I know that 2 qubits are entangled if it is impossible to represent their joint state as a tensor product. But when we are given a joint state, how can we tell if it is possible to represent it as a tensor product? For example, I am asked to tell if the qubits are entangled for each of the following situations:

$$\begin{align} \left| 01 \right>\\ \frac 12(\left| 00 \right> + i\left| 01 \right> - i\left| 10 \right> + i\left| 11 \right> )\\ \frac 12(\left| 00 \right> - \left| 11 \right>)\\ \frac 12(\left| 00 \right> + \left| 01 \right> +i\left| 10 \right> + \left| 11 \right> ) \end{align}$$

  • $\begingroup$ Are you given all the coefficients of $\mid 01 \rangle$ etc? Or are you thinking more as given 2 physical qubits and you can only perform unitaries and measurements? $\endgroup$ – AHusain Jun 10 '18 at 17:44
  • $\begingroup$ For example, I am asked to tell if the qubits are entangled for each of the following situations:| 01> 1/2*(|00> + i*|01> +i*|10> + i*|01> ) 1/2*(|00> - |11>) $\endgroup$ – Archil Zhvania Jun 10 '18 at 17:49
  • $\begingroup$ I will edit my question, thereI will use math notations. $\endgroup$ – Archil Zhvania Jun 10 '18 at 17:53

If you are given a general 2-qubit state $a \mid 00 \rangle + b \mid 01 \rangle + c \mid 10 \rangle + d \mid 11 \rangle$

If it is unentangled, then the coefficients are that of $(\alpha \mid 0 \rangle + \beta \mid 1 \rangle)(\gamma\mid 0 \rangle + \delta \mid 1 \rangle)$ for some $\alpha .. \delta$.

$$ \alpha \gamma = a\\ \alpha \delta = b\\ \beta \gamma = c\\ \beta \delta = d $$

You want to know if those 4 equations are solvable for a given $a,b,c,d$. This question becomes

$$ ad - bc = 0 $$

so if $ad-bc=0$, then you can solve for $\alpha .. \delta$. You don't need to solve for them, you just need to need to know if it is possible.

The generalization for qudits with potentially different values of $d_1$ and $d_2$ are the quadratic polynomials that cut out the Segre embedding as a zero locus.

  • $\begingroup$ Could you please show me how you would solve the third problem from my example? (1/2* (|00> - |11>)) $\endgroup$ – Archil Zhvania Jun 10 '18 at 18:41
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    $\begingroup$ You forgot to normalize, but it doesn't matter for this problem. $a=1/2$, $b=c=0$, $d=-1/2$. $ad-bc=-1/4 \neq 0$ so entangled. $\endgroup$ – AHusain Jun 10 '18 at 18:56

It is done for a specific state (a Bell state) here, and the same procedure can be used for any other two-qubit state.


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