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 $a$.

  • $\begingroup$ entanglement for multipartite states is tricky. There are multiple possible ways to define what you mean. In particular, you might ask about whether any bipartition is entangled (which you can assessed as per linked question). But then a state like $|0\rangle(|00\rangle+|11\rangle)$ would be "entangled", which you might not like b/c one piece is uncorrelated from the rest. Or you might ask whether the state is entangled in a more "genuine" way, that is, whether roughly speaking there's no "uncorrelated part". Your state is in fact "maximally entangled" in this sense. It's called a GHZ state $\endgroup$
    – glS
    Commented Oct 16, 2022 at 19:20

1 Answer 1


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.

  • $\begingroup$ I updated my question add new about what is not entangle state in CMIIW part. What do you think, is that part true? $\endgroup$ Commented Oct 15, 2022 at 13:43
  • $\begingroup$ @MuhammadIkhwanPerwira No. $\frac{1}{\sqrt{2}}\left( |000\rangle +|110\rangle \right) $ cannot be written into $|\psi \rangle =|\psi _{1}\rangle \otimes \cdots \otimes |\psi _{n}\rangle$ form, see this. $\endgroup$
    – narip
    Commented Oct 15, 2022 at 13:49
  • $\begingroup$ What do you mean with pure state? Did you mean product state (something not entangled state). $\endgroup$ Commented Oct 15, 2022 at 13:52
  • $\begingroup$ @MuhammadIkhwanPerwira See this and this for further information. $\endgroup$
    – narip
    Commented Oct 15, 2022 at 13:57

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