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While that question is only asking two-qubit, I'm asking with any size qubit in general about how to check if qubit with any size in entangled state?

For example is this 3-qubit in entangled state? $\frac{1}{\sqrt{2}}(|010\rangle + |101\rangle)$

I think in general, this state is always entangled: $\frac{1}{\sqrt{2}}(|a\rangle \pm |\neg{a}\rangle)$ Where $a$ is any qubit size, for example the size of qubit is 3 like above: $a=010$ therefore $\neg{a}=101$.

So for $\frac{1}{\sqrt{2}}(|p\rangle \pm |q\rangle)$ if $q\neq \neg p$ then it's not entangled (CMIIW).

But I can't prove mathematically that general form above I thought is always correct for every $a$.

glS
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Muhammad Ikhwan Perwira
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1 Answers1

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Yes, your idea is correct. For a pure state, if it is separable, then it can be written as $|\psi \rangle =|\psi _1\rangle \otimes |\psi _2\rangle \otimes ...$. Hence it's easy to notice that after tracing out any qubit, let's say $i$th qubit, we will get $Tr_i\left( |\psi \rangle \langle \psi | \right) $ is still a pure state of form $|\psi _1\rangle \otimes |\psi _2\rangle \otimes ...\otimes |\psi _{i-1}\rangle \otimes |\psi _{i+1}\rangle \otimes ...$ In your question, you can verify that by tracing out any one qubit, you'll get a mixed state which do not satisfy our criterion for separable state above, hence an entangled state.

narip
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