 21 at 21:27
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$\newcommand{\calU}{\mathcal{U}}\newcommand{\ket}[1]{\lvert#1\rangle}$Suppose such a circuit existed, and denote with $\calU$ the unitary describing its overall action. For this to be a "circuit detecting entanglement", there should be two types of output states, one corresponding to the answer "yes, the input was entangled" and the other corresponding to the answer "no, the input was not entangled".

In other words, there should be two orthogonal subspaces $\mathcal V_e$ and $\mathcal V_{ne}$ such that $\calU\lvert\psi\rangle\in\mathcal V_e$ for all entangled states $\lvert\psi\rangle$, and $\calU\lvert\phi\rangle\in\mathcal V_{ne}$ for all separable $\lvert\phi\rangle$.

Let $\ket{\psi_1},\ket{\psi_2}$ be two orthogonal separable states, so that $\calU\ket{\psi_i}\in\mathcal V_{ne}$. Then, by linearity, we must also have $\calU(\alpha\ket{\psi_1}+\beta\ket{\psi_2})\in\mathcal V_{ne}$ for any pair of complex coefficients with $|\alpha|^2+|\beta|^2=1$.

But all entangled states can be written as superpositions of separable states, thus this circuit is bound to incorrectly classify all entangled states as separable as well.

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  • $\begingroup$ I don't get it. Isn't a measurement a POVM $\rho\mapsto \mathrm{tr}[F_k\rho]$? This is linear over mixed states, not pure states. This only works if you insist on unambiguous outcomes, which is not stated in your answer (nor asked for in the question). $\endgroup$ – Norbert Schuch Jul 22 at 18:40
  • $\begingroup$ what do you mean by "unambiguous outcomes"? The way I was imagining it, if the circuit is fixed, that means we are essentially evolving unitarily and then measuring in a fixed basis. Therefore there is a finite number of possible outcomes, and if we are to interpret the outputs as answering the yes/no question, we have to interpret some of the outcomes as the "yes answer" and some as the "no answer". Something like $\mathcal V_e=\text{span}(|00\rangle,|01\rangle)$ etc. $\endgroup$ – glS Jul 22 at 18:56
  • $\begingroup$ Well, it would already be an achievement to be more likely right than wrong. Or to have lots of "don't know outcomes" (I guess there your argument still applies.) -- So let's say that it gives the correct answer with 50.005% probability? That's far from deterministic, but better than guessing. $\endgroup$ – Norbert Schuch Jul 22 at 19:04
  • $\begingroup$ @NorbertSchuch I guess that's possible if we don't require all separable states to be correctly classified. Still, as soon as it works on two separable states $|\psi\rangle,|\phi\rangle$, you will also have to classify as separable everything in $\text{span}(\{|\psi\rangle,|\phi\rangle\})$, which might very well contain entangled states. Unless you are thinking of a different framework or way to interpret the outcomes of the circuit. $\endgroup$ – glS Jul 22 at 19:14
  • $\begingroup$ Unless you do more general types of assignment like "if you get output #1 there is a probability $p_1$ that the input was entangled" etc. But even then a similar argument applies to show that the possible outputs found when a state is entangled are determined by the possible outcomes arising from some separable states whose linear combination gives the entangled one. $\endgroup$ – glS Jul 22 at 19:20

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