Is there a way to construct a quantum circuit/oracle to check if 2 qubits in an unknown pure state are entangled?

Question

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.

Answer 1

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.

Answer 2

$\newcommand{\calU}{\mathcal{U}}\newcommand{\ket}[1]{\lvert#1\rangle}$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.