Is this 3-Qubit state entangled? http://prntscr.com/8vsg0b $|X\rangle=\frac{1}{\sqrt 2}~~|000\rangle+ \frac{i}{\sqrt 2}|111\rangle$

I've worked with 2-Qubit states and you can turn them into bipartite system

A state is said to be not entangled if $|X\rangle$ is part of the Hilbert space $A \otimes B$ and there is a way to rewrite $|X\rangle$ as $|X\rangle=|a\rangle\otimes|b\rangle$

where $\otimes$ is the tensor product, and $X$ is in Hilbert space resulting from the tensor product of the Hilbert spaces $A$ and $B$. $|a\rangle \in A \text{ and } |b\rangle \in B$

MY QUESTION is instead of this bi-partition, is it possible to do a multi-parition and use a similar process to determine if a state is entangled. If this is not required, could someone please show me the process to determining if this state is entangled or not, thank you.


It is an entagled state. We will prove it by contradiction. We will assume that your state is tensor product of 3 qubits.

Take 3 qubits $a_1|0\rangle+b_1|1\rangle,a_2|0\rangle+b_2|1\rangle,a_3|0\rangle+b_3|1\rangle$ and multyply them tensornally. You get that $$(a_1|0\rangle+b_1|1\rangle)\otimes (a_2|0\rangle+b_2|1\rangle)\otimes (a_3|0\rangle+b_3|1\rangle)=\\a_1a_2a_3|000\rangle+a_1a_2b_3|001\rangle+a_1b_2a_3|010\rangle+a_1b_2b_3|011\rangle+\\+b_1a_2a_3|100\rangle+b_1a_2b_3|101\rangle+b_1b_2a_3|110\rangle+b_1b_2b_3|111\rangle.$$ Since we assume that above equals to $\frac{1}{\sqrt 2}~~|000\rangle+ \frac{i}{\sqrt 2}|111\rangle$ we end up with bunch of equations: $$ \left\{ \begin{array}{c} a_1a_2a_3=\frac{1}{\sqrt{2}} \\ a_1a_2b_3=0 \\ a_1b_2a_3=0\\ a_1b_2b_3=0\\ b_1a_2a_3=0\\ b_1a_2b_3=0\\ b_1b_2a_3=0\\ b_1b_2b_3=\frac{i}{\sqrt{2}}\\ \end{array} \right. $$ We will use only the following fact:

$(\star)$ $abc=0$ if and only, if $a=0$ or $b=0$ or $c=0.$ It holds, because $\mathbb{C}$ is a field.

From the first equation we see that $a_1\neq 0$ and $a_2\neq 0$ and $a_3\neq 0.$ From the last we see that $b_1\neq 0$ and $b_2\neq 0$ and $b_3\neq 0.$ But the second equation tells us that $a_1a_2b_3=0.$ It is a contradiction because of $(\star).$


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